expected values, covariance and correlation
TRANSCRIPT
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Expected Values Covariance Examples Correlation Coefficient
Expected Values, Covariance andCorrelation
Bernd Schroder
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition.
If X and Y are jointly distributed random variables,then the expected value of h(X,Y) is
E(h(X,Y)
)={
∑x ∑y h(x,y)p(x,y) if X,Y are discrete,∫∞
−∞
∫∞
−∞h(x,y)f (x,y) dx dy if X,Y are continuous.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. If X and Y are jointly distributed random variables,then the expected value of h(X,Y) is
E(h(X,Y)
)={
∑x ∑y h(x,y)p(x,y) if X,Y are discrete,∫∞
−∞
∫∞
−∞h(x,y)f (x,y) dx dy if X,Y are continuous.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. If X and Y are jointly distributed random variables,then the expected value of h(X,Y) is
E(h(X,Y)
)={
∑x ∑y h(x,y)p(x,y) if X,Y are discrete,∫∞
−∞
∫∞
−∞h(x,y)f (x,y) dx dy if X,Y are continuous.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. To generate energy, a certain house has solar panelsand a wind turbine. Let X be the percentage of time that thesolar panels generate electricity and let Y be the percentage oftime that the wind turbine generates electricity. Assume that thejoint probability density function of X and Y is
p(x,y) =83
(12− x+ y
)for 0≤ x≤ 0.5 and 0≤ y≤ 1.
Compute E(X), E(Y) and E(XY).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. To generate energy, a certain house has solar panelsand a wind turbine. Let X be the percentage of time that thesolar panels generate electricity and let Y be the percentage oftime that the wind turbine generates electricity. Assume that thejoint probability density function of X and Y is
p(x,y) =83
(12− x+ y
)for 0≤ x≤ 0.5 and 0≤ y≤ 1.
Compute E(X), E(Y) and E(XY).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(X)
=∫ 0.5
0
∫ 1
0x
83
(12− x+ y
)dy dx =
83
∫ 0.5
0
∫ 1
0
12
x− x2 + xy dy dx
=83
∫ 0.5
0
12
xy− x2y+12
xy2∣∣∣∣10
dx =83
∫ 0.5
0x− x2 dx
=83
[12
x2− 13
x3]0.5
0=
83
[18− 1
24
]=
83· 1
12=
29
Compare with the result using marginal distributions:
E(X) =∫ 0.5
0x
83
[1− x] dx =83
∫ 0.5
0x− x2 dx = · · · = 2
9
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(X) =∫ 0.5
0
∫ 1
0
x83
(12− x+ y
)dy dx =
83
∫ 0.5
0
∫ 1
0
12
x− x2 + xy dy dx
=83
∫ 0.5
0
12
xy− x2y+12
xy2∣∣∣∣10
dx =83
∫ 0.5
0x− x2 dx
=83
[12
x2− 13
x3]0.5
0=
83
[18− 1
24
]=
83· 1
12=
29
Compare with the result using marginal distributions:
E(X) =∫ 0.5
0x
83
[1− x] dx =83
∫ 0.5
0x− x2 dx = · · · = 2
9
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(X) =∫ 0.5
0
∫ 1
0x
83
(12− x+ y
)dy dx =
83
∫ 0.5
0
∫ 1
0
12
x− x2 + xy dy dx
=83
∫ 0.5
0
12
xy− x2y+12
xy2∣∣∣∣10
dx =83
∫ 0.5
0x− x2 dx
=83
[12
x2− 13
x3]0.5
0=
83
[18− 1
24
]=
83· 1
12=
29
Compare with the result using marginal distributions:
E(X) =∫ 0.5
0x
83
[1− x] dx =83
∫ 0.5
0x− x2 dx = · · · = 2
9
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(X) =∫ 0.5
0
∫ 1
0x
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
x− x2 + xy dy dx
=83
∫ 0.5
0
12
xy− x2y+12
xy2∣∣∣∣10
dx =83
∫ 0.5
0x− x2 dx
=83
[12
x2− 13
x3]0.5
0=
83
[18− 1
24
]=
83· 1
12=
29
Compare with the result using marginal distributions:
E(X) =∫ 0.5
0x
83
[1− x] dx =83
∫ 0.5
0x− x2 dx = · · · = 2
9
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(X) =∫ 0.5
0
∫ 1
0x
83
(12− x+ y
)dy dx =
83
∫ 0.5
0
∫ 1
0
12
x− x2 + xy dy dx
=83
∫ 0.5
0
12
xy− x2y+12
xy2∣∣∣∣10
dx =83
∫ 0.5
0x− x2 dx
=83
[12
x2− 13
x3]0.5
0=
83
[18− 1
24
]=
83· 1
12=
29
Compare with the result using marginal distributions:
E(X) =∫ 0.5
0x
83
[1− x] dx =83
∫ 0.5
0x− x2 dx = · · · = 2
9
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(X) =∫ 0.5
0
∫ 1
0x
83
(12− x+ y
)dy dx =
83
∫ 0.5
0
∫ 1
0
12
x− x2 + xy dy dx
=83
∫ 0.5
0
12
xy− x2y+12
xy2∣∣∣∣10
dx
=83
∫ 0.5
0x− x2 dx
=83
[12
x2− 13
x3]0.5
0=
83
[18− 1
24
]=
83· 1
12=
29
Compare with the result using marginal distributions:
E(X) =∫ 0.5
0x
83
[1− x] dx =83
∫ 0.5
0x− x2 dx = · · · = 2
9
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(X) =∫ 0.5
0
∫ 1
0x
83
(12− x+ y
)dy dx =
83
∫ 0.5
0
∫ 1
0
12
x− x2 + xy dy dx
=83
∫ 0.5
0
12
xy− x2y+12
xy2∣∣∣∣10
dx =83
∫ 0.5
0x− x2 dx
=83
[12
x2− 13
x3]0.5
0=
83
[18− 1
24
]=
83· 1
12=
29
Compare with the result using marginal distributions:
E(X) =∫ 0.5
0x
83
[1− x] dx =83
∫ 0.5
0x− x2 dx = · · · = 2
9
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(X) =∫ 0.5
0
∫ 1
0x
83
(12− x+ y
)dy dx =
83
∫ 0.5
0
∫ 1
0
12
x− x2 + xy dy dx
=83
∫ 0.5
0
12
xy− x2y+12
xy2∣∣∣∣10
dx =83
∫ 0.5
0x− x2 dx
=83
[12
x2− 13
x3]0.5
0
=83
[18− 1
24
]=
83· 1
12=
29
Compare with the result using marginal distributions:
E(X) =∫ 0.5
0x
83
[1− x] dx =83
∫ 0.5
0x− x2 dx = · · · = 2
9
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(X) =∫ 0.5
0
∫ 1
0x
83
(12− x+ y
)dy dx =
83
∫ 0.5
0
∫ 1
0
12
x− x2 + xy dy dx
=83
∫ 0.5
0
12
xy− x2y+12
xy2∣∣∣∣10
dx =83
∫ 0.5
0x− x2 dx
=83
[12
x2− 13
x3]0.5
0=
83
[18− 1
24
]
=83· 1
12=
29
Compare with the result using marginal distributions:
E(X) =∫ 0.5
0x
83
[1− x] dx =83
∫ 0.5
0x− x2 dx = · · · = 2
9
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(X) =∫ 0.5
0
∫ 1
0x
83
(12− x+ y
)dy dx =
83
∫ 0.5
0
∫ 1
0
12
x− x2 + xy dy dx
=83
∫ 0.5
0
12
xy− x2y+12
xy2∣∣∣∣10
dx =83
∫ 0.5
0x− x2 dx
=83
[12
x2− 13
x3]0.5
0=
83
[18− 1
24
]=
83· 1
12
=29
Compare with the result using marginal distributions:
E(X) =∫ 0.5
0x
83
[1− x] dx =83
∫ 0.5
0x− x2 dx = · · · = 2
9
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(X) =∫ 0.5
0
∫ 1
0x
83
(12− x+ y
)dy dx =
83
∫ 0.5
0
∫ 1
0
12
x− x2 + xy dy dx
=83
∫ 0.5
0
12
xy− x2y+12
xy2∣∣∣∣10
dx =83
∫ 0.5
0x− x2 dx
=83
[12
x2− 13
x3]0.5
0=
83
[18− 1
24
]=
83· 1
12=
29
Compare with the result using marginal distributions:
E(X) =∫ 0.5
0x
83
[1− x] dx =83
∫ 0.5
0x− x2 dx = · · · = 2
9
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(X) =∫ 0.5
0
∫ 1
0x
83
(12− x+ y
)dy dx =
83
∫ 0.5
0
∫ 1
0
12
x− x2 + xy dy dx
=83
∫ 0.5
0
12
xy− x2y+12
xy2∣∣∣∣10
dx =83
∫ 0.5
0x− x2 dx
=83
[12
x2− 13
x3]0.5
0=
83
[18− 1
24
]=
83· 1
12=
29
Compare with the result using marginal distributions:
E(X) =∫ 0.5
