Editing Bayes factor (section)

==Example==
Suppose we have a [[random variable]] that produces either a success or a failure.  We want to compare a model ''M''<sub>1</sub> where the probability of success is ''q'' = {{frac|1|2}}, and another model ''M''<sub>2</sub> where ''q'' is unknown and we take a [[prior distribution]] for ''q'' that is [[uniform distribution (continuous)|uniform]] on [0,1].  We take a sample of 200, and find 115 successes and 85 failures.  The likelihood can be calculated according to the [[binomial distribution]]:

:<math>{{200 \choose 115}q^{115}(1-q)^{85}}.</math>

Thus we have for ''M''<sub>1</sub>

:<math>P(X=115 \mid M_1)={200 \choose 115}\left({1 \over 2}\right)^{200} \approx 0.006</math>

whereas for ''M''<sub>2</sub> we have

:<math>P(X=115 \mid M_2) = \int_{0}^1{200 \choose 115}q^{115}(1-q)^{85}dq = {1 \over 201} \approx 0.005 </math>

The ratio is then 1.2, which is "barely worth mentioning" even if it points very slightly towards ''M''<sub>1</sub>.

A [[frequentist]] [[Statistical hypothesis testing|hypothesis test]] of ''M''<sub>1</sub> (here considered as a [[null hypothesis]]) would have produced a very different result.  Such a test says that ''M''<sub>1</sub> should be rejected at the 5% significance level, since the probability of getting 115 or more successes from a sample of 200 if ''q'' = {{frac|1|2}} is 0.02, and as a two-tailed test of getting a figure as extreme as or more extreme than 115 is 0.04.  Note that 115 is more than two standard deviations away from 100. Thus, whereas a [[frequentist]] [[Statistical hypothesis testing|hypothesis test]] would yield [[Statistical significance|significant results]] at the 5% significance level, the Bayes factor hardly considers this to be an extreme result. Note, however, that a non-uniform prior (for example one that reflects the fact that you expect the number of success and failures to be of the same order of magnitude) could result in a Bayes factor that is more in agreement with the frequentist hypothesis test.

A classical [[likelihood-ratio test]] would have found the [[maximum likelihood]] estimate for ''q'', namely <math>\hat q =\frac{115}{200}  = 0.575</math>, whence 
:<math>\textstyle P(X=115 \mid M_2) = {{200 \choose 115}\hat q^{115}(1-\hat q)^{85}} \approx 0.06</math>  
(rather than averaging over all possible ''q'').  That gives a likelihood ratio of 0.1 and points towards ''M''<sub>2</sub>.

''M''<sub>2</sub> is a more complex model than ''M''<sub>1</sub> because it has a free parameter which allows it to model the data more closely.  The ability of Bayes factors to take this into account is a reason why [[Bayesian inference]] has been put forward as a theoretical justification for and generalisation of [[Occam's razor]], reducing [[Type I error]]s.<ref>[http://www.stat.duke.edu/~berger/papers/ockham.html Sharpening Ockham's Razor On a Bayesian Strop]</ref>

On the other hand, the modern method of [[relative likelihood]] takes into account the number of free parameters in the models, unlike the classical likelihood ratio.  The relative likelihood method could be applied as follows.  Model ''M''<sub>1</sub> has 0 parameters, and so its [[Akaike information criterion]] (AIC) value is <math>2\cdot 0 - 2\cdot \ln(0.005956)\approx 10.2467</math>.  Model ''M''<sub>2</sub> has 1 parameter, and so its AIC value is <math>2\cdot 1 - 2\cdot\ln(0.056991)\approx 7.7297</math>.  Hence ''M''<sub>1</sub> is about <math>\exp\left(\frac{7.7297- 10.2467}{2}\right)\approx 0.284</math> times as probable as ''M''<sub>2</sub> to minimize the information loss.  Thus ''M''<sub>2</sub> is slightly preferred, but ''M''<sub>1</sub> cannot be excluded.