# 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 $$\lambda \gt 1$$.