0x
83
[1− x] dx =83
∫ 0.5
0x− x2 dx = · · · = 2
9
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(X) =∫ 0.5
0
∫ 1
0x
83
(12− x+ y
)dy dx =
83
∫ 0.5
0
∫ 1
0
12
x− x2 + xy dy dx
=83
∫ 0.5
0
12
xy− x2y+12
xy2∣∣∣∣10
dx =83
∫ 0.5
0x− x2 dx
=83
[12
x2− 13
x3]0.5
0=
83
[18− 1
24
]=
83· 1
12=
29
Compare with the result using marginal distributions:
E(X)
=∫ 0.5
0x
83
[1− x] dx =83
∫ 0.5
0x− x2 dx = · · · = 2
9
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(X) =∫ 0.5
0
∫ 1
0x
83
(12− x+ y
)dy dx =
83
∫ 0.5
0
∫ 1
0
12
x− x2 + xy dy dx
=83
∫ 0.5
0
12
xy− x2y+12
xy2∣∣∣∣10
dx =83
∫ 0.5
0x− x2 dx
=83
[12
x2− 13
x3]0.5
0=
83
[18− 1
24
]=
83· 1
12=
29
Compare with the result using marginal distributions:
E(X) =∫ 0.5
0x
83
[1− x] dx
=83
∫ 0.5
0x− x2 dx = · · · = 2
9
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(X) =∫ 0.5
0
∫ 1
0x
83
(12− x+ y
)dy dx =
83
∫ 0.5
0
∫ 1
0
12
x− x2 + xy dy dx
=83
∫ 0.5
0
12
xy− x2y+12
xy2∣∣∣∣10
dx =83
∫ 0.5
0x− x2 dx
=83
[12
x2− 13
x3]0.5
0=
83
[18− 1
24
]=
83· 1
12=
29
Compare with the result using marginal distributions:
E(X) =∫ 0.5
0x
83
[1− x] dx =83
∫ 0.5
0x− x2 dx
= · · · = 29
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(X) =∫ 0.5
0
∫ 1
0x
83
(12− x+ y
)dy dx =
83
∫ 0.5
0
∫ 1
0
12
x− x2 + xy dy dx
=83
∫ 0.5
0
12
xy− x2y+12
xy2∣∣∣∣10
dx =83
∫ 0.5
0x− x2 dx
=83
[12
x2− 13
x3]0.5
0=
83
[18− 1
24
]=
83· 1
12=
29
Compare with the result using marginal distributions:
E(X) =∫ 0.5
0x
83
[1− x] dx =83
∫ 0.5
0x− x2 dx = · · ·
=29
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(X) =∫ 0.5
0
∫ 1
0x
83
(12− x+ y
)dy dx =
83
∫ 0.5
0
∫ 1
0
12
x− x2 + xy dy dx
=83
∫ 0.5
0
12
xy− x2y+12
xy2∣∣∣∣10
dx =83
∫ 0.5
0x− x2 dx
=83
[12
x2− 13
x3]0.5
0=
83
[18− 1
24
]=
83· 1
12=
29
Compare with the result using marginal distributions:
E(X) =∫ 0.5
0x
83
[1− x] dx =83
∫ 0.5
0x− x2 dx = · · · = 2
9
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(Y)
=∫ 0.5
0
∫ 1
0y
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
y− xy+ y2 dy dx
=83
∫ 0.5
0
14
y2− 12
xy2 +13
y3∣∣∣∣10
dx
=83
∫ 0.5
0
14− 1
2x+
13
dx
=83
∫ 0.5
0
712− 1
2x dx =
83
[7
12x− 1
4x2]0.5
0
=83
[7
24− 1
16
]=
83
[1448− 3
48
]=
83· 11
48=
1118
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(Y) =∫ 0.5
0
∫ 1
0
y83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
y− xy+ y2 dy dx
=83
∫ 0.5
0
14
y2− 12
xy2 +13
y3∣∣∣∣10
dx
=83
∫ 0.5
0
14− 1
2x+
13
dx
=83
∫ 0.5
0
712− 1
2x dx =
83
[7
12x− 1
4x2]0.5
0
=83
[7
24− 1
16
]=
83
[1448− 3
48
]=
83· 11
48=
1118
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(Y) =∫ 0.5
0
∫ 1
0y
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
y− xy+ y2 dy dx
=83
∫ 0.5
0
14
y2− 12
xy2 +13
y3∣∣∣∣10
dx
=83
∫ 0.5
0
14− 1
2x+
13
dx
=83
∫ 0.5
0
712− 1
2x dx =
83
[7
12x− 1
4x2]0.5
0
=83
[7
24− 1
16
]=
83
[1448− 3
48
]=
83· 11
48=
1118
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(Y) =∫ 0.5
0
∫ 1
0y
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
y− xy+ y2 dy dx
=83
∫ 0.5
0
14
y2− 12
xy2 +13
y3∣∣∣∣10
dx
=83
∫ 0.5
0
14− 1
2x+
13
dx
=83
∫ 0.5
0
712− 1
2x dx =
83
[7
12x− 1
4x2]0.5
0
=83
[7
24− 1
16
]=
83
[1448− 3
48
]=
83· 11
48=
1118
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(Y) =∫ 0.5
0
∫ 1
0y
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
y− xy+ y2 dy dx
=83
∫ 0.5
0
14
y2− 12
xy2 +13
y3∣∣∣∣10
dx
=83
∫ 0.5
0
14− 1
2x+
13
dx
=83
∫ 0.5
0
712− 1
2x dx =
83
[7
12x− 1
4x2]0.5
0
=83
[7
24− 1
16
]=
83
[1448− 3
48
]=
83· 11
48=
1118
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(Y) =∫ 0.5
0
∫ 1
0y
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
y− xy+ y2 dy dx
=83
∫ 0.5
0
14
y2− 12
xy2 +13
y3∣∣∣∣10
dx
=83
∫ 0.5
0
14− 1
2x+
13
dx
=83
∫ 0.5
0
712− 1
2x dx =
83
[7
12x− 1
4x2]0.5
0
=83
[7
24− 1
16
]=
83
[1448− 3
48
]=
83· 11
48=
1118
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(Y) =∫ 0.5
0
∫ 1
0y
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
y− xy+ y2 dy dx
=83
∫ 0.5
0
14
y2− 12
xy2 +13
y3∣∣∣∣10
dx
=83
∫ 0.5
0
14− 1
2x+
13
dx
=83
∫ 0.5
0
712− 1
2x dx =
83
[7
12x− 1
4x2]0.5
0
=83
[7
24− 1
16
]=
83
[1448− 3
48
]=
83· 11
48=
1118
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(Y) =∫ 0.5
0
∫ 1
0y
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
y− xy+ y2 dy dx
=83
∫ 0.5
0
14
y2− 12
xy2 +13
y3∣∣∣∣10
dx
=83
∫ 0.5
0
14− 1
2x+
13
dx
=83
∫ 0.5
0
712− 1
2x dx
=83
[7
12x− 1
4x2]0.5
0
=83
[7
24− 1
16
]=
83
[1448− 3
48
]=
83· 11
48=
1118
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(Y) =∫ 0.5
0
∫ 1
0y
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
y− xy+ y2 dy dx
=83
∫ 0.5
0
14
y2− 12
xy2 +13
y3∣∣∣∣10
dx
=83
∫ 0.5
0
14− 1
2x+
13
dx
=83
∫ 0.5
0
712− 1
2x dx =
83
[7
12x− 1
4x2]0.5
0
=83
[7
24− 1
16
]=
83
[1448− 3
48
]=
83· 11
48=
1118
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(Y) =∫ 0.5
0
∫ 1
0y
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
y− xy+ y2 dy dx
=83
∫ 0.5
0
14
y2− 12
xy2 +13
y3∣∣∣∣10
dx
=83
∫ 0.5
0
14− 1
2x+
13
dx
=83
∫ 0.5
0
712− 1
2x dx =
83
[7
12x− 1
4x2]0.5
0
=83
[7
24− 1
16
]
=83
[1448− 3
48
]=
83· 11
48=
1118
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(Y) =∫ 0.5
0
∫ 1
0y
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
y− xy+ y2 dy dx
=83
∫ 0.5
0
14
y2− 12
xy2 +13
y3∣∣∣∣10
dx
=83
∫ 0.5
0
14− 1
2x+
13
dx
=83
∫ 0.5
0
712− 1
2x dx =
83
[7
12x− 1
4x2]0.5
0
=83
[7
24− 1
16
]=
83
[1448− 3
48
]
=83· 11
48=
1118
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(Y) =∫ 0.5
0
∫ 1
0y
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
y− xy+ y2 dy dx
=83
∫ 0.5
0
14
y2− 12
xy2 +13
y3∣∣∣∣10
dx
=83
∫ 0.5
0
14− 1
2x+
13
dx
=83
∫ 0.5
0
712− 1
2x dx =
83
[7
12x− 1
4x2]0.5
0
=83
[7
24− 1
16
]=
83
[1448− 3
48
]=
83· 11
48
=1118
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(Y) =∫ 0.5
0
∫ 1
0y
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
y− xy+ y2 dy dx
=83
∫ 0.5
0
14
y2− 12
xy2 +13
y3∣∣∣∣10
dx
=83
∫ 0.5
0
14− 1
2x+
13
dx
=83
∫ 0.5
0
712− 1
2x dx =
83
[7
12x− 1
4x2]0.5
0
=83
[7
24− 1
16
]=
83
[1448− 3
48
]=
83· 11
48=
1118
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(XY)
=∫ 0.5
0
∫ 1
0xy
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
xy− x2y+ xy2 dy dx
