I am trying to understand a proof for the Kurtosis of a sum of independent random variables, however, there is one part where I am quite stuck:
Theorem:
for $X_1, X_2, ..., X_n$ independent random variables with means $ \mu_1, ... , \mu_n$ and variance $\sigma_1^2, ... , \sigma_n^2 $ $E(X_i^4) < \infty$
Define $S_n = X_1 + ... + X_n$ and $S_n$ will be appropriately normal
$ kurt(S_n) - 3 = (\sum_{i=1}^n\sigma_i^n)^{-2}\sum_{i=1}^n\sigma_i^4(kurt(X_i)-3)$
Proof (first part):
assume WLOG that $E[X_i] = 0$ for all $i$
$ kurt(Sn) = \frac{E[(X_1 + ... + X_n )^4]}{(\sigma_1^2 + ... +\sigma_n^2)^2 } $
$ E[(X_1 + ... + X_n )^4] = \sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^n\sum_{l=1}^n E(X_iX_jX_kX_l) = \sum_{i=1}^n E(X_i^4) +6 \sum_{i<j}^n\sigma_i^2\sigma_j^2$
NOTE: $E(X_iX_jX_kX_l) = 0 $ unless $ i=j=k=l $ or if combinations of two pairs.
Question: why is the NOTE: true? From my understanding the expected value of INDEPENDENT random variables is equal to the product of the expected values of the random variables. However I intuitively understand this case does not apply, or else, we would not get very far! So what is going on?
Assuming this hickup true, i understand the rest of the proof, which I will not write. (as i am improvising the syntax)
Consider for example the case of $i=j=k=l$ where you are then taking the expectation of $X_i^4$. Since this real number raised to an even power is never negative, and is sometimes positive, its expectation cannot be zero.
On the other hand, consider $X_i^3 X_j$ with $i\neq j$. Now just because $E(X_i) = 0$ that does not allow you to say that $E<X_i^3>$ = 0; consider a discrete random that is $-1$ with probability $\frac23$ and $1$ with probability $\frac13$. But by the multiplication of independent expectations, you get zero since the expectation of $X_j$ is zero.
Only if all the independent variates are raised to powers greater than $1$ can the expectation be non-zero. For four variables, the only such cases are four of a kind and two pairs.