Show $P(A|B) > P(A|B^c)$ is equivalent to $P(B|A) > P(B|A^c)$

Question

This is a problem from Blitzstein and Hwang Introduction to Probability.

Consider the following scenario, from Tversky and Kahneman [30]: Let A be the event that before the end of next year, Peter will have installed a burglar alarm system in his home. Let B denote the event that Peter’s home will be burglarized before the end of next year.

(c) Show that for any events $A$ and $B$ (with probabilities not equal to 0 or 1), $P(A|B) > P(A|B^c )$ is equivalent to $P(B|A) > P(B|A^c)$.

Blitzstein, Joseph K.. Introduction to Probability (Chapman & Hall/CRC Texts in Statistical Science) (p. 76). Chapman and Hall/CRC. Kindle Edition.

Using Bayes: $$ \frac{P(A|B)}{P(A|B^c)}=\frac{P(B|A)P(B^c)}{P(B^c|A)P(B)} $$

Not sure how to go beyond this step.

Answer 1

Let $$a = P(A^c \cap B)$$ $$b = P(A \cap B)$$ $$c = P(A \cap B^c)$$ $$d = P(A^c \cap B^c)$$ These denote the 4 regions of the Venn Diagram for events $A$ and $B$.

By assumption : $$P(A|B) > P(A|B^c) \Leftrightarrow$$ $$\frac{P(A \cap B)}{P(B)} > \frac{P(A \cap B^c)}{P(B^c)}$$ We substitute to get $$\frac{b}{a+b} > \frac{c}{c+d}$$ We multiply by the denominators, and simplify to get $$bd > ac \Leftrightarrow$$ $$ab + bd > ab + ac \Leftrightarrow$$ $$b(a+d) > a(b+c) \Leftrightarrow$$ $$\frac{b}{b+c} > \frac{a}{a+d}$$ Substitue back to get $$\frac{P(B \cap A)}{P(A)} > \frac{ P(B \cap A^c)}{P(A^c)} \Leftrightarrow$$ $$P(B|A) > P(B|A^c)$$

Answer 2

Hint: show that

$$P(B)+P(B^c)\frac{P(A|B^c)}{P(A|B)}=\frac{P(A)P(B)}{P(A\cap B)}=P(A)+P(A^c)\frac{P(B|A^c)}{P(B|A)}$$