=83
∫ 0.5
0
14
xy2− 12
x2y2 +13
xy3∣∣∣∣10
dx
=83
∫ 0.5
0
712
x− 12
x2 dx
=83
[7
24x2− 1
6x3]0.5
0
=83
[7
96− 1
48
]=
83· 5
96=
536
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(XY) =∫ 0.5
0
∫ 1
0
xy83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
xy− x2y+ xy2 dy dx
=83
∫ 0.5
0
14
xy2− 12
x2y2 +13
xy3∣∣∣∣10
dx
=83
∫ 0.5
0
712
x− 12
x2 dx
=83
[7
24x2− 1
6x3]0.5
0
=83
[7
96− 1
48
]=
83· 5
96=
536
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(XY) =∫ 0.5
0
∫ 1
0xy
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
xy− x2y+ xy2 dy dx
=83
∫ 0.5
0
14
xy2− 12
x2y2 +13
xy3∣∣∣∣10
dx
=83
∫ 0.5
0
712
x− 12
x2 dx
=83
[7
24x2− 1
6x3]0.5
0
=83
[7
96− 1
48
]=
83· 5
96=
536
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(XY) =∫ 0.5
0
∫ 1
0xy
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
xy− x2y+ xy2 dy dx
=83
∫ 0.5
0
14
xy2− 12
x2y2 +13
xy3∣∣∣∣10
dx
=83
∫ 0.5
0
712
x− 12
x2 dx
=83
[7
24x2− 1
6x3]0.5
0
=83
[7
96− 1
48
]=
83· 5
96=
536
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(XY) =∫ 0.5
0
∫ 1
0xy
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
xy− x2y+ xy2 dy dx
=83
∫ 0.5
0
14
xy2− 12
x2y2 +13
xy3∣∣∣∣10
dx
=83
∫ 0.5
0
712
x− 12
x2 dx
=83
[7
24x2− 1
6x3]0.5
0
=83
[7
96− 1
48
]=
83· 5
96=
536
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(XY) =∫ 0.5
0
∫ 1
0xy
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
xy− x2y+ xy2 dy dx
=83
∫ 0.5
0
14
xy2− 12
x2y2 +13
xy3∣∣∣∣10
dx
=83
∫ 0.5
0
712
x− 12
x2 dx
=83
[7
24x2− 1
6x3]0.5
0
=83
[7
96− 1
48
]=
83· 5
96=
536
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(XY) =∫ 0.5
0
∫ 1
0xy
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
xy− x2y+ xy2 dy dx
=83
∫ 0.5
0
14
xy2− 12
x2y2 +13
xy3∣∣∣∣10
dx
=83
∫ 0.5
0
712
x− 12
x2 dx
=83
[7
24x2− 1
6x3]0.5
0
=83
[7
96− 1
48
]=
83· 5
96=
536
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(XY) =∫ 0.5
0
∫ 1
0xy
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
xy− x2y+ xy2 dy dx
=83
∫ 0.5
0
14
xy2− 12
x2y2 +13
xy3∣∣∣∣10
dx
=83
∫ 0.5
0
712
x− 12
x2 dx
=83
[7
24x2− 1
6x3]0.5
0
=83
[7
96− 1
48
]=
83· 5
96=
536
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(XY) =∫ 0.5
0
∫ 1
0xy
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
xy− x2y+ xy2 dy dx
=83
∫ 0.5
0
14
xy2− 12
x2y2 +13
xy3∣∣∣∣10
dx
=83
∫ 0.5
0
712
x− 12
x2 dx
=83
[7
24x2− 1
6x3]0.5
0
=83
[7
96− 1
48
]
=83· 5
96=
536
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(XY) =∫ 0.5
0
∫ 1
0xy
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
xy− x2y+ xy2 dy dx
=83
∫ 0.5
0
14
xy2− 12
x2y2 +13
xy3∣∣∣∣10
dx
=83
∫ 0.5
0
712
x− 12
x2 dx
=83
[7
24x2− 1
6x3]0.5
0
=83
[7
96− 1
48
]=
83· 5
96
=5
36
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
E(XY) =∫ 0.5
0
∫ 1
0xy
83
(12− x+ y
)dy dx
=83
∫ 0.5
0
∫ 1
0
12
xy− x2y+ xy2 dy dx
=83
∫ 0.5
0
14
xy2− 12
x2y2 +13
xy3∣∣∣∣10
dx
=83
∫ 0.5
0
712
x− 12
x2 dx
=83
[7
24x2− 1
6x3]0.5
0
=83
[7
96− 1
48
]=
83· 5
96=
536
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition.
If X and Y are jointly distributed randomvariables, then the expected value of X +Y is
E(X +Y) = E(X)+E(Y).
Proof (discrete case only).E(X +Y) = ∑
x∑y
(x+ y)p(x,y)
= ∑x
∑y
xp(x,y)+ yp(x,y)
= ∑x
∑y
xp(x,y)+∑y
∑x
yp(x,y)
= ∑x
x∑y
p(x,y)+∑y
y∑x
p(x,y)
= ∑x
xpX(x)+∑y
ypY(y) = E(X)+E(Y)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed randomvariables, then the expected value of X +Y is
E(X +Y) = E(X)+E(Y).
Proof (discrete case only).E(X +Y) = ∑
x∑y
(x+ y)p(x,y)
= ∑x
∑y
xp(x,y)+ yp(x,y)
= ∑x
∑y
xp(x,y)+∑y
∑x
yp(x,y)
= ∑x
x∑y
p(x,y)+∑y
y∑x
p(x,y)
= ∑x
xpX(x)+∑y
ypY(y) = E(X)+E(Y)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed randomvariables, then the expected value of X +Y is
E(X +Y) = E(X)+E(Y).
Proof (discrete case only).E(X +Y) = ∑
x∑y
(x+ y)p(x,y)
= ∑x
∑y
xp(x,y)+ yp(x,y)
= ∑x
∑y
xp(x,y)+∑y
∑x
yp(x,y)
= ∑x
x∑y
p(x,y)+∑y
y∑x
p(x,y)
= ∑x
xpX(x)+∑y
ypY(y) = E(X)+E(Y)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed randomvariables, then the expected value of X +Y is
E(X +Y) = E(X)+E(Y).
Proof (discrete case only).
E(X +Y) = ∑x
∑y
(x+ y)p(x,y)
= ∑x
∑y
xp(x,y)+ yp(x,y)
= ∑x
∑y
xp(x,y)+∑y
∑x
yp(x,y)
= ∑x
x∑y
p(x,y)+∑y
y∑x
p(x,y)
= ∑x
xpX(x)+∑y
ypY(y) = E(X)+E(Y)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed randomvariables, then the expected value of X +Y is
E(X +Y) = E(X)+E(Y).
Proof (discrete case only).E(X +Y)
= ∑x
∑y
(x+ y)p(x,y)
= ∑x
∑y
xp(x,y)+ yp(x,y)
= ∑x
∑y
xp(x,y)+∑y
∑x
yp(x,y)
= ∑x
x∑y
p(x,y)+∑y
y∑x
p(x,y)
= ∑x
xpX(x)+∑y
ypY(y) = E(X)+E(Y)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed randomvariables, then the expected value of X +Y is
E(X +Y) = E(X)+E(Y).
Proof (discrete case only).E(X +Y) = ∑
x∑y
(x+ y)p(x,y)
= ∑x
∑y
xp(x,y)+ yp(x,y)
= ∑x
∑y
xp(x,y)+∑y
∑x
yp(x,y)
= ∑x
x∑y
p(x,y)+∑y
y∑x
p(x,y)
= ∑x
xpX(x)+∑y
ypY(y) = E(X)+E(Y)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed randomvariables, then the expected value of X +Y is
E(X +Y) = E(X)+E(Y).
Proof (discrete case only).E(X +Y) = ∑
x∑y
(x+ y)p(x,y)
= ∑x
∑y
xp(x,y)+ yp(x,y)
= ∑x
∑y
xp(x,y)+∑y
∑x
yp(x,y)
= ∑x
x∑y
p(x,y)+∑y
y∑x
p(x,y)
= ∑x
xpX(x)+∑y
ypY(y) = E(X)+E(Y)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed randomvariables, then the expected value of X +Y is
E(X +Y) = E(X)+E(Y).
Proof (discrete case only).E(X +Y) = ∑
x∑y
(x+ y)p(x,y)
= ∑x
∑y
xp(x,y)+ yp(x,y)
= ∑x
∑y
xp(x,y)+∑y
∑x
yp(x,y)
= ∑x
x∑y
p(x,y)+∑y
y∑x
p(x,y)
= ∑x
xpX(x)+∑y
ypY(y) = E(X)+E(Y)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed randomvariables, then the expected value of X +Y is
E(X +Y) = E(X)+E(Y).
Proof (discrete case only).E(X +Y) = ∑
x∑y
(x+ y)p(x,y)
= ∑x
∑y
xp(x,y)+ yp(x,y)
= ∑x
∑y
xp(x,y)+∑y
∑x
yp(x,y)
= ∑x
x∑y
p(x,y)+∑y
y∑x
p(x,y)
= ∑x
xpX(x)+∑y
ypY(y) = E(X)+E(Y)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed randomvariables, then the expected value of X +Y is
E(X +Y) = E(X)+E(Y).
Proof (discrete case only).E(X +Y) = ∑
x∑y
(x+ y)p(x,y)
= ∑x
∑y
xp(x,y)+ yp(x,y)
= ∑x
∑y
xp(x,y)+∑y
∑x
yp(x,y)
= ∑x
x∑y
p(x,y)+∑y
y∑x
p(x,y)
= ∑x
xpX(x)+∑y
ypY(y)
= E(X)+E(Y)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed randomvariables, then the expected value of X +Y is
E(X +Y) = E(X)+E(Y).
