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

Theorem: Let $X$ be a random matrix following a matrix-normal distribution:

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

Then, the transpose of $X$ also has a matrix-normal distribution:

\[\label{eq:matn-trans} X^\mathrm{T} \sim \mathcal{MN}(M^\mathrm{T}, V, U) \; .\]

Proof: The probability density function of the matrix-normal distribution is:

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

Define $Y = X^\mathrm{T}$. Then, $X = Y^\mathrm{T}$ and we can substitute:

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

Using $(A+B)^\mathrm{T} = (A^\mathrm{T} + B^\mathrm{T})$, we have:

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

Using $\mathrm{tr}(ABC) = \mathrm{tr}(CAB)$, we obtain

\[\label{eq:matn-pdf-Y-s3} f(Y) = \frac{1}{\sqrt{(2\pi)^{np} |V|^n |U|^p}} \cdot \exp\left[-\frac{1}{2} \mathrm{tr}\left( U^{-1} (Y-M^\mathrm{T})^\mathrm{T} \, V^{-1} (Y-M^\mathrm{T}) \right) \right]\]

which is the probability density function of a matrix-normal distribution with mean $M^T$, covariance across rows $V$ and covariance across columns $U$.

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Metadata: ID: P144 | shortcut: matn-trans | author: JoramSoch | date: 2020-08-03, 22:21.