Covariance of functions of a random variable

**gziv** · June 14th 2011, 12:18 PM

Let X be a random variable in the range (0,1), i.e. P(X)=0 for X<=0 or X>=1. Is there a way to prove that Cov(X,X/(1-X)) > 0 ? Is it true for any pdf f(X) whose support is in (0,1)?

**theodds** · June 14th 2011, 05:42 PM

Wow, this took me a surprisingly long time to show. Yes, it is true, provided that you allow the inequality to be attained (if X is a point mass you get a covariance of 0).

Let $f(x) = \frac x {1 - x}$ . Let $\mu = \mbox{E}[X]$ and $\tau = \mbox{E}[f(X)]$ . First, note that $f(x)$ is convex, and hence by Jensen's Inequality we have $\tau \ge f(\mu)$ .

Next, note that it suffices to show $\mbox{Cov}[1 - X, f(X)] \le 0$ . By definition this is the statement that

$\displaystyle \mbox{E}[(1 - X)f(X)] - \mbox{E}[1 - X]\mbox{E}[f(X)] = \mu - (1 - \mu)\tau \le 0$

and solving for $\tau$ this is the statement $f(\mu) \le \tau$ , which is always true by Jensen's Inequality.

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