Proof (discrete case only).E(X +Y) = ∑
x∑y
(x+ y)p(x,y)
= ∑x
∑y
xp(x,y)+ yp(x,y)
= ∑x
∑y
xp(x,y)+∑y
∑x
yp(x,y)
= ∑x
x∑y
p(x,y)+∑y
y∑x
p(x,y)
= ∑x
xpX(x)+∑y
ypY(y) = E(X)+E(Y)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed randomvariables, then the expected value of X +Y is
E(X +Y) = E(X)+E(Y).
Proof (discrete case only).E(X +Y) = ∑
x∑y
(x+ y)p(x,y)
= ∑x
∑y
xp(x,y)+ yp(x,y)
= ∑x
∑y
xp(x,y)+∑y
∑x
yp(x,y)
= ∑x
x∑y
p(x,y)+∑y
y∑x
p(x,y)
= ∑x
xpX(x)+∑y
ypY(y) = E(X)+E(Y)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition.
If X and Y are jointly distributed independentrandom variables, then the expected value of XY is
E(XY) = E(X)E(Y).
Proof (discrete case only).E(XY) = ∑
x∑y
xyp(x,y) = ∑x
∑y
xypX(x)pY(y)
= ∑x
xpX(x)∑y
ypY(y)
= E(X)E(Y)
Note. For non-independent random variables, the above neednot apply. In the example with the house,
E(XY) =5
366= 2
9· 11
18= E(X)E(Y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed independentrandom variables, then the expected value of XY is
E(XY) = E(X)E(Y).
Proof (discrete case only).E(XY) = ∑
x∑y
xyp(x,y) = ∑x
∑y
xypX(x)pY(y)
= ∑x
xpX(x)∑y
ypY(y)
= E(X)E(Y)
Note. For non-independent random variables, the above neednot apply. In the example with the house,
E(XY) =5
366= 2
9· 11
18= E(X)E(Y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed independentrandom variables, then the expected value of XY is
E(XY) = E(X)E(Y).
Proof (discrete case only).E(XY) = ∑
x∑y
xyp(x,y) = ∑x
∑y
xypX(x)pY(y)
= ∑x
xpX(x)∑y
ypY(y)
= E(X)E(Y)
Note. For non-independent random variables, the above neednot apply. In the example with the house,
E(XY) =5
366= 2
9· 11
18= E(X)E(Y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed independentrandom variables, then the expected value of XY is
E(XY) = E(X)E(Y).
Proof (discrete case only).
E(XY) = ∑x
∑y
xyp(x,y) = ∑x
∑y
xypX(x)pY(y)
= ∑x
xpX(x)∑y
ypY(y)
= E(X)E(Y)
Note. For non-independent random variables, the above neednot apply. In the example with the house,
E(XY) =5
366= 2
9· 11
18= E(X)E(Y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed independentrandom variables, then the expected value of XY is
E(XY) = E(X)E(Y).
Proof (discrete case only).E(XY)
= ∑x
∑y
xyp(x,y) = ∑x
∑y
xypX(x)pY(y)
= ∑x
xpX(x)∑y
ypY(y)
= E(X)E(Y)
Note. For non-independent random variables, the above neednot apply. In the example with the house,
E(XY) =5
366= 2
9· 11
18= E(X)E(Y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed independentrandom variables, then the expected value of XY is
E(XY) = E(X)E(Y).
Proof (discrete case only).E(XY) = ∑
x∑y
xyp(x,y)
= ∑x
∑y
xypX(x)pY(y)
= ∑x
xpX(x)∑y
ypY(y)
= E(X)E(Y)
Note. For non-independent random variables, the above neednot apply. In the example with the house,
E(XY) =5
366= 2
9· 11
18= E(X)E(Y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed independentrandom variables, then the expected value of XY is
E(XY) = E(X)E(Y).
Proof (discrete case only).E(XY) = ∑
x∑y
xyp(x,y) = ∑x
∑y
xypX(x)pY(y)
= ∑x
xpX(x)∑y
ypY(y)
= E(X)E(Y)
Note. For non-independent random variables, the above neednot apply. In the example with the house,
E(XY) =5
366= 2
9· 11
18= E(X)E(Y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed independentrandom variables, then the expected value of XY is
E(XY) = E(X)E(Y).
Proof (discrete case only).E(XY) = ∑
x∑y
xyp(x,y) = ∑x
∑y
xypX(x)pY(y)
= ∑x
xpX(x)∑y
ypY(y)
= E(X)E(Y)
Note. For non-independent random variables, the above neednot apply. In the example with the house,
E(XY) =5
366= 2
9· 11
18= E(X)E(Y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed independentrandom variables, then the expected value of XY is
E(XY) = E(X)E(Y).
Proof (discrete case only).E(XY) = ∑
x∑y
xyp(x,y) = ∑x
∑y
xypX(x)pY(y)
= ∑x
xpX(x)∑y
ypY(y)
= E(X)E(Y)
Note. For non-independent random variables, the above neednot apply. In the example with the house,
E(XY) =5
366= 2
9· 11
18= E(X)E(Y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed independentrandom variables, then the expected value of XY is
E(XY) = E(X)E(Y).
Proof (discrete case only).E(XY) = ∑
x∑y
xyp(x,y) = ∑x
∑y
xypX(x)pY(y)
= ∑x
xpX(x)∑y
ypY(y)
= E(X)E(Y)
Note. For non-independent random variables, the above neednot apply. In the example with the house,
E(XY) =5
366= 2
9· 11
18= E(X)E(Y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed independentrandom variables, then the expected value of XY is
E(XY) = E(X)E(Y).
Proof (discrete case only).E(XY) = ∑
x∑y
xyp(x,y) = ∑x
∑y
xypX(x)pY(y)
= ∑x
xpX(x)∑y
ypY(y)
= E(X)E(Y)
Note.
For non-independent random variables, the above neednot apply. In the example with the house,
E(XY) =5
366= 2
9· 11
18= E(X)E(Y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed independentrandom variables, then the expected value of XY is
E(XY) = E(X)E(Y).
Proof (discrete case only).E(XY) = ∑
x∑y
xyp(x,y) = ∑x
∑y
xypX(x)pY(y)
= ∑x
xpX(x)∑y
ypY(y)
= E(X)E(Y)
Note. For non-independent random variables, the above neednot apply.
In the example with the house,
E(XY) =5
366= 2
9· 11
18= E(X)E(Y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed independentrandom variables, then the expected value of XY is
E(XY) = E(X)E(Y).
Proof (discrete case only).E(XY) = ∑
x∑y
xyp(x,y) = ∑x
∑y
xypX(x)pY(y)
= ∑x
xpX(x)∑y
ypY(y)
= E(X)E(Y)
Note. For non-independent random variables, the above neednot apply. In the example with the house
,
E(XY) =5
366= 2
9· 11
18= E(X)E(Y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed independentrandom variables, then the expected value of XY is
E(XY) = E(X)E(Y).
Proof (discrete case only).E(XY) = ∑
x∑y
xyp(x,y) = ∑x
∑y
xypX(x)pY(y)
= ∑x
xpX(x)∑y
ypY(y)
= E(X)E(Y)
Note. For non-independent random variables, the above neednot apply. In the example with the house,
E(XY) =5
36
6= 29· 11
18= E(X)E(Y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed independentrandom variables, then the expected value of XY is
E(XY) = E(X)E(Y).
Proof (discrete case only).E(XY) = ∑
x∑y
xyp(x,y) = ∑x
∑y
xypX(x)pY(y)
= ∑x
xpX(x)∑y
ypY(y)
= E(X)E(Y)
Note. For non-independent random variables, the above neednot apply. In the example with the house,
E(XY) =5
366= 2
9· 11
18
= E(X)E(Y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Proposition. If X and Y are jointly distributed independentrandom variables, then the expected value of XY is
E(XY) = E(X)E(Y).
Proof (discrete case only).E(XY) = ∑
x∑y
xyp(x,y) = ∑x
∑y
xypX(x)pY(y)
= ∑x
xpX(x)∑y
ypY(y)
= E(X)E(Y)
Note. For non-independent random variables, the above neednot apply. In the example with the house,
E(XY) =5
366= 2
9· 11
18= E(X)E(Y).
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition.
The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX
+
+
-
-
rrr rr
rrrrr
rr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
r
rr rr
rrrrr
rr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rr
r rr
rrrrr
rr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr
rr
rrrrr
rr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr r
rr
rrrrrr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
r
rrrrrr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rr
rrrrr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrr
rrrr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrr
rrr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
r
r
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r
r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r r
r
rrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r r
rr
rr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r r
rrr
r
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r r
rrrr
positive covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
rrr rr
rrrrr
rr
r r
rrrr
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
r r r rr r
r r rr rrr
r rr
r
r
negative covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The covariance between two jointly distributedrandom variables X and Y with expected values µX and µY isCov(X,Y) = E
((X−µX)(Y−µY)
).
y=µY
x=µX+
+
-
-
r r r rr r
r r rr rrr
r rr
r
r
negative covariance
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem.
