I’m hoping, when I get round to it, to give a full explanation of Bayes’ Theorem, it’s use and different forms of it. For now, since I’ve just been formalising the derivations in preparation for a paper I’m writing, I thought I might as well type it up, and no reason not to share in case people want to have a look.

So we begin with a basic axiom of probability theory:

This is to say that the probability of both A and B being the case is the probability of B being the case multiplied by the probability of A given B. One can also put this the other way round:

Since both these latter halves are equal to , it follows that:

Now, if we divide both sides by , we get:

This, in short, is Bayes’ Theorem, which says that the probability of A given B is equal to the probability of A, multiplied by the probability of B given A, divided by the probability of B.

Now, to get to the odds form, we need to do a few more things: firstly, we note that:

And so we can deduce that:

The odds form allows us to compare and directly. To get further towards this, we can go through the whole process again, this time using in place of . This will eventually give us:

From this, we find we can divide by , which gives us the following:

This may look confusing, but we can note that the denominators of both the top half and of the lower half are the same – if we multiply top and bottom by that denominator, we get the much simpler equation:

Separate out some of the terms on the right hand side, and you get:

And you now have the odds form of Bayes’ Theorem! Perfect. As I said, I won’t go into its use or anything here: this is purely to provide the formal derivation for future reference. I hope you won’t be too disappointed, therefore, if you find that there is nothing at all interesting to you in this post.

Calum Miller

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By the way I love Bayes’ theorem and have used it to argue for the resurrection and to show one of Richard Dawkins’s fallacies.

By:

Greg Reeveson January 25, 2012at 11:48 pm

Calum, can you show how you got to your first equation in the derivation of the odds form, the first one with ~A in it? Thanks.

By:

thinkingchristianon January 27, 2012at 6:44 pm

Thanks for the comments. Thinkingchristian, the probability of P(B) can be looked at in terms of a probability tree. This one I’ve found illustrates it quite nicely: http://www.gcsemathstutor.com/images/information/info-tree1.jpg

Suppose P(B) is the probability of getting a blue on the 2nd choice. And suppose P(A) is the probability of getting a blue on the 1st choice. There are two ways we can get a blue on the 2nd choice: by getting a green and then a blue, or by getting a blue and then another blue. The probability of P(B) is the sum of the probabilities of the different ways we can get a blue on the 2nd choice: in this example, we have only these two ways of doing so.

The probability of finding a blue on the 2nd choice via getting a blue on the first choice, is the probability P(A) multiplied by P(B|A).

In contrast, the probability of finding a blue on the 2nd choice via getting a green on the first choice, is the probability P(~A) (i.e. the probability of getting a green), multiplied by the probability of getting a blue after getting a green, which is P(B|~A). Thus, it is equal to P(~A) x P(B|~A).

The sum of these is the total probability of getting a blue on the 2nd choice, and so is equal to [P(A) x P(B|A)] + [P(~A) x P(B|~A)]. And there you have the equation! Hope this helps.

By:

Calum Milleron January 29, 2012at 4:57 pm