A.S. convergence of sum of square-integrable independent random variables with summable variation

I'm working on the following exercise from Achim Klenke's "Probability Theory: A Comprehensive Course" (exercise 6.1.4):

Let $$X_1, X_2, \ldots$$ be independent, square integrable, centered random variables with $$\sum_{i=1}^\infty \mathbf{Var}[X_i] < \infty$$. Show that there exists a square integrable $$X$$ with $$X = \lim_{n \to \infty} \sum_{i=1}^n X_i$$ almost surely.

Chebyshev's inequality gives us $$\mathbf P\left[|S_m - S_n| > \epsilon\right] \leq \epsilon^{-2} \mathbf{Var}\left[ \sum_{i=m+1}^n X_i\right] = \epsilon^{-2} \sum_{i=m+1}^n \mathbf{Var}\left[X_i\right] \xrightarrow{m,n \to \infty} 0.$$ whence $$(S_n)_{n \in \mathbb N}$$ is a Cauchy sequence in probability. Thus $$S_n \xrightarrow{\mathbf P} X$$. Using a similar strategy, we can in fact show that $$S_n \to X$$ in $$L^2$$.

Now, to prove almost sure convergence, I'd like to use the following result (Corollary 6.13 in Klenke):

Let $$(E,d)$$ be a separable metric space. Let $$f, f_1, f_2, \ldots$$ be measurable maps $$\Omega \to E$$. Then the following statements are equivalent.

(i)$$\quad f_n \to f$$ in measure as $$n \to \infty$$.

(ii)$$\quad$$For any subsequence of $$(f_n)_{n \in \mathbb N}$$, there exists a sub-subsequence that converges to $$f$$ almost everywhere.

and somehow use the fact that we're working with a sum of centered random variables to show that in fact every subsequence converges a.s. But I'm not sure how to do this since our $$X_i$$ are not nonnegative. I tried reconstructing the proof of this theorem, but I've only been able to show once again that there are a.e. convergent subsequences.

My other thought was to apply the Borel-Cantelli lemma to the events $$B_n(\epsilon) := \left\{ |X - S_n| > \epsilon\right\}$$ and prove that $$\limsup_{n \to \infty} B_n(\epsilon) =: B(\epsilon)$$ has probability $$0$$, but in the latter case I don't know how to approximate the probability of $$B_n(\epsilon)$$. Chebyshev doesn't seem available to us since strictly speaking we don't know what $$X$$ looks like, only that $$S_n$$ converges in $$L^2$$ to it. Even if we could say $$X - S_n = \sum_{i=n+1}^\infty X_i$$, the above approximation using Chebyshev with $$|X - S_n|$$ instead of $$|S_m - S_n|$$ would work out to $$\mathbf P\left[|X - S_n| > \epsilon\right] \leq \epsilon^{-2} \sum_{i=n+1}^\infty \mathbf{Var}[X_i]$$ which would sum to $$\epsilon^{-2} \sum_{n=1}^\infty n\mathbf{Var}[X_n]$$, but I don't see why this series converges.

Any thoughts on how to prove $$S_n \to X$$ almost surely?

3 Answers

Since $$\lim_{n\to\infty}\mathsf{P}\!\left(\sup_{k\ge n}|S_n-S_k|> \epsilon\right)=0,$$ the set on which the sequence $$\{S_n\}$$ is not Cauchy, $$N=\bigcup_{\epsilon>0}\bigcap_{n\ge 1}\left\{\sup_{j,k\ge n}|S_j-S_k|>\epsilon\right\}$$ is a null set ($$\because \sup_{j,k\ge n}|S_j-S_k|\le 2\sup_{k\ge n}|S_n-S_k|$$). So you define $$X:=\lim_{n\to\infty} S_n1_{N^c}$$.

• This seems a little too good to be true. Why does this not imply, for example, that every sequence of random variables that converges in probability converges almost surely? (This is a false result, e.g. $X_n \sim \mathrm{Ber}_{1/n}$.) Dec 16, 2018 at 14:27
• @DFord "Why does this not imply..."? Why should it?
– user140541
Dec 16, 2018 at 18:20
• Well, replace $S_n$ with any sequence of random variables $X_n$ that converges in probability to $X$. It appears as though the same argument applies: the set on which $\{X_n\}$ is not Cauchy is a null set. Dec 16, 2018 at 18:26
• @DFord Does $\mathsf{P}(\sup_{k\ge n}|X_n-X_k|>\epsilon)$ converge to $0$ in that case?
– user140541
Dec 16, 2018 at 18:28
• Cauchy in prob. means that $\mathsf{P}(|X_j-X_k|>\epsilon)\to 0$ as $j,k\to \infty$. For the a.s. convergence you need a stronger condition.
– user140541
Dec 16, 2018 at 19:00

$$var (\sum _{i=n}^{m} X_i) =\sum _{i=n}^{m} var(X_i) \to 0$$ as $$n,m \to \infty$$so the partial sums of $$\sum X_i$$ form a Cauchy sequence in $$L^{2}$$. Hence there is a square integrable random variable $$X$$ such that $$\sum _1^{n} X_i \to X$$ in $$L^{2}$$. Now convergence in mean square implies convergence in probability and for series of independent random variables convergence in probability implies almost sure convergence.

• For series of independent random variables, convergence in probability implies almost sure convergence." Is this a standard result? Does it have a name? I'm not familiar with this. Dec 16, 2018 at 14:28
• It is a well known result. A proof can be found in Chung's 'A course in probability Theory'. @DFord Dec 16, 2018 at 23:22

Your try is not bad, but I doubt that those methods will give you the result. We should show that the sum of independent random variables $$S_n = \sum_{i=1}^n X_i \to S_\infty$$ almost surely. But we know that in general $$P(\sum_{i=1}^\infty |X_i|<\infty) =1$$ fails. This means that the series is conditionally convergent in most cases, and it is known that some kind of maximal inequality such as $$P(\max_{n\in \mathbb{N}} |S_n|>\lambda) \leq \frac{C}{\lambda^2}\sum_{n=1}^\infty \operatorname{Var}(X_n)$$ provides sufficient and necessary condition for the almost sure convergence. It is necessary since we need to control the oscillation of the sequence $$n\mapsto S_n(\omega)$$ for almost all $$\omega\in \Omega$$. Fortunately there are several known maximal inequalities such as Kolmogorov's maximal inequality, Etemadi's inequality or martingale maximal inequalities. In particular, Kolmogorov's inequality can establish that $$S_n \text{ converges a.s.} \iff S_n \text{ converges in probability.}$$ (Or you can see this: https://en.wikipedia.org/wiki/Kolmogorov%27s_two-series_theorem.) If you are allowed to use more powerful tools such as martingale convergence theorem, then $$S_n = \sum_{i=1}^n X_i \to S_\infty$$ almost surely follows immediately.

• Is there a clear reason why $S_n$ converges a.s. $\iff$ $S_n$ converges in probability that follows from Kolmogorov's inequality? I'm having trouble seeing it. Dec 16, 2018 at 18:48
• It seems that there are (at least) two Kolmogorov's maximal inequalities. The version I refered to is this one: Let $x>a>0$ and $p = \max_{j\leq n} P(|S_n-S_j|>a).$ Then $P(\max_{j\leq n}|S_j|>x) \leq \frac{1}{1-p}P(|S_n|>x-a)$. What this inequality can show is @d.k.o's argument below. Of course, $S_n=X_1+X_2+\cdots +X_n$ and $X_i$ are independent (need not be identical). Dec 16, 2018 at 18:53