Exercise - Martingales - Convergence problem

Question

I have a problem considering the following exercise.

Let $(\Omega,\mathfrak{F},\mathbb{P},\mathfrak{(F_n})_{n\in\mathbb{N}})$ be a filtered probability space on which we consider a bounded martingale $(M_n)_{n\in\mathbb{N}}$ (i.e. $|M_n|\leq K <\infty$ for every $n$). Define $$X_n:=\sum_{k=1}^n\frac{1}{k}(M_k-M_{k-1}).$$ Show that $(X_n)_{n\geq 1}$ is an $(\mathfrak{F})_{n\geq 1}$-martingale converging $a.s.$ and in $L^2$.

To show that it is a martingale is easy, however I do not know to to justify the two convergences. I tried to apply a result about $L^p$-convergence of martingales, but unfortunately I was not successful.

In fact, I am stuck at $|X_n|\leq4K^2 \big(\sum_{k=1}^n\frac{1}{k})^2$. I would like to take $n$ to $\infty$ in order to have $|X_n|$ bounded by something independent of n, but the harmonic series does not converge, perhaps somebody knows how to deal with the square of it.

Thanks in advance for your help!

Answer 1

Notice that $$ X_n = \sum_{k=1}^{n-1}\frac{M_k}{k(k+1)} + \frac{M_n}{n} - M_0. $$ Hence, $(X_n)$ is a bounded martingale: $$ |X_n| \leq 2K + K\sum_{k=1}^\infty\frac{1}{k(k+1)} = 3K. $$ As a consequence, it is trivially bounded in $L^p$ for every $p \geq 1$. We conclude that $X_n$ converges a.s. and in $L^p$ for every $p \geq 1$.

Answer 2

We have $X_n = M_n - M_0$ as a telescopic sum, and thus $|X_n|\leq 2K$ uniformly in $n$. So now you can apply any martingale convergence theorem.

Answer 3

The random variables $Y_k=M_k-M_{k-1}$ for $k\geqslant1$ are uncorrelated and $|Y_k|\leqslant2K$ almost surely hence the identity $X_n=Y_1+\frac12Y_2+\cdots+\frac1nY_n$ implies that $$ \mathbb E(X_n^2)=\sum_{k=1}^n\frac1{k^2}\mathbb E(Y_k^2)\leqslant(2K)^2\sum_{k\geqslant1}\frac1{k^2}, $$ is bounded uniformly in $n$.