Index: The Book of Statistical ProofsProbability Distributions ▷ Multivariate continuous distributions ▷ Normal-gamma distribution ▷ Special case of normal-Wishart distribution

Theorem: The normal-gamma distribution is a special case of the normal-Wishart distribution where the number of columns of the random matrices is $p = 1$.

Proof: Let $X$ be an $n \times p$ real matrix and let $Y$ be a $p \times p$ positive-definite symmetric matrix, such that $X$ and $Y$ jointly follow a normal-Wishart distribution:

$\label{eq:nw} X,Y \sim \mathrm{NW}(M, U, V, \nu) \; .$

Then, $X$ and $Y$ are described by the probability density function

$\label{eq:nw-pdf} \begin{split} p(X,Y) = \; & \frac{1}{\sqrt{(2\pi)^{np} |U|^p |V|^{\nu}}} \cdot \frac{\sqrt{2^{-\nu p}}}{\Gamma_p \left( \frac{\nu}{2} \right)} \cdot |Y|^{(\nu+n-p-1)/2} \cdot \\ & \exp\left[-\frac{1}{2} \mathrm{tr}\left( Y \left[ (X-M)^\mathrm{T} \, U^{-1} (X-M) + V^{-1} \right] \right) \right] \end{split}$

where $\lvert A \rvert$ is a matrix determinant, $A^{-1}$ is a matrix inverse and $\Gamma_p(x)$ is the multivariate gamma function of order $p$. If $p = 1$, then $\Gamma_p(x) = \Gamma(x)$ is the ordinary gamma function, $x = X$ is a column vector and $y = Y$ is a real number. Thus, the probability density function of $x$ and $y$ can be developed as

$\label{eq:ng-pdf-s1} \begin{split} p(x,y) = \; & \frac{1}{\sqrt{(2\pi)^n |U| |V|^{\nu}}} \cdot \frac{\sqrt{2^{-\nu}}}{\Gamma \left( \frac{\nu}{2} \right)} \cdot y^{(\nu+n-2)/2} \cdot \\ & \exp\left[-\frac{1}{2} \mathrm{tr}\left( y \left[ (x-M)^\mathrm{T} \, U^{-1} (x-M) + V^{-1} \right] \right) \right] \\ = \; & \sqrt{\frac{|U^{-1}|}{(2\pi)^n}} \cdot \frac{\sqrt{\left(2 |V|\right)^{-\nu}}}{\Gamma \left( \frac{\nu}{2} \right)} \cdot y^{\frac{\nu}{2}+\frac{n}{2}-1} \cdot \\ & \exp\left[-\frac{1}{2} \left( y \left[ (x-M)^\mathrm{T} \, U^{-1} (x-M) + 2 \left(2 V\right)^{-1} \right] \right) \right] \\ = \; & \sqrt{\frac{|U^{-1}|}{(2\pi)^n}} \cdot \frac{\left(\frac{1}{2 |V|}\right)^{\frac{\nu}{2}}}{\Gamma \left( \frac{\nu}{2} \right)} \cdot y^{\frac{\nu}{2}+\frac{n}{2}-1} \cdot \\ & \exp\left[-\frac{y}{2} \left( (x-M)^\mathrm{T} \, U^{-1} (x-M) + 2 \left(\frac{1}{2 V}\right) \right) \right] \\ \end{split}$

In the matrix-normal distribution, we have $M \in \mathbb{R}^{n \times p}$, $U \in \mathbb{R}^{n \times n}$, $V \in \mathbb{R}^{p \times p}$ and $\nu \in \mathbb{R}$. Thus, with $p = 1$, $M$ becomes a column vector and $V$ becomes a real number, such that $V = \lvert V \rvert = 1/V^{-1}$. Finally, substituting $\mu = M$, $\Lambda = U^{-1}$, $a = \frac{\nu}{2}$ and $b = \frac{1}{2V}$, we get

$\label{eq:ng-pdf-s2} p(x,y) = \sqrt{\frac{|\Lambda|}{(2 \pi)^n}} \frac{b^a}{\Gamma(a)} \cdot y^{a+\frac{n}{2}-1} \exp \left[ -\frac{y}{2} \left( (x-\mu)^\mathrm{T} \Lambda (x-\mu) + 2b \right) \right]$
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Metadata: ID: P324 | shortcut: ng-nw | author: JoramSoch | date: 2022-05-20, 18:23.