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).
 A: Break the problem into two parts.

*

*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).


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