Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It only takes a minute to sign up.
Sign up to join this community
Is there a way to resolve probability of an event, given another event that never happens? Mathematically speaking the problem is:
Given that $P(B) = 0$,
$$P(A|B)=\frac{P(A \cap B)}{P(B)} = \frac{0}{0}$$
Is this probability vacuously $0$ of $1$? Can we show that it's one or the other?
The comment by Dilip Sarwate points to conditioning on the level of densities which can be interpreted as conditioning on a family of events of probability zero. It is a probabilistic version of Radon-Nikodym derivative.
One can also condition on an individual event of probability zero, if that event admits a natural approximation by events of positive probability. For example, begin with a probability measure $\nu$ on the space $C[0,1]$ of continuous functions such that $\nu(\{f(0)=0\})=0$. Restrict it to the set of functions $f$ such that $|f(0)| \le \epsilon$. Normalize this restriction to a probability measure. If these normalized restrictions converge (in some sense) to a probability measure $\mu$ on $C[0,1]$, we can regard $\mu$ as conditional distribution on the zero-probability event $f(0)=0$.
The definition of conditional probability $$P(A|B)=\frac{P(A \cap B)}{P(B)}$$ means that we can't condition on an event with zero probability; if $P(B)=0$, then $P(A|B)$ is undefined for any event $A$.
Sometimes it makes sense to define conditional probability differently. Let $X$ and $Y$ be jointly continuous random variables; then we can define conditional PDF of $X$ given $Y$ $$f_{X|Y}(x|y)=\frac{f_{X,Y}(x,y)}{f_Y(y)}$$ for all $y$ such that $f_Y(y)>0$, and using this definition we can define $$P(X\in A|Y=y)=\int_Af_{X|Y}(x|y)dx$$ which implies conditioning on zero probability event $\{Y=y\}$.
The answer to the general OP question is: if $P(B)=0$, the conditional probabilities given event $B$ are undefined, unless a different definition of conditional probability is used.
This argument applies only to probability of events and not for probability with random variables.
Firstly the definition, $$P(A|B)=\dfrac{P(A \cap B)}{P(B)}$$ applies only when $P(B) \neq 0$.
What is it when $P(B)=0$?
Well, $P(A|B)$ is the probability of $A$ happening given that $B$ happened is ridiculous to even ask. How can you be given that an event that cannot happen has happened? So, it shall remain indeterminate in my opinion.
However, some texts, do give that $P(A|B)=0$ and some others assign $1$, or so my teacher has told us that he has seen one book doing the latter (asssigning $1$).
But, it just does not matter.
