$E(X_i \cdot I(X_j>\theta))$, where theta is the median? 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...I'm looking for an expression in function of $\mu$ if possible, or something similarly friendly. 
Any help would be appreciated. 
Edit: This question came up as I was trying to find the $Cov\left(\sum^n_{i=1}X_i,\sum^n_{i=1}sgn(X_i-\theta)\right)$, where $sgn(u)=1$ if $u>0$, $sgn(u)=-1$ if $u<0$, and zero otherwise. If you know the answer, please feel free to share your knowledge with me. :)
 A: If $i\neq j$, you are right. Formally, the expected value here is taken with respect to the joint distribution of $X_i$ and $X_j$,
$$ E[X_i \cdot I(X_j>\theta)]  = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f_{ij}(x_i,x_j)x_i\cdot I(x_j>\theta)dx_idx_j$$
where due to independence
$$=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f_{i}(x_i)f_{j}(x_j)x_i\cdot I(x_j>\theta)dx_idx_j$$
$$=\int_{-\infty}^{\infty}f_{j}(x_j) I(x_j>\theta)\int_{-\infty}^{\infty}f_{i}(x_i)x_idx_idx_j$$
$$=\int_{-\infty}^{\infty}f_{j}(x_j) I(x_j>\theta)E(X_i)dx_j$$
$$=E(X_i)\int_{-\infty}^{\infty}f_{j}(x_j) I(x_j>\theta)dx_j = E(X_i)\int_{\theta}^{\infty}f_{j}(x_j) dx_j$$
$$=E(X_i)\cdot[1-F_X(\theta)] = \frac 12E(X_i)$$
If $i=j$, by the Law of Total Expectation
$$E[X_i \cdot I(X_i>\theta)] = E[X_i \cdot I(X_i>\theta)\mid X_i>\theta]P( X_i>\theta) + E[X_i \cdot I(X_i>\theta)\mid X_i\leq \theta]P( X_i\leq\theta)$$
$$=E(X_i \cdot 1\mid X_i>\theta)\frac 12 + E(X_i \cdot 0\mid X_i\leq \theta)\frac 12$$
$$=\frac 12E(X_i \mid X_i>\theta)=\frac 12\int_{\theta}^{\infty}\frac {x f_X(x)}{1-F_X(\theta)}dx = \int_{\theta}^{\infty}x f_X(x)dx$$
i.e. here we have "half the value of the truncated from below at $\theta$" distribution, or, the expected value of the "restricted from below at $\theta$" distribution.
One could also obtain this immediately by using the so-called "Law of the Unconscious Statistician", 
$$E[X_i \cdot I(X_i>\theta)] = \int_{-\infty}^{\infty}f_X(x)[x I(x>\theta)]dx = \int_{\theta}^{\infty}x f_X(x)dx$$
