Convexity of a function

Question

Let: $F_{j+1,y}(s)$ be the cumulative distribution function of a binomial distribution with mean $y$, $j+1$ independent trials considered for $s$ successes. Is it possible to show in any way that:

$\frac{1}{F_{j+1,y}(s+2)}-\frac{2}{F_{j+1,y}(s+1)}+\frac{1}{F_{j+1,y}(s)}>0, 0\leq s\leq j-1$

?

This can be helpful when comparing the expected value of random processes governed by poisson-binomial distributions against ones governed by binomial distribution. And in infer statistical dominance of a process over the other, indeed is a frequent term in bandit algorithms problems Thank you for your kind attention

Answer 1

The answer is yes. Indeed, the cdf $F$ of the binomial distribution is log concave -- see e.g. Theorem 2 on p. 152, used with $\alpha=1$, $r=\infty$, and $q$ being the probability mass function of the binomial distribution. So, $1/F$ is log convex on the set $S:=\{0,\dots,j+1\}$ of all atoms of the distribution. So, $1/F$ is strictly convex on $S$.

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Detail on the last statement: Let $g:=1/F$, $i\in\{1,\dots,j\}$, $x:=g(i-1)$, $y:=g(i)$, $z:=g(i+1)$. We have $x>z$ and $(\ln x+\ln z)/2\ge\ln y$. We want to show that then $(x+z)/2>y$. But $(\ln x+\ln z)/2\ge\ln y$ means that $y\le\sqrt{xz}$ and hence, by the inequality of arithmetic and geometric means, $(x+z)/2\ge\sqrt{xz}\ge y$, and the first of the latter two inequalities is strict, because $x>z$ and hence $x\ne z$. Thus, $(x+z)/2>y$.