Index: The Book of Statistical ProofsProbability Distributions ▷ Matrix-variate continuous distributions ▷ Matrix-normal distribution ▷ Drawing samples

Theorem: Let $X \in \mathbb{R}^{n \times p}$ be a random matrix with all entries independently following a standard normal distribution. Moreover, let $A \in \mathbb{R}^{n \times n}$ and $B \in \mathbb{R}^{p \times p}$, such that $A A^\mathrm{T} = U$ and $B^\mathrm{T} B = V$.

Then, $Y = M + A X B$ follows a matrix-normal distribution with mean $M$, covariance across rows $U$ and covariance across rows $U$:

$\label{eq:matn-samp} Y = M + A X B \sim \mathcal{MN}(M, U, V) \; .$

Proof: If all entries of $X$ are independent and standard normally distributed

$\label{eq:xij-dist} x_{ij} \overset{\mathrm{i.i.d.}}{\sim} \mathcal{N}(0, 1) \quad \text{for all} \quad i = 1,\ldots,n \quad \text{and} \quad j = 1,\ldots,p \; ,$ $\label{eq:vecX-dist} \begin{split} \mathrm{vec}(X) &\sim \mathcal{N}\left(\mathrm{vec}(0_{np}), I_{np} \right) \\ &\sim \mathcal{N}\left(\mathrm{vec}(0_{np}), I_p \otimes I_n \right) \end{split}$

where $0_{np}$ is an $n \times p$ matrix of zeros and $I_n$ is the $n \times n$ identity matrix.

Due to the relationship between multivariate and matrix-normal distribution, we have:

$\label{eq:X-dist} X \sim \mathcal{MN}(0_{np}, I_n, I_p) \; .$

Thus, with the linear transformation theorem for the matrix-normal distribution, it follows that

$\label{eq:matn-samp-qed} \begin{split} Y = M + AXB &\sim \mathcal{MN}\left(M + A 0_{np} B, A I_n A^\mathrm{T}, B^\mathrm{T} I_p B \right) \\ &\sim \mathcal{MN}\left(M, A A^\mathrm{T}, B^\mathrm{T} B \right) \\ &\sim \mathcal{MN}\left(M, U, V \right) \; . \end{split}$

Thus, given $X$ defined by \eqref{eq:xij-dist}, $Y$ defined by \eqref{eq:matn-samp} is a sample from $\mathcal{MN}\left(M, U, V \right)$.

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Metadata: ID: P297 | shortcut: matn-samp | author: JoramSoch | date: 2021-12-07, 08:43.