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Question

# In deterministic logic, the statement “$A$ implies $B$” is equivalent to its contrapositive, “not $B$ implies not $A$”. In this problem we will consider analogous statements in probability, the logic of uncertainty. Let $A$ and $B$ be events with probabilities not equal to 0 or 1. (a) Show that if $P(B|A) = 1$, then $P(A^c|B^c) = 1$. Hint: Apply Bayes’ rule and LOTP. (b) Show however that the result in (a) does not hold in general if $=$ is replaced by $\approx$. In particular, find an example where $P(B|A)$ is very close to 1 but $P(A^c|B^c)$ is very close to 0. Hint: What happens if A and B are independent?

Solution

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#### (a)

Let $P(B|A) = 1$.

\begin{align*} &P(B|A) = \dfrac{P(B)P(A|B)}{P(A|B)P(B)+P(A|B^c)P(B^c)} = 1 \\ &\Rightarrow P(B)P(A|B) = P(A|B)P(B)+P(A|B^c)P(B^c) \Rightarrow P(A|B^c)P(B^c) = 0 \end{align*}

Thus $P(A|B^c) = 0$ or $P(B^c) = 0$. If $P(A|B^c) = 0 \Rightarrow P(A^c|B^c) = 1$, so we are done in this case. If $P(B^c) = 0$ then $P(A^c|B^c)$ does not have a sense.

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