section 4.4 suppose x and y are two random variables such that (1) (2) (3)
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Section 4.4 Suppose X and Y are two random variables such that (1) (2) (3). E( Y | X = x ) is a linear function of x ,. Var( Y | X = x ) is a constant for each x. Y | X = x has a normal distribution for any real number x. Then we know that (1) (2). - PowerPoint PPT PresentationTRANSCRIPT
Section 4.4
Suppose X and Y are two random variables such that
(1)
(2)
(3) Y | X = x has a normal distribution for any real number x.
E(Y | X = x) is a linear function of x,
Var(Y | X = x) is a constant for each x.
Then we know that
(1)
(2)
E(Y | X = x) =Y
Y + — (x – X)X
(because when a conditional mean is a linear function, that linear function must be the least squares line).
Var(Y | X = x) =
–
2Y|x f1(x) dx =
Var(Y | X = x) =
–
2Y|x f1(x) dx =
–
y – h(y | x) dy f1(x) dx =
–
Y
Y + — (x – X)X
2
–
f(x,y) dy dx =
–
Y
(y – Y) – — (x – X) X
2
E Y
(Y – Y) – — (X – X) = X
2
E Y
(Y – Y) – — (X – X) = X
2
E Y 2
Y
(Y – Y)2 – 2 — (X – X)(Y – Y) + 2 — (X – X)2 = X 2
X
Y 2Y
E[(Y – Y)2] – 2 — E[(X – X)(Y – Y)] + 2 — E[(X – X)2] = X 2
X
2Y –
Y2 — Cov(X,Y) + X
22Y = 2
Y – 222Y + 22
Y = 2Y (1 – 2)
Suppose X and Y are two random variables such that
(1)
(2)
(3)
Then we know that
(1)
(2)
E(Y | X = x) =Y
Y + — (x – X)X
Var(Y | X = x) = 2Y (1 – 2)
h(y | x) =
Y | X = x has a normal distribution for any real number x,
E(Y | X = x) is a linear function of x,
Var(Y | X = x) is a constant for each x.
exp
Y
– y – Y + — (x – X)X
2
22Y (1 – 2)
(2)1/2Y (1 – 2)1/2
– < y <
(3) For all x,
Now, suppose that X has a normal distribution. Then, the joint p.d.f. of (X,Y) is f(x,y) = f1(x) h(y | x) =
exp– (x – X)2
22X
(2)1/2X
exp
Y
– y – Y + — (x – X)X
2
22Y (1 – 2)
(2)1/2Y (1 – 2)1/2
exp
Y 2Y
(y – Y)2 – 2 — (x – X)(y – Y) + 2 — (x – X)2
X 2X
22Y (1 – 2)
1 x – X– — ——— – 2 X
2
2X Y (1 – 2)1/2
=
=
exp
Y 2Y
(y – Y)2 – 2 — (x – X)(y – Y) + 2 — (x – X)2
X 2X
22Y (1 – 2)
1 x – X– — ——— – 2 X
2
2X Y (1 – 2)1/2=
exp
y – Y x – X y – Y x – X ——— – 2 ——— ——— + 2 ——— Y X Y X
2 (1 – 2)
1 x – X– — ——— – 2 X
2
2X Y (1 – 2)1/2=
2 2
exp
y – Y x – X y – Y x – X ——— – 2 ——— ——— + 2 ——— Y X Y X
2 (1 – 2)
1 x – X– — ——— – 2 X
2
2X Y (1 – 2)1/2=
2 2
exp
2X Y (1 – 2)1/2
2 (1 – 2)
2 x – X——— – X
x – X y – Y y – Y 2 ——— ——— + ——— X Y Y
2
This is called a bivariate normal p.d.f., and (X, Y) are said to have a bivariate normal distribution with correlation coefficient .
(Note: Completing this derivation is Text Exercise 4.4-2.)
– < x <
– < y <
exp
2X Y (1 – 2)1/2
2 (1 – 2)
2 x – X——— – X
x – X y – Y y – Y 2 ——— ——— + ——— X Y Y
2
– < x <
– < y <
From our derivation of the bivariate normal p.d.f., we know that
(1) X has a N( X , 2X ) distribution,
(2) Y | X = x has a N( , ) distribution.Y
Y + — (x – X)X
2Y (1 – 2)
From the symmetry of the bivariate normal p.d.f., we find that
(1) Y has a distribution,
(2) X | Y = y has a N( , ) distribution.X
X + — (y – Y)Y
2X (1 – 2)
N( Y , 2Y )
Important Theorem in the Text:
If (X, Y) have a bivariate normal distribution with correlation coefficient , then X and Y are independent if and only if = 0.
Theorem 4.4-1
1.
(a)
Random variables X and Y are respectively the height (inches) and weight (lbs.) of adult males in a particular population. It is known that X and Y have a bivariate normal distribution with
X = 70, Y = 150, X2 = 9, Y
2 = 225, and = +0.4 .
