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Set up

Consider a collection of random events $A_1, \dots, A_n$ and $B_1, \dots, B_n$. It is clear from Bayes rule that $$ P(A_i \mid B_i) = \frac{P(B_i \mid A_i) P(A_i)}{P(B_i)}, $$ for each $i=1,\dots,n$. Therefore by summing we get $$ \sum_{i=1}^n P(A_i \mid B_i) = \sum_{i=1}^n \frac{P(B_i \mid A_i) P(A_i)}{P(B_i)}. \tag{1} $$

Now, we establish another fact. Define $I \sim \mathrm{unif}\{1, \dots, n\}$ independent of all events. Then by Bayes rule $$ P(A_I \mid B_I) = \frac{P(B_I \mid A_I) P(A_I)}{P(B_I)}. $$ Expanding each of these terms using the law of total probability, we see e.g. that $P(B_I) = \frac{1}{n} \sum_{i=1}^n P(B_i)$ and $$ P(A_I \mid B_I) = \sum_{i=1}^n P(A_i \mid B_i, I=i) P(I=i) = \frac{1}{n} \sum_{i=1}^n P(A_i \mid B_i) \tag{?} $$ and likewise for $P(B_I \mid A_I)$ and $P(A_I)$. Plugging this into the previous display, we get $$ \sum_{i=1}^n P(A_i \mid B_i) = \frac{\sum_{i=1}^n P(B_i \mid A_i) \sum_{i=1}^n P(A_i)}{\sum_{i=1}^n P(B_i)}. \tag{2} $$

Consequence and confusion

By equating the RHS terms in eqs. (1) and (2), we get $$ \frac{\sum_{i=1}^n P(B_i \mid A_i) \sum_{i=1}^n P(A_i)}{\sum_{i=1}^n P(B_i)} = \sum_{i=1}^n \frac{P(B_i \mid A_i) P(A_i)}{P(B_i)}. $$ These two expressions in eq. (3) are clearly not always equal, indicating an error in the previous development. I can't find it, however I am suspicious of eq. (?). Would someone please point out the error? Thanks.

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    $\begingroup$ Such a well elaborated question for a new member. Congratulations! $\endgroup$
    – Dominique
    Commented Jul 10 at 9:37

1 Answer 1

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The problem is indeed in (?). Here you are applying the law of total probability to a conditional probability space, but not quite correctly. It should be $$P(A_I\mid B_I)=\sum_{i=1}^nP(A_I\mid B_I,I=i)P(I=i\mid B_I).$$

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  • $\begingroup$ Earlier, I made an answer claiming that this was a common mistake made in probability. Seems like I wasn't wrong! $\endgroup$ Commented Jul 10 at 17:45

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