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Conditional probability $P(A|B)$ is defined as: $P(A|B) = \frac{P(A\cap B )}{P(B)}$ when $P(B) > 0$, where $A$ and $B$ are events in the sample space.

Is $P(A|B)$ not defined for $P(B)=0$?

Consider the following problem.

The experiment is an infinite sequence of coin tosses. Suppose the coin lands $Heads$ at each toss with a probability $p \in (0,1)$.

Suppose $B = \{Heads, Heads, Heads,..., Heads,...\}$ i.e, coin lands $Heads$ after every toss,

and

$A$ is the event in which the 1st coin toss is Heads i.e., $A=\{Heads, H/T, H/T,...\}$.

We have, $P(A)=p$ and $P(B)=0$

I am interested in $P(A|B)$.

Consider the following attempt.

Suppose we consider the first $n$ coin toss.

Let $B_n = \{Heads, Heads,...,Heads\}$.

Let $A_n$ be the event in which the first coin toss is Heads.

We have: $ A_n \cap B_n = B_n$

$ \displaystyle P(A_n|B_n) = \frac{P(A_n \cap B_n)}{P(B_n)} $

$ \implies P(A_n|B_n) = \frac{P(B_n)}{P(B_n)} $

$ \implies P(A_n|B_n) =1. $

$ \implies \displaystyle \lim_{n\to\infty} P(A_n|B_n) = \lim_{n\to\infty} 1 $

$ \implies P(A|B) = 1 $

This result matches our intuition. But, $P(A|B)$ is not defined by the definition when $P(B)=0$.

Where am I getting it wrong?

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The equality: $$P(A\mid B)P(B)=P(A\cap B)\tag1$$ characterizes conditional probability.

If $P(B)=0$ then also $P(A\cap B)=0$ and we get:$$P(A\mid B)0=0$$

This is true for every value given to $P(A\mid B)$ and gives us the liberty to fill in a value that agrees with our intuition.

I admit that then a rigorous definition of $P(A\mid B)$ lacks and care is needed to avoid accidents.

Nevertheless, personally I prefer $(1)$ and nothing more than $(1)$ above the "handicapped" expression $P(A\cap B)/P(B)$.

Another example (next to yours): let it be that $X$ has binomial distribution, that $Y$ has normal distribution and that $X$ and $Y$ are independent.

Then how about expressions like $P(X=4\mid Y=2)$?

Following my intuition I choose for $P(X=4\mid Y=2)=P(X=4)$ on base of the independence.

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  • $\begingroup$ Well, in the case of $P(X=4|Y=2) $, its hybrid joint distributions (X: discrete r.v and Y: continuous r.v). $P(X=4|Y=2)$ is defined as $P(X=4|Y=2) = \frac{p_Y(2|X=4) P(X=4)}{p_{Y}(y)}$. But, I understand your point $\endgroup$ – Suhan Shetty Aug 19 '18 at 16:13
  • $\begingroup$ Any ways, $P(A|B)P(B) = P(A \cap B)$ still does not define $P(A|B)$ in the case $P(B)=0$ $\endgroup$ – Suhan Shetty Aug 19 '18 at 16:20

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