Timeline for Bayesian Hypothesis Testing: Bayes Factor vs Comparing the Posterior Distribution
Current License: CC BY-SA 4.0
8 events
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| Oct 5, 2021 at 23:22 | comment | added | Tom Bennett | Good point. I should have said that the hypothesis priors are indirectly determined by the parameter priors $\pi(.)$. Yes, I've seen people doing a composite $\pi(.)$ with $P(H_i)$ chosen first, but don't know how to effectively use it and its full implications. | |
| Oct 5, 2021 at 22:38 | comment | added | Christian Hennig | @TomBennett "we can't arbitrarily choose $P(H_0)$ and $P(H_1)$." Well, we can choose $\pi$ and the hypotheses probabilities by implication. We can also choose $\pi$ to make $P(H_0)=P(H_1)$, or even specify $\pi$ conditionally on $H_0, H_1$ with $P(H_0)$ and $P(H_1)$ chosen first. | |
| Oct 5, 2021 at 22:12 | comment | added | Tom Bennett | Yes, one has to do something special for point hypotheses. Typically people redefine the prior as $\pi(\theta) =\pi_0 \delta(\theta_0) + (1-\pi_0) \pi'(\theta)$. | |
| Oct 5, 2021 at 20:53 | comment | added | bdeonovic | I've just haven't seen prior odds for hypotheses defined in this way. I guess it makes sense, but it breaks down for point hypotheses e.g. $H_0: \theta=0$ which seems to be a big issue | |
| S Oct 5, 2021 at 20:51 | history | suggested | Tom Bennett | CC BY-SA 4.0 |
fixed url of the R package
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| Oct 5, 2021 at 20:03 | review | Suggested edits | |||
| S Oct 5, 2021 at 20:51 | |||||
| Oct 5, 2021 at 20:02 | comment | added | Tom Bennett | I think this is a bit different from model selection, in that the priors cannot be chosen, but instead are determined by the priors of the model. For example, let $\theta$ be the parameter that we are interested in and let it be the animal's weights. Let the two hypotheses be H0: $\theta<10$ pounds and H1: $\theta \ge10$. Then $P(H_1)=\int_{\theta<10} \pi(\theta) d\theta$, $P(H_2)=\int_{\theta \ge 10} \pi(\theta) d\theta$ and $P(H_1) + P(H_2)=1$. But we can't arbitrarily choose $P(H_1)$ and $P(H_2)$. By contrast, in model comparison, we often assume $P(M_0)=P(M_1)$. | |
| Oct 5, 2021 at 15:26 | history | answered | bdeonovic | CC BY-SA 4.0 |