Index: The Book of Statistical ProofsStatistical Models ▷ Multivariate normal data ▷ Multivariate Bayesian linear regression ▷ Conjugate prior distribution

Theorem: Let

$\label{eq:GLM} Y = X B + E, \; E \sim \mathcal{MN}(0, V, \Sigma)$

be a general linear model with measured $n \times v$ data matrix $Y$, known $n \times p$ design matrix $X$, known $n \times n$ covariance structure $V$ as well as unknown $p \times v$ regression coefficients $B$ and unknown $v \times v$ noise covariance $\Sigma$.

Then, the conjugate prior for this model is a normal-Wishart distribution

$\label{eq:GLM-NW-prior} p(B,T) = \mathcal{MN}(B; M_0, \Lambda_0^{-1}, T^{-1}) \cdot \mathcal{W}(T; \Omega_0^{-1}, \nu_0)$

where $T = \Sigma^{-1}$ is the inverse noise covariance or noise precision matrix.

Proof: By definition, a conjugate prior is a prior distribution that, when combined with the likelihood function, leads to a posterior distribution that belongs to the same family of probability distributions. This is fulfilled when the prior density and the likelihood function are proportional to the model parameters in the same way, i.e. the model parameters appear in the same functional form in both.

Equation \eqref{eq:GLM} implies the following likelihood function

$\label{eq:GLM-LF-Class} p(Y|B,\Sigma) = \mathcal{MN}(Y; X B, V, \Sigma) = \sqrt{\frac{1}{(2 \pi)^{nv} |\Sigma|^n |V|^v}} \, \exp\left[ -\frac{1}{2} \mathrm{tr}\left( \Sigma^{-1} (Y-XB)^\mathrm{T} V^{-1} (Y-XB) \right) \right]$

which, for mathematical convenience, can also be parametrized as

$\label{eq:GLM-LF-Bayes} p(Y|B,T) = \mathcal{MN}(Y; X B, P, T^{-1}) = \sqrt{\frac{|T|^n |P|^v}{(2 \pi)^{nv}}} \, \exp\left[ -\frac{1}{2} \mathrm{tr}\left( T (Y-XB)^\mathrm{T} P (Y-XB) \right) \right]$

using the $v \times v$ precision matrix $T = \Sigma^{-1}$ and the $n \times n$ precision matrix $P = V^{-1}$.

Seperating constant and variable terms, we have:

$\label{eq:GLM-LF-s1} p(Y|B,T) = \sqrt{\frac{|P|^v}{(2 \pi)^{nv}}} \cdot |T|^{n/2} \cdot \exp\left[ -\frac{1}{2} \mathrm{tr}\left( T (Y-XB)^\mathrm{T} P (Y-XB) \right) \right] \; .$

Expanding the product in the exponent, we have:

$\label{eq:GLM-LF-s2} p(Y|B,T) = \sqrt{\frac{|P|^v}{(2 \pi)^{nv}}} \cdot |T|^{n/2} \cdot \exp\left[ -\frac{1}{2} \mathrm{tr}\left( T \left[ Y^\mathrm{T} P Y - Y^\mathrm{T} P X B - B^\mathrm{T} X^\mathrm{T} P Y + B^\mathrm{T} X^\mathrm{T} P X B \right] \right) \right] \; .$

Completing the square over $\beta$, finally gives

$\label{eq:GLM-LF-s3} p(Y|B,T) = \sqrt{\frac{|P|^v}{(2 \pi)^{nv}}} \cdot |T|^{n/2} \cdot \exp\left[ -\frac{1}{2} \mathrm{tr}\left( T \left[ (B - \tilde{X}Y)^\mathrm{T} X^\mathrm{T} P X (B - \tilde{X}Y) - Y^\mathrm{T} Q Y + Y^\mathrm{T} P Y \right] \right) \right]$

where $\tilde{X} = \left( X^\mathrm{T} P X \right)^{-1} X^\mathrm{T} P$ and $Q = \tilde{X}^\mathrm{T} \left( X^\mathrm{T} P X \right) \tilde{X}$.

In other words, the likelihood function is proportional to a power of the determinant of $T$, times an exponential of the trace of $T$ and an exponential of the trace of a squared form of $B$, weighted by $T$:

$\label{eq:GLM-LF-s4} p(Y|B,T) \propto |T|^{n/2} \cdot \exp\left[ -\frac{1}{2} \mathrm{tr}\left( T \left[ Y^\mathrm{T} P Y - Y^\mathrm{T} Q Y \right] \right) \right] \cdot \exp\left[ -\frac{1}{2} \mathrm{tr}\left( T \left[ (B - \tilde{X}Y)^\mathrm{T} X^\mathrm{T} P X (B - \tilde{X}Y) \right] \right) \right] \; .$

The same is true for a normal-Wishart distribution over $B$ and $T$

$\label{eq:MBLR-prior-s1} p(B,T) = \mathcal{MN}(B; M_0, \Lambda_0^{-1}, T^{-1}) \cdot \mathcal{W}(T; \Omega_0^{-1}, \nu_0)$ $\label{eq:MBLR-prior-s2} p(B,T) = \sqrt{\frac{|T|^p |\Lambda_0|^v}{(2 \pi)^{pv}}} \, \exp\left[ -\frac{1}{2} \mathrm{tr}\left( T (B-M_0)^\mathrm{T} \Lambda_0 (B-M_0) \right) \right] \cdot \frac{1}{\Gamma_v \left( \frac{\nu_0}{2} \right)} \sqrt{\frac{|\Omega_0|^{\nu_0}}{2^{\nu_0 v}}} |T|^{(\nu_0-v-1)/2} \exp\left[ -\frac{1}{2} \mathrm{tr}\left( \Omega_0 T \right) \right]$

exhibits the same proportionality

$\label{eq:MBLR-prior-s3} p(B,T) \propto |T|^{(\nu_0+p-v-1)/2} \cdot \exp\left[ -\frac{1}{2} \mathrm{tr}\left( T \Omega_0 \right) \right] \cdot \exp\left[ -\frac{1}{2} \mathrm{tr}\left( T \left[ (B-M_0)^\mathrm{T} \Lambda_0 (B-M_0) \right] \right) \right]$

and is therefore conjugate relative to the likelihood.

Sources:

Metadata: ID: P159 | shortcut: mblr-prior | author: JoramSoch | date: 2020-09-03, 07:33.