# $X_n$ independent random variables zero one question

For $A := \left\{ \sum X_n \text { converges} \right\}$, I'm wondering how to get started showing $P (A)\in\{0,1\}$.

Is Borel Cantelli a good strategy ?

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Kolmogorov's $0$-$1$ law? –  Dilip Sarwate Feb 6 '12 at 13:31
Maybe Kolmogorov zero-one law? –  Ilya Feb 6 '12 at 13:32
Thanks for the hint, hadn't covered it yet, so it's quite helpful! –  Beltrame Feb 6 '12 at 14:05
What are the tools you're allowed to use? –  Davide Giraudo Feb 6 '12 at 15:54

As mentioned by others, the random variables $(X_n)_{n\geqslant0}$ are independent hence, by Kolmogorov zero-one law, the sigma-algebras $F_k$ generated by $(X_{n+k})_{n\geqslant0}$ have a trivial intersection $F_{\infty}$, in the sense that every $C$ in $F_\infty$ is such that $\mathrm P(C)=0$ or $\mathrm P(C)=1$.

Hence, the task is to prove that the event $A=\big[\sum\limits_{n\geqslant0} X_n\ \text{converges}\big]$ is in $F_\infty$, or, equivalently, that $A$ is in $F_k$ for every $k\geqslant0$.

To this end, fix $k\geqslant0$ and note that $A=A_k$ with $A_k=\big[\sum\limits_{n\geqslant0} X_{n+k}\ \text{converges}\big]$. By definition, $A_k$ depends on $(X_{n+k})_{n\geqslant0}$ only, hence $A=A_k$ is in $F_k$. QED.

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This solution bypasses any precise translation of the convergence of a series in terms of the behaviour of its partial sums, which unnecessarily complicates things when the question is only about the measurability of the event that the series converges. –  Did May 8 '12 at 8:41

In other to expand Dilip Sarwate's and Ilya's comments, let $\mathcal F_n:=\sigma(X_k,k\geq n)$ the smallest $\sigma$-algebra making the random variables $X_k$, for $k\geq n$, measurable. We have to show that $A\in \bigcap_{n\geq 1}\mathcal F_n$ in order to apply Kolmogorov's $0-1$ law. We can write $$A=\bigcap_{p\geq 1}\bigcup_{N\geq p}\bigcap_{N\leq n<m}\left\{\omega:\left|\sum_{j=n+1}^mX_j(\omega)\right|\le \frac 1p\right\}.$$ Since the sequence $\left\{\bigcup_{N\geq p}\bigcap_{N\leq n<m}\left\{\omega:\left|\sum_{j=n+1}^mX_j(\omega)\right|\le \frac 1p\right\}\right\}_p$ is decreasing, we can in fact write for all $p_0$ that $A= \bigcap_{p\geq p_0}\bigcup_{N\geq p}\bigcap_{N\leq n<m}\left\{\omega:\left|\sum_{j=n+1}^mX_j(\omega)\right|\le \frac 1p\right\}$. Since $\bigcup_{N\geq p}\bigcap_{N\leq n<m}\left\{\omega:\left|\sum_{j=n+1}^mX_j(\omega)\right|\le \frac 1p\right\}\in\mathcal F_p\subset \mathcal F_{p_0}$, we get that $A\in\bigcap_{n\geq 1}\mathcal F_n$, so $A$ is a tail event and $P(A)\in \{0,1\}$.

Note that there are cases on which $P(A)=0$, for example $X_n(\omega)=1$ for all $\omega$. We can also have $P(A)=1$ if $X_n(\omega)=0$ for all $n$ and all $\omega$.

Series of independent random variables have nice properties, for example convergence in probability is the same thing as convergence almost everywhere. We also have three series theorem, by Kolmogorov.

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