# Convexity of a function

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

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 $$S$$.
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, $$(x+z)/2>y$$.
• @MarcoMaxFiandri : (i) That the reciprocal of a log concave function is log convex follows immediately from the definitions of such functions -- similarly to the fact that the function $(-f)$ is convex if $f$ is concave. (ii) A better reference than that remark is Theorem 2 of the same linked paper. I have also added details on this. Commented Jan 29 at 17:21
• I see! I understood thank you for your patience! However I don't get how we retrieve the last inequality. Sure we've proved that (calling B_j the reciprocal of the cdf) $\ln(B_{j+2})-2\ln(B_{j+1})+ln{B_{j}}>0$ but this implies that the inequality also hold true for $B_{j}$? Commented Jan 29 at 17:41