Index: The Book of Statistical Proofs ▷ Probability Distributions ▷ Matrix-variate continuous distributions ▷ Matrix-normal distribution ▷ Equivalence to multivariate normal distribution

Theorem: The matrix $X$ is matrix-normally distributed

\[\label{eq:matn} X \sim \mathcal{MN}(M, U, V) \; ,\]

if and only if $\mathrm{vec}(X)$ is multivariate normally distributed

\[\label{eq:mvn} \mathrm{vec}(X) \sim \mathcal{N}(\mathrm{vec}(M), V \otimes U)\]

where $\mathrm{vec}(X)$ is the vectorization operator and $\otimes$ is the Kronecker product.

Proof: The probability density function of the matrix-normal distribution with $n \times p$ mean $M$, $n \times n$ covariance across rows $U$ and $p \times p$ covariance across columns $V$ is

\[\label{eq:matn-pdf} \mathcal{MN}(X; M, U, V) = \frac{1}{\sqrt{(2\pi)^{np} |V|^n |U|^p}} \cdot \exp\left[-\frac{1}{2} \mathrm{tr}\left( V^{-1} (X-M)^\mathrm{T} \, U^{-1} (X-M) \right) \right] \; .\]

Using the trace property $\mathrm{tr}(ABC) = \mathrm{tr}(BCA)$, we have:

\[\label{eq:matn-mvn-s1} \mathcal{MN}(X; M, U, V) = \frac{1}{\sqrt{(2\pi)^{np} |V|^n |U|^p}} \cdot \exp\left[-\frac{1}{2} \mathrm{tr}\left( (X-M)^\mathrm{T} \, U^{-1} (X-M) \, V^{-1} \right) \right] \; .\]

Using the trace-vectorization relation $\mathrm{tr}(A^\mathrm{T} B) = \mathrm{vec}(A)^\mathrm{T} \, \mathrm{vec}(B)$, we have:

\[\label{eq:matn-mvn-s2} \mathcal{MN}(X; M, U, V) = \frac{1}{\sqrt{(2\pi)^{np} |V|^n |U|^p}} \cdot \exp\left[-\frac{1}{2} \mathrm{vec}(X-M)^\mathrm{T} \, \mathrm{vec}\left( U^{-1} (X-M) \, V^{-1} \right) \right] \; .\]

Using the vectorization-Kronecker relation $\mathrm{vec}(ABC) = \left( C^\mathrm{T} \otimes A \right) \mathrm{vec}(B)$, we have:

\[\label{eq:matn-mvn-s3} \mathcal{MN}(X; M, U, V) = \frac{1}{\sqrt{(2\pi)^{np} |V|^n |U|^p}} \cdot \exp\left[-\frac{1}{2} \mathrm{vec}(X-M)^\mathrm{T} \, \left( V^{-1} \otimes U^{-1} \right) \mathrm{vec}(X-M) \right] \; .\]

Using the Kronecker product property $\left( A^{-1} \otimes B^{-1} \right) = \left( A \otimes B \right)^{-1}$, we have:

\[\label{eq:matn-mvn-s4} \mathcal{MN}(X; M, U, V) = \frac{1}{\sqrt{(2\pi)^{np} |V|^n |U|^p}} \cdot \exp\left[-\frac{1}{2} \mathrm{vec}(X-M)^\mathrm{T} \, \left( V \otimes U \right)^{-1} \mathrm{vec}(X-M) \right] \; .\]

Using the vectorization property $\mathrm{vec}(A+B) = \mathrm{vec}(A) + \mathrm{vec}(B)$, we have:

\[\label{eq:matn-mvn-s5} \mathcal{MN}(X; M, U, V) = \frac{1}{\sqrt{(2\pi)^{np} |V|^n |U|^p}} \cdot \exp\left[-\frac{1}{2} \left[ \mathrm{vec}(X) - \mathrm{vec}(M) \right]^\mathrm{T} \, \left( V \otimes U \right)^{-1} \left[ \mathrm{vec}(X) - \mathrm{vec}(M) \right] \right] \; .\]

Using the Kronecker-determinant relation $\lvert A \otimes B \rvert = \lvert A \rvert^m \lvert B \rvert^n$, we have:

\[\label{eq:matn-mvn-s6} \mathcal{MN}(X; M, U, V) = \frac{1}{\sqrt{(2\pi)^{np} |V \otimes U|}} \cdot \exp\left[-\frac{1}{2} \left[ \mathrm{vec}(X) - \mathrm{vec}(M) \right]^\mathrm{T} \, \left( V \otimes U \right)^{-1} \left[ \mathrm{vec}(X) - \mathrm{vec}(M) \right] \right] \; .\]

This is the probability density function of the multivariate normal distribution with the $np \times 1$ mean vector $\mathrm{vec}(M)$ and the $np \times np$ covariance matrix $V \otimes U$:

\[\label{eq:matn-mvn} \mathcal{MN}(X; M, U, V) = \mathcal{N}(\mathrm{vec}(X); \mathrm{vec}(M), V \otimes U) \; .\]

By showing that the probability density functions are identical, it is proven that the associated probability distributions are equivalent.

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Metadata: ID: P26 | shortcut: matn-mvn | author: JoramSoch | date: 2020-01-20, 21:09.