Why, when calculating the conditional probability of A given B, do we assume that the probability of B is greater than zero?

  • 1
    $\begingroup$ think of B containing A. If B had no members, it could not contain a ∈ A ⊂ B. $\endgroup$ – isomorphismes Mar 18 '18 at 3:54
  • 3
    $\begingroup$ If $P(B) = 0$, then you have freedom to choose the value of $P(A\mid B)$, and any such choice never affects the computation of probability under $P$ in view of the law of total probability. In layman's terms, you can assume whatever for something that never happens. This convention is often theoretically useful, especially when you need to consider a family of conditional probabilities (such as transition kernel of a Markov process). $\endgroup$ – Sangchul Lee Mar 18 '18 at 4:14
  • $\begingroup$ @isomorphismes: $P(B)=0$ does not imply $B=\emptyset$. $\endgroup$ – celtschk Mar 18 '18 at 11:24

We have

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

If $P(B)=0$ then the RHS is undefined.

Also, if we're given that $B$ happens then it cannot be the case that $P(B)=0$. That's a contradiction.

Here is an example of when we can still calculate $P(A|B)$ when $P(B)=0$

Let $X\sim N(0,1)$ and $Y\sim N(0,1)$

Let $A$ be the event that $X\lt1$ and let $B$ be the event that $X+Y=2$

Then $P(A)=\Phi(1) \approx 0.8413$ and $P(B)=0$ since the normal distribution is continuous but still $P(A|B)=0.5$

  • $\begingroup$ "Also, if we're given that B happens then it cannot be the case that P(B)=0. That's a contradiction." — Consider an uniform distribution on $[0,1]$. Then $P(\{x\})=0$ for all $x\in [0,1]$. Thus from your claim it would follow that you cannot get any value from $[0,1]$. But a probability distribution of $[0,1]$, including the uniform distribution, exactly means that the random variable takes a values from that interval. A contradiction. $\endgroup$ – celtschk Mar 18 '18 at 11:30
  • $\begingroup$ Your statement in your first section is true in discrete probability spaces but not necessarily in continuous ones. I'm sure you know this but it should be pointed out to others. $\endgroup$ – Rory Daulton Mar 18 '18 at 11:52
  • $\begingroup$ What does the notation N(x,y) stand for? $\endgroup$ – Orphevs Mar 18 '18 at 12:29
  • 1
    $\begingroup$ Normally distributed with mean 0 and variance 1 $\endgroup$ – Remy Mar 18 '18 at 21:09

It is to avoid the divide-by-zero error that occurs when you try to divide by zero.

The definition of a conditional probability mass function, that $\mathsf P(A\mid B):=\mathsf P(A\cap B)\div\mathsf P(B)$ is only viable when $B$ has an non-zero measure.

Still, however, when $\mathsf P(B)=0$ there are other compatible definitions for conditional probability measures that can be used; although they are not necessarily probability mass functions.

  • 1
    $\begingroup$ Could you please elaborate on the other functions that are used for conditional probability, when P(B) = 0? I've never heard of them before. If it's worth it, just add it to your answer :) $\endgroup$ – arya_stark Mar 18 '18 at 3:39

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.