Show that $P(A\mid B) > P(A\mid B^{c}) \implies P(B\mid A) > P(B\mid A^{c})$

Question

My initial thought is to start from $P(A\mid B)$ and $P(A\mid B^{c})$ and transform them to expressions involving $P(B\mid A)$ and $P(B\mid A^{c})$ respectively.

$P(A\mid B) = \dfrac{P(A)P(B\mid A)}{P(B)}$

$P(A\mid B^{c}) = \dfrac{P(A) - P(B) + P(A^{c})P(B\mid A^{c})}{P(B^{c})}$

The expressions get quite messy when I try to compare them, and I am wondering if there is a better a way.

Answer 1

To begin with, notice that \begin{align*} P(A|B) - P(A|B^{c}) > 0 & \Longleftrightarrow \frac{P(A\cap B)}{P(B)} - \frac{P(A\cap B^{c})}{P(B^{c})} > 0\\\\ & \Longleftrightarrow \frac{P(B^{c})P(A\cap B) - P(B)P(A\cap B^{c})}{P(B)P(B^{c})} > 0\\\\ & \Longleftrightarrow P(A\cap B) - P(B)(P(A\cap B) + P(A\cap B^{c})) > 0\\\\ & \Longleftrightarrow P(A\cap B) - P(B)P(A) > 0 \end{align*}

Now we can manipulate the second relation in order to obtain \begin{align*} P(B|A) - P(B|A^{c}) & = \frac{P(A\cap B)}{P(A)} - \frac{P(A^{c}\cap B)}{P(A^{c})}\\\\ & = \frac{P(A^{c})P(A\cap B) - P(A)P(A^{c}\cap B)}{P(A)P(A^{c})}\\\\ & = \frac{P(A\cap B) - P(A)(P(A\cap B) + P(A^{c}\cap B))}{P(A)P(A^{c}))}\\\\ & = \frac{P(A\cap B) - P(A)P(B)}{P(A)P(A^{c})} \end{align*} which is positive due to the first result.

Answer 2

From the given information we can infer from properties of complements that

$$P\left (A^c|B \right ) < P\left (A^c|B^c \right)$$

Thus

$$P\left ( A| B \right ) > P\left( A |B^c \right ) > P \left (A^c |B^c\right )> P \left ( A^c |B\right ) $$

$$\frac{P\left( A \cap B \right )}{P(B)} > \frac{P\left( A \cap B^c \right )}{P(B^c)} >\frac{P\left( A^c \cap B^c \right )}{P(B^c)}> \frac{P\left( A^c \cap B \right )}{P(B)} $$

$$\frac{P(A) P \left (B |A\right )}{P(B)} > \frac{P(A) P \left (B^c |A\right )}{P(B^c)} >\frac{P(A) P \left (B^c |A^c\right )}{P(B^c)}> \frac{P(A^c) P \left (B |A^c\right )}{P(B)} $$

$$\frac{P(A) P \left (B |A\right )}{P(B)} > \frac{P(A^c) P \left (B |A^c\right )}{P(B)}$$

$$P(A) P \left (B |A\right ) > P(A^c) P \left (B |A^c\right )$$

We can also see from the Law of Total Probability that

$$P\left ( A\right ) = P\left (B \right)P\left (A|B\right ) + P\left (B^c \right)P\left (A|B^c\right )$$

$$P\left ( A^c\right ) = P\left (B \right)P\left (A^c|B\right ) + P\left (B^c \right)P\left (A^c|B^c\right )$$

Which means that

$$P(A) > P(A^c)$$

Therefore,

$$ P \left (B |A\right ) > P \left (B |A^c\right )$$