Index: The Book of Statistical ProofsStatistical Models ▷ Multivariate normal data ▷ General linear model ▷ Maximum likelihood estimation

Theorem: Given a general linear model with matrix-normally distributed errors

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

maximum likelihood estimates for the unknown parameters $B$ and $\Sigma$ are given by

$\label{eq:GLM-MLE} \begin{split} \hat{B} &= (X^\mathrm{T} V^{-1} X)^{-1} X^\mathrm{T} V^{-1} Y \\ \hat{\Sigma} &= \frac{1}{n} (Y - X\hat{B})^\mathrm{T} V^{-1} (Y - X\hat{B}) \; . \\ \end{split}$

Proof: In \eqref{eq:GLM}, $Y$ is an $n \times v$ matrix of measurements ($n$ observations, $v$ dependent variables), $X$ is an $n \times p$ design matrix ($n$ observations, $p$ independent variables) and $V$ is an $n \times n$ covariance matrix across observations. This multivariate GLM implies the following likelihood function

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

and the log-likelihood function

$\label{eq:GLM-LL1} \begin{split} \mathrm{LL}(B,\Sigma) = &\log p(Y|B,\Sigma) \\ = &- \frac{nv}{2} \log(2\pi) - \frac{n}{2} \log |\Sigma| - \frac{v}{2} \log |V| \\ &- \frac{1}{2} \, \mathrm{tr}\left[ \Sigma^{-1} (Y - XB)^\mathrm{T} V^{-1} (Y - XB) \right] \; .\\ \end{split}$

Substituting $V^{-1}$ by the precision matrix $P$ to ease notation, we have:

$\label{eq:GLM-LL2} \begin{split} \mathrm{LL}(B,\Sigma) = &- \frac{nv}{2} \log(2\pi) - \frac{n}{2} \log(|\Sigma|) - \frac{v}{2} \log(|V|) \\ &- \frac{1}{2} \, \mathrm{tr}\left[ \Sigma^{-1} \left( Y^\mathrm{T} P Y - Y^\mathrm{T} P X B - B^\mathrm{T} X^\mathrm{T} P Y + B^\mathrm{T} X^\mathrm{T} P X B \right) \right] \; .\\ \end{split}$

The derivative of the log-likelihood function \eqref{eq:GLM-LL2} with respect to $B$ is

$\label{eq:dLL-dB} \begin{split} \frac{\mathrm{d}\mathrm{LL}(B,\Sigma)}{\mathrm{d}B} &= \frac{\mathrm{d}}{\mathrm{d}B} \left( - \frac{1}{2} \, \mathrm{tr}\left[ \Sigma^{-1} \left( Y^\mathrm{T} P Y - Y^\mathrm{T} P X B - B^\mathrm{T} X^\mathrm{T} P Y + B^\mathrm{T} X^\mathrm{T} P X B \right) \right] \right) \\ &= \frac{\mathrm{d}}{\mathrm{d}B} \left( -\frac{1}{2} \, \mathrm{tr}\left[ -2 \Sigma^{-1} Y^\mathrm{T} P X B \right] \right) + \frac{\mathrm{d}}{\mathrm{d}B} \left( -\frac{1}{2} \, \mathrm{tr}\left[ \Sigma^{-1} B^\mathrm{T} X^\mathrm{T} P X B \right] \right) \\ &= - \frac{1}{2} \left( -2 X^\mathrm{T} P Y \Sigma^{-1} \right) - \frac{1}{2} \left( X^\mathrm{T} P X B \Sigma^{-1} + (X^\mathrm{T} P X)^\mathrm{T} B (\Sigma^{-1})^\mathrm{T} \right) \\ &= X^\mathrm{T} P Y \Sigma^{-1} - X^\mathrm{T} P X B \Sigma^{-1} \\ \end{split}$

and setting this derivative to zero gives the MLE for $B$:

$\label{eq:B-MLE} \begin{split} \frac{\mathrm{d}\mathrm{LL}(\hat{B},\Sigma)}{\mathrm{d}B} &= 0 \\ 0 &= X^\mathrm{T} P Y \Sigma^{-1} - X^\mathrm{T} P X \hat{B} \Sigma^{-1} \\ 0 &= X^\mathrm{T} P Y - X^\mathrm{T} P X \hat{B} \\ X^\mathrm{T} P X \hat{B} &= X^\mathrm{T} P Y \\ \hat{B} &= \left( X^\mathrm{T} P X \right)^{-1} X^\mathrm{T} P Y \\ \end{split}$

The derivative of the log-likelihood function \eqref{eq:GLM-LL1} at $\hat{B}$ with respect to $\Sigma$ is

$\label{eq:dLL-dS} \begin{split} \frac{\mathrm{d}\mathrm{LL}(\hat{B},\Sigma)}{\mathrm{d}\Sigma} &= \frac{\mathrm{d}}{\mathrm{d}\Sigma} \left( - \frac{n}{2} \log |\Sigma| - \frac{1}{2} \, \mathrm{tr}\left[ \Sigma^{-1} (Y - X\hat{B})^\mathrm{T} V^{-1} (Y - X\hat{B}) \right] \right) \\ &= - \frac{n}{2} \left( \Sigma^{-1} \right)^\mathrm{T} + \frac{1}{2} \left( \Sigma^{-1} (Y - X\hat{B})^\mathrm{T} V^{-1} (Y - X\hat{B}) \, \Sigma^{-1} \right)^\mathrm{T} \\ &= - \frac{n}{2} \, \Sigma^{-1} + \frac{1}{2} \, \Sigma^{-1} (Y - X\hat{B})^\mathrm{T} V^{-1} (Y - X\hat{B}) \, \Sigma^{-1} \\ \end{split}$

and setting this derivative to zero gives the MLE for $\Sigma$:

$\label{eq:S-MLE} \begin{split} \frac{\mathrm{d}\mathrm{LL}(\hat{B},\hat{\Sigma})}{\mathrm{d}\Sigma} &= 0 \\ 0 &= - \frac{n}{2} \, \hat{\Sigma}^{-1} + \frac{1}{2} \, \hat{\Sigma}^{-1} (Y - X\hat{B})^\mathrm{T} V^{-1} (Y - X\hat{B}) \, \hat{\Sigma}^{-1} \\ \frac{n}{2} \, \hat{\Sigma}^{-1} &= \frac{1}{2} \, \hat{\Sigma}^{-1} (Y - X\hat{B})^\mathrm{T} V^{-1} (Y - X\hat{B}) \, \hat{\Sigma}^{-1} \\ \hat{\Sigma}^{-1} &= \frac{1}{n} \, \hat{\Sigma}^{-1} (Y - X\hat{B})^\mathrm{T} V^{-1} (Y - X\hat{B}) \, \hat{\Sigma}^{-1} \\ I_v &= \frac{1}{n} \, (Y - X\hat{B})^\mathrm{T} V^{-1} (Y - X\hat{B}) \, \hat{\Sigma}^{-1} \\ \hat{\Sigma} &= \frac{1}{n} \, (Y - X\hat{B})^\mathrm{T} V^{-1} (Y - X\hat{B}) \\ \end{split}$

Together, \eqref{eq:B-MLE} and \eqref{eq:S-MLE} constitute the MLE for the GLM.

Sources:

Metadata: ID: P7 | shortcut: glm-mle | author: JoramSoch | date: 2019-12-06, 10:40.