# The ratio of two Bayes factors for two opposite one-tailed hypotheses

I am trying to understand how Bayesian inference works, so this might be a very simple question. I have an experiment where I test two hypotheses predicting opposite results. Let’s say, hypothesis 1 (H1) predicts that x > 0, and hypothesis 2 (H2) predicts that x < 0.

I calculated Bayes factor with informed priors (positive and negative half-normals for H1 and H2 respectively) for two hypotheses. BF10 for the H1 was 0.04, and BF10 for the H2 was 0.13. In other words, both results indicate that I have to believe more in H0 than in any of two alternative hypotheses.

However, if I still want to make some inference on H1 and H2, can I just divide BF10(for H2) by BF10(for H1)? This ratio (it will Bayes Factor too, right?) will be 0.13 / 0.04 = 3.25. Does this result tell me that I have to increase my belief in H2 compared to H1?

• If I understand correctly, Bayes factor for H1 is (prior(H1) * likelihood) / (prior(H0) * likelihood). H0 (x=0) is the same for H1 and H2. Oct 28 '20 at 14:18
• No this is not correct, the likelihood is integrated against the prior, check eg the Wikipedia page Oct 28 '20 at 14:22
• Now, after you asked, i started thinking about it, and it seems that what i did (BF(H1)/BF(H2)) is basically a Bayes factor for these two hypotheses, because the common denominator in two Bayes factors (prior(H0)*lokelihood) is cancelled in division. Would be great if somebody could confirm this. Oct 28 '20 at 14:23
• Thanks @Xi'an ! Oct 28 '20 at 14:30

In short, yes, but your comments indicate you might have gotten the right answer for the wrong reasons.

A Bayes Factor is the ratio of the marginal likelihoods of the data under two different hypotheses.

Let's say $$L_i$$ is the marginal likelihood of the data under hypothesis $$H_i$$, and $$H_0: x = 0$$, $$H_1: x > 0$$, and $$H_2: x < 0$$, where for $$H_1$$ and $$H_2$$ you have some prior over possible values of $$x$$ (positive and negative values, respectively).

Your Bayes Factors are then the following ratios:

• $$BF_{10} = \frac{L_1}{L_0}$$
• $$BF_{20} = \frac{L_2}{L_0}$$
• $$BF_{21} = \frac{L_2}{L_1}$$

If you know $$BF_{20}$$ and $$BF_{10}$$, their ratio is

\begin{align} \frac{BF_{20}}{BF_{10}} &= \frac{\frac{L_2}{L_0}}{\frac{L_1}{L_0}} \\ &= \frac{L_2}{L_0} \times \frac{L_0}{L_1} \\ &= \frac{L_2}{L_1} \\ &= BF_{21} \end{align}

• Indeed, I did not understand Bayes factor calculation well (i.e. marginal likelihood), but I didn't do the calculation myself (I used a script from elsewhere), therefore, the results should be right. So, it's just a BF21. Great, thank you! Oct 28 '20 at 14:29