Talk:Bayes factor

message: Bayesian model comparison should not redirect here. There are many ways to do Bayesian model comparison other than Bayes factors; in fact, Bayes factors are probably one of the least popular ways to do model comparison among Bayesian statisticians. See Practical Bayesian model evaluation using leave-one-out cross-validation and WAIC as an example. Closed Limelike Curves (talk) 20:38, 10 December 2021 (UTC)
thread_title: Bayesian model comparison should not redirect here

message: The claim ... use of Bayes factors is a Bayesian alternative to classical hypothesis testing is not only tired, it is misleading. Most Bayesian inference today calculates an estimate of the entire posterior distribution from an estimate or a decision space, the latter being weighted by a Loss function. For example, a test for the equality of two means calculates the Bayesian posterior of their difference, with the contributions weighted by the respective likelihoods and priors, and if the q-ROPE (or "q-Region Of Practical Equivalence") includes zero, they are considered to be the same. See Bayesian estimation supersedes the t-test as an example.
I don't mean to suggest that Bayes factors ought to be excluded, simply that they are not the only way Bayesian inference is done. This is particularly so, for Wikipedia, since Bayesian model selection now redirects here. 
It's another page, but the Bayesian inference needs updating, too
bayesianlogic.1@gmail.com 
 21:43, 2 September 2019 (UTC)
thread_title: Bayes factors are less and less Bayesian every day

message: What is the proper extension to multiple (more than 2) models?--128.113.89.90 (talk) 15:23, 12 December 2008 (UTC)
thread_title: Extension to multiple models

message: 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)
thread_title:  Bayes information criterion 

message: 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)
replies: text: 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)
thread_title:  Correspondence with p values? 

message: This article and Bayesian model comparison both cover the same material. Opinions as to merging? -3mta3 (talk) 09:55, 27 March 2009 (UTC)
thread_title:  Merger proposal 

message: It seems to me this article could be made more clear by showing how the Bayes factor appears in the posterior odds ratio for the two models. This may seem obvious to those familiar with Bayesian inference but it may be confusing to those who are not. What I mean is, insert
\frac{P(M_1|D)}{P(M_2|D)} = \frac{P(D|M_1)}{P(D|M_2)}\frac{P(M_1)}{P(M_2)}
in there somewhere and discuss it. --- Bjfar (talk) 22:15, 18 December 2012 (UTC)
thread_title:  Odds ratio 

message: Shouldn't the likelihood in the example section be q^{115}(1-q)^{85} instead of {{200 \choose 115}q^{115}(1-q)^{85}}?  However, it won't matter for the ratio since the normalizing constant drops out. --Mguggis (talk) 05:49, 29 July 2013 (UTC)
replies: text: @Mguggis, they are using the Binomial distribution which is appropriate. Monsterman222 (talk) 23:35, 1 August 2014 (UTC)

text: @Monsterman222 , :@Mguggis references to non-publicly (free) available articles should be avoided  — Preceding unsigned comment added by 31.161.255.177 (talk) 12:07, 14 February 2015 (UTC)

text: @Monstermann222 , my confusion is what the distribution of interest is. Is it the sum of bernoullis or the bernoulli?  If it's the sum, then the likelihood is 1 observation from a Binomial(200,p) and the likelihood in the example,  Pr(X_{1}+...+X_{200}|q) = {{200 \choose 115}q^{115}(1-q)^{85}}, is correct.  However if the bernoulli is of interest then the likelihood is the joint distribution of 200 iid observations from a bernoulli, which is  Pr(X_{1},...,X_{200}|q) = \prod_{i=1}^{200} Pr(X_{i}|q) = q^{115}(1-q)^{85} .  I was thinking the second case was the distribution of interest. Mguggis (talk) 22:35, 12 December 2015 (UTC)
thread_title:  Example 