Cov(X,Y) = E(XY)−µXµY
Proof.Cov(X,Y) = E
((X−µX)(Y−µY)
)= E(XY−XµY −µXY + µXµY)= E(XY)−E(XµY)−E(µXY)+E(µXµY)= E(XY)−E(X)µY −µXE(Y)+ µXµY
= E(XY)−µXµY −µXµY + µXµY
= E(XY)−µXµY
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. Cov(X,Y)
= E(XY)−µXµY
Proof.Cov(X,Y) = E
((X−µX)(Y−µY)
)= E(XY−XµY −µXY + µXµY)= E(XY)−E(XµY)−E(µXY)+E(µXµY)= E(XY)−E(X)µY −µXE(Y)+ µXµY
= E(XY)−µXµY −µXµY + µXµY
= E(XY)−µXµY
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. Cov(X,Y) = E(XY)−µXµY
Proof.Cov(X,Y) = E
((X−µX)(Y−µY)
)= E(XY−XµY −µXY + µXµY)= E(XY)−E(XµY)−E(µXY)+E(µXµY)= E(XY)−E(X)µY −µXE(Y)+ µXµY
= E(XY)−µXµY −µXµY + µXµY
= E(XY)−µXµY
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. Cov(X,Y) = E(XY)−µXµY
Proof.
Cov(X,Y) = E((X−µX)(Y−µY)
)= E(XY−XµY −µXY + µXµY)= E(XY)−E(XµY)−E(µXY)+E(µXµY)= E(XY)−E(X)µY −µXE(Y)+ µXµY
= E(XY)−µXµY −µXµY + µXµY
= E(XY)−µXµY
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. Cov(X,Y) = E(XY)−µXµY
Proof.Cov(X,Y)
= E((X−µX)(Y−µY)
)= E(XY−XµY −µXY + µXµY)= E(XY)−E(XµY)−E(µXY)+E(µXµY)= E(XY)−E(X)µY −µXE(Y)+ µXµY
= E(XY)−µXµY −µXµY + µXµY
= E(XY)−µXµY
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. Cov(X,Y) = E(XY)−µXµY
Proof.Cov(X,Y) = E
((X−µX)(Y−µY)
)
= E(XY−XµY −µXY + µXµY)= E(XY)−E(XµY)−E(µXY)+E(µXµY)= E(XY)−E(X)µY −µXE(Y)+ µXµY
= E(XY)−µXµY −µXµY + µXµY
= E(XY)−µXµY
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. Cov(X,Y) = E(XY)−µXµY
Proof.Cov(X,Y) = E
((X−µX)(Y−µY)
)= E(XY−XµY −µXY + µXµY)
= E(XY)−E(XµY)−E(µXY)+E(µXµY)= E(XY)−E(X)µY −µXE(Y)+ µXµY
= E(XY)−µXµY −µXµY + µXµY
= E(XY)−µXµY
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. Cov(X,Y) = E(XY)−µXµY
Proof.Cov(X,Y) = E
((X−µX)(Y−µY)
)= E(XY−XµY −µXY + µXµY)= E(XY)−E(XµY)−E(µXY)+E(µXµY)
= E(XY)−E(X)µY −µXE(Y)+ µXµY
= E(XY)−µXµY −µXµY + µXµY
= E(XY)−µXµY
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. Cov(X,Y) = E(XY)−µXµY
Proof.Cov(X,Y) = E
((X−µX)(Y−µY)
)= E(XY−XµY −µXY + µXµY)= E(XY)−E(XµY)−E(µXY)+E(µXµY)= E(XY)−E(X)µY −µXE(Y)+ µXµY
= E(XY)−µXµY −µXµY + µXµY
= E(XY)−µXµY
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. Cov(X,Y) = E(XY)−µXµY
Proof.Cov(X,Y) = E
((X−µX)(Y−µY)
)= E(XY−XµY −µXY + µXµY)= E(XY)−E(XµY)−E(µXY)+E(µXµY)= E(XY)−E(X)µY −µXE(Y)+ µXµY
= E(XY)−µXµY −µXµY + µXµY
= E(XY)−µXµY
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
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Expected Values Covariance Examples Correlation Coefficient
Theorem. Cov(X,Y) = E(XY)−µXµY
Proof.Cov(X,Y) = E
((X−µX)(Y−µY)
)= E(XY−XµY −µXY + µXµY)= E(XY)−E(XµY)−E(µXY)+E(µXµY)= E(XY)−E(X)µY −µXE(Y)+ µXµY
= E(XY)−µXµY −µXµY + µXµY
= E(XY)−µXµY
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
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Expected Values Covariance Examples Correlation Coefficient
Theorem. Cov(X,Y) = E(XY)−µXµY
Proof.Cov(X,Y) = E
((X−µX)(Y−µY)
)= E(XY−XµY −µXY + µXµY)= E(XY)−E(XµY)−E(µXY)+E(µXµY)= E(XY)−E(X)µY −µXE(Y)+ µXµY
= E(XY)−µXµY −µXµY + µXµY
= E(XY)−µXµY
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
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Expected Values Covariance Examples Correlation Coefficient
Example.
A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
First a remark: The probability p(0,1) is not listed, becauseapparently you need to make a call to place an order.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
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Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
First a remark: The probability p(0,1) is not listed, becauseapparently you need to make a call to place an order.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
First a remark:
The probability p(0,1) is not listed, becauseapparently you need to make a call to place an order.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
First a remark: The probability p(0,1) is not listed, becauseapparently you need to make a call to place an order.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1)
= 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04
+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16
+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10
+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20
+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30
+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20
= 1.66E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20
= 0.8E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20
= 1.5Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2)
= 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04
+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16
+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10
+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20
+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30
+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20
= 0.8E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20
= 1.5Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2)
= 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04
+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16
+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10
+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20
+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30
+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20
= 1.5Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2)
= 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8
= 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
E(X1) = 0 ·0.04+1 ·0.16+1 ·0.10+2 ·0.20+2 ·0.30+2 ·0.20= 1.66
E(X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+1 ·0.30+2 ·0.20= 0.8
E(X1X2) = 0 ·0.04+0 ·0.16+1 ·0.10+0 ·0.20+2 ·0.30+4 ·0.20= 1.5
Cov(X1,X2) = 1.5−1.66 ·0.8 = 0.172
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
A positive covariance makes sense: The more calls you receive,the more orders you should receive. But it is not a no-brainer.Beyond a certain call volume, a negative covariance is at leastconceivable: Stressed sales associates may be less efficient percall.In such a situation, a negative covariance between call volumeand customer satisfaction is a reasonable conjecture. (But thatwas not asked.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
A positive covariance makes sense
: The more calls you receive,the more orders you should receive. But it is not a no-brainer.Beyond a certain call volume, a negative covariance is at leastconceivable: Stressed sales associates may be less efficient percall.In such a situation, a negative covariance between call volumeand customer satisfaction is a reasonable conjecture. (But thatwas not asked.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
A positive covariance makes sense: The more calls you receive,the more orders you should receive.
But it is not a no-brainer.Beyond a certain call volume, a negative covariance is at leastconceivable: Stressed sales associates may be less efficient percall.In such a situation, a negative covariance between call volumeand customer satisfaction is a reasonable conjecture. (But thatwas not asked.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
A positive covariance makes sense: The more calls you receive,the more orders you should receive. But it is not a no-brainer.
Beyond a certain call volume, a negative covariance is at leastconceivable: Stressed sales associates may be less efficient percall.In such a situation, a negative covariance between call volumeand customer satisfaction is a reasonable conjecture. (But thatwas not asked.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
A positive covariance makes sense: The more calls you receive,the more orders you should receive. But it is not a no-brainer.Beyond a certain call volume, a negative covariance is at leastconceivable
: Stressed sales associates may be less efficient percall.In such a situation, a negative covariance between call volumeand customer satisfaction is a reasonable conjecture. (But thatwas not asked.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
A positive covariance makes sense: The more calls you receive,the more orders you should receive. But it is not a no-brainer.Beyond a certain call volume, a negative covariance is at leastconceivable: Stressed sales associates may be less efficient percall.
In such a situation, a negative covariance between call volumeand customer satisfaction is a reasonable conjecture. (But thatwas not asked.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
A positive covariance makes sense: The more calls you receive,the more orders you should receive. But it is not a no-brainer.Beyond a certain call volume, a negative covariance is at leastconceivable: Stressed sales associates may be less efficient percall.In such a situation, a negative covariance between call volumeand customer satisfaction is a reasonable conjecture.
(But thatwas not asked.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the covariance andinterpret the result.
A positive covariance makes sense: The more calls you receive,the more orders you should receive. But it is not a no-brainer.Beyond a certain call volume, a negative covariance is at leastconceivable: Stressed sales associates may be less efficient percall.In such a situation, a negative covariance between call volumeand customer satisfaction is a reasonable conjecture. (But thatwas not asked.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem.
If X and Y are independent jointly distributedrandom variables, then Cov(X,Y) = 0.
Proof.Cov(X,Y) = E(XY)−µXµY
= E(X)E(Y)−µXµY
= µXµY −µXµY = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. If X and Y are independent jointly distributedrandom variables, then Cov(X,Y) = 0.
Proof.Cov(X,Y) = E(XY)−µXµY
= E(X)E(Y)−µXµY
= µXµY −µXµY = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. If X and Y are independent jointly distributedrandom variables, then Cov(X,Y) = 0.
Proof.
Cov(X,Y) = E(XY)−µXµY
= E(X)E(Y)−µXµY
= µXµY −µXµY = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. If X and Y are independent jointly distributedrandom variables, then Cov(X,Y) = 0.
Proof.Cov(X,Y)
= E(XY)−µXµY
= E(X)E(Y)−µXµY
= µXµY −µXµY = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. If X and Y are independent jointly distributedrandom variables, then Cov(X,Y) = 0.
