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.
1 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.
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1$\begingroup$ Thanks, Konstantin. I used a slightly less direct method: $\endgroup$– VitalyCommented Aug 7, 2014 at 19:12
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1$\begingroup$ $$E[B(n,p)^k] \leq (pn)^k + \int_{(pn)^k}^\infty pr[B(n,p)^k \geq t] dt .$$ I then used standard estimates of $pr[B(n,p)^k \geq t]$ (that are based on the moment generating function). I'll try to play with your approach to see what kind of bounds it can give. $\endgroup$– VitalyCommented Aug 7, 2014 at 19:24
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$\begingroup$ @Vitaly I think there is a mistake in your bound. It appears you are assuming $E[X] \le z + \Pr[X\ge z]$, but this is not true in general. (Consider as an example any $X$ that has infinite expectation, like the Cauchy distribution.) Perhaps you are thinking of $E[X] \ge z Pr[X\ge z]$, which holds for positive random variables. $\endgroup$ Commented Dec 27, 2017 at 12:59
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$\begingroup$ Thomas: No I'm not assuming that. I'm just using the trivial fact that for a non-negative r.v. X and any $a>0$, $E[X] = \int_0^\infty pr[X \geq t] dt \leq a + \int_a^\infty pr[X \geq t] dt$ $\endgroup$– VitalyCommented Dec 28, 2017 at 22:49