What is the importance of conditional probability in probability theory?

[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, ... E_n, ...} be a sequence of events, where i≠j implies E_i and E_j 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)}?

 

[mfb]

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

 

[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.

 

[mfb]

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

 

[Mogarrr]

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

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.

 

[scinoob]

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

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