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:


  • $\begingroup$ You cannot expect readers to go through the paper to answer and even understand your question so please produce some context for your question to make sense on its own. $\endgroup$
    – Xi'an
    Commented Jan 15, 2015 at 18:12
  • 2
    $\begingroup$ This question appears to be off-topic because it requires the reader to go through an external link to be understood. $\endgroup$
    – Xi'an
    Commented Jan 15, 2015 at 18:13
  • $\begingroup$ Ok, I will try and edit the question later in the night today. Sorry for it. $\endgroup$
    – Luca
    Commented Jan 15, 2015 at 18:20
  • $\begingroup$ 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?" $\endgroup$
    – whuber
    Commented Jan 15, 2015 at 19:03
  • $\begingroup$ hmmmm... ok, I will need to go away and have a think. I will update the thread as soon as I get the confusions clear in my head. $\endgroup$
    – Luca
    Commented Jan 15, 2015 at 19:06

1 Answer 1


This is just a model assumption the author made. Unfortunately, there aren't standardized procedures to "follow Bayesian statistical modelling", so while you may specify a prior for variance, it isn't a requirement for a linear regression to be bayesian.

  • $\begingroup$ Thanks for the answer. So, I guess the authors choose to only model a few of the unknown variables in the Bayesian setting. However, I guess nothing stops me from also have a prior over the noise precision as well. $\endgroup$
    – Luca
    Commented Jan 15, 2015 at 21:48
  • $\begingroup$ I found the answer in Gelman's book Bayesian Data Analysis (Page 384). He argues that priors for variances are less important for most applications but mentions that a scaled-inverse chi square dist. is a conjugate prior in this setting. $\endgroup$
    – Luca
    Commented Jan 19, 2015 at 12:02

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