Conditional expectation of book Shiryaev page 233

Possible Duplicate:
Help with conditional expectation question

I have problem with exercise, I didn’t solve.

Let $X$ and $Y$ be i.i.d. random variables with $E(X)$ defined. Show that

$$E(X|X+Y)=E(Y|X+Y)= \frac{X+Y}{2}$$ (a.s.)

Thanks very much for your help.

Solutions Collecting From Web of "Conditional expectation of book Shiryaev page 233"

The first part of the equation:

$$E(X|X+Y) = E(Y|X+Y)$$

is true by symmetry. They are independent and identical.

Now, what is $E(X|X+Y)+E(Y|X+Y)$?

Hint:

  • Using the linearity of conditional expectation (and an other property), show that $E(X\mid X+Y)+E(Y\mid X+Y)=X+Y$.
  • Show that $E(X\mid X+Y)=E(Y\mid X+Y)$ by the following argument. Take $B$ a set in the $\sigma$-algebra generated by $X+Y$ ($B=(X+Y)^{-1}(B’)$ for some $B’$), then write
    $$\int_B X \, dP=\int_{\Bbb R}\int_{\mathbb R}x\chi_{x+y\in B’} \, dP_X(x)\, dP_Y(y),$$
    then use independence and a substitution.