# Lower Bound on $E[\frac{1}{X}]$ for positive symmetric distribution

Let $X$ be positive random variable and its distribution is symmetric about its mean value $m$. Then $$E\left[\frac{1}{X}\right] \geq \frac{1}{m} + \frac{\sigma^2}{m^3},$$ where $\sigma^2$ is variance of $X$. I can just prove that $$E\left[\frac{1}{X}\right] \geq \frac{1}{m},$$ using Jensen, but somehow can't incorporate symmetry and get also the second term.

• Do you have a reference for that inequality? Well, probably not ... a positive random variable can be symmetric only if it is bounded, so you can as well assume boundedness too. – kjetil b halvorsen Feb 4 '18 at 13:14
• Unfortunately not, but it is obvious that $X\leq2m,$ but still how to incorporate variance? – Ethan Feb 4 '18 at 14:10

Let's try the usual preliminaries: simplify by choosing appropriate units of measurement and exploiting the symmetry assumption.

### Reframing the question

Change the units of $X$ so that its mean is $m=1$: this will not alter the truth of the inequality. Thus the distribution $F$ of $X$ is symmetric about $1$ and the range of $X$ is within the interval $[0,2]$. Our objective is to prove

$$\int_0^2 \frac{1}{x}dF(x) = E\left[\frac{1}{X}\right] \ge 1 + \sigma^2 = E[X^2] = \int_0^2 x^2 dF(x).$$

Since $dF$ is invariant under the symmetry $x\to 2-x$, break each integral into two integrals over the intervals $[0,1)$ and $(1,2]$ and change the variable from $x$ to $2-x$ over the second interval. We may ignore any probability concentrated at the value $1$ because at that point $1/x = x^2.$ Whence the problem reduces to demonstrating

$$\int_0^1 \left[\left(\frac{1}{x} + \frac{1}{2-x}\right) -(x^2 + (2-x)^2)\right]dF(x) \ge 0.\tag{*}$$

This can happen only if the integrand

$$g(x) = \frac{1}{x} + \frac{1}{2-x} -(x^2 + (2-x)^2)$$

is nonnegative on the interval $(0,1].$ That's what we must show.

### Solution

You could apply differential calculus. Elementary demonstrations are also available. When $0 \le x \le 1$, it will be the case that $0 \le |x-1| \le 1$, whence $1 \le 1/|x-1|$, entailing

$$1 \le \frac{1}{(x-1)^2} \le \frac{1}{(x-1)^4}.$$

This implies

$$0 \le \frac{1}{(x-1)^4} - \frac{1}{(x-1)^2} = \frac{2}{g(x)},$$

showing $g(x) \ge 0$ for $x\in (0,1],$ QED.

The inequality is tight in the sense that when $F$ concentrates its probability closer to $1$, the inequality gets closer to being an equality. Thus, we could not replace $\sigma^2/m^3$ in the original inequality by any multiple $\lambda\sigma^2/m^3$ with $\lambda \gt 1$.