Let $X_i$ be iid with

$$\mathbb{P}(X_i=1)= \mathbb{P}(X_i= -1) = \frac{1}{2i}, \mathbb{P}(X_i=0)=1-\frac{1}{i},$$

where $i=1,2,...$

And define $Y_1=X_1$ and for $k\geq2$

$$Y_k= \begin{cases} X_k, \text{ if } Y_{k-1}=0\\ kY_{k-1}|X_k|, \text { if } Y_{k-1} \neq 0 \end{cases} $$

I have shown that $Y_k$ is a martingale wrt to the natural filtration $\sigma(X_1,...,X_k)$. And also that it converges in probability to zero (by conditioning on the events $(Y_{k-1} = 0 )$ and $(Y_{k-1} \neq0) $). I am however stuck on showing that it doesn't converge almost surely. I guess one has to use the Borell Cantelli Lemma but I don't see how to apply that in this situation.

  • $\begingroup$ What is $M_n$ ? $\endgroup$
    – Surb
    Nov 6, 2021 at 11:42
  • $\begingroup$ Yes sorry, that should be $Y_k$ $\endgroup$ Nov 6, 2021 at 11:52

1 Answer 1


One can show that $$\Bbb P(Y_k\neq0\mid X_0,\ldots,X_{k-1})=\frac1k$$ for all $k\ge1$. Thus $$\sum_{k=1}^\infty\Bbb P(Y_k\neq0\mid X_0,\ldots,X_{k-1})=\infty.$$ By the conditional Borel-Cantelli lemma, this means that the event $\limsup\:\{Y_k\neq0\}=\limsup\:\{|Y_k|\ge1\}$ is almost sure.


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