# $E(X_i \cdot I(X_j>\theta))$, 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. :)

• @user603 , any thoughts on this? :) Jul 28, 2014 at 11:31

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",

$$E[X_i \cdot I(X_i>\theta)] = \int_{-\infty}^{\infty}f_X(x)[x I(x>\theta)]dx = \int_{\theta}^{\infty}x f_X(x)dx$$

• Many thanks for your interest in this question. The last equality, by the law of the unconscious statistician I already had... I was hoping for something as a function of $\mu$. 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. Jul 28, 2014 at 12:33
• Once you truncate the support, you cannot avoiding dealing with the expected value of the truncated distribution. Obviously, $$E[X_i \cdot I(X_i>\theta)] = \mu-E[X_i \cdot I(X_i\leq\theta)]$$ Jul 28, 2014 at 12:41
• I like the last part of this answer best because it most clearly reveals the basic ideas. The first part is a little too restrictive IMHO because it implicitly assumes $F$ is continuous, which is unnecessary (and was not assumed by the OP). Finally, for the case $i=j$ you can obtain the value $\int_0^\infty(1-F(x))dx$.
– whuber
Jul 28, 2014 at 13:43
• @whuber I was under the impression that this specific expression related to non-negative random variables, no? Jul 28, 2014 at 15:20
• And that's exactly the case when a variable is truncated below at zero :-).
– whuber
Jul 28, 2014 at 17:15