Coherence of conditional probabilities

Question

Dennis Lindley's paper The Philosophy of Statistics in 2001 includes the following 'simple' example of statistical coherence: "A set of uncertainty statements is said to be coherent if they satisfy the rules of the probability calculus. Thus, the pair of statements p(A|B) = 0.7 and p(A| ~B) = 0.4 do not cohere with the pair p(B|A)= 0.5 and p(B|~ A) = 0.3. (Here ~B denotes the complement of B.) Think of A as a statement about data x and B as a statement about parameter theta. The first pair refers to uncertainties in the data and coheres with the first parameter statement, p(B|A)= 0.5, for data A. (Take p(B) = 0.4/1.1 = 0.36.) But all three do not cohere with the second parameter statement for data A, that p(B| A) = 0.3. With p(B) = 0.36, the coherent value is 0.22."

How is the coherent value of 0.22 calculated?

Answer 1

$$p(B|A)=\frac{p(A|B)p(B)}{p(A|B)p(B))+p(A|\neg B)p(\neg B)} =\frac{p(A|B)}{p(A|B)+p(A|\neg B)p(\neg B)/p(B)}$$ leads to $$0.5=\frac{0.7}{0.7+0.4p(\neg B)/p(B)}$$ hence to $p(B)=0.4/1.1=0.36$. This means that $$P(B|\neg A)=\frac{p(\neg A|B)p(B)}{p(\neg A|B)p(B)+p(\neg A|\neg B)p(\neg B)}$$ is equal to $$\frac{0.3\cdot 0.36}{0.3\cdot 0.36+0.6\cdot 0.64}\approx 0.22$$ And using instead $$p(B|\neg A)=\frac{p(\neg A|B)p(B)}{p(\neg A|B)p(B))+p(\neg A|\neg B)p(\neg B)} =\frac{p(\neg A|B)}{p(A|B)+p(\neg A|\sim B)p(\neg B)/p(B)}$$ leads to $$0.3=\frac{0.3}{0.3+0.6p(\neg B)/p(B)}$$ hence to $p(B)=0.6/1.3\approx 0.46$.