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Question: I am wondering if there was a way to prove this result without using probability densities:

If $\bf x \sim \mathcal N (m, P)$ and $\bf y \;|\; x \sim \mathcal N (Hx, R)$, then $$\begin{pmatrix} \bf x \\ \bf y \end{pmatrix} \sim \mathcal{N} \left( \begin{pmatrix} \bf m \\ \bf Hm \end{pmatrix}, \begin{pmatrix} \bf P & \bf P H^{\top} \\ \bf HP & \bf H P H^{\top} + R \end{pmatrix} \right)$$

I came across the result here on slide 14 some time ago. A sketch of a proof in the univariate case can be found here.

The author of the slides uses the term "Gaussian densities" so the covariance matrices are non-singular. The author's choice of words may be harmless but it made me wonder about how a proof of the result would be when one cannot use probability densities.

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Because $(\bf y \;|\; x) \sim \mathcal N (Hx, R)$, $\bf y$ can be written as $$\bf y = \bf Hx + \bf \epsilon$$ where $\bf \epsilon \sim \mathcal N (0, R)$ and independent with $\bf x$. It means $$\begin{pmatrix} \bf x \\ \bf \epsilon \end{pmatrix} \sim \mathcal N \left (\begin{pmatrix} \bf m \\ \bf 0 \end{pmatrix}, \begin{pmatrix} \bf P & 0 \\ \bf 0 &\bf R \end{pmatrix} \right ) $$

Then $$ \begin{pmatrix} \bf x \\ \bf y \end{pmatrix} = \begin{pmatrix} \bf I & 0 \\ \bf H &\bf I \end{pmatrix} \begin{pmatrix} \bf x \\ \bf \epsilon \end{pmatrix} $$ Following the fact that $$AY\sim N(A\mu, A\Sigma A') \text { given that } Y\sim N(\mu, \Sigma)$$ The results are there.

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