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In the book an introduction to statistical learning, the following proportionality is given:

$$ p(\beta|X,Y) \propto f(Y|X,\beta)\,p(\beta|X) $$

however, I'm wondering what the proportionality constant is, and how it can be derived.

Frank Vel
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  • The total probability must be $1$. – whuber Jan 19 '18 at 21:22
  • The mathematical points made above are all correct. There's also a practical issue: it is often too difficult to calculate the sum. There are sampling methods that can cope with this. – chrishmorris Jan 19 '18 at 22:35

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Just use the definition of conditional probability: $p(A|B)=\frac{p(A,B)}{p(B)}$.

Left-hand side is $p(\beta|X,Y)=\frac{p(X,Y,\beta)}{p(X,Y)}$

Right-hand side is $p(Y|X,\beta)p(\beta|X)=\frac{p(X,Y,\beta)}{p(X,\beta)}\frac{p(X,\beta)}{p(X)}=\frac{p(X,Y,\beta)}{p(X)}$

Comparing both sides, you can see that the constant of proportionality should be $\frac{p(X)}{p(X,Y)}=\frac{1}{p(Y|X)}$

Moss Murderer
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  • So there is no difference between $f$ and $p$? – Frank Vel Jan 19 '18 at 21:42
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    Given a specific observation $Y$, $p(Y|X,\beta)$ can be viewed as a function of $X$ and $\beta$ instead of as a probability value. So sometimes people use $f$ to emphasize that it is a likelihood function rather than probability distribution. But that is just an interpretation thing, so doesn't matter for the calculation of the constant of proportionality. – Moss Murderer Jan 19 '18 at 22:06