$\newcommand{\P}{\mathbb{P}}$Is it valid to interpret $\P(B|A)$ as "infinite probability" when $\P(A) = 0$?

  • 2
    $\begingroup$ Nope, because Bayes' Theorem aplies when $P(A) > 0$. If $P(A) = 0$ then $B$ is independent of $A$, Hence $P(B) = P(B)$. $\endgroup$
    – Drey
    Commented Nov 15, 2016 at 8:44
  • 2
    $\begingroup$ Conditional probabilities (like all probabilities) must be in the interval $[0,1]$ so cannot be infinite. Your example may or may not be well defined: suppose $X \sim N(0,1)$ and $Y =X^2$ and consider conditional probabilities $P(Y=-3 \mid X=1)$ and $P(X=1 \mid Y=-3)$; the first is zero while the second is not well defined even though both the events $X=1$ and $Y=-3$ have zero probability. $\endgroup$
    – Henry
    Commented Nov 15, 2016 at 9:03
  • $\begingroup$ Negative probability and Negative Probabilities in Financial Modeling $\endgroup$
    – BCLC
    Commented Mar 25, 2021 at 5:10

3 Answers 3


No. First, probability is bounded in $[0, 1]$, so it cannot be infinite.

Notice that there are two cases where probability can be equal to zero:

  1. When you are dealing with empty set $\Pr(\varnothing)=0$. So for example, if you ask "what is the probability that person being -31 years old dies in a car crash?", then it is impossible to answer such question since negative age is impossibility, so the answer for that question is undefined, and basically the question does not make sense. This was noted by  Kolmogorov

    the concept of a conditional probability with regard to an isolated hypothesis whose probability equals 0 is inadmissible

  2. On another hand, as noted by Dilip Sarwate in the comment to this thread and in the following answer by 40 votes, we often want to condition on zero-probability events in case of continuous random variables (where $\Pr(X=x)=0$ for all $x$'s) by approximating it using limits, i.e. by conditioning on probability densities.

Check the referred Probability, conditional on a zero probability event thread on math.stackexchange.com that describes it in greater detail.

  • $\begingroup$ probability can be negative though? $\endgroup$
    – BCLC
    Commented Mar 24, 2021 at 6:56
  • $\begingroup$ @JohnSmithKyon it can't, I'm not sure what you mean. $\endgroup$
    – Tim
    Commented Mar 24, 2021 at 15:55
  • $\begingroup$ Tim, Negative probability and Negative Probabilities in Financial Modeling $\endgroup$
    – BCLC
    Commented Mar 25, 2021 at 5:09
  • $\begingroup$ @JohnSmithKyon that’s not a standard definition of probability. By standard definition they are bounded between 0 and 1 en.wikipedia.org/wiki/Probability_axioms $\endgroup$
    – Tim
    Commented Mar 25, 2021 at 6:08
  • $\begingroup$ well it looks like OP is thinking about something nonstandard since standard probability is $[0,1]$ $\endgroup$
    – BCLC
    Commented Mar 25, 2021 at 6:45

No (no such thing as infinite probability: a certain event has probability 1).

In fact if it is a probability of an event (and not of a continuous random variable) then it is an ill-posed problem: if $P(A)=0$, what is the probability of $B$ given that $A$ happened?

$P(B|A)=\frac{P(B\cap A)}{P(A)}=\frac{0}{0}$

So I think it should just be considered undetermined. (Note that the definition of conditional probability applies only only when the probability of the conditioning event is non-zero, $P(A)\ne0$.)

It is different if you are dealing with the probability of a continuous random variable assuming a particular value. See the comments to this question (which has already been referred to by the other answers).


As far as I know (Stochastic lecture and quick wikipedia lookup), the conditional probability P(B|A) is not defined for P(A)=0. (rule of Bayes)

where A and B are events and P(B) ≠ 0.

So I would say no, not valid.

But if P(A) = 0 then the variables are independent anyways..


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