# Independence for subsets of [0,1]

I'm struggling with this problem:

Let $$\mathbb{P}$$ be the uniform measure on $$[0,1].$$ Define $$A=(a,b)$$ and $$B=(c,d)$$ with $$a State the necessary and sufficient conditions for $$A$$ and $$B$$ to be independent.

What I've done so far is drawn some intervals

From the definition of independence I require that $$\mathbb{P}(A \cap B) = \mathbb{P}(A)\mathbb{P}(B).$$ A sufficient condition is that $$A \cap B = \varnothing.$$ Where I'm getting confused is how to handle non-disjoint sets. I feel like they should be dependent, but another part of me feels like I'm not on the right track. Unless I'm totally off base.

EDIT: After reading the comments I realize that $$A \cap B = \varnothing$$ is not a sufficient condition. I left it in my initial asking in case others thought the same.

• You are on the right track: the next step is just expressing $P(A \cap B)$ in terms of $a, b, c, d$ for all three plots you have drawn, then state $P(A \cap B) = (b - a)(d - c)$. – Zhanxiong May 18 at 15:23
• Do you mean the def of independence is $\mathbb{P}(A \cup B) = \mathbb{P}(A)\mathbb{P}(B)$? Rather than $\mathbb{P}(A \cap B)$? Thatis obviously wrong if you think it through. – AdamO May 18 at 15:27
• If $A\cap B = \emptyset$, then $A$ and $B$ are not independent in general because $P(A\cap B)$ is typically nonzero whille $P(A)$ and $P(B)$ are not. This makes it hard to assert (with a straight face) that the condition $P(A\cap B)=P(A)P(B)$ holds and so $A$ and $B$ are independent events. – Dilip Sarwate May 18 at 15:39
• Drawing pictures is good. You should now be able to see that (except for edge cases) the second and third drawings exemplify non-independence, leaving you to analyze only the first drawing. For that, apply the definition of independence to obtain an algebraic relation among $(a,b,c,d).$ – whuber May 18 at 18:22
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