Proof of $\sigma^2\geq (\mu-m)^2$ without resorting to Jensen's or Chebychev's inequality.

Question

I asked a group of undergrad students (engineering) to prove that $\sigma^2\geq (\mu-m)^2$, where $\sigma^2$, $\mu$ and $m$ are the variance, mean and median of a continuous random variable. For the discrete case it is simple. For the continuous case the easiest way is to use Jensen's inequality. But the students do not know that inequality.

I thought I had an elementary proof, but I was mistaken. Can someone reproduce the proof in this paper http://www.tandfonline.com/doi/abs/10.1080/00031305.1990.10475743?queryID=%24%7BresultBean.queryID%7D#.VCLQ2_ldXfs ? I do not have access to this one.

Thank you in advance, Gustavo.

Answer 1

Suppose $m \le \mu$ and the random variable is $X$. (If $m \gt \mu$, consider $-X$ instead.)

Then $$E[X|X \le m] \le m$$ and $$P(X\le m) \ge \frac12 \ge P(X \gt m)$$ implying $$E[X|X \gt m] \ge 2\mu-m.$$

So $$E[(X-\mu)^2|X \le m] \ge (\mu-m)^2$$ and $$E[(X-\mu)^2|X \gt m] \ge (\mu-m)^2$$ implying $$\sigma^2= E[(X-\mu)^2] \ge (\mu-m)^2.$$

Answer 2

The Cantelli's inequality for the standardized random variable $$ Z=\frac{X-\mu}{\sigma}$$ gives: $$\mathbb{P}[Z\geq k]\leq\frac{1}{1+k^2}.\tag{1}$$ $|m-\mu|\leq\sigma$ is just a straightforward consequence.