message: I taught this material in class and I find that students have difficulty with the abuse of notation of P(D | M) where M is not a random variable. It is better to define a Bayesian model by the likelihood, prior and parameter region of interest:
:M := <P_M(D|\theta), ~P_M(\theta),~\Theta'>
where the default \Theta' is the entire support of the prior. This implies the Bayes Factor can be defined as:
: K = \frac{\Pr(D|M_1)}{\Pr(D|M_2)} = \frac{P_{M_1}(D)}{P_{M_2}(D)}
= \frac{\int_{\Theta'_1}  P_{M_1}(D | \theta_1) P_{M_1}(\theta_1) \,d\theta_1}
{\int_{\Theta'_2}  P_{M_2}(D | \theta_2) P_{M_2}(\Theta'_2) \,d\theta_2}.
And the first expression, \frac{\Pr(D|M_1)}{\Pr(D|M_2)} is admittedly an abuse of notation.
This implies the example becomes:
:M_1 := <P_M(X|\theta) = {{200 \choose 115}\theta^{115}(1-\theta)^{85}}, ~P_M(\theta) = 1/2~ w.p. ~1,~\Theta' = [0, 1]>
:M_2 := <P_M(X|\theta) = {{200 \choose 115}\theta^{115}(1-\theta)^{85}}, ~P_M(\theta) = U(0,1),~\Theta' = [0, 1]>
and thereby,
: K = \frac
{\int_0^1  {{200 \choose 115}\theta^{115}(1-\theta)^{85}} (1) \,d\theta} = \frac{0.005956...}{0.004975...} \approx 1.197
You can also give an example here such as:
:M_1 := H_0 : \theta \leq 0.5 \equiv <P_M(X|\theta) = {{200 \choose 115}\theta^{115}(1-\theta)^{85}}, ~P_M(\theta) = U(0,1),~\Theta' = [0, 0.5]>
:M_2 := H_a : \theta > 0.5 \equiv <P_M(X|\theta) = {{200 \choose 115}\theta^{115}(1-\theta)^{85}}, ~P_M(\theta) = U(0,1),~\Theta' = [0.5, 1]>
and thereby,
: K = \frac{{\int_0^{0.5}  {{200 \choose 115}\theta^{115}(1-\theta)^{85}} (1) \,d\theta}}
{\int_{0.5}^1  {{200 \choose 115}\theta^{115}(1-\theta)^{85}} (1) \,d\theta}
Thoughts?
M is not a random variable under the interpretation of frequency probabilities. However, it is a random variable under the subjective (or evidential) probability interpretation. Which is the interpretation that Bayesian methods use. Mguggis (talk) 18:08, 7 February 2017 (UTC)
thread_title:  Abuse of Notation 

message: Should not every unique idea in the article (definitions and examples included) end with a note? Using this article as an example, there seems to be at least one missing, as there is a footnote saying "[citation needed]" after a sentence in the middle of Definition section of the page. Additionally, while all of the links are in working order, there seems to be a lack of recent source material for this page (only about a third are published after 2000). (I hope to be able to confirm the validity of the sentence in question requiring a citation and perhaps incorporate a recent reference in a future edit of this article.) Llee17 (talk) 03:49, 27 September 2016 (UTC)
thread_title:  Checking citations (Article critique)  

message: In the intro, it is stated that "The aim of the Bayes factor is to quantify the support for a model over another, regardless of whether these models are correct." I think the second part of this sentence is contentious, or at least it should be better explained what is meant by that. Consistency proofs of the BF require the true model to be in the set, which is why many introductory texts drop a note about the BF requiring the true model to be in the set for consistency. It's not clear to me that "support" would even mean for two wrong models, technically (how would I prove the method performs correctly), and epistemologically. In the case of wrong models, AIC has a more straightforward definition via KL divergence.  — Preceding unsigned comment added by 132.199.61.137 (talk) 17:17, 14 November 2017 (UTC)
thread_title:  regardless of whether these models are correct 

message: Hello,
I was reviewing this page, and I noticed what I thought was an error. In the old revision (revision id 880926160), the Definition section included the following text: 
The posterior probability \Pr(M|D) of a model M given data D is given by Bayes' theorem:
:\Pr(M|D) = \frac{\Pr(D|M)\Pr(M)}{\Pr(D)}.
The key data-dependent term \Pr(D|M) is the likelihood of the model M in view of the data D, and represents the probability that some data are produced under the assumption of the model M; evaluating it correctly is the key to Bayesian model comparison. 
However, \Pr(D|M) is not "the likelihood of the model M in view of the data D"; it is instead the likelihood of the Data D in view of the Model M, as is described in the later part of that sentence. The term which is "the likelihood of the model M in view of the data D" is \Pr(M|D). Thus, I edited out the erroneous first part of this sentence. (See the diff here.)
If I am interpreting the page or the probability notation incorrectly, someone can revert this change.
Thebrainkid (talk) 06:21, 18 February 2019 (UTC)
thread_title:  Fixed a error or typo in the Definition section