2. Conditional Expectation

2. Conditional Expectation (9/10/04; cf. Ross) Intro / Definition Examples Conditional Expectation Computing Probabilities by Conditioning 1

2. Conditional Expectation

Intro / Definition Recall conditional probability: Pr(A|B) = Pr(A∩B)/Pr(B) if Pr(B) > 0. Suppose that X and Y are jointly discrete RV’s. Then if Pr(Y = y) > 0, Pr(X = x ∩ Y = y) f (x, y) Pr(X = x|Y = y) = = Pr(Y = y) fY (y) Pr(X = x|Y = 2) defines the probabilities on X given that Y = 2. 2

2. Conditional Expectation

Definition: If fY (y) > 0, then fX|Y (x|y) ≡ ff(x,y) is the (y) Y conditional pmf/pdf of X given Y = y. Remark: Usually just write f (x|y) instead of fX|Y (x|y). Remark: Of course, fY |X (y|x) = f (y|x) = ff (x,y) . (x) X

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2. Conditional Expectation

Old Discrete Example: f (x, y) = Pr(X = x, Y = y).

Y =1 Y =2 Y =3 fX (x)

X = 1 X = 2 X = 3 X = 4 fY (y) .01 .07 .09 .03 .2 .20 .00 .05 .25 .5 .09 .03 .06 .12 .3 .3 .1 .2 .4 1

Find f (x|2).

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2. Conditional Expectation

Then

f (x, 2) f (x, 2) f (x|2) = = = fY (2) 0.5

                  

0.4 if x = 1 0 if x = 2 0.1 if x = 3 0.5 if x = 4 0 otherwise

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2. Conditional Expectation

Old Cts Example: 21 2 x y, f (x, y) = 4

if x2 ≤ y ≤ 1

21 2 fX (x) = x (1 − x4), 8 7 5/2 fY (y) = y , 2

if −1 ≤ x ≤ 1

if 0 ≤ y ≤ 1

Find f (y|X = 1/2). 6

2. Conditional Expectation

, y) f (1 1 2 f (y| ) = 2 fX ( 1 2) = 21 8 ·

21 · 1 y 4 4 1 · (1 − 1 ) , 4 16

32 = y, 15

if 1 4 ≤y≤1

if 1 4 ≤y≤1

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2. Conditional Expectation

More generally, f (x, y) f (y|x) = fX (x) =

21 x2y 4 21 x2(1 − x4) , 8

if x2 ≤ y ≤ 1

2y 2 ≤ y ≤ 1. = if x 1 − x4 Note: 2/(1 − x4) is a constant with respect to y, and we can check to see that f (y|x) is a legit condl pdf: 2y dy = 1. 2 4 x 1−x

Z 1

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2. Conditional Expectation

Typical Problem: Given fX (x) and f (y|x), find fY (y). Steps: (1) f (x, y) = fX (x)f (y|x) (2) fY (y) = < f (x, y) dx. R

Example: fX (x) = 2x, 0 < x < 1. Given X = x, suppose that Y |x ∼ U(0, x). Now find fY (y). 9

2. Conditional Expectation

Solution: Y |x ∼ U(0, x) ⇒ f (y|x) = 1/x, 0 < y < x. So f (x, y) = fX (x)f (y|x) 1 = 2x · , if 0 < x < 1 and 0 < y < x x = 2, if 0 < y < x < 1. Thus, fY (y) =

Z