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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).

Joram Soch
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    Why are you using this strange notation $\langle~\cdot~\rangle$ which is commonly used for inner products ? – JRC Dec 03 '20 at 05:30
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    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. – Xi'an Dec 03 '20 at 08:39
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    The first line of https://stats.stackexchange.com/a/116657/919 obtains the answer for you, because you know $\bar y \sim \mathcal{N}(\mu, \sigma^2/n).$ https://stats.stackexchange.com/a/176814/919 gives another method to obtain this result. – whuber Dec 03 '20 at 12:49
  • @Kolmogorov: Sorry for this. In my field, it is common to do this (but $\mathrm{E}(\cdot)$ is also being used). Wikipedia says that the angle bracket notation is commonly used in physics (not my field). – Joram Soch Dec 03 '20 at 19:13
  • @Xi'an: If I understand correctly, I should apply $\left\langle x^4 \right\rangle = \mu^4 + 6 \mu^2 \sigma^2 + 3 \sigma^4$ to $n \bar{y} \sim \mathcal{N}(n \mu, n \sigma^2)$, right? – Joram Soch Dec 03 '20 at 19:16
  • Since I don't really see how the linked threads answer my question (or that my question is a duplicate of them), I will answer my question in this comment: With $y_i \sim \mathcal{N}(\mu,\sigma^2)$, we have $n\bar{y} \sim \mathcal{N}(n\mu,n\sigma^2)$ because of the linear combination of normal random variables. Thus, we can apply $x \sim \mathcal{N}(\mu,\sigma^2) \Rightarrow \left\langle x^4 \right\rangle = \mu^4 + 6\mu^2\sigma^2 + 3\sigma^4$ which gives $\left\langle (n\bar{y})^4 \right\rangle = n^4\mu^4 + 6n^3\mu^2\sigma^2 + 3n^2\sigma^4$. – Joram Soch Nov 02 '21 at 11:44
  • @Xi'an This is Bra-Ket notation, which is nearly-ubiquitous in Quantum Mechanics. – Galen Nov 03 '21 at 17:42
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    This is a statistics forum, not a quantum mechanics forum! – Xi'an Nov 03 '21 at 20:00
  • @Galen This is not bra-ket notation. QM overloads angle brackets to refer to two distinct operations: Hermitean inner product $\langle\ \mid\ \rangle$ ("bra-ket") and statistical averaging. The vast majority of statistical works use $E[]$ notation for the latter, whereas $E[]$ is rarely seen in the QM literature. – whuber Nov 03 '21 at 20:15
  • @whuber Maybe I am mistaken. Let's try to clarify with an example in 1D. Consider $ \langle x \rangle = \int \langle x| x | x \rangle\ dx = \int \psi (x)^* x \psi (x)\ dx = \int x \psi (x)^* \psi (x)\ dx = \int x |\psi (x)|^2\ dx$. By Born's rule, $\int x |\psi (x)|^2\ dx = \int x Pr(x)\ dx$. What is wrong with this example? – Galen Nov 03 '21 at 20:32
  • @Galen That's a fine example: in your notation, the first angle brackets represent expectations while the angle brackets in the integrand are not, themselves, expectations: as usual, they represent the application of an operator ($\mid x \mid$) to a vector ($\mid x\rangle$) and a covector ($\langle x \mid$). – whuber Nov 03 '21 at 21:15
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    @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. – Galen Nov 03 '21 at 21:30
  • @all: Thanks for your feedback. I just noticed that the confusion was also due to a mistake on my side: I wrote "expectation of $\left\langle (n\bar{y})^4 \right\rangle$", but $\left\langle \cdot \right\rangle$ already means "expectation of". Thus, I should have written "expectation of $(n\bar{y})^4$" or simply "$\mathrm{E}\left[ (n\bar{y})^4 \right]$". – Joram Soch Nov 05 '21 at 09:39

1 Answers1

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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

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

whuber
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    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. – Joram Soch Nov 05 '21 at 09:46