Index: The Book of Statistical ProofsStatistical Models ▷ Univariate normal data ▷ Bayesian linear regression ▷ Accuracy and complexity

Theorem: Let

$\label{eq:GLM} m: \; y = X \beta + \varepsilon, \; \varepsilon \sim \mathcal{N}(0, \sigma^2 V)$

be a linear regression model with measured $n \times 1$ data vector $y$, known $n \times p$ design matrix $X$, known $n \times n$ covariance structure $V$ as well as unknown $p \times 1$ regression coefficients $\beta$ and unknown noise variance $\sigma^2$. Moreover, assume a normal-gamma prior distribution over the model parameters $\beta$ and $\tau = 1/\sigma^2$:

$\label{eq:GLM-NG-prior} p(\beta,\tau) = \mathcal{N}(\beta; \mu_0, (\tau \Lambda_0)^{-1}) \cdot \mathrm{Gam}(\tau; a_0, b_0) \; .$

Then, accuracy and complexity of this model are

$\label{eq:GLM-NG-AnC} \begin{split} \mathrm{Acc}(m) = - &\frac{1}{2} \frac{a_n}{b_n} (y-X\mu_n)^\mathrm{T} P (y-X\mu_n) - \frac{1}{2} \mathrm{tr}(X^\mathrm{T} P X \Lambda_n^{-1}) \\ + &\frac{1}{2} \log |P| - \frac{n}{2} \log (2 \pi) + \frac{n}{2} (\psi(a_n) - \log(b_n)) \\ \mathrm{Com}(m) = \hphantom{+} &\frac{1}{2} \frac{a_n}{b_n} \left[(\mu_0-\mu_n)^\mathrm{T} \Lambda_0 (\mu_0-\mu_n) - 2(b_n-b_0)\right] + \frac{1}{2} \mathrm{tr}(\Lambda_0 \Lambda_n^{-1}) - \frac{1}{2} \log \frac{|\Lambda_0|}{|\Lambda_n|} - \frac{p}{2} \\ + & \, a_0 \log \frac{b_n}{b_0} - \log \frac{\Gamma(a_n)}{\Gamma(a_0)} + (a_n - a_0) \psi(a_n) \; . \end{split}$

where $\mu_n$, $\Lambda_n$, $a_n$ and $b_n$ are the posterior hyperparameters for Bayesian linear regression and $P$ is the data precision matrix: $P = V^{-1}$.

Proof: Model accuracy and complexity are defined as

$\label{eq:lme-anc} \begin{split} \mathrm{LME}(m) &= \mathrm{Acc}(m) - \mathrm{Com}(m) \\ \mathrm{Acc}(m) &= \left\langle \log p(y|\beta,\tau,m) \right\rangle_{p(\beta,\tau|y,m)} \\ \mathrm{Com}(m) &= \mathrm{KL} \left[ p(\beta,\tau|y,m) \, || \, p(\beta,\tau|m) \right] \; . \end{split}$

1) The accuracy term is the expectation of the log-likelihood function $\log p(y|\beta,\tau)$ with respect to the posterior distribution $p(\beta,\tau|y)$. This expectation can be rewritten as:

$\label{eq:GLM-NG-Acc-s1} \begin{split} \mathrm{Acc}(m) &= \iint p(\beta,\tau|y) \, \log p(y|\beta,\tau) \, \mathrm{d}\beta \, \mathrm{d}\tau \\ &= \int p(\tau|y) \int p(\beta|\tau,y) \, \log p(y|\beta,\tau) \, \mathrm{d}\beta \, \mathrm{d}\tau \\ &= \left\langle \left\langle \log p(y|\beta,\tau) \right\rangle_{p(\beta|\tau,y)} \right\rangle_{p(\tau|y)} \; . \end{split}$

With the log-likelihood function for multiple linear regression, we have:

