3
$\begingroup$

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).

$\endgroup$
13
  • 3
    $\begingroup$ Why are you using this strange notation $\langle~\cdot~\rangle$ which is commonly used for inner products ? $\endgroup$
    – JRC
    Dec 3, 2020 at 5:30
  • 2
    $\begingroup$ 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. $\endgroup$
    – Xi'an
    Dec 3, 2020 at 8:39
  • 1
    $\begingroup$ 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. $\endgroup$
    – whuber
    Dec 3, 2020 at 12:49
  • 2
    $\begingroup$ This is a statistics forum, not a quantum mechanics forum! $\endgroup$
    – Xi'an
    Nov 3, 2021 at 20:00
  • 1
    $\begingroup$ @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. $\endgroup$
    – Galen
    Nov 3, 2021 at 21:30

1 Answer 1

4
$\begingroup$

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

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

$\endgroup$
1
  • 1
    $\begingroup$ 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. $\endgroup$
    – Joram Soch
    Nov 5, 2021 at 9:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.