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:(adsbygoogle = window.adsbygoogle || []).push({});

1) Must satisfy properties of probability measures:

a) for any event E, 0≤P(E)≤1;2)P(I|I)=1

b) P(Ω)=1;

c) Sigma-additivity: Let {E_{1}, E_{2}, ... E_{n}, ....} be a sequence of events, where i≠j implies E_{i}and E_{j}are mutually exclusive, then P([itex]\bigcup_{n=1}^∞ E_n[/itex]) = [itex]\sum_{n=1}^∞ P(E_n)[/itex].

3) If [itex]E \subseteq I[/itex] and [itex]F \subseteq I[/itex], and P(I) is greater than 0, then [itex]\frac {P(E|I)}{P(F|I)} = \frac {P(E)}{P(F)}[/itex].

Then a proof is given for the proposition: Whenever P(I) is positive, P(E|I) satisfies the four properties aboveif and only ifP(E|I) = [itex]\frac {P(E \cap I)}{P(I)}[/itex].

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) = [itex]\frac {P(E \cap I)}{P(I)}[/itex].

Here's a quote:

Now we prove the 'only if' part. We prove it by contradiction. Suppose there exists another conditional probability [itex]\bar{P}[/itex] that satifies the four properties. Then There Exists an even E such that:

[itex]\bar{P}(E|I) ≠ P(E|I)[/itex]

It can not be noted that [itex]E \subseteq I[/itex], otherwise we would have:

[itex]\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)} [/itex]

*which would be a contradiction, since if [itex]\bar{P}[/itex] was a conditional probability, it would satisfy:

[itex]\frac {\bar{P}(E|I)}{\bar{P}(I|I)} = \frac {P(E)}{P(I)}[/itex]

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 [itex]\frac {\bar{P}(E|I)}{\bar{P}(I|I)} = \frac {P(E)}{P(I)}[/itex]?

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# Conditional Probability Proof

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