Faktor Bayes

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Dina statistik, the use of Bayes factors is a Bayesian alternative to classical hypothesis testing.

Given a model selection problem in which we have to choose between two models M1 and M2, on the basis of a data vector x. The Bayes factor K is given by

K = \frac{p(x|M_1)}{p(x|M_2)}.

In a sense this is a likelihood-ratio test. Generally, the models M1 and M2 will be parametrised by vectors of parameters θ1 and θ2; thus K is given by

K = \frac{p(x|M_1)}{p(x|M_2)} = \frac{\int \,p(\theta_1|M_1)p(x|\theta_1 M_1)d\theta_1}{\int \,p(\theta_2|M_2)p(x|\theta_2 M_2)d\theta_2}.

A value of K > 1 means that the data indicate that M1 is more likely than M2 and vice versa. Harold Jeffreys gave a scale for interpretation of K:

K Strength of evidence
< 1 Negative (supports M2)
1 to 3 Barely worth mentioning
3 to 12 Positive
12 to 150 Strong
> 150 Very strong

Many Bayesian statisticians would use a Bayes factor as part of making a choice, but would also combine it with their estimates of the prior probability of each of the models and the loss functions associated with making the wrong choice.

[édit] Conto

Suppose we have a variabel acak which produces either a success or a failure. We want to consider a model M1 where the probability of success is q=1/2, and another model M2 where q is completely unknown and we take a prior distribution for q which is uniform on [0,1]. We take a sample of 200, and find 115 success and 85 failures. The likelihood is

{200 \choose 115}q^{115}(1-q)^{85}.

So we have

P(X=115|M_1)={200 \choose 115}\left({1 \over 2}\right)^{200}=0.00595...

but

P(X=115|M_2)=\int_{q=0}^1 1{200 \choose 115}q^{115}(1-q)^{85}dq = {1 \over 201} = 0.00497...

The ratio is then 1.197..., which is "barely worth mentioning" even if it points very slightly towards M1.

This is not the same as a classical likelihood ratio test, which would have found the maximum likelihood estimate for q, namely 115/200=0.575, and from that get a ratio of 0.1045..., and so pointing towards M2. A frequentist hypothesis test would have produced an even more dramatic result, saying that that M1 could be rejected at the 5% confidence level, since the probability of getting 115 or more successes from a sample of 200 if q=1/2 is 0.0200..., and as a two-tailed test of getting a figure as extreme as or more extreme than 115 is 0.0400... Note that 115 is more than two standard deviations away from 100.

M2 is a more complex model than M1 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 errors.

[édit] Tempo oge