Probability of $A|B|C$

Question

Because $P$ can only be aplied to an event, $A|B$ must be an event, but I can not see the way to go from $P(A|B|C)$ to conclude that this is the same thing as $P(A|B,C)$, I am sorry if this is an absurd question, but I, even though I understand the logic behind the statement, I believe it should be possible to use the definition to show one can derive that conclussion and say the conditional probability is consistent with that intuiton.

Answer 1

If you have a probability space $(\Omega, \mathcal{F}, P)$ so that $A$ and $B$ are events, then they are subsets of $\Omega$ and elements of $(\Omega, \mathcal{F}, P)$. You cannot then say that $A\mid B$ is an event in that probability space.

But you can assign a conditional probability value, providing that $P(B)>0$, namely $P(A \mid B)=\frac{P(A \cap B)}{P(B)}$. This uses a different probability space, and in particular this new probability measure.

You can keep $\Omega$ and $\mathcal{F}$, or you can reduce $\Omega$ to $B$ and adjust the elements of $\mathcal{F}$ to their intersections with $B$. If you use the former, then $A$ is still an event with new probability $P(A \mid B)$ though $A \cap B$ has the same probability with this new measure; if you use the latter, $A$ will not be an event in the new probability space if it is not a subset of $B$, though $A \cap B$ is an event with probability $P(A \mid B)$. So how you define the new probability space affects how you might interpret $A \mid B$ as either $A$ or $A \cap B$ with this new probability measure, either having new probability $P(A \mid B)$, and this is why many people reject describing $A \mid B$ as an event in the original probability space.

Moving on to your question about $P(A \mid B \mid C)$, it seems that the conditional probability you are looking for uses two steps of conditionality and will be $$\frac{P(A \cap B \mid C)}{P(B \mid C)} = \frac{\frac{P(A \cap B \cap C)}{P(C)}}{\frac{P(B \cap C)}{P(C)}} =\frac{P(A \cap B \cap C)}{P(B \cap C)}=P(A \mid B \cap C)$$ also written as $P(A \mid B , C)$, and again well defined so long as $P(B \cap C)>0$. But you are changing the probability space again, and whether the corresponding event is $A$ or $A \cap B \cap C$ is unclear: both have the same probability with the newest probability measure.

In summary, you can find these conditional probabilities so long as you do not divide by zero, but trying to specify conditional events is ambiguous and unclear so best avoided.