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

Theorem: Let

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

be a univariate Gaussian data set with unknown mean $\mu$ and known variance $\sigma^2$. Moreover, assume two statistical models, one assuming that $\mu$ is zero (null model), the other imposing a normal distribution as the prior distribution on the model parameter $\mu$ (alternative):

$\label{eq:UGkv-m01} \begin{split} m_0&: \; y_i \sim \mathcal{N}(\mu, \sigma^2), \; \mu = 0 \\ m_1&: \; y_i \sim \mathcal{N}(\mu, \sigma^2), \; \mu \sim \mathcal{N}(\mu_0, \lambda_0^{-1}) \; . \end{split}$

Then, the log Bayes factor in favor of $m_1$ against $m_0$ is

$\label{eq:UGkv-LBF} \mathrm{LBF}_{10} = \frac{1}{2} \log\left( \frac{\lambda_0}{\lambda_n} \right) - \frac{1}{2} \left( \lambda_0 \mu_0^2 - \lambda_n \mu_n^2 \right)$

where $\mu_n$ and $\lambda_n$ are the posterior hyperparameters for the univariate Gaussian with known variance which are functions of the inverse variance or precision $\tau = 1/\sigma^2$ and the sample mean $\bar{y}$.

$\label{eq:LBF-LME} \mathrm{LBF}_{12} = \mathrm{LME}(m_1) - \mathrm{LME}(m_2) \; .$

The LME of the alternative $m_1$ is equal to the log model evidence for the univariate Gaussian with known variance:

$\label{eq:UGkv-LME-m1} \mathrm{LME}(m_1) = \frac{n}{2} \log\left( \frac{\tau}{2 \pi} \right) + \frac{1}{2} \log\left( \frac{\lambda_0}{\lambda_n} \right) - \frac{1}{2} \left( \tau y^\mathrm{T} y + \lambda_0 \mu_0^2 - \lambda_n \mu_n^2 \right) \; .$

Because the null model $m_0$ has no free parameter, its log model evidence (logarithmized marginal likelihood) is equal to the log-likelihood function for the univariate Gaussian with known variance at the value $\mu = 0$:

$\label{eq:UGkv-LME-m0} \mathrm{LME}(m_0) = \log p(y|\mu=0) = \frac{n}{2} \log\left( \frac{\tau}{2 \pi} \right) - \frac{1}{2} \left( \tau y^\mathrm{T} y \right) \; .$

Subtracting the two LMEs from each other, the LBF emerges as

$\label{eq:UGkv-LBF-m10} \mathrm{LBF}_{10} = \mathrm{LME}(m_1) - \mathrm{LME}(m_0) = \frac{1}{2} \log\left( \frac{\lambda_0}{\lambda_n} \right) - \frac{1}{2} \left( \lambda_0 \mu_0^2 - \lambda_n \mu_n^2 \right)$ $\label{eq:UGkv-post-par} \begin{split} \mu_n &= \frac{\lambda_0 \mu_0 + \tau n \bar{y}}{\lambda_0 + \tau n} \\ \lambda_n &= \lambda_0 + \tau n \end{split}$

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

Sources:

Metadata: ID: P215 | shortcut: ugkv-lbf | author: JoramSoch | date: 2021-03-24, 09:05.