Talk:Bayes factor

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[edit] Prior distributions

How do you achieve this result (solving the integral)? \int_{q=0}^1{200 \choose 115}q^{115}(1-q)^{85}dq = {1 \over 201}

Any guidance would be greatly appreciated OrangeDog 20:20, 1 September 2006 (UTC)

Try integration by parts 85 times and note that
\int_a^b x^{n}(1-x)^{m}\,dx = \left[ {\frac{1}{n+1}} x^{n+1}(1-x)^{m}\right]_{a}^{b} + \int_a^b {\frac{m}{n+1}} x^{n+1}(1-x)^{m-1}\,dx
--Henrygb 21:19, 1 September 2006 (UTC)
Sorry, I'm just not seing it. Can you just give me the general result, as I really don't have time to integrate by parts 85 times: I'm on a pencil and paper here. OrangeDog 21:57, 1 September 2006 (UTC)
Try doing it once, put in the numbers (look at those integration limits), and then see if you can see how things are going to turn out... Jheald 20:54, 2 September 2006 (UTC)

Ah, trapesium rules are a wonderful thing. As is the real world when you only need 8 sig. fig. OrangeDog 16:29, 3 September 2006 (UTC)

You don't need any approximation here. See what \left[ {\frac{1}{n+1}} x^{n+1}(1-x)^{m}\right]_{a}^{b} is when a=0 and b=1. A pencil can work out that \int_0^1 x^{n}(1-x)^{m}\,dx = \int_0^1 {\frac{x^{n+m}}{{n+m \choose n}}}\,dx = {\frac{1}{(n+m+1){n+m \choose n}}} --Henrygb 10:08, 4 September 2006 (UTC)
Yeah, but I had lots of similar problems (with various limits) and was looking for an exact general solution. Anyway, I'm done now OrangeDog 18:14, 5 September 2006 (UTC)

[edit] Bayes information criterion

Consider mentioning approximations of integrated likelihoods and Bayes factors, and in particular linking to the article on Bayesian information criterion.

Dfarrar 02:21, 15 March 2007 (UTC)

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