# Bayesian Weighted Linear regression

I am currently reading the following paper which formulates the weighted linear regression in a Bayesian setting. In classic weighted LS, we minimise the following:

$$\sum_{i=1}^{N} w_i (\beta^Tx_i - y_i)$$

In this paper, they try and have a Bayesian formulation of the WLS. So, it makes the following modelling choices about the probability distributions of the random variables:

$$y_i \sim N(\beta^tx_i, \sigma^2/w_i)$$

So, here we are modelling each of the $y_i$ to have variance which can be weighted by their individual weight. There is a normal prior also over the regression parameters $\beta$.

$$\beta \sim N(\beta_0, \Sigma_{\beta, 0})$$

There is a Gamma prior over the weights $w_i$.

$$w_i \sim Gamma(a_i, b_i)$$

Now, my question is that the regression problem is basically:

$$y_i = \beta^T x_i + \epsilon_i$$

My question is why is there no prior on $\epsilon$? In this paper, they estimate $\sigma^2$ through some standard regression formula (Apologies as I have not gone far to derive it yet). However, to me it seems that $\sigma^2$ is also an unknown parameter in the model and if we follow Bayesian statistical modelling, we should specify a prior for it.

If anyone is curious, the paper is here:

• Between the start and end of your question you changed the formulation of the model: you converted "$y_i \sim N(\beta^tx_i, \sigma^2/w_i)$" into "$y_i = \beta^T x_i + \epsilon_i$". That changed the model. (The role played by $\epsilon_i$ in the latter is replaced by the distributional assumptions on $y_i$ in the former.) So basically it looks like you're telling us "the paper assumes this, but I assume something else, so why doesn't the paper deal with my assumption?" – whuber Jan 15 '15 at 19:03