Uniformly integrable local martingale

Question

Can someone give me an example of a uniformly integrable local martingale that is not a martingale? Or are all U.I. local martingales true martingales (continuous, of course).

Answer 1

It is not the case that all uniformly integrable local martingales are true martingales. In fact, it is not even true that $L^2$-bounded local martingales must be true martingales. Since a family of $L^p$-bounded random variable (for $p>1$) must be uniformly integrable it is sufficient to establish this second claim.

What follows is almost identical to my answer here.

Let $B_t$ be a standard Brownian motion in $\mathbb{R}^3$ and define $X_t = |B_t|^{-1}$ for $t \in (\varepsilon, \infty)$ where $0<\varepsilon<1$. Let $\mathcal{F}_t$ be the filtration generated by $B$.

We consider the process $Y_t = X_{1+t}$ and the filtration $\mathcal{G}_t = \mathcal{F}_{1+t}$. We have that $Y_t$ is adapted to $\mathcal{G}_t$ and by Ito's formula that $Y$ is a continuous local martingale since $x \mapsto |x|^{-1}$ is harmonic away from $0$ and $B$ doesn't visit $0$.

An explicit calculation gives that $\mathbb{E}[X_t^2] = t^{-1}$ so that $Y$ is $L^2$-bounded. In particular, as I noted earlier, this implies that $Y$ is a uniformly integrable continuous local martingale.

However, since Brownian motion is transient in dimension $3$, $Y_t \to 0$ almost surely as $t \to \infty$. Since $Y$ is uniformly integrable, if $Y$ were a true martingale we would have $Y_t = \mathbb{E}[Y_\infty \mid \mathcal{G}_t] = 0$ almost surely and it is clear this is not the case, so $Y$ is not a martingale.