1
$\begingroup$

How we can prove the following: $X_1, X_2, ..., X_n$ are $i.i.d$ random variables having all the moments of order less than 4, so \begin{eqnarray*} \sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^n\sum_{l=1}^n \mathbb{E}[X_i X_j X_k X_l] = \sum_{i=1}^n \mathbb{E}[X_i^4] + 3 n(n-1) \big(\mathbb{E}[X_1^2]\big)^2. \end{eqnarray*}

Thank you in advance.

$\endgroup$
4
  • $\begingroup$ thank you Davide Giraudo for your solution, But the coefficient is not equal to 3. $\endgroup$
    – Emera
    Nov 2, 2013 at 19:33
  • $\begingroup$ Which coefficient do you mean? $\endgroup$ Nov 3, 2013 at 9:33
  • $\begingroup$ The coefficient $\binom 42$ is equal to 6 not 3. Can you explain more please. Than you so much Davide Giraudo. $\endgroup$
    – Emera
    Nov 3, 2013 at 15:17
  • $\begingroup$ I agree that $\binom 42=6$, but the sum I wrote is $\frac{n(n-1)}2, which is what is written. $\endgroup$ Nov 3, 2013 at 15:27

1 Answer 1

1
$\begingroup$

We have to assume that $\mathbb EX_0=0$, otherwise the formula doesn't hold true (tale $X_i=1$). Let $S:=\{(i_1,i_2,i_3,i_4)\in [n]\}$. If $\mathbf i=(i_1,i_2,i_3,i_4)\in S$, denote by $S_k:=\{\mathbb i\in S, \mbox{ the maximum of the }i_j \mbox{'s is reached }k\mbox{ times}\}$.

Then $$\sum_{\mathbf i\in S}\mathbb E(X_{i_1}X_{i_2}X_{i_3}X_{i_4})=\sum_{\mathbf i\in S_1}+\sum_{\mathbf i\in S_2}+\sum_{\mathbf i\in S_3}+\sum_{\mathbf i\in S_4}.$$ The first sum of the RHS vanishes by independence and because the random variables are zero-mean. The same argument shows that the third sum vanishes.

The second one is subdivided into two ones: the non-maximals indexes are the same or not. If not, the expectation is $0$. If yes, we get the terms of the RHS (the coefficient is $\binom 42\times\sum_{1\leqslant i\lt j\leqslant n}1$).

The fourth sum is $n\mathbb EX_1^4$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .