I'm trying to solve the following problem:
Let $X_1, X_2, \ldots$ be independent, square integrable random variables with $\sum_{n=1}^\infty \frac 1{n^2}\mathbf{Var}\left[X_n\right] < \infty$. Use the martingale convergence theorem to show the strong law of large numbers for $\left(X_n\right)_{n \in \mathbb N}$.
So I need to show $$\displaystyle \mathbf P \left[ \limsup_{n \to \infty} \left| \frac 1 n \tilde S_n\right|=0\right] = 1,$$
where $\tilde S_n = \sum_{i=1}^n \left(X_i – \mathbf E\left[X_i\right]\right)$. Since $X = (X_n)_{n \in \mathbb N}$ is independent, $\tilde S = \left(\tilde S_n\right)_{n \in \mathbb N_0}$ can be shown to be a square-integrable martingale with respect to the filtration $\mathbb F = \left(\mathcal F_n\right)_{n \in \mathbb N}$, $\mathcal F_n = \sigma(X_1, \ldots, X_n)$. If I can show that $\tilde S$ converges (which seems unlikely a priori), then it follows that $\frac 1 n \tilde S_n \to 0$.
I'm trying to use the following convergence theorem for square-integrable martingales:
Let $M$ be a square integrable martingale with square variation process $\langle M \rangle$. Then the following four statements are equivalent:
- $\sup_{n \in \mathbb N} \mathbf E\left[M_n^2\right]<\infty$,
- $\lim_{n \to \infty} \mathbf E\left[\langle M\rangle_n\right]<\infty,$
- $M$ converges in $L^2$,
- $M$ converges almost surely and in $L^2$.
The square variation process is
\begin{align*}
\left\langle \tilde S\right\rangle_n &= \sum_{i=1}^n \mathbf E\left[\left(\tilde S_i – \tilde S_{i-1}\right)^2\big| \mathcal F_{i-1}\right] = \sum_{i=1}^n \mathbf E\left[\left(X_i – \mathbf E\left[X_i\right]\right)^2\big| \mathcal F_{i-1}\right] = \sum_{i=1}^n \left( \mathbf E\left[E_i^2\right] – \mathbf E\left[X_i\right]^2\right) \\
&= \sum_{i=1}^n \mathbf{Var}\left[X_i\right].
\end{align*}
If I can show this is uniformly bounded, we're done, but it's not clear to me why it should be true. I'm sure it has something to do with the assumption that $\sum_{n=1}^\infty \frac 1{n^2}\mathbf{Var}\left[X_n\right] < \infty$, but in principle $\displaystyle \langle \tilde S \rangle$ could still be unbounded. Any thoughts? Is there another consequence of the martingale convergence theorem I'm overlooking?
Best Answer
W.l.o.g. we may assume $EX_i = 0$. Consider $M_k = \sum_{i=1}^k \frac{X_i}i$. It's easy to see that the process $M_k$ is a square-integrble martingale w.r.t. $(\mathcal{F}_n)_{n \in \mathbb{N}}$.
$$ \langle M \rangle_{\infty} = \lim_k \langle M \rangle_k= \lim_k \sum_{i=1}^k E( (M_i - M_{i-1})^2 | \mathcal{F}_{i-1})= \sum_{i=1}^{\infty} \frac{DX_i}{i^2} < \infty. $$
Hence $M_k$ converges a.s.
It's sufficient to apply Kronecker's lemma.
We know that if $\sum_k a_k < \infty$, $b_k = k$, then $\lim_n \frac{\sum_{k=1}^n k a_k}{n} \to 0.$
Taking $a_k = \frac{X_k}k$ we get $\frac{\sum_{k=1}^n X_k}n \to 0$, q.e.d.