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The marginal of a multivariate Gaussian can be computed in closed form, i.e.,

$p(x) = \int_y \mathcal{N}((x,y);\mu,\Sigma)\ dy$

is simple. But what I need is

$L(x) = \int_y \mathcal{N}((x\mid y); \mu(y),\Sigma_{\mid y})\ dy$,

i.e., the integral over the variable that is being conditioned on. I know that's not a distribution in $x$, but can the integral be computed in closed form? Or approximated efficiently without sampling?

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    $\begingroup$ Note that this is equivalent to $L(x) = p(x)\int_y \frac{p(y\mid x)}{p(y)} dy$, where all distributions are marginals or conditionals of a joint Gaussian distribution $p(x,y)$. This might be easier to tackle. $\endgroup$
    – ASML
    Jan 23, 2015 at 20:37
  • $\begingroup$ $p(y|x)$ and $p(y)$ are both densities of Gaussians, so you should be able to "complete the square" in the exponent and figure out what the normalization constant should be. I.e., it should be the case that $\frac{p(y|x)}{p(y)}=f(x) g(y;x)$, where $g(y;x)$ is a (properly normalized) Gaussian density with respect to y, and $f(x)$ is some function. In that case, $L(x)=p(x)f(x)$, since $\int g(y;x)dy=1$. $\endgroup$
    – martin
    Jan 31, 2015 at 14:15

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