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?
  • 6
    $\begingroup$ 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? $\endgroup$ Jun 26, 2010 at 1:46
  • $\begingroup$ 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 $\endgroup$ Jun 29, 2010 at 18:50
  • $\begingroup$ 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. $\endgroup$ Oct 20, 2010 at 22:15

2 Answers 2


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



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...


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.