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

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).

• Why are you using this strange notation $\langle~\cdot~\rangle$ which is commonly used for inner products ? Dec 3 '20 at 5:30
• You are basically asking for the fourth moment of a Normal random variable, whose value can be found on the Wikipedia page for the Normal distribution. Dec 3 '20 at 8:39
• The first line of stats.stackexchange.com/a/116657/919 obtains the answer for you, because you know $\bar y \sim \mathcal{N}(\mu, \sigma^2/n).$ stats.stackexchange.com/a/176814/919 gives another method to obtain this result.
– whuber
Dec 3 '20 at 12:49
• This is a statistics forum, not a quantum mechanics forum! Nov 3 at 20:00
• @whuber Okay, I understand. I've realized I tacitly have an opinion that $\langle x \rangle$ always entails the use of Bra-Kets, so they go hand-in-hand for me. If I'm not implying the existence of a wave function, I use $\mathbb{E}[]$. Thanks for your explanation. Nov 3 at 21:30

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}

• Thanks for the very nice and very general answer! Yes, the example was intended to include that the $y_i$'s are independent, i.e. $\Sigma_{ij} = 0$ for all $i \neq j$, but you're right, I didn't say that explicitly. Nov 5 at 9:46