Proof.Cov(X,Y) = E(XY)−µXµY
= E(X)E(Y)−µXµY
= µXµY −µXµY = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. If X and Y are independent jointly distributedrandom variables, then Cov(X,Y) = 0.
Proof.Cov(X,Y) = E(XY)−µXµY
= E(X)E(Y)−µXµY
= µXµY −µXµY = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. If X and Y are independent jointly distributedrandom variables, then Cov(X,Y) = 0.
Proof.Cov(X,Y) = E(XY)−µXµY
= E(X)E(Y)−µXµY
= µXµY −µXµY
= 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. If X and Y are independent jointly distributedrandom variables, then Cov(X,Y) = 0.
Proof.Cov(X,Y) = E(XY)−µXµY
= E(X)E(Y)−µXµY
= µXµY −µXµY = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. If X and Y are independent jointly distributedrandom variables, then Cov(X,Y) = 0.
Proof.Cov(X,Y) = E(XY)−µXµY
= E(X)E(Y)−µXµY
= µXµY −µXµY = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Warning.
A covariance of zero does not imply that the randomvariables are independent.Consider the following joint density function for X (across) andY (down).
−1 0 1−1 0.1 0 0.1
0 0.25 0.1 0.251 0.1 0 0.1
1. X and Y are not independent:0 = P(X = 0,Y =−1) = P(X = 0)P(Y =−1) wouldimply P(X = 0) = 0 or P(Y =−1) = 0 and then we wouldhave a row or a column of zeros, which we don’t.
2. µX = µY = 0 (Marginal distributions are symmetric withrespect to zero.)
3. Cov(X,Y) = E(XY) = 0.1−0.1−0.1+0.1 = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Warning. A covariance of zero does not imply that the randomvariables are independent.
Consider the following joint density function for X (across) andY (down).
−1 0 1−1 0.1 0 0.1
0 0.25 0.1 0.251 0.1 0 0.1
1. X and Y are not independent:0 = P(X = 0,Y =−1) = P(X = 0)P(Y =−1) wouldimply P(X = 0) = 0 or P(Y =−1) = 0 and then we wouldhave a row or a column of zeros, which we don’t.
2. µX = µY = 0 (Marginal distributions are symmetric withrespect to zero.)
3. Cov(X,Y) = E(XY) = 0.1−0.1−0.1+0.1 = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Warning. A covariance of zero does not imply that the randomvariables are independent.Consider the following joint density function for X (across) andY (down).
−1 0 1−1 0.1 0 0.1
0 0.25 0.1 0.251 0.1 0 0.1
1. X and Y are not independent:0 = P(X = 0,Y =−1) = P(X = 0)P(Y =−1) wouldimply P(X = 0) = 0 or P(Y =−1) = 0 and then we wouldhave a row or a column of zeros, which we don’t.
2. µX = µY = 0 (Marginal distributions are symmetric withrespect to zero.)
3. Cov(X,Y) = E(XY) = 0.1−0.1−0.1+0.1 = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Warning. A covariance of zero does not imply that the randomvariables are independent.Consider the following joint density function for X (across) andY (down).
−1 0 1−1 0.1 0 0.1
0 0.25 0.1 0.251 0.1 0 0.1
1. X and Y are not independent:
0 = P(X = 0,Y =−1) = P(X = 0)P(Y =−1) wouldimply P(X = 0) = 0 or P(Y =−1) = 0 and then we wouldhave a row or a column of zeros, which we don’t.
2. µX = µY = 0 (Marginal distributions are symmetric withrespect to zero.)
3. Cov(X,Y) = E(XY) = 0.1−0.1−0.1+0.1 = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Warning. A covariance of zero does not imply that the randomvariables are independent.Consider the following joint density function for X (across) andY (down).
−1 0 1−1 0.1 0 0.1
0 0.25 0.1 0.251 0.1 0 0.1
1. X and Y are not independent:0 = P(X = 0,Y =−1)
= P(X = 0)P(Y =−1) wouldimply P(X = 0) = 0 or P(Y =−1) = 0 and then we wouldhave a row or a column of zeros, which we don’t.
2. µX = µY = 0 (Marginal distributions are symmetric withrespect to zero.)
3. Cov(X,Y) = E(XY) = 0.1−0.1−0.1+0.1 = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Warning. A covariance of zero does not imply that the randomvariables are independent.Consider the following joint density function for X (across) andY (down).
−1 0 1−1 0.1 0 0.1
0 0.25 0.1 0.251 0.1 0 0.1
1. X and Y are not independent:0 = P(X = 0,Y =−1) = P(X = 0)P(Y =−1)
wouldimply P(X = 0) = 0 or P(Y =−1) = 0 and then we wouldhave a row or a column of zeros, which we don’t.
2. µX = µY = 0 (Marginal distributions are symmetric withrespect to zero.)
3. Cov(X,Y) = E(XY) = 0.1−0.1−0.1+0.1 = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Warning. A covariance of zero does not imply that the randomvariables are independent.Consider the following joint density function for X (across) andY (down).
−1 0 1−1 0.1 0 0.1
0 0.25 0.1 0.251 0.1 0 0.1
1. X and Y are not independent:0 = P(X = 0,Y =−1) = P(X = 0)P(Y =−1) wouldimply P(X = 0) = 0 or P(Y =−1) = 0
and then we wouldhave a row or a column of zeros, which we don’t.
2. µX = µY = 0 (Marginal distributions are symmetric withrespect to zero.)
3. Cov(X,Y) = E(XY) = 0.1−0.1−0.1+0.1 = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Warning. A covariance of zero does not imply that the randomvariables are independent.Consider the following joint density function for X (across) andY (down).
−1 0 1−1 0.1 0 0.1
0 0.25 0.1 0.251 0.1 0 0.1
1. X and Y are not independent:0 = P(X = 0,Y =−1) = P(X = 0)P(Y =−1) wouldimply P(X = 0) = 0 or P(Y =−1) = 0 and then we wouldhave a row or a column of zeros
, which we don’t.2. µX = µY = 0 (Marginal distributions are symmetric with
respect to zero.)3. Cov(X,Y) = E(XY) = 0.1−0.1−0.1+0.1 = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Warning. A covariance of zero does not imply that the randomvariables are independent.Consider the following joint density function for X (across) andY (down).
−1 0 1−1 0.1 0 0.1
0 0.25 0.1 0.251 0.1 0 0.1
1. X and Y are not independent:0 = P(X = 0,Y =−1) = P(X = 0)P(Y =−1) wouldimply P(X = 0) = 0 or P(Y =−1) = 0 and then we wouldhave a row or a column of zeros, which we don’t.
2. µX = µY = 0 (Marginal distributions are symmetric withrespect to zero.)
3. Cov(X,Y) = E(XY) = 0.1−0.1−0.1+0.1 = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Warning. A covariance of zero does not imply that the randomvariables are independent.Consider the following joint density function for X (across) andY (down).
−1 0 1−1 0.1 0 0.1
0 0.25 0.1 0.251 0.1 0 0.1
1. X and Y are not independent:0 = P(X = 0,Y =−1) = P(X = 0)P(Y =−1) wouldimply P(X = 0) = 0 or P(Y =−1) = 0 and then we wouldhave a row or a column of zeros, which we don’t.
2. µX = µY = 0
(Marginal distributions are symmetric withrespect to zero.)
3. Cov(X,Y) = E(XY) = 0.1−0.1−0.1+0.1 = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Warning. A covariance of zero does not imply that the randomvariables are independent.Consider the following joint density function for X (across) andY (down).
−1 0 1−1 0.1 0 0.1
0 0.25 0.1 0.251 0.1 0 0.1
1. X and Y are not independent:0 = P(X = 0,Y =−1) = P(X = 0)P(Y =−1) wouldimply P(X = 0) = 0 or P(Y =−1) = 0 and then we wouldhave a row or a column of zeros, which we don’t.
2. µX = µY = 0 (Marginal distributions are symmetric withrespect to zero.)
3. Cov(X,Y) = E(XY) = 0.1−0.1−0.1+0.1 = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Warning. A covariance of zero does not imply that the randomvariables are independent.Consider the following joint density function for X (across) andY (down).
−1 0 1−1 0.1 0 0.1
0 0.25 0.1 0.251 0.1 0 0.1
1. X and Y are not independent:0 = P(X = 0,Y =−1) = P(X = 0)P(Y =−1) wouldimply P(X = 0) = 0 or P(Y =−1) = 0 and then we wouldhave a row or a column of zeros, which we don’t.
2. µX = µY = 0 (Marginal distributions are symmetric withrespect to zero.)
3. Cov(X,Y) = E(XY)
= 0.1−0.1−0.1+0.1 = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
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Expected Values Covariance Examples Correlation Coefficient
Warning. A covariance of zero does not imply that the randomvariables are independent.Consider the following joint density function for X (across) andY (down).
−1 0 1−1 0.1 0 0.1
0 0.25 0.1 0.251 0.1 0 0.1
1. X and Y are not independent:0 = P(X = 0,Y =−1) = P(X = 0)P(Y =−1) wouldimply P(X = 0) = 0 or P(Y =−1) = 0 and then we wouldhave a row or a column of zeros, which we don’t.
2. µX = µY = 0 (Marginal distributions are symmetric withrespect to zero.)
3. Cov(X,Y) = E(XY) = 0.1
−0.1−0.1+0.1 = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Warning. A covariance of zero does not imply that the randomvariables are independent.Consider the following joint density function for X (across) andY (down).
