Almost sure convergence of a martingale sum

Question

Consider the senquence of iid r.v. $(Y_k)_{k\geq1}$ such that $\mathbb{P}(Y_k=1)=\mathbb{P}(Y_k=-1)=\frac{1}{2}$ and then consider the process $X=(X_k)_{n\geq1}$ such that $X_n=\sum_{k=1}^n\frac{Y_k}{k}$. It's very easy to see that $X$ is a martingale w.r.t. the filtration that it generate itself. However i tried proving that it is almost surely convergent using the convergence theorem for martingales but i failed. Is this really almost surely convergent? if yes is possible to prove this using martingales theory or is better to use other techniques like Borel-Cantelli lemmas?

Answer 1

$EX_n^{2}=\sum_{k=1}^{n} var (\frac {Y_k} k)=\sum_{k=1}^{n} \frac 1 {k^{2}}$ which is bounded. It is well known that this implies uniform integrability and any uniformly integrable martingale converges almost surely.

Answer 2

For a « martingale » proof, the convergence theorem for martingales states that if $(X_n)$ is a martingale wrt some filtration, and $\mathbb{E}[|X_n|]$ is bounded, then $X_n$ converges pointwise.

Now note that $\mathbb{E}[|X_n|^2]$ is clearly bounded and is not lower than $\mathbb{E}[|X_n|]^2$.

Answer 3

One can see that $$ \mathbb E\left[X_n^2\right]=\sum_{k=1}^n\frac 1{k^2}\mathbb E\left[Y_k^2\right] $$ hence that the martingale $\left(X_n\right)_{n\geqslant 1}$ is bounded in $\mathbb L^2$.