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If we have that $\mathbf{Y \mid \theta} \sim \mathcal{N}_k \left(\mathbf{\theta}, A_1\right)$ and that $\mathbf{\theta} \sim \mathcal{N}_k \left(\mathbf{\mu}, A_2\right)$, where both of these are multivariate normal distributions of dimension $k$, how can I prove that the joint distribution is normal as well?

user136503
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    Just a guess. There might be some conditional expectation involved http://math.stackexchange.com/questions/581229/finding-the-moment-generating-function-of-the-product-of-two-standard-normal-dis – BCLC Dec 12 '15 at 15:27
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    Hint: Let $V = Y - \Theta$, and $Z = \begin{pmatrix}V \ \Theta\end{pmatrix}$. Note that $V|\Theta \sim \mathcal{N}(0, A_1)$. Establish that $Z$ is a jointly gaussian RV using the chain rule. Now use the fact that $\begin{pmatrix}Y \ \Theta\end{pmatrix}$ is a linear transformation of $\begin{pmatrix}V \ \Theta\end{pmatrix}$. – stochasticboy321 Dec 13 '15 at 04:00
  • @stochasticboy321 How exactly do you establish that $Z$ is gaussian using the chain rule? Do you mean a jacobian here is used? – user136503 Dec 14 '15 at 12:18
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    Added an answer, take a look and ask me if you don't understand something. – stochasticboy321 Dec 14 '15 at 19:03

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Let $V=Y−\Theta$, and $Z=\begin{pmatrix} V \\ Θ\end{pmatrix}$. Note that $V|Θ \sim \mathcal{N}(0,A_1)$

\begin{align}P_{V\Theta}(v,\theta) &= P_{V|\Theta}(v|\theta)P_\Theta(\theta)\\ &= \frac{1}{\sqrt{(2\pi)^n|A_1|}} \exp\left(\frac{1}{2} v^TA_1^{-1}v\right)\frac{1}{\sqrt{(2\pi)^n|A_2|}} \exp\left(\frac{1}{2} (\theta - \mu)^TA_2^{-1}(\theta-\mu)\right) \\ \implies P_Z(z) &= \frac{1}{\sqrt{(2\pi)^{2n} |B|}} \exp\left(\frac{1}{2} (z-\hat{\mu})^TB^{-1}(z-\hat{\mu})\right)\end{align} with $\hat{\mu} = \begin{pmatrix} 0 \\ \mu\end{pmatrix}, B = \begin{pmatrix} A_1 & 0 \\ 0 & A_2 \end{pmatrix}$

Thus, $Z$ is a gaussian random vector. Finally, $\begin{pmatrix} Y \\ Θ\end{pmatrix} = \begin{pmatrix} 1 & 1\\0 & 1 \end{pmatrix}Z$.

stochasticboy321
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