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"$A$" is an event that has probability somewhere in interval $(0,1)$. "$B$" is an event that has probability equal to zero.

Given that, I will prove that $\mathbb{P}(A \mid B)$ doesn't exist.

The first proof is based on reasoning alone: $\mathbb{P}(A \mid B)$ doesn't exist because it's logically contradictory. It basically says "probability of event $A$, given that event $B$, that couldn't happen, - happened". Such assumption is contradiction in terms. If event B is really impossible, then it means that we assumed it to be possible. But in our definition of $B$ we already established that B is impossible, thus we contradict ourselves.

The second proof is based on our attempt to calculate value of $\mathbb{P}(A \mid B)$. Remember that $$ \mathbb{P}(A \mid B) = \frac{\mathbb{P}(A \cap B)}{\mathbb{P}(B)}. $$ And we are in trouble because we already established that $\mathbb{P}(B)=0$, meaning that we will divide by zero. In other words, our answer is undefined (because division by zero itself is undefined), what implies that $\mathbb{P}(A \mid B)$ for impossible $B$ is undefined, which for our purposes is equal to it not existing at all.

P.S. The same logic works for cases when $\mathbb{P}(A) \in \{0,1\}$, I restricted $A$ to set of events that can either happen or not, in order to make the first proof less confusing.

KarmaPeasant
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  • @ViktorGlombik Wait, why does it make sense to define P(A|B) as 0 for impossible B? After all, as I shown, it has no value. And no value isn't equal to zero because lack of value and zero as value are two different things – KarmaPeasant Jun 04 '19 at 06:56
  • @ViktorGlombik My bad, by possible event I meant event that can either happen on not. – KarmaPeasant Jun 04 '19 at 07:03
  • @ViktorGlombik Are you sure you don't confuse it with P(AB), which would be zero for impossible B? We already have formula for calculating P(A|B) and it disagrees with your conclusion that the value would be zero. – KarmaPeasant Jun 04 '19 at 07:08
  • I replaced words "set of possible events" with "set of events that can either happen or not" in order to avoid confusion. – KarmaPeasant Jun 04 '19 at 07:10

1 Answers1

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It depends on your definition of $\Bbb P(A\mid B)$. I usually just have the definition

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

According to that definition, it is clear that $\Bbb P(A\mid B)$ is not defined if $\Bbb P(B)=0$. However, everyone is free to define $\Bbb P(A\mid B)$ to be whatever they want in that case. Some people will then define $\Bbb P(A\mid B)=0$ if $B$ happens almost never.

Your first argument is a little dangerous, because you argue based on the "meaning" of $\Bbb P(A\mid B)$, which might be different for different people. It is safer to simply argue based on the formal definition that we both agree on.

Maximilian Janisch
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