In bayes rule, P(th|D)=P(D|th)P(th)/P(D), what are the significant of each of the terms? Thanks

Question

anonymous · Answer

$$P(\theta|D)=\frac{P(D|\theta)P(\theta)}{P(D)}$$

anonymous · Answer

$P(D)$ is the probability of D being true. As an example, D might be obtaining a positive result on a lab test for a disease, and $P(D)$ would be the probability that a random person taking the test would get a positive result.

$P(\theta)$ is the probability of $\theta$ being true. As an example, $\theta$ might be having a disease, and $P(\theta)$ would be the probability that a random person would be suffering from the disease.

$P(D|\theta)$ is the conditional probability of $D$ being true given that $\theta$ is true. In our example this would be the probability that a person suffering from the disease would actually test positive for the disease if they took a lab test. In essence this is a measure of how effective the test is; if the probability $P(D|\theta)$ is close to 1 then the test is very effective.

$P(\theta|D)$ is the conditional probability of $\theta$ being true given that $D$ is true. In our example this would be the probability that a person taking a lab test for a disease and testing positive for it is in fact actually suffering from the disease. (If they're not suffering from the disease then the test has produced a false positive.)

anonymous · Answer

Hi frankhecker. Thank you for your reply. I should've made it clear that the context actually refers to $$posterior = \frac{likelihood \times prior} {evidence}$$The theta in the original equation actually denotes a random variable rather than an event. And that's where my confusion lies.