Let $x_1$ and $x_2$ be IID with mean $\mu$. (a) Show that E($X_1$| $X_1+X_2$) =$\dfrac {(X_1 +X_2)}{2}$.

Question

Let $x_1$ and $x_2$ be IID with mean $\mu$.

(a) Show that E($X_1$| $X_1+X_2$) =$\dfrac {(X_1 +X_2)}{2}$.

(b) Show that in general E($X_1$| $X_1X_2$)$\neq$ $\sqrt {X_1X_2}$

Answer 1

(a) By symmetry $$\alpha = E(X_1|X_1+X_2)=E(X_2|X_1+X_2).$$

$$2\alpha = E(X_1|X_1+X_2)+ E(X_2|X_1+X_2) = E(X_1+X_2|X_1+X_2) = X_1+X_2,$$

$$\alpha = \frac{1}{2} (X_1+X_2).$$

(b) Again, invoking symmetry, we want to show that

$$E(X_1|X_1X_2) E(X_2|X_1X_2) \ne E(X_1X_2|X_1 X_2)=X_1X_2.$$

Well, since we are showing an inequality, a counterexample should be sufficient.

Let $X_1$ and $X_2$ be independent standard normal random variables (mean $\mu=0$, variance $\sigma^2 =1$). If the product $X_1X_2$ is positive, are we expecting $X_1$ and $X_2$ to both be positive. Or are they both negative? The conditional distribution of $X_1$ may no longer be gaussian, but it will be symmetric, and thus $E(X_1|X_1X_2) = 0 \ne X_1X_2$ (unless $X_1X_2=0$).