# Why does the expected value of $\left[{1\over2}(X_i-X_j)^2-\sigma^2\right] \left[{1\over2}(X_i-X_k)^2-\sigma^2\right]$ equal $(\mu_4-\sigma^4)/4$?

Question

Let $X_i,X_j,X_k$ be IID random variables with finite moments. In particular we denote $E[X_i] := \mu$, $E[(X_i - \mu)^4] := \mu_4$ and $Var[X_i] := \sigma^2$.

Why does the expected value of $$\left[{1\over2}(X_i-X_j)^2-\sigma^2\right] \left[{1\over2}(X_i-X_k)^2-\sigma^2\right]$$

equal $(\mu_4-\sigma^4)/4$? This question comes from this answer (point 2) to a question about the variance of the sample variance.

My attempt

The only route I see is to expand the product, I obtain $$\frac{1}{4} E[X_i^4] + E[X_i^2]^2 - E[X_i^3]E[X_i] -\frac{5}{2} \sigma^2 E[X_i^2] +3\sigma^2E[X_i^2] + \sigma^4$$ which is not correct.

Calculating it along your route leads to:$$\frac14[\mu_4-4\mu_1\mu_3+3\mu_2^2]-\frac12\sigma^2[4\mu_2-4\mu_1^2]+\sigma^4\tag1$$
WLOG you may assume that $\mu_1=0$ and consequently $\mu_2=\sigma^2$.
This because the expression does not change if $X_i,X_j,X_k$ are replaced by $X_i-\mu_1,X_j-\mu_1,X_k-\mu_1$ respectively.
Substituting this in $(1)$ leads to: $$\frac14\mu_4-\frac14\sigma^4$$