On martingale convergence

Question

Let $(X_t)_{t\ge0}$ be a martingale with continuous paths. It was previously shown here and here that then it is impossible that $X_t\to\infty$ almost surely as $t\to\infty$.

Is it possible that there exist a nonrandom sequence $(t_k)$ in $[0,\infty)$ converging to $\infty$ such that $X_{t_k}\to\infty$ almost surely as $k\to\infty$? That is,

do there exist a martingale $(X_t)_{t\ge0}$ with continuous paths and a nonrandom sequence $(t_k)$ in $[0,\infty)$ converging to $\infty$ such that $X_{t_k}\to\infty$ almost surely as $k\to\infty$?

Answer 1

You can construct a continuous local martingale $X$ such that $X(n) = n$ almost surely. Indeed, for $t < 1$, set $Y(t) = B(t/(1-t))$ for $B$ a Brownian motion and then $X(t) = Y(t \wedge \tau)$ with $\tau = \inf\{t>0:Y(t) = 1\}$. Since $Y$ is a continuous local martingale and $\tau$ is a stopping time, $X$ is a continuous local martingale such that $X(1) = 1$ almost surely. It then suffices to concatenate independent copies of that process.

One can tweak the construction to get a genuine martingale by replacing $\tau$ by $\tau_n = \inf \{t>0:Y(t) \in \{1,-n^2\}\}$ over the $n$th interval. The fact that one still has $\lim_{n \to \infty} X(n) = +\infty$ in this case is a consequence of the fact that, by the martingale property, $\mathbb{P}(X(n+1) \neq X(n)+1) = 1/(1+n^2)$ which only happens finitely many times by Borel-Cantelli.

Answer 2

If the answer to Nate Rivers' last question, and sorry that im not just linking to it, is yes, and people seem to think it is, then you can have convergence in probability to infinity, from which a sequence converging pointwise can be extracted $$$$ I am editing this comment because the machine suggested doing it rather than posting another answer. $$$$ On the interval (n-1,n) represent an r.v. using the incremental brownian motion that has the distribution 1 with prob $1-2^{-n}$ and A with prob $2^{-n}$ where A is chosen to give the r.v mean 0. Then, stringing them together, the martingale converges to infinity at integer points because negative values occur only finitely often.