I do not know which of the two distributions is being sampled. My goal is to determine this distribution. I have a prior belief that F is the distribution that is being sampled. This prior is called p.

According to Bayes Rule, given a sample of length N, (s1, s2, s3,...,sN), the probability that I am sampling from distribution F, is given by

(p*f(s1)*f(s2)*...*f(sN))/{p*f(s1)*f(s2)*...*f(sN) (1-p)}

Intuitively it'seems that, as N approaches infinity, this quantity should approach one with probability one if the sample is being drawn from F and 0 with probability 1 if the sample is being drawn from the uniform.

I am a little stumped on how to prove this. Can anyone offer some help or a reference?

Many thanks.

(p*f(s1)*f(s2)*...*f(sN))/{p*f(s1)*f(s2)*...*f(sN) + aN(1-p)}

where 0 < a < 1. Denote this P. Then

1/P = 1 + (1 - p)/p · aN / f(s1)·f(s2)·...·f(sN))

If you assume that the sample is distributed according to f, there is a better chance for f(s) to be above a, i.e. nearer the right end where f exceeds uniform density. In which case, indeed, I would expect the second addend to fo to 0 as N grows. On the other hand, if the sample is distributed uniformly on some interval A, B (so that a = 1/(B - A), give or take ...) and f is, say, a straight line equal a at the midpoint (A+B)/2, then for two picks at the same distance from the midpoint but on different sides would be f1 = a - x, f2 = a + x so that

a² > a² - x² = (a - x)(a + x)

and I would expected the second addend grow without bound. For, as N grows, there would be a high probability of being able to combine the samepls into pairs, more or less on the same distance from (A+B)/2. And then you would have a growing number of factors

a² / (a² - x²) > 1

How to do that formally I am afraid I do not know.