Is a local martingale locally uniformly integrable martingale ?
Here I define a local martingale to be the process with a localizing sequence $\tau_n$ such that the stopped process is martingale.
But how can we find a localizing sequence such that the stopped process is a uniformly integrable martingale ?
The solution I gave is $\min (\tau_n , n)$, could somebody please confirm ?
Thanks in advance !
Best Answer
First, convince yourself that the following claim is true.
Let $X \in L^1(P)$ be a random variable defined on a probability space $(\Omega,\mathcal{F},P)$ . Then, the collection $C = \{E[X\mid\mathcal{G}]: \mathcal{G} \subset \mathcal{F}]\}$ is uniformly integrable.
Now let $M$ denote a local martingale with localizing sequence $(\tau_n)$. We want to show $\{M^{\tau_n\wedge n}_t:t \in \mathbf{R}_+\}$ is uniformly integrable. $$M^{\tau_n\wedge n}_t = M^{\tau_n}_{t\wedge n} = E[M^{\tau_n}_n \mid \mathcal{F}_t]$$
Observe that $M^{\tau_n}_n \in L^1(P)$ by definition of a martingale and $\mathcal{F}_t \subset \mathcal{F}$. Here $(\mathcal{F}_t)_{t\geq 0}$ is some appropriate filtration.