Suppose that $A$ and $B$ are disjoint, so $P (A \cap B) = 0$. It follows that $P(A \mid B) = P(B \mid A) = 0$?

Question

My textbook, Statistical Inference by Casella and Berger, says the following:

Suppose that $A$ and $B$ are disjoint, so $P (A \cap B) = 0$. It follows that $P(A \mid B) = P(B \mid A) = 0$.

Intuitively, I don't see how this makes sense. We know that, if two events are disjoint, then the probability of them both occurring at the same time is $0$. And my understanding is that the converse is true; that is, if the probability of two events occurring is $0$, then the two events are disjoint. With all of that said, if the probability that both events $A$ and $B$ occur is equal to $0$ (that is, the events are independent), then I do not see how that necessarily implies that $P(A \mid B) = P(B \mid A) = 0$?

I would greatly appreciate it if people would please take the time to clarify this.

if $p(A\cap B)=0$ they are mutually exclusive not independent — David Quinn, Feb 06 '20 at 08:41
@DavidQuinn Whoops, you're right; I made an elementary error: https://math.stackexchange.com/a/941158/356308 — The Pointer, Feb 06 '20 at 08:43

score 3 · Accepted Answer · edited Jun 15 '21 at 04:08

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If the two events are disjoint, if one occurs, the other does not. So, the probability of one of them occurring given that the other has occurred, is in fact zero. Also, you know that $$ P(A \mid B) = \frac{P(A \cap B)}{P(B)}, \qquad P(B \mid A) = \frac{P(A \cap B)}{P(A)} $$ and the results comes directly out of the formula.

edited Jun 15 '21 at 04:08

The Pointer

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answered Feb 06 '20 at 08:43

PierreCarre

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score 2 · Answer 2 · edited Jun 15 '21 at 04:09

2

Okay: First of all, $\mathbb{P}(A\cap B) = 0$ does not imply that $A\cap B = \emptyset$. This depends on whether your probability measure allows non-trivial null sets (this is for instance not the case for the counting measure, so there your conclusion would hold).

Furthermore, $\mathbb{P}(A\cap B) = 0$ certainly does not imply that $A$ and $B$ are independent. In fact, if $\mathbb{P}(A),\mathbb{P}(B) > 0$, then this implies the opposite. On the other hand, if, say, $A$ were itself a null-set, then it would be independent of anything, so the statement becomes vacuous.

That being said, by definition, if $\mathbb{P}(B)>0$

$$ \mathbb{P}(A \mid B) = \frac{\mathbb{P}(A\cap B)}{\mathbb{P}(B)} = \frac{0}{\mathbb{P}(B)}=0 $$ and likewise for the other conditioning.

edited Jun 15 '21 at 04:09

The Pointer

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answered Feb 06 '20 at 08:42

WoolierThanThou

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I haven't studied measure theory yet. Out of curiosity, what is a "non-trivial null set"? – The Pointer Feb 06 '20 at 08:47
Furthermore, can you please clarify why $\mathbb{P}(A),\mathbb{P}(B)>0$ implies that $A$ and $B$ are dependent? – The Pointer Feb 06 '20 at 08:48
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If $\mathbb{P}(A\cap B)=0$ and $\mathbb{P}(A),\mathbb{P}(B)>0$, then $\mathbb{P}(A\cap B)\neq \mathbb{P}(A)\mathbb{P}(B)$, so the events aren't independent. – WoolierThanThou Feb 06 '20 at 08:50
Ahh, yes, of course. – The Pointer Feb 06 '20 at 08:50
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As for what a non-trivial null-set is, imagine you're observing the distribution of the sum of two dice on ${1,...,12}$, then $\mathbb{P}({1})=0$, but ${1}$ is, obviously, not empty. – WoolierThanThou Feb 06 '20 at 08:50

score 0 · Answer 3 · edited Jun 15 '21 at 04:08

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"$P(A \mid B)$" is the probability that $A$ occurs given that $B$ occurs. That is, with the given conditions, the probability that both occur, which is the probability of their intersection, which is zero. And likewise with $B$ and $A$ swapped.

edited Jun 15 '21 at 04:08

The Pointer

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answered Feb 06 '20 at 08:41

Eric Towers

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Suppose that $A$ and $B$ are disjoint, so $P (A \cap B) = 0$. It follows that $P(A \mid B) = P(B \mid A) = 0$?

3 Answers3