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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 !

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  • $\begingroup$ hint think about incresingly bounding your process regards $\endgroup$
    – TheBridge
    Jan 12, 2014 at 22:02
  • $\begingroup$ I assume the local martingale to be cadlag. So the jumps may be unbounded. I don't know how to bound it. could you please explain it more clearly ? I thank you for your help very much. $\endgroup$ Jan 12, 2014 at 22:39
  • $\begingroup$ Hi the fact that the jumps are not bounded doesn't mean for a càdlàg process that a path is not bounded by a constant $n$ up to some time getting the sequence of stopping time of interest. (increasing $n$ to infinity you get your full process almost surely). Then the process stopped both by those st and the localization sequence gives you a sequence of bounded martingales which are then uniformly integrable. You have to know that some authors include the uniform integrability condition to the definition of local martingality. Best regards. $\endgroup$
    – TheBridge
    Jan 13, 2014 at 8:19

2 Answers 2

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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.

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The result is true, is given as e.g. Theorem I.50 of Protter's book "Stochastic integration and differential equations", and your argument is correct. In detail, to obtain the result, assume that $M$ is a local martingale. Let $(T_n)$ be a localising sequence, such that $M^{T_n}$ is a martingale for each $n$. Then $M^{T_n\land n}$ is a uniformly integrable martingale, since $M^{T_n\land n}_t = M^{T_n}_{n\land t} = E(M^{T_n}_n | \mathcal{F}_{n\land t})$ almost surely for $t\ge0$.

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