−1 0 1−1 0.1 0 0.1
0 0.25 0.1 0.251 0.1 0 0.1
1. X and Y are not independent:0 = P(X = 0,Y =−1) = P(X = 0)P(Y =−1) wouldimply P(X = 0) = 0 or P(Y =−1) = 0 and then we wouldhave a row or a column of zeros, which we don’t.
2. µX = µY = 0 (Marginal distributions are symmetric withrespect to zero.)
3. Cov(X,Y) = E(XY) = 0.1−0.1
−0.1+0.1 = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Warning. A covariance of zero does not imply that the randomvariables are independent.Consider the following joint density function for X (across) andY (down).
−1 0 1−1 0.1 0 0.1
0 0.25 0.1 0.251 0.1 0 0.1
1. X and Y are not independent:0 = P(X = 0,Y =−1) = P(X = 0)P(Y =−1) wouldimply P(X = 0) = 0 or P(Y =−1) = 0 and then we wouldhave a row or a column of zeros, which we don’t.
2. µX = µY = 0 (Marginal distributions are symmetric withrespect to zero.)
3. Cov(X,Y) = E(XY) = 0.1−0.1−0.1
+0.1 = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Warning. A covariance of zero does not imply that the randomvariables are independent.Consider the following joint density function for X (across) andY (down).
−1 0 1−1 0.1 0 0.1
0 0.25 0.1 0.251 0.1 0 0.1
1. X and Y are not independent:0 = P(X = 0,Y =−1) = P(X = 0)P(Y =−1) wouldimply P(X = 0) = 0 or P(Y =−1) = 0 and then we wouldhave a row or a column of zeros, which we don’t.
2. µX = µY = 0 (Marginal distributions are symmetric withrespect to zero.)
3. Cov(X,Y) = E(XY) = 0.1−0.1−0.1+0.1
= 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Warning. A covariance of zero does not imply that the randomvariables are independent.Consider the following joint density function for X (across) andY (down).
−1 0 1−1 0.1 0 0.1
0 0.25 0.1 0.251 0.1 0 0.1
1. X and Y are not independent:0 = P(X = 0,Y =−1) = P(X = 0)P(Y =−1) wouldimply P(X = 0) = 0 or P(Y =−1) = 0 and then we wouldhave a row or a column of zeros, which we don’t.
2. µX = µY = 0 (Marginal distributions are symmetric withrespect to zero.)
3. Cov(X,Y) = E(XY) = 0.1−0.1−0.1+0.1 = 0.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
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Expected Values Covariance Examples Correlation Coefficient
Definition.
The correlation coefficient of the jointlydistributed random variables X and Y is
Corr(X,Y) := ρX,Y :=Cov(X,Y)
σXσY
The covariance can be large if both variables have largevariance, and yet, the correlation between the variables couldstill be “large” or “small.” That is, although the sign of thecovariance tells us something about how the variables correlate,the numerical value of the covariance has limited usefulness.By dividing by the variances, we remove the influence ofvariability in one variable and obtain a standardized measure ofcorrelation.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The correlation coefficient of the jointlydistributed random variables X and Y is
Corr(X,Y) := ρX,Y :=Cov(X,Y)
σXσY
The covariance can be large if both variables have largevariance, and yet, the correlation between the variables couldstill be “large” or “small.” That is, although the sign of thecovariance tells us something about how the variables correlate,the numerical value of the covariance has limited usefulness.By dividing by the variances, we remove the influence ofvariability in one variable and obtain a standardized measure ofcorrelation.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The correlation coefficient of the jointlydistributed random variables X and Y is
Corr(X,Y) := ρX,Y :=Cov(X,Y)
σXσY
The covariance can be large if both variables have largevariance
, and yet, the correlation between the variables couldstill be “large” or “small.” That is, although the sign of thecovariance tells us something about how the variables correlate,the numerical value of the covariance has limited usefulness.By dividing by the variances, we remove the influence ofvariability in one variable and obtain a standardized measure ofcorrelation.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The correlation coefficient of the jointlydistributed random variables X and Y is
Corr(X,Y) := ρX,Y :=Cov(X,Y)
σXσY
The covariance can be large if both variables have largevariance, and yet, the correlation between the variables couldstill be “large” or “small.”
That is, although the sign of thecovariance tells us something about how the variables correlate,the numerical value of the covariance has limited usefulness.By dividing by the variances, we remove the influence ofvariability in one variable and obtain a standardized measure ofcorrelation.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The correlation coefficient of the jointlydistributed random variables X and Y is
Corr(X,Y) := ρX,Y :=Cov(X,Y)
σXσY
The covariance can be large if both variables have largevariance, and yet, the correlation between the variables couldstill be “large” or “small.” That is, although the sign of thecovariance tells us something about how the variables correlate
,the numerical value of the covariance has limited usefulness.By dividing by the variances, we remove the influence ofvariability in one variable and obtain a standardized measure ofcorrelation.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The correlation coefficient of the jointlydistributed random variables X and Y is
Corr(X,Y) := ρX,Y :=Cov(X,Y)
σXσY
The covariance can be large if both variables have largevariance, and yet, the correlation between the variables couldstill be “large” or “small.” That is, although the sign of thecovariance tells us something about how the variables correlate,the numerical value of the covariance has limited usefulness.
By dividing by the variances, we remove the influence ofvariability in one variable and obtain a standardized measure ofcorrelation.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The correlation coefficient of the jointlydistributed random variables X and Y is
Corr(X,Y) := ρX,Y :=Cov(X,Y)
σXσY
The covariance can be large if both variables have largevariance, and yet, the correlation between the variables couldstill be “large” or “small.” That is, although the sign of thecovariance tells us something about how the variables correlate,the numerical value of the covariance has limited usefulness.By dividing by the variances, we remove the influence ofvariability in one variable
and obtain a standardized measure ofcorrelation.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The correlation coefficient of the jointlydistributed random variables X and Y is
Corr(X,Y) := ρX,Y :=Cov(X,Y)
σXσY
The covariance can be large if both variables have largevariance, and yet, the correlation between the variables couldstill be “large” or “small.” That is, although the sign of thecovariance tells us something about how the variables correlate,the numerical value of the covariance has limited usefulness.By dividing by the variances, we remove the influence ofvariability in one variable and obtain a standardized measure ofcorrelation.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The correlation coefficient of the jointlydistributed random variables X and Y is
Corr(X,Y) := ρX,Y :=Cov(X,Y)
σXσY
1. The correlation coefficient is always between −1 and 1.2. Strong correlation: |ρ| ≥ 0.8.3. Moderate correlation: 0.5 < |ρ|< 0.8.4. Weak correlation: |ρ| ≤ 0.5.5. Uncorrelated: ρ = 0. (Remember that this does not mean
“independent.”)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The correlation coefficient of the jointlydistributed random variables X and Y is
Corr(X,Y) := ρX,Y :=Cov(X,Y)
σXσY
1. The correlation coefficient is always between −1 and 1.
2. Strong correlation: |ρ| ≥ 0.8.3. Moderate correlation: 0.5 < |ρ|< 0.8.4. Weak correlation: |ρ| ≤ 0.5.5. Uncorrelated: ρ = 0. (Remember that this does not mean
“independent.”)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The correlation coefficient of the jointlydistributed random variables X and Y is
Corr(X,Y) := ρX,Y :=Cov(X,Y)
σXσY
1. The correlation coefficient is always between −1 and 1.2. Strong correlation: |ρ| ≥ 0.8.
3. Moderate correlation: 0.5 < |ρ|< 0.8.4. Weak correlation: |ρ| ≤ 0.5.5. Uncorrelated: ρ = 0. (Remember that this does not mean
“independent.”)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The correlation coefficient of the jointlydistributed random variables X and Y is
Corr(X,Y) := ρX,Y :=Cov(X,Y)
σXσY
1. The correlation coefficient is always between −1 and 1.2. Strong correlation: |ρ| ≥ 0.8.3. Moderate correlation: 0.5 < |ρ|< 0.8.
4. Weak correlation: |ρ| ≤ 0.5.5. Uncorrelated: ρ = 0. (Remember that this does not mean
“independent.”)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The correlation coefficient of the jointlydistributed random variables X and Y is
Corr(X,Y) := ρX,Y :=Cov(X,Y)
σXσY
1. The correlation coefficient is always between −1 and 1.2. Strong correlation: |ρ| ≥ 0.8.3. Moderate correlation: 0.5 < |ρ|< 0.8.4. Weak correlation: |ρ| ≤ 0.5.
5. Uncorrelated: ρ = 0. (Remember that this does not mean“independent.”)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The correlation coefficient of the jointlydistributed random variables X and Y is
Corr(X,Y) := ρX,Y :=Cov(X,Y)
σXσY
1. The correlation coefficient is always between −1 and 1.2. Strong correlation: |ρ| ≥ 0.8.3. Moderate correlation: 0.5 < |ρ|< 0.8.4. Weak correlation: |ρ| ≤ 0.5.5. Uncorrelated: ρ = 0.
(Remember that this does not mean“independent.”)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Definition. The correlation coefficient of the jointlydistributed random variables X and Y is
Corr(X,Y) := ρX,Y :=Cov(X,Y)
σXσY
1. The correlation coefficient is always between −1 and 1.2. Strong correlation: |ρ| ≥ 0.8.3. Moderate correlation: 0.5 < |ρ|< 0.8.4. Weak correlation: |ρ| ≤ 0.5.5. Uncorrelated: ρ = 0. (Remember that this does not mean
“independent.”)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
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Expected Values Covariance Examples Correlation Coefficient
Example.
