An inequality for the cumulants of a Bernoulli variable

by passerby51   Last Updated June 30, 2020 02:20 AM

The cumulants of a Bernoulli variable, with success probability $p$, satisfy $$\kappa_{n+1} = \kappa_2 \frac{d \kappa_n}{dp}$$ as functions of $p \in [0,1]$. Here is a conjecture: $|\kappa_n| \le \kappa_2$ for all $n \ge 2$.

It seems to hold at least for $2 \le k \le 6$. Is it true for all $n \ge 2$? We have $\kappa_2 = p(1-p)$ to start the recursion.



Answers 1


I think the answer is "no".

My computer tells me that the cumulant generating function $f(x)=\log(\frac{1+e^x}2)$ for a $p=1/2$ Bernoulli random variable has the expansion $$f(x)=\frac 1 2 x + \frac 1 8 x^2 - \frac{1}{192}x^4+\frac{1}{2880} x^6-\frac{17}{645120}x^8+\frac{31}{14515200}x^{10}+\cdots.$$ Recall that $$f(x)=\sum_{n\ge1}\frac {\kappa_n}{n!}x^n$$ so the cumulants are, $$ \kappa_1=\frac 1 2, \kappa_2=\frac 1 4, \kappa_4=-\frac 1 8, \kappa_6=\frac 1 4, \kappa_8=\frac{17}{16}, \kappa_{10}=\frac{31}4, \ldots $$ So the conjecture $|\kappa_n|\le\kappa_2$ does not seem to hold for $n=8$ or $n=10$.

kimchi lover
kimchi lover
June 30, 2020 00:57 AM

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