In Bayesian Essentials, page 40, there's the following formula:

enter image description here

I've tried to derive it.

$P(M=k|D=\frac{P(D|M=k)P(M=k)}{\sum_jP(D|M=j)P(M=j)}=\frac{\int P(D|\theta,M=k)P(\theta|M=k)\ d\theta \ \ P(M=k)}{\sum_jP(D|M=j)P(M=j)}=\frac{\int l_k(\theta)\pi_k(\theta)\ d\theta \ \ P(M=k)}{\sum_jP(D|M=j)P(M=j)}$

However, I'm stuck at this last step...

Any help would be appreciated.


1 Answer 1


Sorry about this terrible typo!!!

$$ \mathbb{P}^\pi(\mathfrak{M}=k|\mathscr{D}_n) = \dfrac{\mathbb{P}^\pi(\mathfrak{M}=k) \int \ell(\theta_k|\mathscr{D}_n)\pi_k(\theta_k)\,\text{d}\theta_k}{ \sum_{j=1}^J \mathbb{P}^\pi(\mathfrak{M}=j)\pi_j(\theta_j)\,\text{d}\theta_j}\,. $$ should read as $$ \mathbb{P}^\pi(\mathfrak{M}=k|\mathscr{D}_n) = \dfrac{\mathbb{P}^\pi(\mathfrak{M}=k) \int \ell(\theta_k|\mathscr{D}_n)\pi_k(\theta_k)\,\text{d}\theta_k}{ \sum_{j=1}^J \mathbb{P}^\pi(\mathfrak{M}=j)\int \ell(\theta_j|\mathscr{D}_n)\pi_j(\theta_j)\,\text{d}\theta_j}\,. $$ which is also what you propose as the last equation in your question. For some reason the cut-and-paste between numerator and denominator did not work completely and we failed to spot the mistake. Apologies!

  • 1
    $\begingroup$ Thanks for the help. By the way, do you know of the existence of an errata for this book? $\endgroup$ Jan 15, 2017 at 19:45
  • $\begingroup$ @Anoldmaninthesea. There is no errata list so far but you demonstrated the need for one!!! $\endgroup$
    – Xi'an
    Jan 26, 2017 at 11:03

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