$\label{eq:GLM-NG-Acc-s2} \begin{split} \mathrm{Acc}(m) &= \left\langle \left\langle \log \left( \sqrt{\frac{1}{(2\pi)^n |\sigma^2 V|}} \cdot \exp\left[ -\frac{1}{2} (y - X\beta)^\mathrm{T} (\sigma^2 V)^{-1} (y - X\beta) \right] \right) \right\rangle_{p(\beta|\tau,y)} \right\rangle_{p(\tau|y)} \\ &= \left\langle \left\langle \log \left( \sqrt{\frac{\tau^n |P|}{(2\pi)^n}} \cdot \exp\left[ -\frac{1}{2} (y - X\beta)^\mathrm{T} (\tau P) (y - X\beta) \right] \right) \right\rangle_{p(\beta|\tau,y)} \right\rangle_{p(\tau|y)} \\ &= \left\langle \left\langle \frac{1}{2} \log |P| + \frac{n}{2} \log \tau - \frac{n}{2} \log (2 \pi) - \frac{1}{2} (y-X\beta)^\mathrm{T} (\tau P) (y-X\beta) \right\rangle_{p(\beta|\tau,y)} \right\rangle_{p(\tau|y)} \\ &= \left\langle \left\langle \frac{1}{2} \log |P| + \frac{n}{2} \log \tau - \frac{n}{2} \log (2 \pi) - \frac{\tau}{2} \left[ y^\mathrm{T} P y - 2 y^\mathrm{T} P X \beta + \beta^\mathrm{T} X^\mathrm{T} P X \beta \right] \right\rangle_{p(\beta|\tau,y)} \right\rangle_{p(\tau|y)} \; . \end{split}$

With the posterior distribution for Bayesian linear regression, this becomes:

$\label{eq:GLM-NG-Acc-s3} \mathrm{Acc}(m) = \left\langle \left\langle \frac{1}{2} \log |P| + \frac{n}{2} \log \tau - \frac{n}{2} \log (2 \pi) - \frac{\tau}{2} \left[ y^\mathrm{T} P y - 2 y^\mathrm{T} P X \beta + \beta^\mathrm{T} X^\mathrm{T} P X \beta \right] \right\rangle_{\mathcal{N}(\beta; \mu_n, (\tau \Lambda_n)^{-1})} \right\rangle_{\mathrm{Gam}(\tau; a_n, b_n)} \; .$

If $x \sim \mathrm{N}(\mu, \Sigma)$, then its expected value is

$\label{eq:mvn-mean} \left\langle x \right\rangle = \mu$ $\label{eq:mvn-meansqr} \left\langle x^\mathrm{T} A x \right\rangle = \mu^\mathrm{T} A \mu + \mathrm{tr}(A \Sigma) \; .$

Thus, the model accuracy of $m$ evaluates to:

$\label{eq:GLM-NG-Acc-s4} \begin{split} \mathrm{Acc}(m) &= \left\langle \frac{1}{2} \log |P| + \frac{n}{2} \log \tau - \frac{n}{2} \log (2 \pi) - \right. \\ &\hphantom{= -} \left. \frac{\tau}{2} \left[ y^\mathrm{T} P y - 2 y^\mathrm{T} P X \mu_n + \mu_n^\mathrm{T} X^\mathrm{T} P X \mu_n + \frac{1}{\tau} \mathrm{tr}(X^\mathrm{T} P X \Lambda_n^{-1}) \right] \right\rangle_{\mathrm{Gam}(\tau; a_n, b_n)} \\ &= \left\langle \frac{1}{2} \log |P| + \frac{n}{2} \log \tau - \frac{n}{2} \log (2 \pi) - \frac{\tau}{2} (y-X\mu_n)^\mathrm{T} P (y-X\mu_n) - \frac{1}{2} \mathrm{tr}(X^\mathrm{T} P X \Lambda_n^{-1}) \right\rangle_{\mathrm{Gam}(\tau; a_n, b_n)} \; . \end{split}$

If $x \sim \mathrm{Gam}(a, b)$, then its expected value is

$\label{eq:gam-mean} \left\langle x \right\rangle = \frac{a}{b}$ $\label{eq:gan-logmean} \left\langle \log x \right\rangle = \psi(a) - \log(b) \; .$

