I'm self-studying Martingales. I came accross the following exercise (exercise 4.3.1.) in Durrett's Probability Theory and Examples (5th Edition).

Exercise. Give an example of a martingale $X_n$ with $\sup_n|X_n|<\infty$ and $\mathbb P(X_n = a \text{ i. o. } )=1$ for $a=-1,0,1$.

Attempt 1.

I think that something in the following lines works.

Fix the probability space $(\Omega,\mathcal F,\mathbb P)$. Define the independent sequence of random variables $\xi_k$ such that

$$\mathbb P(\xi_k= 0) = \frac 1{k^2}, \ \ \ \ \mathbb P(\xi_k = 1) = 1-\frac{1}{k^2} $$ Then I set \begin{align*} X_n = \sum_{k=1}^n (-1)^k (\xi_k-\mathbb E[\xi_k]) \end{align*} This $X_n$ is a martingale with respect to its natural filtration. I know from the First Borel Cantelli that for $\mathbb P$-a.s. $\omega \in \Omega$ after some index $K$ we have $\xi_k(\omega)=1$ for all $k>K$. So I guess that I can say that $X_k$ is almost surely oscillating. I think it is very clear that this does not mean that it oscillates between the three values $-1,0$ and $1$.

I think that something like that works, but I am at the same time skeptic about that because $$ |X_{n+1}-X_n| = |\xi_{n+1}-\mathbb E[\xi_{n+1}]| \leq 2$$ But then from a previous theorem (in the same book) I know that $X_n$ either converges or oscillates between $-\infty$ and $\infty$ which makes the confusion only worse.

This means that if I take $X_n= \sum_{k=1}^n \eta_k$ with $\eta_k$ independent random variables, then we should have that $|\eta_k|$ is not bounded by a real number.

Attempt 2.

I thought maybe three values for $a$ is a little difficult. I tried to construct one martingale oscillating between two values. Let $U_n$ and $V_n$ be two Martingales w.r.t. some filtration $\mathcal F_n$ that converge to $0$ and $1$ respectively. Let $A_n$ be a Bernouilli random variable that is predictable. Then I take $X_n$ as $$X_n = A_n U_n + (1-A_n)V_n$$ This $X_n$ is clearly a Martingale, but I don't know how to proceed rigorously or if it even works. How can I make sure that for almost surely $\omega\in\Omega$ the sequence $A_n(\omega)$ is oscillating?


1 Answer 1


Let $(Y_n)_{n \in \mathbb{N}}$ be a sequence of independent random variables such that

$$\mathbb{P}(Y_n = 1) = \mathbb{P}(Y_n=-1) = \frac{1}{2n} \qquad \mathbb{P}(Y_n=0) = 1- \frac{1}{n}.$$

If we define

$$X_n := \begin{cases} Y_n, & X_{n-1} = 0, \\ n X_{n-1} |Y_n|, & X_{n-1} \neq 0 \end{cases} \qquad X_0 := 0$$

then the process $(X_n)_{n \in \mathbb{N}_0}$ is a martingale with respect to $\mathcal{F}_n := \sigma(Y_k; k \leq n)$. Indeed:

$$\begin{align*} \mathbb{E}(X_n \mid \mathcal{F}_{n-1}) &= 1_{\{X_{n-1}=0\}} \underbrace{\mathbb{E}(Y_n \mid \mathcal{F}_{n-1})}_{=\mathbb{E}(Y_n)=0} + n 1_{\{X_{n-1} \neq 0\}} X_{n-1} \underbrace{\mathbb{E}(|Y_n| \mid \mathcal{F}_{n-1})}_{=\mathbb{E}(|Y_n|) = 1/n} \\ &= 0 \cdot 1_{\{X_{n-1}=0\}} + 1_{\{X_{n-1} \neq 0\}} X_{n-1} = X_{n-1}. \end{align*}$$

For any fixed $a \in \{-1,0,1\}$ we have

$$\begin{align*} \sum_{n \geq 1} \mathbb{P}(Y_{2n}=0, Y_{2n+1}=a) &= \sum_{n \geq 1} \mathbb{P}(Y_{2n}=0) \mathbb{P}(Y_{2n+1}=a) \\ &\geq \sum_{n \geq 1} \left(1-\frac{1}{2n} \right) \frac{1}{2(2n+1)} = \infty, \end{align*}$$

and therefore the Borel-Cantelli lemma shows that for almost all $\omega$ it happens for infinitely many $n \in \mathbb{N}$ that $Y_{2n}(\omega)=0$, $Y_{2n+1}(\omega)=a$. By the very definition, this implies that $X_{2n}(\omega)=0$ and $$X_{2n+1}(\omega)=Y_{2n+1}(\omega)=a$$ for any such $n \in \mathbb{N}$. Consequently, we have shown that $$\mathbb{P}(X_k = a \, \, \text{infinitely often})=1$$ for any $a \in \{-1,0,1\}$. It remains to prove that $$\sup_{n \in \mathbb{N}} |X_n(\omega)| < \infty \quad \text{a.s.}$$ To this end, we note that $$\sum_{n \geq 1} \mathbb{P}(Y_n \neq 0, Y_{n+1} \neq 0) = \sum_{n \geq 1} \mathbb{P}(Y_n \neq 0) \mathbb{P}(Y_{n+1} \neq 0) \leq \sum_{n \geq 1} \frac{1}{n^2} < \infty,$$ applying the Borel-Cantelli lemma we find that for almost all $\omega$ we can choose $N=N(\omega)$ such that $$Y_{n}(\omega) \neq 0 \implies Y_{n+1}(\omega)=0 \quad \text{for all $n \geq N$.}$$ As $$X_n(\omega) \neq 0 \implies Y_n(\omega) \neq 0 \quad \text{and} \quad Y_{n+1}(\omega) = 0 \implies X_{n+1}(\omega)=0$$ this means that $$X_n(\omega) \neq 0 \implies X_{n+1}(\omega)=0 \quad \text{for all $n \geq N$.}$$ By the definition of $X_n$, this implies that $|X_n(\omega)| \leq |Y_n(\omega)| \leq 1$ for all $n \geq N$. Thus, $$\sup_{n \in \mathbb{N}} |X_n(\omega)| \leq \sup_{n \leq N} |X_n(\omega)| + 1<\infty.$$

  • $\begingroup$ I liked it! Many thanks! I would never construct such martingale. So I have one last question, is it just experience or is this example based on something one could easily think of? $\endgroup$
    – Shashi
    Commented Oct 22, 2018 at 17:20
  • 1
    $\begingroup$ @Shashi With increasing experience/practice with martingales it certainly becomes easier to construct counterexamples to certain statements, but this particular (counter)example took me quite a while to figure out. Roughly, the idea is that $(Y_n)_{n \in \mathbb{N}}$ has all the desired properties except that it is not a martingale... so we have to "perturb" it a bit to get a martingale. $\endgroup$
    – saz
    Commented Oct 22, 2018 at 17:31
  • $\begingroup$ yes thanks again!! $\endgroup$
    – Shashi
    Commented Oct 22, 2018 at 18:03
  • $\begingroup$ it's worth mentioning that this is an example of a martingale that converges to zero in probability, but not almost surely. $\endgroup$
    – Alan
    Commented Jul 3, 2021 at 21:51

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