What does it mean for one event to cause another event — for example, smoking () to cause cancer ()? There is a long history in philosophy, statistics, and the sciences of trying to clearly analyze the concept of a cause. One tradition says that causes raise the probability of their effects; we may write this symbolically is (|)>().(1)
Another way to formulate a probabilistic theory of causation is to say that (|)>(|^).(2)
I know that equation (1) implies that (|)>().
Show that equation (1) implies equation (2)
By the law of total probability, we can relate $P(E)$ to $P(E\mid C)$ and $P(E\mid C^c)$
$$P(E) = P(E\mid C)P(C) + P(E\mid C^c)(1-P(C)) = P(C)[P(E\mid C) - P(E\mid C^c)] + P(E\mid C^c) $$
Using our assumption $P(E\mid C) > P(E)$ we get:
$$P(C)[P(E\mid C) - P(E\mid C^c)] + P(E\mid C^c) = P(E) < P(E\mid C)$$
Collecting conditional probabilities on either side of the inequality we get
$$ P(E\mid C^c)- P(C)P(E\mid C^c) < P(E\mid C) - P(C)P(E\mid C) $$
Finally, we factor out $1-P(C)$ on both sides:
$$ (1-P(C))P(E\mid C^c) < (1-P(C))P(E\mid C)$$
Assuming $P(C) < 1$ ($P(C)=1$ would make it omnipresent and not a typical "cause"), we can divide by the strictly positive value $1-P(C)$ to get our implication:
$$P(C) < 1 \text{ and } (1-P(C))P(E\mid C^c) < (1-P(C))P(E\mid C) \implies P(E\mid C^c) < P(E\mid C)\;\; \square$$