stat 712 fa 2021 lec 9 slides conditional distributions

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STAT 712 fa 2021 Lec 9 slides

Conditional distributions, independence

Karl B. Gregory

University of South Carolina

These slides are an instructional aid; their sole purpose is to display, during the lecture,definitions, plots, results, etc. which take too much time to write by hand on the blackboard.

They are not intended to explain or expound on any material.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 1 / 39

Conditional pmfs and pdfs

1 Conditional pmfs and pdfs

2 Conditional pmfs and pdfs

3 Independence of random variables

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 2 / 39

Conditional pmfs and pdfs

Conditional probability mass functionsLet (X ,Y ) be discrete rvs with joint pmf p and marginal pmfs pX and pY .

For any y such that pY (y) > 0, the conditional pmf of X |Y = y is

p(x |y) = p(x , y)

pY (y), x 2 R.

Likewise the conditional pmf of Y |X = x .

Exercise: Show that p(x |y) is a valid pmf.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 3 / 39

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Conditional pmfs and pdfs

Let X = sum of two dice rolls, Y = max.

Y

X

1 2 3 4 5 6 pX (x)2 1/36 1/36

3 2/36 2/36

4 1/36 2/36 3/36

5 2/36 2/36 4/36

6 1/36 2/36 2/36 5/36

7 2/36 2/36 2/36 6/36

8 1/36 2/36 2/36 5/36

9 2/36 2/36 4/36

10 1/36 2/36 3/36

11 2/36 2/36

12 1/36 1/36

pY (y) 1/36 3/36 5/36 7/36 9/36 11/36

Exercise: Tabulate p(x |y = 4) and p(y |x = 7).

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 4 / 39

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Conditional pmfs and pdfs

Conditional probability density functionsLet (X ,Y ) be continuous rvs with joint pdf f and marginal pdfs fX and fY .

For any y such that fY (y) > 0, the conditional pdf of X |Y = y is

f (x |y) = f (x , y)

fY (y), x 2 R.

Likewise the conditional pdf of Y |X = x .

Exercise: Show that f (x |y) is a valid pdf.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 5 / 39

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Conditional pmfs and pdfs

Exercise: Let (X ,Y ) ⇠ f (x , y) = y(1 � x)y�1e�y1(0 < x < 1, 0 < y < 1).

1 Find the pdf f (x |y) of X given Y = y .

2 Find P(X < 1/2|Y = 2).

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 6 / 39

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Conditional pmfs and pdfs

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Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 7 / 39

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Conditional pmfs and pdfs

Exercise: Let (X ,Y ) have joint pdf given by

f (x , y) =1

2⇡x�3/2 exp

� 1

2x(y2 + 1)

�1(0 < x < 1,�1 < y < 1).

1 Find the conditional pdf f (y |x) of Y given X = x .

2 Find P(Y > 1|X = 4).

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 8 / 39

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Conditional pmfs and pdfs

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Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 9 / 39

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Conditional pmfs and pdfs

Exercise: Let (U,V ) be a pair of rvs with joint pdf given by

f (u, v) = 6(v � u)1(0 < u < v < 1).

1 Find the conditional pdf f (u|v) of U given V = v .

2 Find P(U < 1/4|V = 1/2).

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 10 / 39

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Conditional pmfs and pdfs

Conditional expectationLet (X ,Y ) be discrete or continuous rvs on X and Y and g : R2 ! R.

For any y 2 Y, the conditional expectation of g(X ) given Y = y is

E[g(X )|Y = y ] =

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x2Xg(x) · p(x |y) for (X ,Y ) discrete

Z

Rg(x) · f (x |y) for (X ,Y ) continuous

The conditional expectation E[g(Y )|X = x ] is likewise defined.

Note that E[g(X )|Y = y ] is a function of y , since X is summed/integrated out.

We often have g(x) = x , so that we consider E[X |Y = y ].

