# High order central moments of a symmetric binomial variable

Consider a random variable $X\sim B(n,\frac 12)$. I'm trying to estimate the asymptotic behaviour of its central moments $E((X-\frac n2)^r)$, where $r$ is even and in the range $\Omega(1)\leq r\leq O(n)$, and $n$ goes to $\infty$.

I've looked at inequalities for central moments of sums of independent variables, but they seem too general. I'd be very grateful if someone can point me in the right direction.

• You can use the binomial formula to get exact expression. Better suited for math exchange. Mar 17 '16 at 17:13
• Thank you. Indeed, the binomial formula yields an exact expression, but what I'm looking for is the asymptotic growth of this expression. I will edit the question to reflect this. Mar 17 '16 at 17:35
• I think it is a reasonable question. To clarify, do you want a strict upper bound or instead the asymptotic behaviour as $n \rightarrow \infty$? A normal approximation of the Binomial random variable quickly gives you $E((X-n/2)^r) \approx (\sqrt{n}/2)^r (r-1)!!$, where $(r-1)!! = (r-1)(r-3) ... 1$. Mar 17 '16 at 17:45
• By exact I actually meant you can write the expectation as a closed form function of N, r. It will involve some special functions, but their asymptotic behavior is pretty well undetstood. My mathematica is broken right now but I will try to do it once it's reinstalled. Mar 17 '16 at 22:31
• Surely it has been done before, but I'd do it thus: the mgf is $$E e^{tX} = 2^{-n}e^{-tn/2}(e^t+1)^n.$$ Now estimate the coefficient of $t^r$ using the saddle-point method (or otherwise). Mar 18 '16 at 0:25

Let $M:=(E(X-n/2)^r)^{1/r}$. By Corollary 2 in [Latala], $M\sim S$, where $A\sim B$ means that $\frac1C\,B\le A\le C B$ for some universal positive constant $C$ and \begin{equation} S:=r\sup\{t(2n/r)^t\colon1/r\le t\le t_*\}, \end{equation} where $t_*:=\frac12\wedge\frac nr$. It is not hard to see that $t(2n/r)^t$ increases in $t\in[1/r,t_*]$, and so, $S=rt_*(2n/r)^{t_*}$. Thus, $M\sim\sqrt{nr}$ if $r\le2n$ and $M\sim n(n/r)^{n/r}\sim n$ if $r\ge2n$. That is, $M\sim\sqrt{n(r\wedge n)}$.

• Thanks! This reference is pretty much what I was looking for. Mar 24 '16 at 13:10
• Since I'm interested in $M^r$, this gives an estimate tight up to a factor exponential in $r$. Will update if I find anything better. Again, thank you very much. Mar 24 '16 at 13:17

I wonder if this perhaps-naive approach can help you. Let's shift by the mean, so consider $X = \sum_{j=1}^n X_j$ where each $X_j = 0.5$ or $-0.5$ independently with one-half probability each. So $\mathbb{E} X^r$ should be the value you're looking for. Now,

\begin{align} X^r &= \left(X_1 + \dots + X_n\right)^r \\ &= \sum_{\text{$j_1,\dots,j_r$}} ~~ \prod_{s=1}^r X_{j_s} \end{align} where the sum is over all vectors $(j_1,\dots,j_r)$ of indices in $\{1,\dots,n\}$. Now the expectation of the sum is the sum of the expectations, so let's look at the expectation of one term: \begin{align} \mathbb{E} \prod_{s=1}^r X_{j_s} &= \left(0.5^r\right) \mathbb{E} \prod_{s=1}^r \begin{cases} 1 & X_{j_s} > 0 \\ -1 & \text{otherwise} \end{cases} \\ &= \left(0.5^r\right) \begin{cases} 1 & \text{each index appears in $\vec{j}$ an even number of times} \\ 0 & \text{otherwise}. \end{cases} \end{align} (Why is this true? We split the product into a product over distinct indices $j$ of $X_j^{m(j)}$, where $m(j)$ is the multiplicity of $j$ in the multiset. The indices are independent Bernoullis, so the expectation of the product is the product of the expectations. Each distinct index's expected product is 1 if that index has even multiplicity and zero if odd.)

So if I've made no mistakes, the expectation you're looking for is equal to $0.5^r$ times the number of vectors in $\{1,\dots,n\}^r$ where each $j \in \{1,\dots,n\}$ appears in the vector an even number of times.

Example: For $r=2$, there are $n$ vectors of all-even multiplicity (one for each index, where that index appears twice), so we get $0.5^2*n = n/4$. For $r=4$, there are $n$ vectors containing just $1$ element and $6{n\choose 2}$ vectors containing two elements twice each (because there are six ways to arrange 2 indices into 4 slots, each taking two). So we get $0.5^4 \left(n + 6{n \choose 2}\right) = \frac{1}{16}\frac{2n + 6n^2 - 6n}{2} = \frac{3n^2}{16} - \frac{n}{8}$ (which is also correct -- it's nice to check).

I didn't get to think about how hard this is to estimate in general (I hope someone much more knowledgable in combinatorics can comment).

• From this it is easy to recover the exact formula $2^{-n}\sum_j\binom{n}{n/2+j} j^r$, but I don't know to get asymptotics from it directly. Mar 18 '16 at 3:14
• @usul, you've actually given my motivation for asking this question. I started by looking for the asymptotics of the number of sequences of vectors in $\{1\ldots n\}^r$ in which each element appears an even number of times. :-) Mar 18 '16 at 9:40
• @Yonatan, aww, well, it's nice to see the connection but sorry it wasn't useful!
– usul
Mar 18 '16 at 12:18

Symmetric Binomial $$B(n,1/2)$$ has sub-gaussian tails with variance-proxy $$\sigma^2 = n/4$$. Thus the growth of of the $$r$$-th moment is at most $$(\mathbf{E}\|S\|^r)^{1/r} = O(\sqrt{r}\cdot \sigma) = O(\sqrt{r n})$$.

The constant can be calculated by integrating the tails, see for example https://www.hse.ru/data/2016/11/24/1113029206/Concentration%20inequalities.pdf