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Let us have two events, $A$ and $B$ whose probabilities are $P(A)$ and $P(B)$. In the absence of any other information, what would be a reasonable probability to assign to $AB$, that is, $A$ and $B$ being true simultaneously?

The first thing that inevitably comes to mind is $P(AB):= P(A)P(B)$. That is, if we have no reason to assume any of the many possible ways in which $A$ and $B$ are dependent, let us assume that they are independent. However, an "independent" seems to stand out as very special among all the possible "dependent" s. Since $0 \leq P(AB) \leq \min(P(A), P(B))$ it seems somehow odd, that one point in this interval would be the "default" somehow.

In which situation would we use one intersection prior over another?

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If we assume that $A$ and $B$ are independent then the probability that they occur at the same time would be the same as the probability of their intersection. $P(A\cap B)=P(A)P(B)$. This scenario could be represented as rolling two dice at the same time. And if we further assume that $P(A)\neq \{0,1\},\, P(B)\neq \{0,1\}$ then $0<P(A\cap B)<min\{P(A),\,P(B)\}$.

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  • $\begingroup$ Thank you for your answer. I see what follows when you assume independence, but I am more curious whether or not it is a reasonable assumption in the absence of any evidence. Also, I am interested in some alternatives to the independence prior and in their justifications. $\endgroup$ – Martin Drozdik Feb 10 at 22:25

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