Replacing Intersection With Union in Bayes' Theorem

Question

In this question, I asked about deriving the formula $$Pr(R\ |\ (B\ \cap\ E)) = \frac{Pr(R\ |\ B)\ Pr(E\ |\ (B\ \cap\ R))}{Pr(R\ |\ B)\ Pr(E\ |\ (B\ \cap\ R)) + Pr(R^c\ |\ B)\ Pr(E\ |\ (B\ \cap\ R^c))}.$$ However, I don't think I got the operator on the LHS correct; it only makes sense to consider the probability of $R$ given the union of $B$ and $E$, not the intersection. Is it valid to correct this by flipping the intersections on both sides to unions (and vice-versa, if there were any unions), yielding $$Pr(R\ |\ (B\ \cup\ E)) = \frac{Pr(R\ |\ B)\ Pr(E\ |\ (B\ \cup\ R))}{Pr(R\ |\ B)\ Pr(E\ |\ (B\ \cup\ R)) + Pr(R^c\ |\ B)\ Pr(E\ |\ (B\ \cup\ R^c))},$$ or would the entire formula need to be derived from scratch. If the latter, how could this be achieved, since we no longer have the conditional probability rule to start from?

Answer 1

Here is a counter-example for your second equation, where you replace $\cap$s with $\cup$s.

Suppose that $Pr(R \cup E) = 1/2$, with $Pr(R \cap E) = 1/6$, and $Pr(R) = Pr(E) = 1/3$. And suppose that $B = (R \cup E)^c$, so $Pr(B) = 1/2$, with $Pr(B \cap R) = Pr(B \cap E) = 0$.

Calculating all the terms in the second formula, we have:

• $Pr(R \mid B \cup E) = (1/6)/(5/6) = 1/5$,
• $Pr(R \mid B) = 0$,
• $Pr(E \mid B \cup R) = (1/6)/(5/6) = 1/5$,
• $Pr(R^c \mid B) = (1/2)/(2/3) = 3/4$, and
• $Pr(E \mid B \cup R^c) = (1/6)/(2/3) = 1/4$.

Putting this all together, the conjectured second formula gives $$1/5 = \frac{0 \times 1/5}{(0 \times 1/5) + (3/4 \times 1/4)} = 0,$$ so it is not correct.

Given my understanding of the context of the original question, the original operator on the LHS, $Pr(R \mid B \cap E)$, is correct. You want the probability of $R$ (resurrection) given $B$ (background), and $E$ (evidence). Why would you want to condition on $B$ or $E$? Only knowing ''at least one of $B$ or $E$'' will give you less information about $R$ than knowing ''both $B$ and $E$''. In another context, if $R$ is the event of rolling a 6 on a die, $B$ was the event that the roll is bigger than 4, and $E$ is the event that the roll is even. If you know that both $B$ and $E$ are true, then you have more information about $R$ by conditioning on $B \cap E$ than conditioning on $B \cup E$.