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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:

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

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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

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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, $(x+z)/2>y$.

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  • $\begingroup$ Thank you @IosifPinelis ! Where do I find the reference that the reciprocal of a log concave function is log convex? It appears to me that the remark states for the pdf to be logconcave not for the cdf, am I missing someething? $\endgroup$
    – Marco Max Fiandri
    Commented Jan 29 at 15:05
  • $\begingroup$ @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. $\endgroup$
    – Iosif Pinelis
    Commented Jan 29 at 17:21
  • $\begingroup$ 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 $B_{j}$? $\endgroup$
    – Marco Max Fiandri
    Commented Jan 29 at 17:41
  • $\begingroup$ @MarcoMaxFiandri : I have added a detail on this as well -- this is just the AM-GM inequality. $\endgroup$
    – Iosif Pinelis
    Commented Jan 29 at 18:21
  • $\begingroup$ thank you so much I will acknowledge your priceless contribution in the article $\endgroup$
    – Marco Max Fiandri
    Commented Jan 29 at 18:41

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