Examples of martingales that oscillate between finite values infinitely often.

[meiji1]

Homework Statement

Find an example of a martingale [tex] (X_n, (\mathcal{F}_n)_{n=0}^\infty, P) [/tex] such that [tex] P(X_n = a \mbox{ } i.o.) = 1 [/tex] for [tex] a = -1, 0, 1 [/tex] and
[tex] sup_n X_n < \infty [/tex]. (i.o. = infinitely often)

Homework Equations

We must have [tex] sup_n X_n < \infty [/tex].

The Attempt at a Solution

I've made a number of attempts at very different sorts of solutions. The most straightforward construction consists of the following steps:

1) Choose an integer-valued [tex] X_0 [/tex] with distribution [tex] P(X_0 = k) = 2^{-k}[/tex] for each [tex] k \in \mathbb{N} [/tex].

2) For [tex] X_1 [/tex], take the conditional distribution to be

[tex] P(X_1 = a | X_0 = k) = \frac{1}{3k} [/tex]

for each [tex]a \in \{-1,0,1\}[/tex] and decide the remaining parts of the conditional distribution to satisfy the martingale relation

[tex]
E(X_1 | \mathcal{F}_0)|_{\{X_0 = k\}} = \sum_{m = -1}^k mP(X_1 = m | X_0 = k) = k
[/tex]

so that all values assumed by [tex]X_1[/tex] are bounded in absolute value by [tex]X_0[/tex]. For subsequent [tex] X_n [/tex], the same approach is assumed, but with conditioning on [tex] X_0 [/tex], so that [tex] sup_n X_n < \infty [/tex] is maintained.

First of all, I don't think this approach can possibly work, since it seems we would always have [tex] sup_n EX_n^{+} < \infty [/tex], which implies convergence to a finite limit almost surely, by the Martingale Convergence Theorem. If we were to circumvent that, we'd end up with [tex] E|X_n| = \infty [/tex], in which case we no longer have a martingale.

I'm told this problem can be solved with a finite state Markov chain / martingale, which is similar to the unworkable approach I took above.

 

[lippyka]

I would like to know how one might construct such a finite state Markov chain / martingale. Any help would be appreciated.