How to find E(X|Y=y) and Var(Y|X=x)?

2016-12-05 5:05 pm
Let (X,Y) have the joint p.d.f, f(x,y)=2 if 0<x,0<y, x+y<1 and f(x,y)=0 for elsewhere.

How to find E(X|Y=y) and Var(Y|X=x)?
Thanks!

回答 (1)

2016-12-05 11:40 pm
✔ 最佳答案
First of all, we compute marginal pdfs.
fx(x) = ∫(y = 0 to 1-x) 2 dy = 2 - 2x for x in (0, 1), and 0 otherwise.

Similarly, fy(y) = 2 - 2y for y in (0, 1), and 0 otherwise.
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Next, we compute the marginal pdfs.
fx|y (x|y) = f(x,y)/fy(y)
..............= 2/(2 - 2y)
..............= 1/(1 - y) for y in (0, 1) (and 0 otherwise).

Similarly, fy|x (y|x) = 1/(1 - x) for x in (0, 1) (and 0 otherwise).
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Hence,
E(X|Y = y) = ∫ x fx|y (x|y) dx
..................= ∫(x = 0 to 1-y) x * (1/(1 - y)) dx
..................= (1/2)(1 - y) for y in (0, 1), and 0 otherwise.

Similarly,
E(Y|X = x) = ∫ y fy|x (y|x) dy
..................= ∫(y = 0 to 1-y) y * (1/(1 - x)) dy
..................= (1/2)(1 - x) for x in (0, 1), and 0 otherwise.

E(Y^2|X = x) = ∫ y^2 fy|x (y|x) dy
..................= ∫(y = 0 to 1-c) y^2 * (1/(1 - x)) dy
..................= (1/3)(1 - x)^2 for x in (0, 1), and 0 otherwise.

Thus, Var(Y|X = x) = (1/3)(1 - x)^2 - [(1/2)(1 - x)]^2 = (1/12)(1 - x)^2 for x in (0, 1), and 0 otherwise.
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I hope this helps!


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