2
$\begingroup$

If I have some data that I believe is Normally distributed and I just want to test the hypotheses that the mean is equal to 1 of 3 values, my understanding is that the Bayes Factor is the ratio of marginal likelihoods. So

BF = $\frac{p(x_i|H_0)}{p(x_i|H_1)}$.

Furthermore, from the texts I read, the marginal likelihood for the null is equal to

$\int(p(\theta_0|H_0)p(x_i|\theta_0,H_0)d\theta_0$

where $\theta_i$ is your parameter, in this case the mean.

My question is, since the alternate hypothesis is that the mean is equal to one of the two values other than the null, is the marginal likelihood $p(x_i|H_1)$ equal to

$\int(p(\theta_{1,1}|H_1)p(x_i|\theta_{1,1},H_0)d\theta_{1,1}$$\int(p(\theta_{1,2}|H_1)p(x_i|\theta_{1,2},H_1)d\theta_{1,2}$.

$\endgroup$
2
$\begingroup$

You are almost right. Use the sum of the two integrals rather than the product.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.