# Correlation between probability of events

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.

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

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}

• So if $0 <= P(A) <= 1$ and $0 <= P(B) <= 1$, is it safe to conclude that they are independent? – Arthur L Feb 27 '16 at 16:35
• No. Where did you get that? – Graham Kemp Feb 27 '16 at 23:39
• @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.... – Graham Kemp Feb 29 '16 at 21:42