I have the following distributional assumptions on some on my RV and model parameters:

$$ y_i \sim N(\beta x_i, w_i^{-1}\Sigma_y) $$

There is a normal prior on the parameters $\beta$ as well:

$$ \beta \sim N(\beta_0, \Sigma_{\beta}) $$

AFAIK, the conditional distribution for the mean parameter $\beta$, given by $P(\beta|y, w) should also be normally distributed due to conjugate relationship. I have been struggling to derive that.

So, I want to get the expression for the conditional distribution of $\beta$, which in this case is given by $p(y|\beta, w) P(\beta)$ and this is proportional to:

$$ |\Sigma_y|^\frac{-N}{2}\Bigg(\exp \sum_{i=1}^N(y_i - \beta x_i)^T\Sigma_y^{-1}(y-\beta x_i) \Bigg)|\Sigma_{\beta}|^{-0.5} \exp (\beta-\beta_0)^T\Sigma_{\beta}^{-1}(\beta -\beta_0) $$

Here I am dropping the terms depending on $w$. This becomes $$ |\Sigma_y|^\frac{-N}{2} |\Sigma_{\beta}|^{-0.5} \exp \big((\beta-\beta_0)^T\Sigma_{\beta}^{-1}(\beta -\beta_0) + \sum_{i=1}^N(y_i - \beta x_i)^T\Sigma_y^{-1}(y-\beta x_i)\big) $$

However, after this I am quite stuck. I am not sure how I can express this in the form of a normal distribution.


1 Answer 1


First note that the availability of a closed-form posterior, which essentially amounts to conjugacy as you notice, depends also on $\sigma^2$ and on what prior distribution it is assigned. If $\sigma^2$ is known, then your prior $\beta \sim N(\beta_0, \Sigma_0)$ is indeed conjugate, otherwise if $\sigma^2$ is random, you need an inverse gamma prior on it, and you also need to change your prior on $\beta$ as $$\beta\vert\sigma^2 \sim N(\beta_0, \sigma^2\Sigma_0).$$

Showing conjugacy requires some tedious algebra, a crucial point in the algebra is to use the least squares estimate $$\hat \beta = (XX^T)^{-1}X^Ty$$ in order to center the posterior distribution of $\beta$. You can show that the posterior is $$\beta\vert \sigma^2, Y, X\sim N(\beta_n, \sigma^2\Sigma_n)$$ where $$\Sigma_n = (X^TX+\Sigma_0^{-1})^{-1},\quad \beta_n = \Sigma_n((X^TX)\hat\beta+\Sigma_0^{-1}\beta_0).$$

The steps are well detailed on the Wikipedia page of Bayesian linear regression, see here: http://en.wikipedia.org/wiki/Bayesian_linear_regression#Posterior_distribution

PS: mind your expression of the posterior distribution for $\beta$ where you are missing some terms. It should read $$\Bigg(\prod_{i=1}^N \exp -\frac{w_i}{2 \sigma^2}\big(y_i - \beta^Tx_i\big)^T\big(y_i - \beta^Tx_i\big)\Bigg)\exp -\frac{1}{2 \sigma^2}(\beta-\beta_0)^T\Sigma_0^{-1}(\beta -\beta_0)$$

  • $\begingroup$ Thanks for your reply. May I ask what this $\sigma^2$ term is. I do not see it. Is the same as the variace weights $w_i$ in my case? In my case, the variance for $y_i$ is given by $w_i^{-1} \Sigma_y$ as $y$ is a vector quantity. $\endgroup$
    – Luca
    Feb 26, 2015 at 15:57
  • 1
    $\begingroup$ the $w_i$ is like a concentration parameter while the inverse $\sigma^2$ is the variance. I was meaning the parameter that you had in the model equation in the previous version of your question. $\endgroup$
    – julyan
    Feb 26, 2015 at 16:02
  • $\begingroup$ Ah ok. Sorry, I reformulated it to use a multivariate form. So, if I understand you correctly, I need to change the prior on $\beta$ as $(\beta_0, \Sigma_{y} \Sigma_{\beta})$. I was wondering if you would be kind enough to elaborate on why this form is necessary? $\endgroup$
    – Luca
    Feb 26, 2015 at 16:06

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