Let the probability distribution for $X=x$ have a binomial distribution with parameters $n$ and $p$: $$\mathrm{P}(X=x) = \binom{n}{x} p^x\, (1-p)^{n-x}, \quad x \in \{0,\dotsc,n\}.$$

I'm looking for upper and lower bounds of its raw $k$th moment, possibly as functions of $k$, $n$, $p$: $$L(k,n,p) \leqslant \mathrm{E}(X^k) \leqslant U(k,n,p).$$

In particular I am interested in the case where $k$, $n$, $k/n$ are large and $p$, $kp$, $np$ are small. (Concrete values: $n\approx 10^3$, $k\approx 5\times 10^5$, $p\approx 10^{-16}$). Probably this translates into the limit $k\to\infty$, $kp\to 0$, $n\to \infty$, $k/n\to \infty$.

I have tried playing around with the recursion formula in Johnson, Kotz, Balakrishnan: Univariate Discrete Distributions (3rd ed. Wiley 2005), §3.3: $$\mathrm{E}(X^{k+1}) = \biggl[np + (1-p)p \frac{\mathrm{d}}{\mathrm{d}p}\biggr] \mathrm{E}(X^k),$$ but to no avail.

I have read this answer, based on Jensen's inequality, and it seems to suggest that $\mathrm{E}(X^k)\simeq n^k$ for $k$ large; in that case I'd need some extra terms in this asymptotic expansion. I have also read this answer, but I'm not sure that its series expansion is valid in my case.

Updated context: What I'm actually interested in is the ratio $\frac{\mathrm{E}(X^{k+1})}{\mathrm{E}(X^k)}$ in the asymptotic case described above. The first answer I mentioned seems to imply that this ratio is asymptotically equal to $n^{k+1}/n^k \equiv n$, but I'd like to find out the next correction to this approximation.

It turns out that the ratio $\frac{\mathrm{E}(X^{k+1})}{n\mathrm{E}(X^k)}$ is the posterior distribution of an exchangeable model for two events (e.g. coin tossing), with a binomial hyperparameter prior, conditioned on the observation of $k$ instances of the first event (say, Heads). When $k$ is large this posterior should give probability almost 1 to the first event, which is in accord with common sense and with the result above from Jensen's inequality. What interests me is the "almost", that is, the next correction to $\mathrm{E}(X^k)\simeq n^k$. I'm looking up more about the proof of Jensen's inequality, maybe I'll find something there.

  • $\begingroup$ When you ask about whether it is asymptotically like $n$, do you mean in the limit $k \to \infty$ at fixed $n$ and $p$? $\endgroup$
    – Ian
    Jun 17, 2017 at 14:48
  • $\begingroup$ @Ian: It's a tricky question, because I'm not considering sequences. I have definite values of $k$, $n$, $p$. As I explain above, $k$ is large, $k/n$ is large, $kp$ is small. Your "fixed $n$, $p$" should translate to $kn$ large, $kp$ large, so that's not the limit I'm looking for. Thank you! $\endgroup$
    – pglpm
    Jun 17, 2017 at 15:07
  • $\begingroup$ Consider if the nonasymptotic bound here math.stackexchange.com/a/2581762/7072 is useful for your applications. Also, have a look at the great notes by Jelani Nelson people.seas.harvard.edu/~minilek/madalgo2015/notes1.pdf with lots of tricks for similar problems. $\endgroup$ Dec 27, 2017 at 12:42
  • $\begingroup$ @ThomasAhle: It looks useful indeed. And great reference! Thank you, Thomas $\endgroup$
    – pglpm
    Dec 28, 2017 at 14:51

1 Answer 1


Normalized moments of IID sums are known up to a constant factor.


This will give $(\mathbf{E} X^k)^{1/k}$ up to a constant factor.

This is more relevant than Khincthine's inequalities developed for symmetric variables.


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