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Edwards, Lindman and Savage propose the MBF as the value that minimizes the Bayes Factor, where the alternate hypothesis has all its prior density concentrated at the maximum likelihood estimate. The MBF in this case is exp(-0.5z^2), so for a z of 2, corresponding to p-value (or alpha) of 5%, this would be approximately 1/7.

However, if you imagine another scenario, where the alternate hypothesis is not a number, but rather a binary state e.g. has breast cancer versus the null of no breast cancer, the Bayes Factor in this case would be alpha/beta, where alpha is the p-value and beta is the power of the test. As above, if p-value (or alpha) is 5%, the MBF would be 5/100, or 1/20, assuming your test has maximum power.

Could someone please explain why the MBF is 1/7 in one case and 1/20 in the other? I get that the alternate hypothesis can take on a range of values in the first case, while it's a binary state in the second case. But in both scenarios, you're lining up the entire prior density of the alternate hypothesis behind a single value, which should be the equivalent of saying your test or condition has 100% power. I'm searching for an intuitive explanation for why one method yields numerically different results from the other. Thanks.

Please refer to pages 18, 22, 27-28 for more details: https://papers.ssrn.com/sol3/papers.cfm?abstract_id=2893930

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  • $\begingroup$ I am a little late to the party but you might want to check the paper Held, L., & Ott, M. (2018). On p-values and Bayes factors. and the related pCalibrate R package, as they studied lots of different calculation methods for the minimum bayes factor and the influence of various parameters such as effect size and sample size. $\endgroup$ – gaborous Sep 4 at 5:31

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