# When using Bayes Rule, what are the rules for flipping the conditions and the event of interest?

Here is Bayes Rule:

$$P(A\mid B) = \frac{P(B\mid A) P(A)}{P(B)}$$

This paper (http://www.cogsci.northwestern.edu/Bayes/Sivia_1996.pdf) uses Bayes rule on page 21 in the context of model selection where $H$ is the single model parameter, $D_1$ and $D_2$ are two sets of data from different experiments, and $I$ represent all information known prior to the two experiments:

$$P(H\mid D_2, D_1, I) \propto P(D_2, D_1\mid H, I) P(H\mid I)$$

Is valid to select a subset of conditions (in this case $D_1$ and $D_2$) to swap with the event of interest? Or is this just allowed due to the special case where $I$ actually represents all info prior to the experiment?

Thank you!

Yes, as you say $I$ represents all info prior to the experiment: effectively it restricts the sample space. When working with expressions like this I find it convenient to manipulate them without the $I$ and then add it in back afterwards. So you have $$P(H|D_2,D_1) \propto P(D_2,D_1|H)P(H)$$ Then write $D=D_2\wedge D_1$ and you have Bayes Rule again $$P(H|D) = \propto P(D|H)P(H)$$
(Because it's using $\propto$ rather than having $P(D)$ on the denominator on the RHS, the probabilities will have to be normalized.)
• for example in your case $P(H\mid D_2, D_1, I) \propto P(D_2, D_1\mid H, I) P(H\mid I)$ I'd say "conditioning everything on $I$, we have $P(H\mid D_2, D_1) \propto P(D_2, D_1\mid H) P(H)$ hence $P(H\mid D_2, D_1) / P(H) \propto P(D_2, D_1\mid H)$ and now writing in the conditioning again explicitly $P(H\mid D_2, D_1, I) / P(H|I) \propto P(D_2, D_1\mid H,I)$" – TooTone Mar 21 '14 at 17:47