3
$\begingroup$

While studying the following theorem from Loeve's book on probability:

Let $X_n$ be uniformly bounded random variables. If $\sum X_n$ converges a.s. then $\sum \sigma^{2}X_n$ and $\sum E X_n$ converges

He writes the proof as follows:

To the random variables $X_n$ we associate random variables $X_n'$ such that $x_n$ and $X_n'$ are identically distributed for every n and $X_1,X_1',X_2,X_2'...$ is a sequence of independent random variables. We form the "symmetrized" sequence $X_{n}^{s} = X_n - X_n' $ of independent random variables.

So I checked the section on symmetrization but I cannot figure out the following:

-How can such a $X_n'$ be built in general? (they need to be independent and identically distributed) -How can I proof that the "symmetrized" sequence is made up with independent random variables?

$\endgroup$
  • $\begingroup$ apparently it has to do with kolmogorov-daniell theorem $\endgroup$ – Javier Jan 14 '16 at 11:25
  • 1
    $\begingroup$ This is always doable, possibly enlarging the sample space. If the random variables $X_n$ are independent then the random variables $X_n^s$ are independent as well. $\endgroup$ – Did Jan 14 '16 at 11:37
  • $\begingroup$ how do you know the symmetrised random variables are independent? $\endgroup$ – BCLC Jan 17 '16 at 18:56
  • $\begingroup$ It is implicit in the quote i made from Loève: " the symmetrized sequence of independent random variables" so that's why I ask how can i build them $\endgroup$ – Javier Jan 17 '16 at 19:00
1
$\begingroup$

Not a mathematician so please let me know if this is completely off base, and I will delete this answer. The way I understand symmetrization, as used in the proof, is the subtraction, from each $X_n$, of an independent RV $X_n'$ with the same distribution as $X_n$. Assuming $X_n$ were independent (not sure if this was supposed to be the case, but is suggested by the proof wherein the sequence $X_n$,$X_n'$ is independent), $X_n'$ will also be independent because they are identically distributed. By the same reason, for all $n$ and $m$, $X_n'$ will be independent of $X_m$, because otherwise $X_n$ would not be independent of $X_m$. Independence of $X_n^s$ follows from 3 independent RVs A,B,C satisfying

$$ P_{A-B|C}(a-b<x) = \int_{-\infty}^\infty dP_{B|C}(b) P_{A|C}(a<x+b) = \int_{-\infty}^\infty dP_{B}(b) P_{A}(a<x+b) = P_{A-B}(a-b<x) $$

Then if we have another independent RV D, we have both C and D independent of A-B, so C-D is independent of A-B. This is a special case of this general proof.

To use this symmetrized series to show $\sum\sigma^2X_n$ and $\sum EX_n$ converge, note convergence of $\sum\sigma^2X_n$ implies convergence of the latter, and $\sum\sigma^2X_n=1/2\sum\sigma^2X_n^s$, so it suffices to show convergence of $\sum\sigma^2X_n^s$. I will omit this complicated proof, as it seems beyond the scope of the question.

$\endgroup$
  • $\begingroup$ Indeed this is way offbase. There is no way to prove that $(X'_n)$ and $(X_n)$ are independent, actually one must assume it. I fail to understand your "By the same reason, $X_n'$ will be independent of $X_n$, because otherwise $X_n$ would not be independent of itself." (Maybe the upvoter can explain?) $\endgroup$ – Did Aug 10 '16 at 10:04
  • $\begingroup$ @Did sorry for the lack of clarity. I did start with the assumption $X'_n$ and $X_n$ were independent for all $n$, and what I meant in the later statement was for $X'_n$ to be independent of $X_m$ for all n and m. Is that reasonable? I will edit. $\endgroup$ – obsolesced Aug 10 '16 at 10:45

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.