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 $X$?

1$\begingroup$ fixed.........! $\endgroup$ – Nate Eldredge Nov 7 '10 at 3:47

1$\begingroup$ Out of interest, where does this inequality arise? $\endgroup$ – Yemon Choi Nov 7 '10 at 5:04
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 $\mathbb{P}(X=0)=1a$ which has variance of $aa^2$. So we get the inequality $$16(aa^2) \le 8a+1$$ for any $0\le a \le 1$ which is equivalent to $$0\le (4a1)^2 \ .$$

$\begingroup$ Would you like to call $\mathbb{E}(X)$ something other than $e$? I got confused at first thinking you meant Euler's constant. $\endgroup$ – Nate Eldredge Nov 7 '10 at 12:09
