Index: The Book of Statistical ProofsStatistical Models ▷ Univariate normal data ▷ Univariate Gaussian with known variance ▷ Log model evidence

Theorem: Let

$\label{eq:ug} m: \; y = \left\lbrace y_1, \ldots, y_n \right\rbrace, \quad y_i \sim \mathcal{N}(\mu, \sigma^2), \quad i = 1, \ldots, n$

be a univariate Gaussian data set with unknown mean $\mu$ and known variance $\sigma^2$. Moreover, assume a normal distribution over the model parameter $\mu$:

$\label{eq:UGkv-prior} p(\mu) = \mathcal{N}(\mu; \mu_0, \lambda_0^{-1}) \; .$

Then, the log model evidence for this model is

$\label{eq:UGkv-LME} \log p(y|m) = \frac{n}{2} \log\left( \frac{\tau}{2 \pi} \right) + \frac{1}{2} \log\left( \frac{\lambda_0}{\lambda_n} \right) - \frac{1}{2} \left( \tau y^T y + \lambda_0 \mu_0^2 - \lambda_n \mu_n^2 \right) \; .$

where the posterior hyperparameters are given by

$\label{eq:UGkv-post-par} \begin{split} \mu_n &= \frac{\lambda_0 \mu_0 + \tau n \bar{y}}{\lambda_0 + \tau n} \\ \lambda_n &= \lambda_0 + \tau n \end{split}$

with the sample mean $\bar{y}$ and the inverse variance or precision $\tau = 1/\sigma^2$.

Proof: According to the law of marginal probability, the model evidence for this model is:

$\label{eq:UGkv-ME-s1} p(y|m) = \int p(y|\mu) \, p(\mu) \, \mathrm{d}\mu \; .$

According to the law of conditional probability, the integrand is equivalent to the joint likelihood:

$\label{eq:UGkv-ME-s2} p(y|m) = \int p(y,\mu) \, \mathrm{d}\mu \; .$

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

$\label{eq:UG-LF-class} \begin{split} p(y|\mu,\sigma^2) &= \prod_{i=1}^{n} \mathcal{N}(y_i; \mu, \sigma^2) \\ &= \prod_{i=1}^{n} \frac{1}{\sqrt{2 \pi} \sigma} \cdot \exp\left[ -\frac{1}{2} \left( \frac{y_i-\mu}{\sigma} \right)^2 \right] \\ &= \left( \sqrt{\frac{1}{2 \pi \sigma^2}} \right)^n \cdot \exp\left[ -\frac{1}{2 \sigma^2} \sum_{i=1}^{n} \left( y_i-\mu \right)^2 \right] \end{split}$

which, for mathematical convenience, can also be parametrized as

$\label{eq:UG-LF-Bayes} \begin{split} p(y|\mu,\tau) &= \prod_{i=1}^{n} \mathcal{N}(y_i; \mu, \tau^{-1}) \\ &= \prod_{i=1}^{n} \sqrt{\frac{\tau}{2 \pi}} \cdot \exp\left[ -\frac{\tau}{2} \left( y_i-\mu \right)^2 \right] \\ &= \left( \sqrt{\frac{\tau}{2 \pi}} \right)^n \cdot \exp\left[ -\frac{\tau}{2} \sum_{i=1}^{n} \left( y_i-\mu \right)^2 \right] \end{split}$

using the inverse variance or precision $\tau = 1/\sigma^2$.

When deriving the posterior distribution $p(\mu \vert y)$, the joint likelihood $p(y,\mu)$ is obtained as

$\label{eq:UGkv-LME-s1} p(y,\mu) = \left( \frac{\tau}{2 \pi} \right)^\frac{n}{2} \cdot \sqrt{\frac{\lambda_0}{2 \pi}} \cdot \exp \left[ -\frac{\lambda_n}{2} (\mu - \mu_n)^2 -\frac{1}{2} \left( \tau y^\mathrm{T} y + \lambda_0 \mu_0^2 - \lambda_n \mu_n^2 \right) \right] \; .$

Using the probability density function of the normal distribution, we can rewrite this as

$\label{eq:UGkv-LME-s2} p(y,\mu) = \left( \frac{\tau}{2 \pi} \right)^\frac{n}{2} \cdot \sqrt{\frac{\lambda_0}{2 \pi}} \cdot \sqrt{\frac{2 \pi}{\lambda_n}} \cdot \mathcal{N}(\mu; \lambda_n^{-1}) \cdot \exp \left[ -\frac{1}{2} \left( \tau y^\mathrm{T} y + \lambda_0 \mu_0^2 - \lambda_n \mu_n^2 \right) \right] \; .$

Now, $\mu$ can be integrated out using the properties of the probability density function:

$\label{eq:UGkv-LME-s3} p(y|m) = \int p(y,\mu) \, \mathrm{d}\mu = \left( \frac{\tau}{2 \pi} \right)^\frac{n}{2} \cdot \sqrt{\frac{\lambda_0}{\lambda_n}} \cdot \exp \left[ -\frac{1}{2} \left( \tau y^\mathrm{T} y + \lambda_0 \mu_0^2 - \lambda_n \mu_n^2 \right) \right] \; .$

Thus, the log model evidence of this model is given by

$\label{eq:UGkv-LME-s4} \log p(y|m) = \frac{n}{2} \log\left( \frac{\tau}{2 \pi} \right) + \frac{1}{2} \log\left( \frac{\lambda_0}{\lambda_n} \right) - \frac{1}{2} \left( \tau y^T y + \lambda_0 \mu_0^2 - \lambda_n \mu_n^2 \right) \; .$
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Metadata: ID: P213 | shortcut: ugkv-lme | author: JoramSoch | date: 2021-03-24, 06:45.