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I have a multivariately distributed random 3-vector $$\mathbf{x}=\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}\sim\mathbf{N}\left(\begin{pmatrix}2\\-1\\3\end{pmatrix},\begin{pmatrix}4&1&0\\1&2&1\\0&1&30\end{pmatrix}\right)$$ and I'd like to compute the covariance between two products of the $x_i$, in particular $$\text{cov}(x_1x_2,x_1x_3)$$ I don't think the fact that $\text{cov}(x_1,x_3)=0$, i.e. the fact that $x_1$ and $x_3$ are independent, implies the covariance I want also vanishes, but I'm stuck because of the non-linearity of the transformation. Suggestions? I assume the problem simplifies immensely because of the normality assumption and that the general question (without the distribution assumption) would have no closed form solution, but I still don't know how to proceed.

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up vote 1 down vote accepted

By the definition of covariance, $\text{cov}(x_1 x_2, x_1 x_3) = E[x_1^2 x_2 x_3] - E[x_1 x_2] E[x_1 x_3]$. You might use Isserlis's theorem to compute $E[x_1^2 x_2 x_3]$, after first expressing it in terms of $y_i = x_i - E[x_i]$. Or use the joint MGF.

EDIT: Let ${\bf x}$ be the column vector $[x_1, x_2, x_3]^T$, and ${\bf t} = [t_1, t_2, t_3]^T$. The joint MGF is $$E[e^{{\bf t} \cdot {\bf x}}] = e^{{\bf \mu} \cdot {\bf t} + {\bf t}^T \Sigma {\bf t}/2}$$ where ${\bf \mu}$ is the vector of means and $\Sigma$ is the covariance matrix. Expanding this out, the coefficient of $t_1^{i_1} t_2^{i_2} t_3^{i_3}$ is $E[x_1^{i_1} x_2^{i_2} x_3^{i_3}]/(i_1! i_2! i_3!)$. In your case, $$ E[e^{{\bf t} \cdot {\bf x}}] = e^{2 t_1 - t_2 + 3 t_3 + 2 t_1^2 + t_1 t_2 + t_2^2 + t_2 t_3 + 15 t_3^2}$$ Expanding this out, the coefficient of $t_1^2 t_2 t_3$ seems to be $-2$, which makes $E[x_1^2 x_2 x_3] = -4$.

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Thanks Robert. Can you elaborate on the second approach ("or use the joint MGF")? I don't have a good sense of how to do things with MGFs; I realize this is the problem. – symplectomorphic Feb 7 '13 at 7:01

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