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X and Y are independent random variables. f is a function with E|f(X,Y)| < ∞. show E(f(X,Y)|X) = E(f(X,Y))

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a bit too homeworky? –  Mitch Mar 28 '11 at 23:57

2 Answers 2

up vote 3 down vote accepted

I think you want to prove $E(f(X,Y)|X=x) = E(f(x,Y))$. This follows immediately from the independence of $X$ and $Y$.

Let $f_{X,Y}(x,y)$ be the joint probability density function for $X$ and $Y$. Since $X$ and $Y$ are independent, we have $f_{XY}(x,y) = f_X(x) \times f_Y(y)$. And hence $f_{X,Y|X}(y) = f_Y(y)$.

Therefore, $$E(f(X,Y)|X=x) = \int f(x,y) f_{X,Y|X}(y) dy = \int f(x,y) f_{Y}(y) dy$$

$$E(f(x,Y)) = \int f(x,y) f_{Y}(y) dy$$

Hence, $$E(f(X,Y)|X=x) = E(f(x,Y))$$ whenever $X$ and $Y$ are independent.

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This is false. Take $X$ and $Y$ to be any nonconstant random variables with finite means and $f(x,y) = x+y$. Then $E(f(X,Y) | X) = X + EY$ and $E(f(X,Y)) = EX + EY$.

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