If we write E[X |Y ], without specifying a value y for Y , then E[X |Y ] is a rv.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 12 / 39

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Conditional pmfs and pdfs

Exercise: Let (U,V ) be a pair of rvs with joint pdf given by

f (u, v) = 6(v � u)1(0 < u < v < 1).

1 Find E[U|V = v ].2 Find E[U|V = 1/2].

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 13 / 39

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Conditional pmfs and pdfs

Conditional varianceThe conditional variance of X given that Y = y is

Var[X |Y = y ] = E[(X � E[X |Y = y ])2|Y = y ].

If we write Var[X |Y ], without specifying a value y for Y , then Var[X |Y ] is a rv.

Useful expression: Var[X |Y ] = E[X 2|Y ]� (E[X |Y ])2

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 14 / 39

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Conditional pmfs and pdfs

Exercise: Let (X ,Y ) ⇠ f (x , y) = y(1 � x)y�1e�y1(0 < x < 1, 0 < y < 1).

1 Find E[X |Y = y ].2 Find Var[X |Y = y ].3 Find Var[X |Y = 2].

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 15 / 39

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Conditional pmfs and pdfs

Exercise: Let (X ,Y ) have joint pdf given by

f (x , y) =1

2⇡x�3/2 exp

� 1

2x(y2 + 1)

�1(0 < x < 1,�1 < y < 1).

1 Find E[Y |X = x ].2 Find Var[Y |X = x ].3 Find Var[Y |X = 2].

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 16 / 39

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Independence of random variables

1 Conditional pmfs and pdfs

2 Conditional pmfs and pdfs

3 Independence of random variables

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 17 / 39

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Independence of random variables

On (S ,B,P), independence of events A,B 2 B is the property that

P(A \ B) = P(A)P(B).

Independence of random variablesTwo rvs X and Y are independent if

P(X 2 A,Y 2 B) = P(X 2 A)P(Y 2 B) for all A,B 2 B(R).

We sometimes write X ?? Y to express that X and Y are independent.

Basically: If all pairs of events concerning X and Y are independent, then X ?? Y .

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 18 / 39

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Independence of random variables

Theorem (Independence iff joint is product of marginals)

If (X ,Y ) discrete with joint pmf p and marginal pmfs pX and pY , then

X ?? Y () p(x , y) = pX (x)pY (y) for all (x , y) 2 R2.

If (X ,Y ) continuous with joint pdf f and marginal pdfs fX and fY , then

X ?? Y () f (x , y) = fX (x)fY (y) for all (x , y) 2 R2.

Check independence by checking whether the joint is the product of the marginals.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 19 / 39

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Independence of random variables

Exercise: Suppose there are six chairs in a circle numbered 1, . . . , 6. Then:

1 Let X be the roll of a die and sit in chair X .

2 Roll die again and move that many chairs clockwise.

3 Let Y be the number of the chair in which you now sit.

Are X and Y independent?

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 20 / 39

Independence of random variables

Exercise: Let X and Y be independent rvs with marginal pmfs given by

px(x) = px(1 � p)1�x · 1(x 2 {0, 1})

pY (y) =

✓3

y

◆⌘y (1 � ⌘)3�y · 1(y 2 {0, 1, 2, 3})

Give the joint pmf of the rv pair (X ,Y ).

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 21 / 39

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Independence of random variables

Exercise: Let (X ,Y ) be a pair of rvs with joint pdf given by

f (x , y) =6

5[1 � (x � y)2] · 1(0 < x < 1, 0 < y < 1).

Check whether X and Y are independent.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 22 / 39

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Independence of random variables

Exercise: Let X1 and X2 be independent rvs with the Exponential(1) distribution.

1 Give the joint pdf of (X1,X2)2 Find P(X2 < X1 < 2X2).

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 23 / 39

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Independence of random variables

Theorem (Easier independence check)Let (X ,Y ) be discrete or continuous rvs with joint pmf p or joint pdf f .

Then X ?? Y () there exist functions g and h such that

p(x , y) = g(x)h(y) for all (x , y) 2 R2 or

f (x , y) = g(x)h(y) for all (x , y) 2 R2, respectively.

