Mrthdrb

I am trying to understand Bayes Factor (BF). I believe they are like likelihood ratio of 2 hypotheses. So if BF is 5, it means H1 is 5 times more likely than H0. And value of 3-10 indicates moderate evidence, while >10 indicates strong evidence.

However, for P-value, traditionally 0.05 is taken as cut-off. At this P value, H1/H0 likelihood ratio should be about 95/5 or 19.

So why a cut-off of >3 is taken for BF while a cut-off of >19 is taken for P values? These values are not anywhere close either.

A few things:

The BF gives you evidence in favor of a hypothesis, while a frequentist hypothesis test gives you evidence against a (null) hypothesis. So it's kind of "apples to oranges."

These two procedures, despite the difference in interpretations, may lead to different decisions. For example, a BF might reject while a frequentist hypothesis test doesn't, or vice versa. This problem is often referred to as the Jeffreys-Lindley's paradox. There have been many posts on this site about this; see e.g. here, and here.

"At this P value, H1/H0 likelihood should be 95/5 or 19." No, this isn't true because, roughly $p(y mid H_1) neq 1- p(y mid H_0)$. Computing a p-value and performing a frequentist test, at a minimum, does not require you to have any idea about $p(y mid H_1)$. Also, p-values are often integrals/sums of densities/pmfs, while a BF doesn't integrate over the data sample space.

The Bayes factor $B_01$ can be turned into a probability under equal weights as
$$P_01=frac11+frac1large B_01$$but this does not make them comparable with a $p$-value since

1. 
   $P_01$ is a probability in the parameter space, not in the sampling space


2. its value and range depend on the choice of the prior measure, they are thus relative rather than absolute (and Taylor's mention of the Lindley-Jeffreys paradox is appropriate at this stage)


3. both $B_01$ and $P_01$ contain a penalty for complexity (Occam's razor) by integrating out over the parameter space

If you wish to consider a Bayesian equivalent to the $p$-value, the posterior predictive $p$-value (Meng, 1994) should be investigated
$$Q_01=mathbb P(B_01(X)le B_01(x^textobs))$$
where $x^textobs$ denotes the observation and $X$ is distributed from the posterior predictive
$$Xsim int_Theta f(x|theta) pi(theta|x^textobs),textdtheta$$
but this does not imply that the same "default" criteria for rejection and significance should apply to this object.

Some of your confusion might stem from taking the number 95/5 directly from the fact that the p value is 0.05 - is this what you are doing? I do not believe this is correct. The p value for a t-test, for example, reflects the chance of getting the observed difference between means or a more extreme difference if the null hypothesis is in fact true. If you get a p value of 0.02, you say 'ah, there is only a 2% chance of getting a difference like this, or a greater difference, if the null is true. That seems very improbable, so I propose that the null is not true!'. These numbers are just not the same thing that goes into the Bayes factor, which is the ratio of the posterior probabilities given to each competing hypothesis. These posterior probabilities are not computed in the same way as the p-value, and so thinking of 95/5 as being like posterior probabilities that would give a BF of 19 is not correct.

As a side note, I would suggest strongly guarding against thinking of different BF values as meaning particular things. These assignments are completely arbitrary, just like the .05 significance level. Problems such as p-hacking will occur just as readily with Bayes Factors if people start to believe that only particular numbers warrant consideration. Try to understand them for what they are, which are something like relative probabilities, and use your own sense to determine whether you find a BF number convincing evidence or not.