Statistical inequality

Question

Let $X$ be a finite discrete variable and $X\ge0$. Is it true that $$16\operatorname{Var}(X) \le \left[8{\mathbb E}(X) + \operatorname{Range}(X)\right]\operatorname{Range}(X)$$ where $\operatorname{Var}(X)$, $\mathbb{E}(X)$ and $\operatorname{Range}(X)$ are respectively the variance, mean, and range of $X$?

Answer 1

I'm not sure what you mean by $Range(X)$, but assuming you mean the difference between the (essential) sup and inf then the answer is yes. The requirement that $X$ be discrete is also unnecessary.

By translating $X$ we can make the infimum be 0. This doesn't change the variance or range but decreases $\mathbb{E}(X)$ so proving this case is enough. Multiplying then by $\max^{-1}(X)$ we get a r.v. with range of 1 (this multiply both sides of the inequality by a factor of $\max^{-2}(X)$. We now get: $$16var(X)\le 8\mathbb{E}(X)+1 \ .$$ Now, among all r.v. in $[0,1]$ with $\mathbb{E}(X)=a$ the one which maximizes the variance is the one with $\mathbb{P}(X=1)=a$ and $\mathbb{P}(X=0)=1-a$ which has variance of $a-a^2$. So we get the inequality $$16(a-a^2) \le 8a+1$$ for any $0\le a \le 1$ which is equivalent to $$0\le (4a-1)^2 \ .$$