Check if the joint is the product of a function of just x and a function of just y .

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 24 / 39

Independence of random variables

Exercise: Let (X ,Y ) have the joint pdf

f (x , y) = 48xy(y � xy) · 1(0 < x < 1, 0 < y < 1).

Check whether X and Y are independent.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 25 / 39

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Independence of random variables

Exercise: Let (X ,Y ) be a pair of rvs with joint pdf given by

f (x , y) =6

5[1 � (x � y)2] · 1(0 < x < 1, 0 < y < 1).

Check whether X and Y are independent.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 26 / 39

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Independence of random variables

Exercise: Let (X ,Y ) be a pair of rvs with joint pdf given by

f (x , y) =1

8(x + y) · 1(0 < x < 2, 0 < y < 2).

Check whether X and Y are independent.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 27 / 39

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Independence of random variables

Exercise: Let (X ,Y ) have the joint pdf

f (x , y) =1

4⇡exp

�x2 + y2 � 2x + 1

4

�for all x , y 2 R.

Check whether X and Y are independent.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 28 / 39

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Independence of random variables

Exercise: Let (X ,Y ) be a pair of rvs with joint pdf given by

f (x , y) = y(1 � x)y�1e�y1(0 < x < 1, 0 < y < 1)

Check whether X and Y are independent.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 29 / 39

Independence of random variables

Exercise: Let (X ,Y ) be a pair of rvs with joint pdf given by

f (x , y) =1

2⇡x�3/2 exp

� 1

2x(y2 + 1)

�1(0 < x < 1,�1 < y < 1)

Check whether X and Y are independent.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 30 / 39

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Independence of random variables

Exercise: Let (X ,Y ) be a pair of rvs with joint pdf given by

f (x , y) = 4xy · 1(0 < x < 1, 0 < y < 1).

Check whether X and Y are independent.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 31 / 39

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Independence of random variables

Exercise: Let (U,V ) be a pair of rvs with joint pdf given by

f (u, v) = 6(v � u)1(0 < u < v < 1).

Check whether U and V are independent.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 32 / 39

9

Independence of random variables

Theorem (Expectation of the product of independent rvs)Let X and Y be independent rvs. Then

EXY = EXEY .

Moreover, for any functions g : R ! R and h : R ! R,

Eg(X )h(Y ) = Eg(X )Eh(Y ).

Exercise: Prove the result.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 33 / 39

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Independence of random variables

Exercise: Let X and Y be independent rvs with marginal pdfs

fX (x) = 2e�2x1(x > 0)

fY (y) = e�2|y |1(�1 < y < 1)

Find EXY 2.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 34 / 39

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Independence of random variables

Exercise: Let X and Y be independent rvs such that

X ⇠ Beta(3, 4)

Y ⇠ Beta(1, 2)

Find E[XY ].

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 35 / 39

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Independence of random variables

Theorem (mgf of a sum of independent random variables)If X and Y are indep. rvs with mgfs MX and MY , the mgf of V = X + Y is

MV (t) = MX (t)MY (t)

for all t at which MX (t) and MY (t) are defined.

Exercise: Prove the result.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 36 / 39

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Independence of random variables

Exercise: Let X and Y be independent rvs such that

X ⇠ Binomial(n, p)

Y ⇠ Binomial(m, p)

Find the distribution of U = X + Y .

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 37 / 39

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Independence of random variables

Exercise: Let X and Y be independent rvs such that

X ⇠ Normal(µX ,�2X )

Y ⇠ Normal(µY ,�2Y )

Find the distribution of U = X + Y .

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 38 / 39

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Independence of random variables

Exercise: Let X1 and X2 be independent rvs such that

X1 ⇠ Poisson(�1)

X2 ⇠ Poisson(�2)

Find the distribution of Y = X1 + X2.

Karl B. Gregory (U. of South Carolina) STAT 712 fa 2021 Lec 9 slides 39 / 39

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