I'm told that$\mathbf{\Sigma}$ is a positive definite matrix and also that it's the variance of$\mathbf{X}$ where$\mathbf{X}=\bf{\mu}+BZ$ such that$\mathbf{\mu},\mathbf{Z}\in\mathbb{R}^n$ and$Z_1,\dots,Z_n\sim_{\mathrm{iid}}\mathcal{N}(0,1)$. In addition, I'm to partition the matrix$\bf\Sigma$ in the following way:
$$ \bf \Sigma= \begin{pmatrix} \bf \Sigma_p & \bf \Sigma_r \\ \bf \Sigma_r^{\top} & \bf \Sigma_q \end{pmatrix}. $$
So the question is to prove that the marginal vectors$\mathbf{X}_p$ and$\mathbf{X}_q$ are also multivariate normal, with$ \mathbf{X}_p\sim\mathcal{N}(\mu_p,\mathbf{\Sigma}_p)$ and$ \mathbf{X}_q\sim\mathcal{N}(\mu_q,\mathbf{\Sigma}_q)$.
I've tried to simply rewrite the expression for$\bf X$ in terms of the marginal vectors and use matrix multiplication to equate components but it seems to be a dead end because I can't simplify anything down or at least I dont know how to simplify everything. In addition I've tried showing that the MGF of the marginal vectors will have be the one for the multivariate gaussian but my working unfortunately seems to not go anywhere. Any help would be appreciated, thank you.
1 Answer1
$$X_p=AX=A\mu+(AB)Z,$$where$$A=\begin{bmatrix}I_{p\times p} & 0_{p\times q}\end{bmatrix}.$$Therefore,$X_p$ is normally distributed with$\mathsf{E}X_p=A\mu=\mu_p$ and$\operatorname{Var}(X_p)=A\Sigma A^{\top}=\Sigma_p.$
- $\begingroup$I like this proof it's really slick. I think I was overthinking on how to go about it. In saying that, it seems like you jump over the way $\mathbf{\Sigma}$ was supposed to be partitioned? How did you know to ignore the way it was partitioned (at least indirectly)? Thanks!$\endgroup$Alias K– Alias K2020-09-19 09:55:14 +00:00CommentedSep 19, 2020 at 9:55
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