Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

My earlier question became too long so succintly:

What are $P(T|A)=P(T\cap A)/P(A)$ and $P(T|B)=P(T\cap B)/P(B)$ if $P(A)=0$ and $P(B)=0$?

I think they are undefined because of the division by zero. How can I specify the conditional probabilities now? Please, note that the basic events $A$ and $B$ depend on $T$ because $T$ consists of them, namely $T=A \cup B$.

share|improve this question

2 Answers 2

Yes, it is undefined in general. It is generally pointless to ask for the conditional probability of $T$ when $A$ occurs when it is known that $A$ almost surely never happens.

But a meaningful specification in your particular case that $T = A\cup B$ is by some intuitive notion of continuity. For $P(A) \neq 0$, if $C\supseteq A$ we must have $P(C|A) = 1$. Hence one can argue from a subjective interpretation of probability that $P(T|A) = 1$ since $T\supseteq A$. And similarly for $P(T|B)$.

share|improve this answer
    
I cannot understand the point "by some intuitive notion of continuity". –  hhh Feb 26 '13 at 20:22
    
If we make the "intuitive" assumption that $P(T|A)$ for $T = A\cup B$ is a continuous function of $P(A)$, then by taking the limit $P(A)\to 0$ we have that $P(T|A) = 1$. It is in analogy to the fact that the function $x/x$ is not defined at $x = 0$. But there exists a unique continuous function on $\mathbb{R}$ that equals $x/x$ away from $x=0$. –  Willie Wong Feb 26 '13 at 23:31
    
The situation is perhaps better illuminated by the even specialer case of $B = \emptyset\implies T = A$. I think it is more than reasonable to assume $P(A|A) = 1$ even for probability zero events $A$. –  Willie Wong Feb 26 '13 at 23:34

I discussed with my teacher and he said that the conditional probability formula is a tautology. If you assume zero probability in some input of the OR, then you have different topology and hence you cannot use the formula like that -- you need reform the problem -- because OR port can be simplified with certain zero-probabilities.

The key is to differentiate the different layers: logic, probability and real life. You can claim whatever you want in the logical layer before you think about probabilities -- even though you may get undefined in probabilistic layer. The real-world can be then anything between the logic and probability.

Example with interval probability

Suppose $T=A\cup B$. Now if $P(B)=0$, then $T=A$. Now if $P(B)\in [0,0.2]$, then $T=A$ if $P(B)=0$ and $T=A\cup B$ if $P(B)\not = 0$. So interval probabilities become pretty messy if you assume extreme probabilities such as $0$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.