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I have a sequence Xj of random variables, each of which individually is uniformly distributed on the unit circle in the complex plane, and a corresponding sequence cj of positive coefficients. My sequence of coefficients has the property that $\sum_{j=1}^\infty c_j^2$ converges but $\sum_{j=1}^\infty c_j$ diverges.

Note that I am not assuming that the Xj are independent. What I do know about the variables is this: any odd-index one $X_{2j-1}$ is independent from any finite collection of other $X_i$. In other words, all the odd-index ones are independent of one another and of the even-index ones, but there might be dependences among the even-index ones.

I want to write down three other random variables: $$ A = \sum_{j=1}^\infty c_j X_j, \quad B_1 = \sum_{j=1}^\infty c_{2j-1} X_{2j-1}, \quad B_2 = \sum_{j=1}^\infty c_{2j} X_{2j}. $$ That B1 is a well-defined random variable is no problem: since the X2j-1 are all independent, the sum defining B1 exists almost surely thanks to the $\ell^2$-convergence of the cj.

There is no immediate reason to think that A itself is a well-defined random variable; however, in my situation, I have extra information that ensures that A really is well-defined.

So my two questions (finally) are these:

  1. Is the above information enough to prove that B2 is a well-defined random variable?
  2. Is the above information enough to prove that B1 and B2 are independent?
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I'm assuming "well-defined" means "finite a.s.". We have $B_2=A-B_1$, so since $A$ and $B_1$ are finite a.s., so is $B_2$. And since the variables making up $B_1$ are independent of those making up $B_2$, it must be that $B_1$ and $B_2$ are independent. Is there some subtlety to the problem that I'm missing? –  Kevin O'Bryant Jun 26 '10 at 1:46
    
Thanks - "well-defined" would indeed be better worded as "finite almost surely". Is there a subtlety? Well, maybe not, but this is what I'm worried about: I agree that the $B_2 = A - B_1$ argument shows that $B_2$ is finite a.s. But the lack of a priori a.s. convergence of $\sum_{j=1}^\infty c_{2j}X_{2j}$ makes me leery of applying reasoning based on properties of the $X_{2j}$. – Greg Martin 0 secs ago –  Greg Martin Jun 29 '10 at 18:50
    
You definitely need more info here. If the $X_{2j}$ are not just identically distributed but actually the same random variable Z which is uniform on the circle, then 1 is false. –  Abdelmalek Abdesselam Oct 20 '10 at 22:15
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2 Answers

For the First one I don't know sorry.

For the second one the answer is Yes

You might consider the Sigma-algebra generated by your family $X_{2j-1}$ and the see that it is independent from the Sigma-algebra generated by your family $X_{2j}$ by recurrence.

I can add that if you cannot see it you can have a look at the 0's Chapter of Itô's lectures at Aarhus University viewable here

Regards

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To get around the worry about using properties of the $X_{2j}$ without a priori convergence of the series defining $B_2$, you might use a relative of the following result.

If the $\pi$-systems of events $H_{\alpha,\beta}$, $(\alpha,\beta) \in \mathcal{J}$, are mutually independent, then the $\sigma$-algebras $$G_\alpha =\sigma \left(\bigcup_\beta H_{\alpha,\beta} \right)$$ are mutually independent.

I got this out of Amir Dembo's notes, but I imagine the result is in many books (see, e.g. Durrett's Probability: Theory and Examples, Third Edition, Chapter 1, (4,4).)

To apply the proposition, we would let $\alpha$ run over $1,2$. Let $H_{1,\beta}=\sigma(X_{2 \beta-1})$ and let $H_{2,\beta} = \sigma(X_{2\beta})$. The $\beta$, of course, range over the natural numbers. Unfortunately, the hypothesis doesn't quite hold due to the dependence of the $X_{2j}$.

But perhaps it might help to look at the proof and see if we can weaken the assumption of mutual independence of all the $H_{\alpha,\beta}$? Essentially we might define $K_{\alpha,\ell} = \sigma(H_{\alpha,1},H_{\alpha,2},\ldots,H_{\alpha,\ell})$ and assume $K_{1,\ell}$ is independent of $K_{2,m}$ for every $m$ and $n$. Can we conclude then then the limits are independent by reasoning similar to that of the above proposition? I suspect not exactly -- we need some finer control than that, but maybe it can be extended...

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