It is a Bayesian Regression, y and X are observed data, the response and the feature matrix respectively. \begin{eqnarray*} \mathbf y \mid \mathbf X,\mathbf w,\sigma^2 &\sim& N (\mathbf X \mathbf w, \sigma^2 \mathbf I ) \\ \mathbf w\mid \lambda &\sim &N(\mathbf 0, \lambda \mathbf I) \\ \sigma, \lambda >0 \end{eqnarray*}

How could I show that $\mathbf w \mid \mathbf X , \mathbf y , \lambda , \sigma^2$ is multivariate Gaussian? And what is the mean vector and variance-covariance matrix? I have been looking into this for many days and dont know how to solve it at all. Thanks!

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1 Answer 1


The posterior distribution for $\textbf{w}$ will come from the product

$$p(\textbf{w}|\textbf{y},\textbf{X},\lambda,\sigma^{2}) \propto N(\textbf{X} \textbf{w},\sigma^{2}I)N(0,\lambda I)$$

by discarding all the terms that do not include $\textbf{w}$.

If you do that you will end up with

$$e^{-\frac{1}{2\sigma^{2}}(-2\textbf{w}^{T}\textbf{X}\textbf{y}+\textbf{w}^{T}\textbf{X}^{T}\textbf{X}\textbf{w})-\frac{1}{2\lambda}\textbf{w}^{T}\textbf{w}} \ \ \ \ (*)$$

Now let's assume that we have a multivariate normal distribution $\textbf{z}\sim N(\textbf{m},\Sigma)$, if we expand each pdf and keep only terms that have $\textbf{z}$, we will end up with

$$e^{-\frac{1}{2}(\textbf{z}^{T}\Sigma^{-1}\textbf{z}-2\textbf{z}\Sigma^{-1}\textbf{m})} \ \ \ \ \ (**)$$

Hence, in order to identify the mean and covariance of $\textbf{w}$ try to bring $(*)$ into the form $(**)$ and identify which terms correspond to $\Sigma^{-1}$ and $\textbf{m}$


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