Thus, the model accuracy of $m$ evaluates to

$\label{eq:GLM-NG-Acc-s5} \begin{split} \mathrm{Acc}(m) = & - \frac{1}{2} \frac{a_n}{b_n} (y-X\mu_n)^\mathrm{T} P (y-X\mu_n) - \frac{1}{2} \mathrm{tr}(X^\mathrm{T} P X \Lambda_n^{-1}) \\ & + \frac{1}{2} \log |P| - \frac{n}{2} \log (2 \pi) + \frac{n}{2} (\psi(a_n) - \log(b_n)) \end{split}$

which proofs the first part of \eqref{eq:GLM-NG-AnC}.

2) The complexity penalty is the Kullback-Leibler divergence of the posterior distribution $p(\beta,\tau|y)$ from the prior distribution $p(\beta,\tau)$. This can be rewritten as follows:

$\label{eq:GLM-NG-Com-s1} \begin{split} \mathrm{Com}(m) &= \iint p(\beta,\tau|y) \, \log \frac{p(\beta,\tau|y)}{p(\beta,\tau)} \, \mathrm{d}\beta \, \mathrm{d}\tau \\ &= \iint p(\beta|\tau,y) \, p(\tau|y) \, \log \left[ \frac{p(\beta|\tau,y)}{p(\beta|\tau)} \, \frac{p(\tau|y)}{p(\tau)} \right] \, \mathrm{d}\beta \, \mathrm{d}\tau \\ &= \int p(\tau|y) \int p(\beta|\tau,y) \, \log \frac{p(\beta|\tau,y)}{p(\beta|\tau)} \, \mathrm{d}\beta \, \mathrm{d}\tau + \int p(\tau|y) \, \log \frac{p(\tau|y)}{p(\tau)} \int p(\beta|\tau,y) \, \mathrm{d}\beta \, \mathrm{d}\tau \\ &= \left\langle \mathrm{KL} \left[ p(\beta|\tau,y)\,||\,p(\beta|\tau) \right] \right\rangle_{p(\tau|y)} + \mathrm{KL} \left[ p(\tau|y)\,||\,p(\tau) \right] \; . \end{split}$

With the prior distribution given by \eqref{eq:GLM-NG-prior} and the posterior distribution for Bayesian linear regression, this becomes:

$\label{eq:GLM-NG-Com-s2} \begin{split} \mathrm{Com}(m) &= \left\langle \mathrm{KL} \left[ \mathcal{N}(\beta; \mu_n, (\tau \Lambda_n)^{-1})\,||\,\mathcal{N}(\beta; \mu_0, (\tau \Lambda_0)^{-1}) \right] \right\rangle_{\mathrm{Gam}(\tau; a_n, b_n)} \\ &+ \mathrm{KL} \left[ \mathrm{Gam}(\tau; a_n, b_n)\,||\,\mathrm{Gam}(\tau; a_0, b_0) \right] \; . \end{split}$ $\label{eq:mvn-kl} \mathrm{KL}[\mathcal{N}(\mu_1, \Sigma_1)\,||\,\mathcal{N}(\mu_2, \Sigma_2)] = \frac{1}{2} \left[ (\mu_2 - \mu_1)^\mathrm{T} \Sigma_2^{-1} (\mu_2 - \mu_1) + \mathrm{tr}(\Sigma_2^{-1} \Sigma_1) - \ln \frac{|\Sigma_1|}{|\Sigma_2|} - n \right]$ $\label{eq:gam-kl} \mathrm{KL}[\mathrm{Gam}(a_1, b_1)\,||\,\mathrm{Gam}(a_2, b_2)] = a_2 \, \ln \frac{b_1}{b_2} - \ln \frac{\Gamma(a_1)}{\Gamma(a_2)} + (a_1 - a_2) \, \psi(a_1) - (b_1 - b_2) \, \frac{a_1}{b_1} \; ,$

the model complexity of $m$ evaluates to:

