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)?

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- Jun 14th 2011, 12:18 PMgzivCovariance of functions of a random variable
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)?

- Jun 14th 2011, 05:42 PMtheoddsRe: Covariance of functions of a random variable
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 $\displaystyle f(x) = \frac x {1 - x}$. Let $\displaystyle \mu = \mbox{E}[X]$ and $\displaystyle \tau = \mbox{E}[f(X)]$. First, note that $\displaystyle f(x)$ is convex, and hence by Jensen's Inequality we have $\displaystyle \tau \ge f(\mu)$.

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

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

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