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Following paragraph comes from Page 18 of E. T. Jaynes's Probability Theory: The Logic of Sciencehttp://bayes.wustl.edu/etj/prob/book.pdf

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I really have no idea how (1.36) works given (1.34) and (1.35)..

I tried Bayes' theorem as well as the Definition of Conditional Probability, but still can't get (1.36) from (1.34) and (1.35).

Does anyone have any ideas about this?

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Note: here I use $'$ to denote the other event as in the question, and am not using it to denote the complementary event.

Suppose $Pr(A|C')>Pr(A|C)$ and that $Pr(B|A\cap C) = Pr(B|A\cap C')$

Rewording this information, we have then as hypotheses

$\frac{Pr(A\cap C')}{Pr(C')}>\frac{Pr(A\cap C)}{Pr(C)}$ and $\frac{Pr(A\cap B\cap C)}{Pr(A\cap C)}=\frac{Pr(A\cap B\cap C')}{Pr(A\cap C')}$

The first is impossible if $Pr(A\cap C')=0$ as it could not be strictly larger than any other probability, so we may assume that it is positive.

Then $Pr(A\cap B|C') = \frac{Pr(A\cap B\cap C')}{Pr(C')}=\frac{Pr(A\cap B\cap C')}{Pr(C')}\cdot \frac{Pr(A\cap C')}{Pr(A\cap C')}=\frac{Pr(A\cap B\cap C)}{Pr(A\cap C)}\frac{Pr(A\cap C')}{Pr(C')}$

$=\frac{Pr(A\cap B\cap C)}{Pr(A\cap C)}\frac{Pr(A\cap C')}{Pr(C')}\frac{Pr(C)}{Pr(C)}=Pr(A\cap B|C)\cdot \frac{Pr(A|C')}{Pr(A|C)}\geq Pr(A\cap B|C)$ since $Pr(A|C')>Pr(A|C)$

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