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| Bayes: A man of the cloth |
There is a long tradition of Bayes theorem being used in discussions about the probability of God. I've never been very keen on using Bayes to "prove" God's existence: I expressed my reservations in this short paper where I discussed the use of Bayes Theorem by Christians Roger Forster and Paul Marsden in their book Reason and Faith. In this connection, however, I noticed this post by Christian Blogger James Knight where once again we see God and Bayes appearing together.
Well, in this instance I didn't want to get embroiled with the subject of God and Bayes, but in my correspondence with James I picked up on a technical issue which obliquely impinged upon his post.
The theorem that interested me can be expressed as follows....
If
P(A) < P(A|B)
....then it follows that....
P(B) < P(B|A)
....where P(A) and P(B) are the unconditional probabilities of A and B respectively and P(A|B) and P(B|A) are the respective conditional probabilities of A and B.
As per my practice in my paper on randomness I'm going to use Venn diagrams. But Such an approach implicitly assumes my frequentist interpretation of probability, an interoperation I won't attempt to justify here.
In terms of a Venn diagram the relationship of A and B will in general look something like this....
Here the area labelled A represents the set of possible cases with property A and the area B represents the set of possible cases with property B. This Venn diagram is imagined to reside in a large domain of a total number of possible cases of T.
Now, if N(A) = number of cases with property A, then the unconditional probability of A is given by P(A) where...
P(A) = N(A) / T
If the number of cases with property B is N(B) and the number of cases where B and A overlap is expressed as N(A|B) = N(B|A), then the probability of A given B, P(A|B), equates to....
P(A|B) = N(A|B) / N(B)
Now we postulate that:
P(A) < P(A|B)
Expressed in frequentist terms we can write that as.....
N(A) / T < N(A|B) / N(B).
We now multiply both sides of this inequality by N(B) and this gives......
N(A) N(B) / T < N(A|B)
Now divide both sides of the latter inequality by N(A) and this returns.
N(B) / T < N(A|B) / N(A)
But N(A|B) = N(B|A) and so the above inequality becomes....
N(B) / T < N(B|A) / N(A)
Expressed in terms of probabilities the latter inequality can be written as.....
P(B) < P(B|A)
....and this inequality has thus been proved from our first postulate which was...
P(A) < P(A|B)
In other words:
P(B) < P(B|A) => P(A) < P(A|B).
*****
James was concerned that the apparent symmetry of this result is contrary to his intuition that the general case is far from symmetric. However this intuition of asymmetry is backed up by the following special case where we have....
From this diagram we see that B=>A (i.e. B logically implies A). But clearly given A the probability of B, depending the relatives sizes of the two sets A and B, may be quite low. This may be the kind of asymmetry that James is thinking of.
