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P (B)P (A|B) = P (A ∩ B) (2.2)

But, we can simply make the same statement about P (B|A), which is akin to asking "What

is the probability of seeing clouds, given that it is raining?":

P (B|A) =

P (B ∩ A)

P (A)

(2.3)

Note that P (A ∩ B) = P (B ∩ A) and so by substituting the above and multiplying by P (A),

we get:

P (A)P (B|A) = P (A ∩ B) (2.4)

We are now able to set the two expressions for P (A ∩ B) equal to each other:

P (B)P (A|B) = P (A)P (B|A) (2.5)

If we now divide both sides by P (B) we arrive at the celebrated Bayes’ rule:

P (A|B) =

P (B|A)P (A)

P (B)

(2.6)

However, it will be helpful for later usage of Bayes’ rule to modify the denominator, P (B)

on the right hand side of the above relation to be written in terms of P (B|A). We can actually

write:

P (B) = ∑ a∈A

P (B ∩ A) (2.7)

This is possible because the events A are an exhaustive partition of the sample space.

So that by substituting the defintion of conditional probability we get:

P (B) = ∑ a∈A

P (B ∩ A) = ∑ a∈A

P (B|A)P (A) (2.8)

Finally, we can substitute this into Bayes’ rule from above to obtain an alternative version of

Bayes’ rule, which is used heavily in Bayesian inference:

P (A|B) =

P (B|A)P (A)

∑a∈A P (B|A)P (A) (2.9)

Now that we have derived Bayes’ rule we are able to apply it to statistical inference.

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