stat 712 fa 2021 lec 9 slides conditional distributions

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


3 Independence of random variables



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.


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4 1/36 2/36 3/36

5 2/36 2/36 4/36

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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).


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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.


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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).


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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).


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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).


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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.


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f (u, v) = 6(v � u)1(0 < u < v < 1).

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


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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].


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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].


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Independence of random variables



3 Independence of random variables


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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 .


II


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.


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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?



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) =

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y

◆⌘y (1 � ⌘)3�y · 1(y 2 {0, 1, 2, 3})

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


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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.


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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).


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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 .



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

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



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f (x , y) =6

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



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f (x , y) =1

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

f (x , y) =1

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�x2 + y2 � 2x + 1

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�for all x , y 2 R.



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�1(0 < x < 1,�1 < y < 1)



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



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f (u, v) = 6(v � u)1(0 < u < v < 1).

Check whether U and V are independent.


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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.


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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.


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Exercise: Let X and Y be independent rvs such that

X ⇠ Beta(3, 4)

Y ⇠ Beta(1, 2)

Find E[XY ].


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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.


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X ⇠ Binomial(n, p)

Y ⇠ Binomial(m, p)

Find the distribution of U = X + Y .


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X ⇠ Normal(µX ,�2X )

Y ⇠ Normal(µY ,�2Y )

Find the distribution of U = X + Y .


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Exercise: Let X1 and X2 be independent rvs such that

X1 ⇠ Poisson(�1)

X2 ⇠ Poisson(�2)

Find the distribution of Y = X1 + X2.


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stat 712 fa 2021 lec 9 slides conditional distributions

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