Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
up vote 1 down vote accepted

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 $p \geq 1$.

share|cite|improve this answer

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 $n$.

share|cite|improve this answer
Thank you very much! – Mathoman Jan 26 '13 at 17:04
A possible typo: $X_n = Y_1+\frac12Y_2+\dots +\frac1nY_n$ – Ilya Jan 26 '13 at 17:44

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.

share|cite|improve this answer
I am sorry for the inconvenience, but I made a mistake while copying. The definition of $X_n$ was false, I changed it. I added also one more comment, perhaps you still know the correct answer? Thanks! – Mathoman Jan 26 '13 at 15:02
@Mathoman: sorry, I haven't seen that. Anyway, you've received an answer. – Ilya Jan 26 '13 at 17:45
No problem, it was my fault! – Mathoman Jan 26 '13 at 21:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.