2
$\begingroup$

Suppose there are two events $A$ and $B$ and that $P(A|A\cup B)P(B|A\cup B) = P(A\cap B | A \cup B)$. Then I am asked to find if $A$ and $B$ are independent, positively or negatively correlated.

My intuition is that $A$ and $B$ are independent. If they are,$P(A|A\cap B)P(B|A\cup B)$ would yield only $P(A)P(B)$, and the right side would be equal. But I am not sure if this is a rigorous proof.

$\endgroup$
  • 1
    $\begingroup$ "But I am not sure if this is a rigorous proof." What is "this"? $\endgroup$ – Did Feb 27 '16 at 9:47
2
$\begingroup$

Suppose there are two events $A$ and $B$ and that $\mathsf P(A\mid A\cup B)~\mathsf P(B\mid A\cup B) ~=~ \mathsf P(A\cap B \mid A \cup B)$. Then I am asked to find if $A$ and $B$ are independent, positively or negatively correlated.

What you have is conditional independence given the union of events.   (Think about what that says†.)

My intuition is that $A$ and $B$ are independent. If they are, $\mathsf P(A\mid A\cap B)~\mathsf P(B\mid A\cup B)$ would yield only $\mathsf P(A)~\mathsf P(B)$, and the right side would be equal. But I am not sure if this is a rigorous proof.

Independence requires $\mathsf P(A\cap B)=\mathsf P(A)~\mathsf P(B)$ That is all; nothing else.

Total Probability says: $$\begin{align}\mathsf P(A\cap B)~=& ~\mathsf P(A\cap B\mid A\cup B)~\mathsf P(A\cup B) + \mathsf P(A\cap B\mid (A\cup B)^\complement)~\mathsf P((A\cup B)^\complement) \\[1ex] = & ~ \mathsf P(A\cap B\mid A\cup B)~\mathsf P(A\cup B) + 0 \end{align}$$

With what we are given, that means:

$$\begin{align} \mathsf P(A\cap B) ~ = & ~ \mathsf P(A \mid A\cup B)~\mathsf P(B\mid A\cup B)~\mathsf P(A\cup B) \\[1ex] = & ~ \mathsf P(A\mid A\cup B)~\mathsf P(B) \\[1ex] = & ~ \frac{\mathsf P(A)}{\mathsf P(A\cup B)}~\mathsf P(B) \end{align}$$

So your given sets will only be independent if: $\mathsf P(A\cup B)=1$, otherwise $\mathsf P(A\cap B)> \mathsf P(A)~\mathsf P(B)$


† Also consider that $\mathsf P(A\cup B\mid A\cup B) = \mathsf P(A\mid A\cup B)+\mathsf P(B\mid a\cup B)-\mathsf P(A\cap B\mid A\cup B)$

So $$\begin{align}1 ~= ~& \mathsf P(A\mid A\cup B)+\mathsf P(B\mid A\cup B)-\mathsf P(A\mid A\cup B)~\mathsf P(B\mid A\cup B) \\[3ex] \therefore \mathsf P(A\mid A\cup B)~ = & ~ \mathsf P(B\mid A\cup B) ~=~ 1\end{align}$$

$\endgroup$
  • $\begingroup$ So if $0 <= P(A) <= 1$ and $0 <= P(B) <= 1$, is it safe to conclude that they are independent? $\endgroup$ – Arthur L Feb 27 '16 at 16:35
  • $\begingroup$ No. Where did you get that? $\endgroup$ – Graham Kemp Feb 27 '16 at 23:39
  • $\begingroup$ @LeBronJames Because, like all probability measures, $0~\leq ~\mathsf P(A\cup B)~\leq~ 1$ so if it is not one (nor zero either, it should be added) then $\mathsf P(A\cap B) ~=~ \mathsf P(A)~\mathsf P(B)\big/\mathsf P(A\cup B)$ means.... $\endgroup$ – Graham Kemp Feb 29 '16 at 21:42

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.