I am learning probability and statistics from the book Mathematical Statistics and Data Analysis, 3rd Edition by Rice. However just couple pages reading I think his definition on conditional probability is incomplete and misleading.
Notice that function P only takes subset of Ω, nothing else.
Now here is his definition on conditional probability:
The problem is that the P in left is not the same P as in right. Firstly lets ask what is A|B. If A|B is a subset of Ω, then what is the exact components of that? Its awkward to assign anythings into it because really you can't if you use the old P. P on Ω is fundamentally updated by knowing that B is true. So in old P P(B)=something, in new P P(B)=1, and all elements associated with B got updated value. In new P we can say that A|B = A ∩ B or A|B = A.
But all those are not clearly stated in his definition (in his following explanation he said the sample space becomes B rather than Ω, which is even more misleading because Ω doesn't need to change).
Is my understanding correct? Thanks for any help.
A following question can be found in here:
Fallacy on using interpretation instead of definition in computing conditional probability? (using multiplication law circularly?)
(also I think its best not to define conditional probability this way. This should be a result instead of a definition.)
Best Answer
You definitely have a point if you observe an old and a new $P$.
Actually if $\langle\Omega,\mathcal A,P\rangle$ is a probability space then every $B\in\mathcal A$ with $P(B)>0$ somehow induces a new probability space $\langle\Omega,\mathcal A,P_B\rangle$ where $P_B$ is defined by:$$P_B(A)=\frac{P(A\cap B)}{P(B)}$$
Subscript $B$ emphasizes that we are dealing with a probability that depends on $B$.
If you are dealing with probabilities conditional with respect to $B$ then in fact tacitly you have stepped over to that space.
A nicer and generally accepted notation for $P_B(A)$ is $P(A\mid B)$.
I think it is the best here if you interpret $P(A\mid B)$ as nothing else but an abbreviation of $\frac{P(A\cap B}{P(B)}$.
Keeping in mind of course that you dealing with conditional probabilities.