Find each of the following:
P(140 < Y < 170)
Y has a distribution.N( , )150 225
P(140 < Y < 170) = 140 – 150 Y – 150 170 – 150P( ———— < ——— < ———— ) = 15 15 15
P(– 0.67 < Z < 1.33) = (1.33) – (– 0.67) =
(1.33) – (1 – (0.67)) = 0. 9082 – (1 – 0.7486) = 0.6568
1.-continued
(b) P(X > 65)
X has a distribution.N( , ) 70 9
P(X > 65) = X – 70 65 – 70P( ——— > ———) = 3 3
P(Z > –1.67) =
1 – (– 1.67) = 1 – (1 – (1.67)) = 1 – (1 – 0.9525) = 0.9525
(c)
(d)
E(Y | X = x)
Var(Y | X = x)
=Y
Y + — (x – X) =X
15150 + 0.4 — (x – 70) =
310 + 2x
= 2Y (1 – 2) = 225(1 – 0.42) = 189
1.-continued
(e) P(140 < Y < 170 | X = 72)
Y | X = 72 has a distribution.N( , )154 189
P(140 < Y < 170 | X = 72 ) = 140 – 154 Y – 154 170 – 154P( ———— < ——— < ———— ) = 189 189 189
P(– 1.02 < Z < 1.16) = (1.16) – (– 1.02) =
(1.16) – (1 – (1.02)) = 0. 8770 – (1 – 0.8461) = 0.7231
Y | X = 68 has a distribution.N( , )146 189
P(140 < Y < 170 | X = 68 ) = 140 – 146 Y – 146 170 – 146P( ———— < ——— < ———— ) = 189 189 189
P(– 0.44 < Z < 1.75) = (1.75) – (– 0.44) =
(1.75) – (1 – (0.44)) = 0. 9599 – (1 – 0.6700) = 0.6299
(f) P(140 < Y < 170 | X = 68)
1.-continued
(g) P(140 < Y < 170 | X = 70)
Y | X = 70 has a distribution.N( , )150 189
P(140 < Y < 170 | X = 70 ) = 140 – 150 Y – 150 170 – 150P( ———— < ——— < ———— ) = 189 189 189
P(– 0.73 < Z < 1.45) = (1.45) – (– 0.73) =
(1.45) – (1 – (0.73)) = 0.9265 – (1 – 0.7673) = 0.6938
(h)
(i)
E(X | Y = y)
Var(X | Y = y)
=X
X + — (y – Y) =Y
370 + 0.4 — (y – 150) =
1558 + 0.08y
= 2X (1 – 2) = 9(1 – 0.42) = 7.56
1.-continued
(j) P(X > 65 | Y = 140)
X | Y = 140 has a distribution.N( , )69.2 7.56
P(X > 65 | Y = 140 ) = X – 69.2 65 – 69.2P( ———— > ———— ) = 7.56 7.56
P(Z > – 1.53) = 1 – (– 1.53) = 1 – (1 – (1.53)) =
1 – (1 – 0.9370) = 0.9370
2.
(a)
(b)
(c)
(d)
(e)
(f)
The random variables X and Y have a bivariate normal distribution with X = 10, Y = 5, X
2 = 16, Y2 = 36, and = –0.4 .
Find each of the following:
E(X + Y) = 10 + 5 = 15
Var(X + Y) = 16 + 36 + 2(–0.4)(16)1/2(36)1/2 = 32.8
E(X – Y) = 10 – 5 = 5
Var(X – Y) = 16 + 36 – 2(–0.4)(16)1/2(36)1/2 = 71.2
E(X – 3Y) = 10 – 3(5) = – 5
Var(X – 3Y) =16 + (–3)2(36) + (2)(–3)(–0.4)(16)1/2(36)1/2 = 397.6
Complete #2 for homework!
3.
(a)
(b)
Suppose the random variables X and Y have joint p.d.f.
2e + xy e
– (x2 + y2)/2 1 – (x2 + y2)/2
f(x,y) = – < x < if – < y <
Are X and Y independent? Why or why not?
Since f(x,y) cannot be factored into the product of a function of x alone and a function of y alone, then X and Y cannot possibly be independent.
f1(x) =
–
dy =2
e + xy e– (x2 + y2)/2 1 – (x2 + y2)/2
Find the marginal p.d.f. of X and the marginal p.d.f. of Y.
–
2dy +
e– (x2 + y2)/2
–
2dy
xy e1 – (x2 + y2)/2
This is a bivariate normal p.d.f. with
X = , 2X = , Y = , 2
Y = , and = .This is an odd function, which implies that the integral of this function is .Consequently, f1(x) is a N(0,1) p.d.f., and
similarly we see that f2(y) is a N(0,1) p.d.f.
0 1 0 1 0
Decide whether each of the following statements is true or false: (c)
If (X, Y) have a bivariate normal distribution, then the marginal distribution for each of X and Y must be a normal distribution.
If the marginal distribution for each of X and Y is a normal distribution, then (X, Y) must have a bivariate normal distribution.
False
True
0