# Conditional Probability Proof

1. Jul 1, 2014

### Mogarrr

I'd like some help understanding a proof, from http://www.statlect.com/cndprb1.htm. Properties are introduced, which a conditional probability ought to have:
1) Must satisfy properties of probability measures:
a) for any event E, 0≤P(E)≤1;
b) P(Ω)=1;
c) Sigma-additivity: Let {E1, E2, ... En, ....} be a sequence of events, where i≠j implies Ei and Ej are mutually exclusive, then P($\bigcup_{n=1}^∞ E_n$) = $\sum_{n=1}^∞ P(E_n)$.​
2)P(I|I)=1
3) If $E \subseteq I$ and $F \subseteq I$, and P(I) is greater than 0, then $\frac {P(E|I)}{P(F|I)} = \frac {P(E)}{P(F)}$.​

Then a proof is given for the proposition: Whenever P(I) is positive, P(E|I) satisfies the four properties above if and only if P(E|I) = $\frac {P(E \cap I)}{P(I)}$.

I'm having a hard time following the proof of the "only if" part. That is, if P(E|I) satisfies the four properties above, then P(E|I) = $\frac {P(E \cap I)}{P(I)}$.

Here's a quote:
Now we prove the 'only if' part. We prove it by contradiction. Suppose there exists another conditional probability $\bar{P}$ that satifies the four properties. Then There Exists an even E such that:
$\bar{P}(E|I) ≠ P(E|I)$

It can not be noted that $E \subseteq I$, otherwise we would have:
$\frac {\bar{P}(E|I)}{\bar{P}(I|I)} = \frac {\bar{P}(E|I)}1 ≠ \frac {P(E|I)}1 = \frac {P(E \cap I)}{P(I)} = \frac {P(E)}{P(I)}$

*which would be a contradiction, since if $\bar{P}$ was a conditional probability, it would satisfy:
$\frac {\bar{P}(E|I)}{\bar{P}(I|I)} = \frac {P(E)}{P(I)}$​

The proof by contradiction, seems more like a proof of the uniqueness of a conditional probability.

Anyways, I'm not really seeing the statement, *. How is it that $\frac {\bar{P}(E|I)}{\bar{P}(I|I)} = \frac {P(E)}{P(I)}$?

2. Jul 1, 2014

### Staff: Mentor

For "only if", you assume that $\bar P(E|I)$ satisfies the four properties. In particular, it satisfies the third one with F=I.

3. Jul 1, 2014

### Mogarrr

OH....

So Property 3 asserts that: If $E \subseteq I$ and $F \subseteq I$, where P(F) is positive, then for any probability A, $\frac {A(E|I)}{A(F|I)} = \frac {P(E)}{P(F)}$?

BTW, the other property was this: If $E \subseteq I^C$ then P(E|I)=0.

4. Jul 2, 2014

### Staff: Mentor

Oh, I thought the RHS would have bars as well, they are hard to see. Maybe a typo there and it should be $\bar P$.

5. Jul 3, 2014

### Mogarrr

Yes, I agree, but...

If $\frac {\bar{P}(E|I)}{\bar{P}(I|I)} = \frac {\bar{P}(E)}{\bar{P}(I)}$, then there is no contradiction.

So I'm looking for a justification for the statement *.

I'm fairly certain that I quoted the author correctly, but if there is doubt, you can always visit statlect.com. The notes on Conditional probability are in the Fundamentals of probability theory section.

Last edited: Jul 3, 2014
6. Jul 9, 2014

### scinoob

I think the notation for probabilities and conditional probabilities is not the same and therefore there is no typo.