Probability Proof(Finan #5.1.15)

Question

Show that $P(A|B)>P(A)$ if and only if $P(A^c |B)<P(A^c)$. We assume that $0<P(A)<1$ and $0<P(B)<1$ ( Finan #5.1.5)

Taking the complement of probabilities reverses the sign of the inequality (as shown below): $$P(X)>P(Y)\Rightarrow-P(X)<P(Y)\Rightarrow1-P(X)<1-P(Y)\Rightarrow P(X^c)<P(Y^c) $$

Hence, take the complement of both sides of $P(A|B)>P(A)$ giving us the desired result $$P(A^c |B)<P(A^c)$$

Is this correct?

Notes:

1. Finan is a prep guide for Exam Probability of SOA-USA
2. The answer key at the back of the book has a different, lengthier solution.

Answer 1

$P(A) = 1 - P(A^C)$
$P(A|B) = 1 - P(A^C|B)$

If $P(A|B) > P(A)$ then ...

$1 - P(A^C|B) > 1 - P(A^C)$
$-P(A^C|B) > -P(A^C)$
$P(A^C|B) < P(A^C)$

Answer 2

Your proof is correct but it is only halfway complete. Note the instructions say show $P(A|B)>P(A)$ if and only if $P(AC|B)<P(AC)$. Formally speaking, they are asking you to prove $P(A|B)>P(A)⇔P(AC|B)<P(AC)$. This means you must first prove $P(A|B)>P(A)⇒P(AC|B)<P(AC)$, which you do, but then you must also prove $P(AC|B)<P(AC)⇒P(A|B)>P(A)$, which you do not