What is the expectation of $\left\langle (n \bar{y})^4 \right\rangle$, if $y_i \sim \mathcal{N}(\mu,\sigma^2)$?

Question

Let $y_i \sim \mathcal{N}(\mu,\sigma^2), \; i = 1,\ldots,n$ and $\bar{y} = \frac{1}{n} \sum_{i=1}^n y_i$, such that $n \bar{y} = y_1 + \ldots + y_n$.

Then, we want to know what the expectation of $(n \bar{y})^4$ is.

As an inspiration, here's my derivation of the expectation $(n \bar{y})^2$:

$$ \begin{split} \left\langle (n \bar{y})^2 \right\rangle &= \left\langle (y_1 + \ldots + y_n) (y_1 + \ldots + y_n) \right\rangle \\ &= \left\langle \left( \sum_{i=1}^n y_i \right) \left( \sum_{j=1}^n y_j \right) \right\rangle \\ &= \left\langle \sum_{i=1}^n \sum_{j=1}^n y_i y_j \right\rangle \\ &= \left\langle n y_i^2 + (n^2-n) y_i y_j \right\rangle \\ &= n (\mu^2 + \sigma^2) + (n^2-n) (\mu \cdot \mu) \\ &= n (\mu^2 + \sigma^2) + (n^2-n) \mu^2 \\ &= n^2 \mu^2 + n \sigma^2 \end{split} $$

However, the expectation $(n \bar{y})^4$ is not as easy, because the combinatorics are much more complicated (products of different numbers of independent or non-independent random variables).

Answer 1

Break the problem into two parts.

1. Work out the distribution of $n\bar y.$ Assume $(y_1,\ldots, y_n)$ has an $n$-variate Normal distribution. (Without this assumption, or some specific assumption about the joint distribution, the problem is insoluble.) Under this assumption $n\bar y,$ being a linear combination of the $y_i,$ has a Normal distribution.

   Let $\Sigma = (\operatorname{Cov}(y_i,y_j))$ be the covariance matrix. (We are told that $\Sigma_{ii}=\sigma^2$ but are given no information about its other entries.) Linearity of expectation yields $$E[n\bar y] = E[y_1] + \cdots + E[y_n] = n\mu$$ and bilinearity of covariance gives $$\operatorname{Var}(n\bar y) = \sum_{i,j} \operatorname{Cov}(y_i,y_j) = \sum_{i,j} \Sigma_{ij} = n\sigma^2 + 2\sum_{j\gt i}\Sigma_{ij}.$$

   Thus, the distribution of $n\bar y$ is completely determined (because it's Normal and we have worked out its mean and variance).

2. Use the central moments of that distribution to find the answer algebraically. Let $Z$ be any standard Normal variable. Clearly $n\bar y$ has the same distribution as $X = n\mu + Z\sqrt{\sum_{ij}\Sigma_{ij}}$ because both are Normally distributed with the same means and variances. Letting $k=4$ and writing $X=\nu + \tau Z$ (to simplify the notation), apply the Binomial Theorem to compute

   $$\begin{aligned} E\left[\left(n\bar y\right)^k\right] &= E\left[\left(\nu+\tau Z\right)^k\right]\\ &= \sum_{i=0}^k \binom{k}{i} \nu^{k-i} \tau^{i} E[Z^i]. \end{aligned}$$

   The odd terms are zero (because $Z$ is symmetric about $0$ and has finite moments of all orders) while the expectations of the even terms are $$E[Z^{2j}]=\frac{(2j)!}{2^jj!}.$$ Plugging these in yields

   $$\begin{aligned}E\left[\left(y_1+y_2+\cdots+y_n\right)^k\right] &= E\left[\left(n\bar y\right)^k\right]\\&= \sum_{j=0}^{\lfloor k/2\rfloor}\binom{k}{2j} (n\mu)^{k-2j} \left(\sum_{ij}\Sigma_{ij}\right)^j\frac{(2j)!}{2^jj!}.\end{aligned}$$

As an example, consider the case where the $y_i$ are uncorrelated, which with the assumption $\operatorname{Var}(y_i)=\sigma^2$ implies $\sum_{ij}\Sigma_{ij}=n\sigma^2.$ With $k=4$ (as in the question) the preceding formula reduces to

$$\begin{aligned} E\left[\left(n\bar y\right)^4\right] &= \sum_{j=0}^{2}\binom{4}{2j} (n\mu)^{4-2j} \left(n\sigma^2\right)^j\frac{(2j)!}{2^jj!}\\ &= \left(n\mu\right)^4\frac{(0)!}{2^00!} + 6\left(n\mu\right)^2\left(n\sigma^2\right)\frac{(2)!}{2^11!} + \left(n\sigma^2\right)^2\frac{(4)!}{2^22!}\\ &= \mu^4n^4\, + \, 6\mu^2\sigma^2n^3\, + \, 3\sigma^4n^2. \end{aligned}$$