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Suppose I have a mutivariate Gaussian distributed variable $u\sim\mathcal{N}(\mu,\Sigma)$, where $\Sigma$ is a dense matrix. I wish to calculate the expectation of $f(u_i)$.

Is $$E(f(u_i))=\int f(u_i)\mathcal{N}(u_i|\mu_i,\Sigma_{ii})\,du_i$$ If so provide a simple proof? Also does this apply to other multivariate distributions? eg. Multivariate T.

Otherwise do I have to consider that $$E(f(u_i))=\int f(u_i)P(u_i|u_{\backslash i})\,du_i \,P(u_{\backslash i})\,du_{\backslash i}$$ where, $u_{\backslash i}$ indicates variable $u$ without the i-th dimension.

Aside: in my particular case $f(u_i)=\exp(-u_i)$, however I'm more concerned of the general case.

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  • $\begingroup$ What is the subscript $_i$ for? $\endgroup$
    – AdamO
    Commented Dec 5, 2013 at 17:33

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Yes because the marginal distribution of $u_i$ is $N(\mu_i,\Sigma_{ii})$.

The marginal distribution in this case is defined as the following:

$\int ... \int\int...\int N(\mu,\Sigma) du_1 ... du_{i-1} du_{i+1} ... du_n$.

Which can be shown to equal the density of a $N(\mu_i,\Sigma_{ii})$ distribution.

Once you have the marginal, you can apply the law of the unconscious statistician which says that if X is a random variables with density $f_X$ then $E[g(X)] = \int g(x) f_X(x) dx$.

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    $\begingroup$ You can integrate over "$\mu_i$" if you find the correct distribution of $\mu_i$. Of course finding such a distribution is far more difficult than calculating the simple conditional mean: $\mbox{E}(\mu_i) = \mbox{E}(\mbox{E}(\mu_i| X_1, X_2, \ldots, X_n))$. $\endgroup$
    – AdamO
    Commented Dec 5, 2013 at 17:34
  • $\begingroup$ Right, I'll edit it for clarity. $\endgroup$ Commented Dec 5, 2013 at 17:38
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    $\begingroup$ The marginal distribution of a multivariate normal is a univariate normal with corresponding $\mu_i$ and $\Sigma_{ii}$ $\endgroup$
    – sachinruk
    Commented Dec 5, 2013 at 20:31

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