)}{P(\textbf{y})}\nonumber\\
    &amp;=&amp;\frac{ N(\textbf{X}\beta, \sigma^2\textbf{I})N(0, \tau\textbf{I}) }{P(\textbf{y})}.\nonumber
\end{eqnarray}\]</div>
<p>Taking logarithm on both sides we get</p>
<div class="arithmatex">\[\begin{eqnarray}
    \ln(P(\beta|\textbf{y})) = -\frac{1}{2}\left(\frac{(\textbf{y}-\textbf{X}\beta)^T(\textbf{y}-\textbf{X}\beta)}{\sigma^2} + \frac{\beta^T\beta}{\tau}\right) + C,\nonumber
\end{eqnarray}\]</div>
<p>where <span class="arithmatex">\(C\)</span> is a constant indepe