Given parameter/s $\theta$, data $X$ and prior on the parameter/s $p(\theta)$, Bayes' theorem allows us to estimate the posterior distribution $p(\theta | X)$:

$p(\theta | X) = \frac{p(\theta) p(X | \theta)}{p(X)}$

$\to p(\theta | X) \ \propto \ p(\theta) p(X | \theta)$

From the Bayesian linear regression Wiki page:

$p(\beta, \sigma^2 | y, X) \ \propto \ p(y | X, \beta, \sigma^2) p(\beta | \sigma^2) p(\sigma^2)$

I was expecting something like:

$p(\beta, \sigma^2 | y, X) \ \propto p(y | X, \beta, \sigma^2) \color{red}{p(X | \beta, \sigma^2)} p(\beta | \sigma^2) p(\sigma^2)$ or

since I guess

$p(\beta, \sigma^2, y, X) = p(y | X, \beta, \sigma^2) p(X, \beta, \sigma^2)$

Without $p(X | \beta, \sigma^2)$, I guess it's still true, but why not include it?


1 Answer 1


$X$ does not depend on $\beta$, $\sigma$. These values have to do with getting $y$ from $x$. So $P(X| \beta, \sigma)=P(X)$. Since we are maximizing with respect to $\theta$ we don't care about this constant term.


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