Forth Moment of Sum of Normal with Equal Correlation

Question

I have $X_1,\dots,X_n$ identically normal distributed $N(0,\sigma^2)$ and $\operatorname{corr}(X_i,X_j)=\rho $ for all $i\neq j$. I'd like to compute \begin{equation} E\left(\sum_{i=1}^nX_i\right)^4. \end{equation} I have tried to expand $\left(\sum_{i=1}^nX_i\right)^4$ using mutinomial expansion, but it become difficult to handle the number of terms having form $X_i^3X_j$, $X_i^2X_j^2$ and $X_i^2X_jX_k$ and also $X_iX_jX_kX_l$. Could anyone help me? Thanks for any help and suggestion.

Answer 1

If the variables are jointly normal, then their sum, denote it $S_n$ will be a normal random variable, with mean zero and a variance, denote it simply by $v_s$. Then the random variable $Z = S_n^2/v_s$ will follow a chi-square distribution with one degree of freedom. Then, the random variable $W = v_sZ =S^2_n$ follows a Gamma distribution with shape parameter $k=1/2$ and scale parameter $\theta=2v_s$, and we have $$E(W) = k\theta = v_s,\;\;\operatorname {Var}(W) = k\theta^2=2v_s^2$$

Now, we want to calculate $$E[S_n^4] = E[(S_n^2)^2] = E[W^2] = \operatorname{Var}(W) + \left(E(W)\right)^2$$

$$\Rightarrow E[S_n^4] = 2v_s^2 + \left(v_s\right)^2 = 3v_s^2 = 3\cdot\left[\operatorname{Var}(S_n)\right]^2$$

For the variance of $S_n$ we have

$$\operatorname{Var}(S_n) = \sum_{i=1}^n\sigma^2 + 2\sum_{i\neq j}\operatorname{Cov}(X_i,X_j)$$

The variables are identical in distribution and equicorrelated, so

$$\forall\; i\neq j,\;\operatorname{Cov}(X_i,X_j)=\rho\sigma^2 $$

Then

$$\operatorname{Var}(S_n) = n\sigma^2 + n(n-1)\rho\sigma^2 = [1+(n-1)\rho]n\sigma^2$$

As noted in a comment, this variance could become negative, placing a restriction on $\rho$, $\rho > -\frac{1}{n-1}$ which is a well known restriction for equicorrelated and identical random variables (in other words, if we want $\rho$ to be even more negative, then the equicorrelated variables cannot be identically distributed, i.e. cannot have the same variance).

Assuming the restriction holds, substituting we finally obtain

$$E[S_n^4] = 3\cdot\left[[1+(n-1)\rho]n\sigma^2\right]^2 = 3\cdot[1+(n-1)\rho]^2n^2\sigma^4$$

So we see that the answer given through the MGF approach was correct (always assuming that the variables are jointly normal, and that the restriction on the correlation coefficient is respected).

Answer 2

I've not worked out the details but I am sure it will go through. Calculate the moment generating function $E[exp(s \sum X_i)]$. Differentiate this enough times and you get your answer.

Taking the covariance matrix to have compound symmetry then I get the generating function for $\sum X_i$ to be $exp(n(1+(n-1)\rho)\sigma^2 s^2 /2)$. This is at least right for $n=1$ and also right for $\rho=0$.

It gives $Var[\bar{X}] = \frac{1+(n-1)\rho}{n}\sigma^2$ which looks right for $\rho=1$ so I am hopeful that it is correct.

If it is, then $E[(\sum X_i)^4]=3 n^2 (1+ (n-1)\rho)^2 \sigma^4$