Timeline for Integral over conditioning variable of a Gaussian

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Jan 31, 2015 at 14:15	comment	added	martin		$p(y\|x)$ 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 $\int g(y;x)dy=1$.
Jan 23, 2015 at 20:38	history	edited	ASML	CC BY-SA 3.0	added 8 characters in body
Jan 23, 2015 at 20:37	comment	added	ASML		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.
Jan 23, 2015 at 20:13	review	First posts
Jan 23, 2015 at 20:21
Jan 23, 2015 at 20:09	history	asked	ASML	CC BY-SA 3.0