(related to Proposition: Characterization of Independent Events)
Assuming \(0 < p(B) < 1\), we have to show the equivalence
\[p(A|B)=p(A|\overline{B})\Longleftrightarrow p(A)=p(A|B).\]
Assume the events \(A\) and \(B\) are independent. Then the equation \(p(A|B)=p(A|\overline{B})\) holds (hypothesis). Because \(A=(A\cap B )\cup(A\cap\overline{B})\), it follows from the
\[\begin{array}{rcll} p(A)&=&p((A\cap B )\cup(A\cap\overline{B}))&\text{because }A=(A\cap B )\cup(A\cap\overline{B})\\ &=&p(A\cap B )+p(A\cap\overline{B})-p(A\cap B\cap A\cap\overline{B})&\text{probability of event union}\\ &=&p(A\cap B )+p(A\cap\overline{B})-p(\emptyset)&\text{because }B\cap \overline{B}=\emptyset\text{ and }A\cap \emptyset=\emptyset\\ &=&p(A\cap B )+p(A\cap\overline{B})&\text{probability of impossible event}\\ &=&p(A|B)p(B)+p(A|\overline{B})p(\overline{B})&\text{probability of joint events}\\ &=&p(A|B)p(B)+p(A|B)p(\overline{B})&\text{by hypothesis}\\ &=&p(A|B)(p(B)+p(\overline{B}))&\text{distributivity law of real numbers}\\ &=&p(A|B)(p(B)+1-p(B))&\text{probability of complement event}\\ &=&p(A|B)&\text{cancellation of }p(B)\text{ and multiplication by }1\\ \end{array}\]
Now assume that \(p(A|B)=p(A)\) (hypothesis). Because \(0 < p(B) < 1\), it follows from the probability of complement event that \(1 > p(\overline{B}) > 0\).
By applying
\[\begin{array}{rcll} p(A|\overline{B})&=&\frac{p(A\cap \overline{B})}{p(\overline{B})}&\text{conditional probability and }p(\overline{B}) > 0\\ &=&\frac{p(A)- p(A\cap B)}{p(\overline{B})}&\text{probability of event difference}\\ &=&\frac{p(A)- p(A|B)p(B)}{p(\overline{B})}&\text{probability of joint events}\\ &=&\frac{p(A)- p(A)p(B)}{p(\overline{B})}&\text{hypothesis}\\ &=&\frac{p(A)(1-p(B))}{p(\overline{B})}&\text{distributivity of real numbers}\\ &=&\frac{p(A)(p(\overline{B}))}{p(\overline{B})}&\text{probability of complement event}\\ &=&p(A)&\text{cancelling }p(\overline{B}) \end{array}\]
