2
$\begingroup$

I am asked to prove the following claim:

If $A$ and $B$ are separate events then $A$ and $B$ cannot be independent.

Intuitively, it is easy to see that the claim is true, since if $A$ occurs then $B$ must not have occurred (hence knowledge about $A$ gives complete knowledge about $B$, which is very strong form dependence).

Mathematically, $A$ and $B$ are independent if $P(A\cap B) = P(A)P(B)$. However, if $A$ and $B$ are separate, then there is nothing in the intersection of $A$ and $B$. Hence $P(A\cap B)=0$. However, $P(A)P(B)>0$ if $A$ and $B$ are possible events... So I have arrived in a contradiction? Is this sufficient to prove the claim?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
0
$\begingroup$

What is meant by separate? Disjoint? Do you mean $A \cap B = \emptyset$?

If so, then your claim is equivalent to:

If $A \cap B = \emptyset$, then $P(A \cap B) \ne P(A) P(B)$

To try to prove this, let us suppose $P(A \cap B) = P(A) P(B)$ but $A \cap B = \emptyset$.

As you pointed out, $P(A \cap B) = P(\emptyset) = 0$.

Then $P(A) P(B) = 0$ which means $P(A) = 0$ or $P(B) = 0$.

I don't really see any contradiction here unless $P(A) > 0$ and $P(B) > 0$.

You said $P(A)P(B) > 0$ if they are possible events. Are they? If not, I think your claim is false.

Improve this answer
$\endgroup$
3
  • $\begingroup$ It is said they are possible events in a probability space $(\Omega, \mathcal{F})$. With separate I meant disjoint, yes. $\endgroup$
    – lisa
    Commented Sep 18, 2015 at 11:17
  • 1
    $\begingroup$ Is separate not a commonly used term? If one would draw a Venn diagram, I would find calling two non-overlapping blobs separate somehow intuitive. $\endgroup$
    – lisa
    Commented Sep 18, 2015 at 11:20
  • $\begingroup$ Probability space is $(\Omega, \mathscr{F}, \mathbb{P})$ :P Anyway, where did you say that they are possible events? Okay $A, B \subset \Omega$, $A, B \in \mathscr{F}$, but does either mean $P(A), P(B) > 0$ ? Even if they are jointly exhaustive (perhaps that is an assumption you forgot to state?) ie$A \cup B = \Omega$, I don't see how that means $P(A), P(B) > 0$. I think separated means something different in math. I think the terminology is that the events are mutually exclusive (i.e. disjoint) $\endgroup$
    – BCLC
    Commented Sep 18, 2015 at 11:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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