• $(\Omega,\mathcal A,\operatorname P)$ be a probability space
  • $(\mathcal F_t)_{t\ge0}$ be a filtration on $(\Omega,\mathcal A)$
  • $M$ be a continuous local $\mathcal F$-martingale on $(\Omega,\mathcal A,\operatorname P)$

If $M$ is $L^2$-bounded, i.e. $$\sup_{t\ge0}\left\|M_t\right\|_{L^2(\operatorname P)}<\infty\tag1\;,$$ are we able to show that $M$ is an $\mathcal F$-martingale?

Let $(\tau_n)_{n\in\mathbb N}$ be an $\mathcal F$-localizing sequence for $M$. By definition, $$M_s^{\tau_n}=\operatorname E\left[M_t^{\tau_n}\mid\mathcal F_s\right]\;\;\;\text{for all }t\ge s\ge0\tag2$$ for all $n\in\mathbb N$. Since $M$ is continuous and $\tau_n\xrightarrow{n\to\infty}\infty$ almost surely, $$M_s^{\tau_n}\xrightarrow{n\to\infty}M_s\;\;\;\text{almost surely}\tag3$$ for all $s\ge0$. So, the desired result would follow, if we would be able to apply the dominated convergence theorem for the conditional expectation. However, I don't find an uniform integrable bound for the $M^{\tau_n}$.


1 Answer 1


It needn't be the case that an $L^2$-bounded continuous local martingale is a true martingale. Below I outline an example that can be found in "Diffusions, Markov Processes and Martingales: Volume 2" by Rogers and Williams.

Let $B_t$ be a standard Brownian motion in $\mathbb{R}^3$ and consider the process $X_t = |B_t|^{-1}$. Let $\mathcal{F}_t$ be the (augmented) filtration generated by $B$ and let $Y_t = X_{1+t}$ and $\mathcal{G}_t = \mathcal{F}_{1+t}$ so $Y$ is adapted to $\mathcal{G}_t$. We can see 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. By standard properties of Brownian motion, $Y_t \to 0$ almost surely as $t \to \infty$. So since $Y$ is uniformly integrable (since it is $L^2$-bounded) 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.


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