Question

What does it mean for one event 𝐶 to cause another event 𝐸 – for example, smoking (𝐶) to cause cancer (𝐸)? 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 as

\( 𝑃(𝐸|𝐶) > 𝑃(𝐸)  \) –  –  –  –  –  –  –  –  –  – (1)

a) Does equation (1) imply that 𝑃(𝐶|𝐸) > 𝑃(𝐶)?

If so, prove it. If not, give a counter-example.

b) Another way to formulate a probabilistic theory of causation is to say that

\( P (E | C) > P(E | C^c)    \)   –  –  –  –  –  –  –  –  –  – (2)

Show that equation (1) implies equation (2).

SOLUTION

(a)

Given,

\( P(E|C) > P(E)  \)

Also, by definition, \( \ P(E|C)\ =\ \ \dfrac{P(E|C)}{P\left(C\right)} \)

\( \Rightarrow\ \dfrac{P(E|C)}{P\left(C\right)}\ >\ P\left(E\right) \)

\( ⇒ P(E∩C) > P(E). P(C)  \)  —– (a)

Again,
\( P(E∩C) = P(C|E).P(E)  \)  —– (b)

from (a) and (b), it follows that

\( P(C|E).P(E) > P(E). P(C) \)

\( or, P(C|E) > P(C) \)

Hence Proved

SOLUTION

(b)

Given,

\( P(E|C) > P(E) —– (a) \)

By definition,

\( P (E | C^C) = \dfrac{P(E∩C^C)}{P(C^C)} \)

\( P(E) = P(E|C).P(C) + P(E|C^C).P(C^C) —– (a) \)

From (a) and (b), we have

\( P(E|C) > P(E|C).P(C) + P(E|C^C).P(C^C) \)

or

\( P(E|C) – P(E|C).P(C) > P(E|C^C).P(C^C) \)

\( P(E|C)( 1 – P(C) ) > P(E|C^C).P(C^C) —– (c) \)

Also,

\( 1 – P(C) = P(C^C) —– (d) \)

From (c) and (d), we have

\( P(E|C)P(C^C) > P(E|C^C).P(C^C) \)

or

\( P(E|C) > P(E|C^C) \)

Hence Shown.