Show that sum of independent but not identically distributed r.v. does not converge a.s.

Question

Let $X_k$, $k\geq 1$, be independent with the following distribution: $P(X_k=\frac{1}{\sqrt{k}})=\frac12=P(X_k=-\frac{1}{\sqrt{k}})$. Define $S_n:=\sum_{k=1}^{n} X_k$. Show that $S_n$ does not converge a.s.

So far I have explored (without success) the following ideas: showing that $(S_n)_n$ is a.s. not a Cauchy sequence and looking at the limiting characteristic function.

Any other ideas? Or maybe followups to mine? Than you!

Later edit: Thank you for the ideas so far. Any other ones using less "heavy machinery" than Lindeberg-Feller or Kolmogorov's three series theorem?

Answer 1

With Kolmogorov's three-series theorem:

Note that $$\operatorname{var} X_k = \mathbb{E}[X_k^2] = \frac{1}{k}$$ and therefore $$ \sum_{k=1}^\infty \operatorname{var} X_k = \infty\,. $$ Therefore, condition (iii) of Kolmogorov's three-series theorem is violated: the series $\sum_{k=1}^\infty X_k$ does not converge almost surely.

Answer 2

There is a martingale approach to Kolmogorov's three series theorem which can be invoked to produce a self-contained (modulo standard facts on martingales) proof for the current problem (see D. Williams, Probability with Martingales, chapter 12 for more details). So in a sense, what we present below is a fragment toward martingale proof of Kolmogorov's three series theorem.

Let $a_n^2 = Var(S_n)$, then it is straightforward to see (in view of the independence condition) that $M_n := S_n^2 - a_n^2 $ is a martingale (with respect to filtration $\mathcal{F}_n = \sigma(X_1,...,X_n)$).

Now assume for contradiction that $S_n$ converges almost surely. Then $S_n$ is bounded almost surely, and for some (large) constant $0<c<\infty$ we must have \begin{equation} \mathbb{P}(|S_n| \leq c \ \text{ for all } \ n \in \mathbb{N} ) > 0 . \tag{1} \end{equation}

Consider the stopping time $T = \inf\{ n \in \mathbb{N}: \ |S_n| > c \}$. Since $M_n$ is a martingale, it follows that so is the stopped process $M_{n \wedge T}$, in particular $\mathbb{E} M_{n \wedge T} =0$, which gives $$ \mathbb{E} a_{n\wedge T}^2 = \mathbb{E} S_{n\wedge T}^2. $$ From here we get \begin{equation} a_n^2 \mathbb{P}(T = \infty) \leq \mathbb{E} a_{n\wedge T}^2 = \mathbb{E}( S_{n}^2 \mathbb{I}_{T = \infty}) + \mathbb{E}( S_{n\wedge T}^2 \mathbb{I}_{T < \infty}) \leq c^2 + (c + 1)^2, \tag{2} \end{equation} where for the last inequality we used the fact that $|S_n| \leq c$ as long as $n \leq T$, and that $|S_T - S_{T-1} | = |X_T| \leq 1$.

Finally, observe that $(1)$ implies $\mathbb{P}(T = \infty) > 0$, which combined with $(2)$ shows that $a_n^2 $ is bounded above. This is a contradiction, as $a_n \sim \log n $, and the proof is complete.