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Consider the integral $$ p(Y) = \int \mathcal{N}(HX, I \sigma_e^2) \mathcal{N}(0, FF^\top\sigma_w^2) dX $$

I'm trying to evaluate $p(Y)$ by solving this integral. However, so far I'm not able to do it. We know that $p(Y)$ is Gaussian. So we just need to work with the argument of the exponential function. $$p(Y) \propto \int \exp \left(-\frac{1}{2\sigma_e^2} (Y - HX)^\top(Y-HX)\;\;-\frac{1}{2} X^\top(FF^\top\sigma_w^2)^{-1}X\right) dX.$$ It is not clear for me how we can go from this, to $$ p(Y) \propto\exp \left(-\frac{1}{2} Y^\top(HFF^\top H^\top\sigma_w^2 + I \sigma_e^2)^{-1}Y\right) $$

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This can be done by completing the square in $X$ then using the Schur complement. We end up writing $$p(Y) = \int p(Y)p(X|Y)dX = p(Y) \int p(X|Y)dX = p(Y) $$. That is, the trick is to separate $X$ in a quadratic form that corresponds to the posterior distribution. What remains will be independent of X and will give the required PDF.

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