1
$\begingroup$

I feel like this should be obvious, but I don't see the approach. This question comes from the philosophy of science, so I will pose it first in that context and then in a probability context.

In the philosophy of science, often multiple hypotheses are needed to make a prediction. One hypothesis might be the hypothesis being investigated, while auxiliary hypotheses (such as how the instrument works, etc.) are also needed for a prediction to be made. Each hypothesis is either true or false, but these are unknown. Thus in the Bayesian sense, each hypothesis is assigned a belief $p(H_i)$. Assuming we have priors for each hypothesis, how should the belief in each hypothesis change if an observation (either confirmatory or disconfirmatory) is made?

In probability terms suppose $p(A|H_1,H_2,H_3) = a$, $p(A)=b$, that $p(H_1)=h_1$, $p(H_2)=h_2$, and $p(H_3)=h_3$, and that each $H_i$ is independent, what should the probabilities $p(H_i|A)$ be? It is not clear to me that there's a unique answer; obviously answers have to be constrained such that $p(H_1,H_2,H_3|A) = ah_1h_2h_3/b$, but I don't see any further constraint. Am I missing something obvious here?

Improve this question
$\endgroup$
3
  • $\begingroup$ What is the meaning of $p(A|H_1,H_2,H_3)$? what is the meaning of "those probabilities are independent"? $\endgroup$
    – Xi'an
    Commented Dec 6, 2015 at 15:31
  • $\begingroup$ We say that events are independent, or that random variables are independent. Unless the probabilities are themselves random variables, we wouldn't normally speak of them being independent of other probabilities. So for example you could say that "the events '$H_1$ is true' and '$H_2$ is true' are independent" or you could define $H_i$ to be a random variable that takes the value $1$ when it's true and $0$ otherwise, and then say "the variables $H_1$ and $H_2$ are independent". Can you also explain how you arrived at the thing you say is "obvious" in your second-last sentence? $\endgroup$
    – Glen_b
    Commented Dec 6, 2015 at 15:42
  • $\begingroup$ Thanks for the reply! Yes, I would represent $p(H_1)=1$ if $H_1$ is true, and 0 if false. In the Bayesian framework, as I understand it, we could take $p(H_1)$ as our belief in the truth in $H_1$. I should have said $H_i$ are independent, not $p(H_i)$; the meaning is just $p(H_i,H_j) = p(H_i)p(H_j)$. Sorry for the confusion; I will edit it. The "obvious" constraint is Bayes theorem: $p(A|B) = p(B|A)p(A)/p(B)$ $\endgroup$
    – Hans
    Commented Dec 6, 2015 at 15:48

1 Answer 1

Reset to default
0
$\begingroup$

Using Bayes rule, law of total probability and chain rule, and treating $H_i$ as binary variables: $$ p(H_1 | A) = \frac{p(H_1, A)}{p(A)} = \frac{\sum_{H_2, H_3 \in \{0, 1\}} p(H_1, H_2, H_3, A)}{p(A)} = \frac{\sum_{H_2, H_3 \in \{0, 1\}} p(H_1, A | H_2, H_3) p(H_2) p(H_3)}{p(A)} = \frac{\sum_{H_2, H_3 \in \{0, 1\}} p(A | H_1, H_2, H_3) p(H_1) p(H_2) p(H_3)}{p(A)} $$

In order to compute that you'll need to specify probability of $A$ given all 8 possible configurations of $H_1, H_2, H_3$.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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