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)

Wow. Working hard in these answers. This is an integral that you memorize in the theory of probability/statistics because it is reoccuring. Thats why people prattle off the solution without to much thought. It is called "the beta function". http://en.wikipedia.org/wiki/Beta_function. It is closely related to the beta density. Which is closely related to bernoulli R.V.s Which are closely related to ... lots... Which is why it should be memorized or atleast known to exist. Jeremiahrounds 14:33, 20 June 2007 (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)

[edit] Correspondence with p values?

If a scientist requests a 95% statistical significance level is there a corresponding value for K? More generally, is it meaningful to compare p-values with K values? Pgr94 (talk) 15:43, 4 February 2008 (UTC)

There is a way to do that, but you also need to know the number of degrees of freedom that have gone into computing the aggregate "K value" (discriminating information). It turns out that 2 times the discriminating (Kullback) information is asymptotically distributed as chi-square with that many degrees of freedom. Hence you can use an inverse chi-square table or algorithm to convert it to a "probability" (that the aggregate evidence would a priori have produced at least that much weight in support of the hypothesis if it were true). If you search the pre-Web net archives for an "i-hat" archive you may find C code I contributed long ago that computes this, along with Unix manual pages documenting its use. — DAGwyn (talk) 00:45, 16 February 2008 (UTC)