$E(X_i \cdot I(X_i>\theta)$ expected value of when X is greater than the median.

Question

Let $X_1, ..., X_n$ be iid with a distribution F.

Let $\theta$ be the median of F.

What is the value of $E(X_i \cdot I(X_j>\theta))$?

If $i\neq j$, then $E(X_i \cdot I(X_j>\theta))= 1/2 \cdot \mu$, right?

When $i=j$, I don't seem to find it...

Any help would be appreciated.

Answer 1

Break it up into two using conditional expectation:

$E(X_i \cdot I(X_i>\theta))=P(X_i\leq\theta)E(X_i \cdot I(X_i>\theta)|X_i\leq\theta)+P(X_i>\theta)E(X_i \cdot I(X_i>\theta)|X_i>\theta)=\frac{1}{2}[0+ E[X_i|X_i>\theta]]=\frac{E[X_i|X_i>\theta]}{2}$

That's as far as you can go without knowing the specific distribution.