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Let X : $\Omega \mapsto \mathcal{R}^4$ be a Gaussian vector with ${E}(X) = 0. $

Express: $$ \int_\Omega X_1(\omega)X_2(\omega)X_3(\omega)X_4(\omega) \mathcal{P}(d\omega) $$ as a function of $a_{ij}$ = Cov($X_i,X_j)$ with 1 $\leq i < j \leq 4$.

Denote by $\phi_X(t)$ = ${E}[e^{i⟨t,X⟩}]$the characteristic function of X.

How should I proceed?

Ricter
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1 Answers1

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Let's denote $t = (t_1,t_2,t_3,t_4)$

We have $\phi_X(t) = E[e^{i(t_1X_1 + t_2X_2 + t_3X_3 + t_4X_4)}]$

We see that : $\partial_{t_1} \phi_X(t) = E[iX_1e^{i(t_1X_1 + t_2X_2 + t_3X_3 + t_4X_4)}]$

and :

$f(t) = \partial_{t_1}\partial_{t_2}\partial_{t_3}\partial_{t_4} \phi_X(t) = E[X_1X_2X_3X_4e^{i(t_1X_1 + t_2X_2 + t_3X_3 + t_4X_4)}]$

So $f(0)$ is the required value.

Now, you have to explicit $\phi_X(t)$ in terms of $a_{i,j}$

The characteristic function of $X$ is given by : https://en.wikipedia.org/wiki/Multivariate_normal_distribution

$\phi_X(t) = e^{-\frac{1}{2}t^T \Sigma t}$ with $\Sigma = (a_{i,j})_{i,j\in[[1,4]]}$

Now you can explicit $t^T \Sigma t$ and differentiate 4 times with respect to $t_1,t_2,t_3,t_4$

Good luck...

badinmaths
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  • Thanks! Is it correct that the explicit form for $\phi_x(t)$ is $E[X_1X_2X_3X_4]=E[X_1X_2]E[X_3X_4]+E[X_1X_3]E[X_2X_4]+E[X_1X_4]E[X_2X_3]$ ? – Ricter Jan 03 '23 at 16:31