I am interested in the sign of odd, central moments of a binomial distribution. From DOI. 10.1137/070700024 I have the formulae:


$ E\left[\left(X-\mu\right)^d\right]= \sum\_{i=0}^n\binom{n}{i}\left(-p\right)^i\sum\_{l=0}^{i}\left(-1\right)^l\binom{i}{l}\left(l-\mu\right)^d$


where i can see that in the odd case the sign of $\left(l-\mu\right)^n$ changes if $l > pn$. But I get no control about all the other stuff, not even the inner sum.


Another way to approach that, also from this paper mentioned above, would be to use the identity:
$E\left[\left(X-\mu\right)^d\right]= \sum\_{i=0}^n\binom{n}{i}\left(-\mu\right)^{d-i}E\left[X^d\right]$


and take advantage of the fact that:


$\left(np\right)^d \leq E\left[X^d\right] \leq n^d$


I would highly appreciate to derive the sign of an odd, central moment in dependence of $p$ and $n$, where $p \neq 0.5$.



sign of odd central moments of binomial distribution
Let $X$ have the binomial distribution with parameters $n,p$. Then for any natural $d$
$$E(X-np)^d=E\Big(\sum\_1^n Y\_i\Big)^d,$$
where $Y,Y\_1,\dots,Y\_n$ are iid random variables such that $P(Y=q)=p=1-P(Y=-p)$, with $q:=1-p$. Let $d$ be a natural number.


Suppose now that $p\le1/2$. Then for all natural $k$
$$m\_k:=EY^k=pq(q^{k-1}-(-p)^{k-1})\ge0.$$
Also, $E\Big(\sum\_1^n Y\_i\Big)^d$ is a linear combination with positive coefficients of terms of the form $m\_{k\_1}\cdots m\_{k\_j}\ge0$ with natural $k\_1,\dots,k\_j$.


Thus, $E(X-np)^d\ge0$ if $p\le1/2$. By symmetry, $E(X-np)^d\le0$ if $p\ge1/2$ and $d$ is odd.


Working slightly harder, we can see that, if $d$ is an odd number $\ge3$, then
$$
E(X-np)^d\begin{cases}
>0&\text{ if }0<p<1/2,\\
<0&\text{ if }1/2<p<1,\\
=0&\text{ if }p=1/2.
\end{cases}
$$


