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Question

In deterministic logic, the statement “AA implies BB” is equivalent to its contrapositive, “not BB implies not AA”. In this problem we will consider analogous statements in probability, the logic of uncertainty. Let AA and BB be events with probabilities not equal to 0 or 1. (a) Show that if P(BA)=1P(B|A) = 1, then P(AcBc)=1P(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(BA)P(B|A) is very close to 1 but P(AcBc)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(BA)=1P(B|A) = 1.

P(BA)=P(B)P(AB)P(AB)P(B)+P(ABc)P(Bc)=1P(B)P(AB)=P(AB)P(B)+P(ABc)P(Bc)P(ABc)P(Bc)=0\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(ABc)=0P(A|B^c) = 0 or P(Bc)=0P(B^c) = 0. If P(ABc)=0P(AcBc)=1P(A|B^c) = 0 \Rightarrow P(A^c|B^c) = 1, so we are done in this case. If P(Bc)=0P(B^c) = 0 then P(AcBc)P(A^c|B^c) does not have a sense.

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