1
$\begingroup$

I'm trying to understand the Kolmogorov's three series theorem. Studying the proof I bump into the following proposition:

Let $\{X_n\}_{n\in \mathbb{N}}, \{Y_n\}_{n\in \mathbb{N}}$ be two sequences of real independent r.v. defined on the same probability space (the variables $\{Y_n\}$ are also independent from the $\{X_n\}$) and uniformly bounded (i.e. $\exists \lambda>0 \ s.t. |X_n(\omega)|\leq \lambda$ for all $\omega \in \Omega, n\in \mathbb{N}$). Suppose that $Y_n$ has the same probability distribution of $X_n$, and that $\sum_{n=1}^{+\infty} X_n$ convergers almost surely, then also $\sum_{n=1}^{+\infty} Y_n $ converges almost surely.

How can I prove this?

I understand that since $X_n$ and $Y_n$ have the same distribution and $\sum_{n=1}^{+\infty} X_n$ converges a.s. then it must converge also in probability and then $\sum_{n=1}^{+\infty} Y_n$ must converge in probability but this does not imply, in general, that $\sum_{n=1}^{+\infty} Y_n$ converges a.s.

Thanks for the help.

$\endgroup$
  • $\begingroup$ It may be worth noting a theorem of P. Lévy: for a series of independent random variables, convergence in probability implies almost sure convergence. $\endgroup$ – John Dawkins Sep 13 '16 at 17:57
1
$\begingroup$

Note that the event $\left\{\omega\in\Omega, \sum_{l=1}^{+\infty}X_l\right\}$ is convergent may be written as $$\bigcap_{i\geqslant 1}\bigcup_{N\geqslant 1}\bigcap_{N\leqslant m\leqslant n}\left\{\left|\sum_{l=m}^n X_l\right|\leqslant\frac 1i \right\} .$$ Therefore, the probability of $\left\{\omega\in\Omega, \sum_{l=1}^{+\infty}X_l\right\}$ depends only on the distribution of the sequence $\left(X_l\right)_{l\geqslant 1}$.

In the context of the question, the sequences $\left(X_l\right)_{l\geqslant 1}$ and $\left(Y_l\right)_{l\geqslant 1}$ have the same distribution.

$\endgroup$

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.