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From 5.31 we want \(x>y\) but
\[\log \frac{x}{1-x} +\log \frac{a_x}{b_x}< \log \frac{y}{1-y}+ \log \frac{a_y}{b_y} \]
i.e.
\[\log \frac{x}{1-x} - \log \frac{y}{1-y} <\log \frac{a_y}{b_y} -\log \frac{a_x}{b_x} \]
or, multiplicatively
\[\frac{x(1-y)}{y(1-x)}< \frac{a_y b_x}{a_x b_y}.\]
This is of course very possible. It means simply that the level of trust of N is sufficiently different between X and Y that the the difference of evidence perceived by X and Y from the announcement is bigger than their prior difference of evidences.
Everything in this exercise is conditional on \(I\).
If the “graphical model” assumption \(C\to B\to A\), is true (meaning that \(A\) and \(C\) are independent conditional on \(B\)), then 5.43 becomes
\[P(A|CI)=qP(A|BI)+(1-q)P(A|\overline{B}I)\]
and since \(P(A|\overline{B}I)\) is bounded by \(1\), as \(q\to 1\) we do have \(P(A|CI) \to P(A|BI).\)
In general, however, even \(q=1\) does not imply closeness of \(P(A|BI)\) and \(P(A|CI)\). Suppose we have a fair 4 sided die. Let \(C\) be the event “\(4\) is rolled”, \(B\) the event “the result is even”, \(A\) the event “the result is 1 or 2”. Then \(q=P(B|C)=1\), but \(P(A|C)=0\) while \(P(A|B)=1/2\).
By taking more-sided dice we can even make \(P(A|B)\) arbitrarily small while keeping the other implications.
An example “from life”: \(B=\)“I’m in San Francisco” implies that the probability that \(A=\)“It is snowing around me” conditional on \(B\) is very low. However, if I know \(C=\)“I’m in San Francisco and the date is February 5, 1976” then I am certain of \(B|C\) (so \(q=1\)), but also certain of \(A|C\).