Derivation using complements on conditional probabilities.

Consider two events E and F that satisfy P(E|F) = P(E).

Show that $$P(E^{c}|F^{c}) = P(E^{c})$$ assuming $$P(F^{c}) \neq 0$$.

At first sight this seems like a simple derivation. In my attempt I've tried using the multiplication rule: $$P(A \cap B) = P(A) \cdot P(B|A)$$ and the identity $$P(E) =1 - P(E^{c} | F)$$ but I seem to be running in circles:

This is where I've gone so far:

$$P (E|F) = P(E)$$

$$\therefore 1-P(E^{c}|F) = P(E)$$

$$\therefore P(E^{c}|F) = P(E) + 1$$

$$\therefore \frac{P(E^{c} \cap F)}{P(F)} = P(E) + 1$$

$$\therefore P(E^{c} \cap F) = P(F) P(E) + P(F)$$

$$\therefore P(E^{c} \cap F) = P(E) P(E | F) + 1 \ \ \$$ since $$P(E) = P(E|F)$$

$$\therefore P(E^{c} \cap F) = P(E \cap F) + 1 \ \ \$$ by using the multiplication rule.

The intuition I was following was to introduce a $$F^{c}$$ into this equality and then eventually reduce it to the statement we have to prove.

I'd by grateful if someone could tell me if I'm going in the right direction. If someone could find a quicker way to make the derivation, it would go a long way.

• The third conclusion is incorrect. In fact, you would get $P(E^c|F) =1-P(E)=P(E^c)$. – zhoraster Sep 27 '15 at 5:49

We are told that $\Pr(E\mid F)=\Pr(E)$. It follows by multiplication by $\Pr(F)$ that $\Pr(E\cap F)=\Pr(E)\Pr(F)$, that is, $E$ and $F$ are independent.

We want to show that $\Pr(E^c\cap F^c)=\Pr(E^c)\Pr(F^c)$, for then division by $\Pr(F^c)$ will give us the desired result. Informally, this is true, since $E$ and $F$ independent should imply that $E^c$ and $F^c$ are independent. But let us compute. We have $$\Pr(E^c\cap F^c)=1-\Pr(E\cup F)=1-(\Pr(E)+\Pr(F)-\Pr(E\cap F))=(1-\Pr(E))(1-\Pr(F))=\Pr(E^c)\Pr(F^c).$$

use

$$P(E^c\cap F^c) = 1 - P(E\cup F)=1- \left[ P(E)+P(F)-P(E \cap F)\right ]$$

$$= 1- P(E)-P(F)+P(E)P(F) = (1-P(E))(1-P(F))$$

so

$$P(E^c\cap F^c|F^c) = \frac {P(E^c\cap F^c) }{P(F^c)} =\frac{(1-P(E))(1-P(F)) }{(1-P(F)) }$$