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Consider the probability space $([0,1]; B[0,1], L)$, where $B[0,1]$ contains the Borel sets intersecting $[0,1]$ and $L$ is the Lebesgue measure. How do I find the sigma algebra generated by a random variable defined on this space, $X = 1_{[0,1/2]}$? Secondly, how do I determine whether random variables defined on this space are independent or not, e.g., are $X = 1_{[0,1/2]}$ and $Y = 1_{[1/4,3/4]}$ independent?

For finding sigma algebra generated by $X$, I find $X^{-1}(1_{[0,1/2]})$ but what will be this inverse in Borel sets? For finding independence, it should be sufficient to show the sigma algebras generated by $X$ and $Y$ are independent right?

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What precisely do you mean by $X^{-1}(1_{[0,1/2]})$? The sigma algebra generated by $X$ consists of $X^{-1}(B)$ where $B$ are Borel sets, not functions... – Byron Schmuland Oct 24 '10 at 21:37
@Byron: You are right. This notation does not make sense. See my answer below for the defition of the sigma algebra generated by $X=1_{[0,1/2]}$. – Rasmus Oct 24 '10 at 21:49
Yes I made a mistake. I meant to take inverse of indicator function. – user957 Oct 24 '10 at 22:00
up vote 8 down vote accepted

The sigma-algebra generated by $1_{[0,1/2]}$ is simply $$ \bigl\{\emptyset,[0,1],[0,1/2],(1/2,1]\bigr\}. $$ It consists of the preimages under the function $1_{[0,1/2]}$ of all Borel sets in the codomain of the function $1_{[0,1/2]}$, namely, $(\mathbb R,B(\mathbb R))$. (Notice that the preimage $1_{[0,1/2]}^{-1}(M)$ is completely determined by the information of whether 0 and 1 do or do not belong to $M$ respectively.)

The situation for $1_{[1/4,3/4]}$ is similar.

The random variables $1_{[0,1/2]}$ and $1_{[1/4,3/4]}$ on $([0,1],B[0,1],L)$ are indeed independent: For this you have to check that $L(A\cap B)=L(A)\cdot L(B)$ for all $A\in 1_{[0,1/2]}^{-1}(B(\mathbb R))$ and $B\in 1_{[1/4,3/4]}^{-1}(B(\mathbb R))$.

The most interesting case is $L([0,1/2]\cap [1/4,3/4])=L([0,1/2])\cdot L([1/4,3/4])$.

Check that both sides are equal!

Also think about the following question: Are the random variables $1_{[0,1/2]}$ and $1_{[1/4,1]}$ on $([0,1],B[0,1],L)$ also independent?

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How is $L([0,1/2]\cap[1/4,3/4])=L([0,1/2]).L([1/4,3/4])$? Lesbegue measure is the length of the interval. How does this extend to independence? – user957 Oct 24 '10 at 22:07
1) Yes, Lesbegue measure is the length of the interval. Plug this in to verify the equation. 2) I explained in my answer how to check the independence. – Rasmus Oct 24 '10 at 22:23
By the way, two random variables of the form $1_A$ and $1_B$ are independent iff the events $A$ and $B$ are independent. – Rasmus Oct 24 '10 at 22:23
Got it. Sorry about the confusion. – user957 Oct 24 '10 at 22:53
@user957: That's not true. Take a look at the example I gave at the end of my answer. – Rasmus Oct 25 '10 at 10:06

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