A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1) = (0−1.66)2 ·0.04+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10
+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20= 0.3044
V(X2) = (0−0.8)2 ·0.04+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20
= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1) = (0−1.66)2 ·0.04+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10
+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20= 0.3044
V(X2) = (0−0.8)2 ·0.04+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20
= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172
V(X1) = (0−1.66)2 ·0.04+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20
= 0.3044V(X2) = (0−0.8)2 ·0.04+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10
+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1)
= (0−1.66)2 ·0.04+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20
= 0.3044V(X2) = (0−0.8)2 ·0.04+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10
+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1) = (0−1.66)2 ·0.04
+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20
= 0.3044V(X2) = (0−0.8)2 ·0.04+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10
+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1) = (0−1.66)2 ·0.04+(1−1.66)2 ·0.16
+(1−1.66)2 ·0.10+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20
= 0.3044V(X2) = (0−0.8)2 ·0.04+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10
+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1) = (0−1.66)2 ·0.04+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10
+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20= 0.3044
V(X2) = (0−0.8)2 ·0.04+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20
= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1) = (0−1.66)2 ·0.04+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10
+(2−1.66)2 ·0.20
+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20= 0.3044
V(X2) = (0−0.8)2 ·0.04+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20
= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1) = (0−1.66)2 ·0.04+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10
+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30
+(2−1.66)2 ·0.20= 0.3044
V(X2) = (0−0.8)2 ·0.04+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20
= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1) = (0−1.66)2 ·0.04+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10
+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20
= 0.3044V(X2) = (0−0.8)2 ·0.04+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10
+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1) = (0−1.66)2 ·0.04+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10
+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20= 0.3044
V(X2) = (0−0.8)2 ·0.04+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20
= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1) = (0−1.66)2 ·0.04+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10
+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20= 0.3044
V(X2)
= (0−0.8)2 ·0.04+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20
= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
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Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1) = (0−1.66)2 ·0.04+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10
+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20= 0.3044
V(X2) = (0−0.8)2 ·0.04
+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20
= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
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Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1) = (0−1.66)2 ·0.04+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10
+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20= 0.3044
V(X2) = (0−0.8)2 ·0.04+(0−0.8)2 ·0.16
+(1−0.8)2 ·0.10+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20
= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1) = (0−1.66)2 ·0.04+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10
+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20= 0.3044
V(X2) = (0−0.8)2 ·0.04+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10
+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1) = (0−1.66)2 ·0.04+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10
+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20= 0.3044
V(X2) = (0−0.8)2 ·0.04+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10+(0−0.8)2 ·0.20
+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1) = (0−1.66)2 ·0.04+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10
+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20= 0.3044
V(X2) = (0−0.8)2 ·0.04+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30
+(2−0.8)2 ·0.20= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1) = (0−1.66)2 ·0.04+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10
+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20= 0.3044
V(X2) = (0−0.8)2 ·0.04+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20
= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Cov(X1,X2) = 0.172V(X1) = (0−1.66)2 ·0.04+(1−1.66)2 ·0.16+(1−1.66)2 ·0.10
+(2−1.66)2 ·0.20+(2−1.66)2 ·0.30+(2−1.66)2 ·0.20= 0.3044
V(X2) = (0−0.8)2 ·0.04+(0−0.8)2 ·0.16+(1−0.8)2 ·0.10+(0−0.8)2 ·0.20+(1−0.8)2 ·0.30+(2−0.8)2 ·0.20
= 0.56
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Corr(X1,X2) =Cov(X1,X2)√V(X1)
√V(X2)
=0.172√
0.3044√
0.56≈ 0.4166.
(Weak correlation.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Corr(X1,X2)
=Cov(X1,X2)√V(X1)
√V(X2)
=0.172√
0.3044√
0.56≈ 0.4166.
(Weak correlation.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Corr(X1,X2) =Cov(X1,X2)√V(X1)
√V(X2)
=0.172√
0.3044√
0.56≈ 0.4166.
(Weak correlation.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Corr(X1,X2) =Cov(X1,X2)√V(X1)
√V(X2)
=0.172√
0.3044√
0.56
≈ 0.4166.
(Weak correlation.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Corr(X1,X2) =Cov(X1,X2)√V(X1)
√V(X2)
=0.172√
0.3044√
0.56≈ 0.4166.
(Weak correlation.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Example. A company keeps track of the number of calls X1 itreceives per day and the number of orders X2 it receives perday. The joint probability mass function p(x1,x2) isp(0,0) = 0.04, p(1,0) = 0.16, p(1,1) = 0.10, p(2,0) = 0.20,p(2,1) = 0.30, p(2,2) = 0.20. Compute the correlationcoefficient.
Corr(X1,X2) =Cov(X1,X2)√V(X1)
√V(X2)
=0.172√
0.3044√
0.56≈ 0.4166.
(Weak correlation.)
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Observation.
Rounding affects correlation.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Observation. Rounding affects correlation.
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem.
The correlation coefficient of the jointly distributedrandom variables X and Y is 1 or −1 if and only if Y = aX +bfor some a 6= 0.
Proof (one direction only).
Corr(X,aX +b) =Cov(X,aX +b)
σXσaX+b
=E(X(aX +b)
)−E(X)E(aX +b)
σX|a|σX
=aE(X2)+E(bX)−aE(X)E(X)−bE(X)
|a|σ2X
=a[E(X2)− (E(X)
)2]
|a|σ2X
=±1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. The correlation coefficient of the jointly distributedrandom variables X and Y is 1 or −1 if and only if Y = aX +bfor some a 6= 0.
Proof (one direction only).
Corr(X,aX +b) =Cov(X,aX +b)
σXσaX+b
=E(X(aX +b)
)−E(X)E(aX +b)
σX|a|σX
=aE(X2)+E(bX)−aE(X)E(X)−bE(X)
|a|σ2X
=a[E(X2)− (E(X)
)2]
|a|σ2X
=±1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. The correlation coefficient of the jointly distributedrandom variables X and Y is 1 or −1 if and only if Y = aX +bfor some a 6= 0.
Proof (one direction only).
Corr(X,aX +b) =Cov(X,aX +b)
σXσaX+b
=E(X(aX +b)
)−E(X)E(aX +b)
σX|a|σX
=aE(X2)+E(bX)−aE(X)E(X)−bE(X)
|a|σ2X
=a[E(X2)− (E(X)
)2]
|a|σ2X
=±1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. The correlation coefficient of the jointly distributedrandom variables X and Y is 1 or −1 if and only if Y = aX +bfor some a 6= 0.
Proof (one direction only).
Corr(X,aX +b)
=Cov(X,aX +b)
σXσaX+b
=E(X(aX +b)
)−E(X)E(aX +b)
σX|a|σX
=aE(X2)+E(bX)−aE(X)E(X)−bE(X)
|a|σ2X
=a[E(X2)− (E(X)
)2]
|a|σ2X
=±1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. The correlation coefficient of the jointly distributedrandom variables X and Y is 1 or −1 if and only if Y = aX +bfor some a 6= 0.
Proof (one direction only).
Corr(X,aX +b) =Cov(X,aX +b)
σXσaX+b
=E(X(aX +b)
)−E(X)E(aX +b)
σX|a|σX
=aE(X2)+E(bX)−aE(X)E(X)−bE(X)
|a|σ2X
=a[E(X2)− (E(X)
)2]
|a|σ2X
=±1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. The correlation coefficient of the jointly distributedrandom variables X and Y is 1 or −1 if and only if Y = aX +bfor some a 6= 0.
Proof (one direction only).
Corr(X,aX +b) =Cov(X,aX +b)
σXσaX+b
=E(X(aX +b)
)−E(X)E(aX +b)
σX|a|σX
=aE(X2)+E(bX)−aE(X)E(X)−bE(X)
|a|σ2X
=a[E(X2)− (E(X)
)2]
|a|σ2X
=±1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. The correlation coefficient of the jointly distributedrandom variables X and Y is 1 or −1 if and only if Y = aX +bfor some a 6= 0.
Proof (one direction only).
Corr(X,aX +b) =Cov(X,aX +b)
σXσaX+b
=E(X(aX +b)
)−E(X)E(aX +b)
σX|a|σX
=aE(X2)+E(bX)−aE(X)E(X)−bE(X)
|a|σ2X
=a[E(X2)− (E(X)
)2]
|a|σ2X
=±1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. The correlation coefficient of the jointly distributedrandom variables X and Y is 1 or −1 if and only if Y = aX +bfor some a 6= 0.
Proof (one direction only).
Corr(X,aX +b) =Cov(X,aX +b)
σXσaX+b
=E(X(aX +b)
)−E(X)E(aX +b)
σX|a|σX
=aE(X2)+E(bX)−aE(X)E(X)−bE(X)
|a|σ2X
=a[E(X2)− (E(X)
)2]
|a|σ2X
=±1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation
logo1
Expected Values Covariance Examples Correlation Coefficient
Theorem. The correlation coefficient of the jointly distributedrandom variables X and Y is 1 or −1 if and only if Y = aX +bfor some a 6= 0.
Proof (one direction only).
Corr(X,aX +b) =Cov(X,aX +b)
σXσaX+b
=E(X(aX +b)
)−E(X)E(aX +b)
σX|a|σX
=aE(X2)+E(bX)−aE(X)E(X)−bE(X)
|a|σ2X
=a[E(X2)− (E(X)
)2]
|a|σ2X
=±1
Bernd Schroder Louisiana Tech University, College of Engineering and Science
Expected Values, Covariance and Correlation