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I'm having a hard time understanding the two different approaches of measuring probability of an event.

Suppose there are two events $A$ and $B$.

1) Event $A$ occurs with probability $p_A$. This $p_A$ itself is random, so it has a pdf $f_A(p_A)$ over $[0,1]$.

2) Likewise, event $B$ occurs with probability $p_B$, and the pdf of $p_B$ is denoted by $f_B$.

3) The joint event $A\cap B$ occurs with probability $p_{A}$. That is, the event $A$ is nested in the event $B$.

What I want to calculate is the "probability of event $A$ when event $B$ has occured". The two approaches I'm confusing is as follows

Approach 1: It's simply the expected value of $p_{A|B}$. So, the prob. is given by $$\int p_{A|B}f_{A|B}dp_{A|B}=\int\frac{p_{A}}{p_B}f_{A|B}(p_A,p_B)dp_Adp_B.$$

Approach 2: Using the formula for a conditional probability, we have $$P[A|B]=\frac{P[A]}{P[B]}=\frac{\int p_Af_A(p_A)dp_A}{\int p_B f_B(p_B)dp_B}.$$

As far as I believe, the two approaches do not have to produce the same answers. Which approach should be the correct one and what are the differences in the interpretation?

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The second approach is correct. In the first approach, the following conclusion is too direct:

It's simply the expected value of $p_{𝐴|𝐵}$

We're saying that $P(A|B)=E[p_{A|B}]$, but at first, $p_{A|B}$ needs a proper definition, which in turn should, maybe, demystify other terms: $f_{A|B}(p_A,p_B)$.

The second approach follows directly from total probability law and Bayes Theorem: $$P(A|B)=\frac{P(A\cap B)}{P(B)},\ \ P(A)=\int \underbrace{P(A|p_A)}_{p_A}f_{p_A}(p_A)dp_A=E[p_A]$$

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  • $\begingroup$ So, it's wrong to use the mean of the probabilities of event to represent the probability of event. and it's because it's just a first moment that does not contain all the information about the probabilities($p_{A|B}$)? $\endgroup$
    – Andeanlll
    Oct 3 '19 at 2:18
  • $\begingroup$ I just have doubts about $p_{A|B}\overbrace{=}^?p_A/p_B$, and the meaning imposed over $p_{A|B}$. How would you describe it? If I'm not mistaken, your sentence "it's wrong to use the mean of the probabilities of event to represent the probability of event" implies that $P(A)\neq E[p_A]$, but it is actually true. $\endgroup$
    – gunes
    Oct 3 '19 at 10:13

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