I'm looking for simple and reasonably tight bounds on the k-th moment of the Binomial distribution $B(n,p)$, namely, $E[B(n,p)^k]$. I'm interested in the case when k is large (say on the order of $\sqrt{n}$) and in exact bounds (not asymptotic).
Using a relatively simple calculation based on the concentration of the binomial sum I got:
$$E[B(n,p)^k] \leq (pn)^k + k \ln n \cdot k^k$$ but I suspect that better bounds are known.
There is an extensive literature with bounds on moments of sums of independent r.v.'s but I could not find something suitable. For example, as far as I understand, in this case Latala's paper gives a bound of:
$\left(O\left(\frac{k}{\ln k}\right)\right)^k \cdot max \{pn,(pn)^k\}$ which is both larger and asymptotic.
Any references or simple alternative bounds would be appreciated.
[Math] Bounds on the moments of the binomial distribution
measure-concentrationpr.probability
Best Answer
For any $\beta>0$, $$\mathbb{E}B(n,p)^k\leq k!\beta^{-k}\mathbb{E}e^{\beta B(n,p)}= k!\beta^{-k}(1-p+pe^{\beta})^n.$$
Now you can plug various $\beta$, e.g. $\beta=\frac{k}{np}$ which yields $$\mathbb{E}B(n,p)^k\leq (np)^k k!k^{-k}\left((1-p)+pe^{\frac{k}{pn}}\right)^n.$$ I assume that you got your estimate by elaborating on this expression, although in the case $1\ll k\ll n$ the above is slightly better.
It feels unlikely that you can get much better universal bounds. Indeed, you can write $$ \mathbb{E}B(n,p)^k=k!\int\frac{(1-p+pe^z)^n}{z^{k+1}}dz, $$ integral being over a contour around zero. When you transform it to pass through a saddle point, you are essentially solving the optimization problem on $\beta$ as above, hence the maximum of the integrand is of the same order. In the limit, Laplace's method gives you at best some polynomial corrections. Another room for small improvements would be a more careful choice of $\beta$; e. g. for $1\ll k\ll n$ you can find a second-order approximation to the extremum.