$\label{eq:GLM-NG-Com-s3} \begin{split} \mathrm{Com}(m) &= \left\langle \frac{1}{2} \left[ (\mu_0 - \mu_n)^\mathrm{T} (\tau \Lambda_0) (\mu_0 - \mu_n) + \mathrm{tr}((\tau \Lambda_0) (\tau \Lambda_n)^{-1}) - \log \frac{|(\tau \Lambda_n)^{-1}|}{|(\tau \Lambda_0)^{-1}|} - p \right] \right\rangle_{p(\tau|y)} \\ &+ a_0 \, \log \frac{b_n}{b_0} - \log \frac{\Gamma(a_n)}{\Gamma(a_0)} + (a_n - a_0) \, \psi(a_n) - (b_n - b_0) \, \frac{a_n}{b_n} \; . \end{split}$

Using $x \sim \mathrm{Gam}(a, b) \Rightarrow \left\langle x \right\rangle = a/b$ from \eqref{eq:gam-mean} again, it follows that

$\label{eq:GLM-NG-Com-s4} \begin{split} \mathrm{Com}(m) &= \frac{1}{2} \frac{a_n}{b_n} \left[ (\mu_0 - \mu_n)^\mathrm{T} \Lambda_0 (\mu_0 - \mu_n) \right] + \frac{1}{2} \mathrm{tr}(\Lambda_0 \Lambda_n^{-1}) - \frac{1}{2} \log \frac{|\Lambda_0|}{|\Lambda_n|} - \frac{p}{2} \\ &+ a_0 \, \log \frac{b_n}{b_0} - \log \frac{\Gamma(a_n)}{\Gamma(a_0)} + (a_n - a_0) \, \psi(a_n) - (b_n - b_0) \, \frac{a_n}{b_n} \; . \end{split}$

Thus, the model complexity of $m$ evaluates to

$\label{eq:GLM-NG-Com-s5} \begin{split} \mathrm{Com}(m) &= \frac{1}{2} \frac{a_n}{b_n} \left[(\mu_0-\mu_n)^\mathrm{T} \Lambda_0 (\mu_0-\mu_n) - 2(b_n-b_0)\right] + \frac{1}{2} \mathrm{tr}(\Lambda_0 \Lambda_n^{-1}) - \frac{1}{2} \log \frac{|\Lambda_0|}{|\Lambda_n|} - \frac{p}{2} \\ &+ a_0 \log \frac{b_n}{b_0} - \log \frac{\Gamma(a_n)}{\Gamma(a_0)} + (a_n - a_0) \psi(a_n) \end{split}$

which proofs the second part of \eqref{eq:GLM-NG-AnC}.

3) A control calculation confirms that

$\label{eq:GLM-NG-AnC-LME} \mathrm{Acc}(m) - \mathrm{Com}(m) = \mathrm{LME}(m)$

where $\mathrm{LME}(m)$ is the log model evidence for Bayesian linear regression:

$\label{eq:GLM-NG-LME} \begin{split} \log p(y|m) = \frac{1}{2} & \log |P| - \frac{n}{2} \log (2 \pi) + \frac{1}{2} \log |\Lambda_0| - \frac{1}{2} \log |\Lambda_n| + \\ & \log \Gamma(a_n) - \log \Gamma(a_0) + a_0 \log b_0 - a_n \log b_n \end{split}$

This requires to recognize that

$\label{eq:GLM-NG-AnC-LME-a1} -\frac{1}{2} \mathrm{tr}(X^\mathrm{T} P X \Lambda_n^{-1}) - \frac{1}{2} \mathrm{tr}(\Lambda_0 \Lambda_n^{-1}) + \frac{p}{2} = 0$

and

$\label{eq:GLM-NG-AnC-LME-a2} \frac{n}{2} (\psi(a_n) - \log(b_n)) - a_0 \log \frac{b_n}{b_0} - (a_n - a_0) \psi(a_n) = a_0 \log b_0 - a_n \log b_n$

thanks to the nature of the posterior hyperparameters for Bayesian linear regression.

Sources:

Metadata: ID: P431 | shortcut: blr-anc | author: JoramSoch | date: 2024-01-12, 14:24.