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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!

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3 Answers 3

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 $p \geq 1$.

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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 $n$.

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Thank you very much! –  Mathoman Jan 26 '13 at 17:04
1  
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.

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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

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