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This is probably an easy question:

A local martingale is an adapted, cadlag process for which there is an increasing sequence of stopping times (going to $\infty$) such that the stopped process is a uniformly integrable martingale for each $n$.

This implies that each component is a local martingale.

Question: Is the converse also true? I.e. you have $M^1$ and $M^2$ as local martingales, is then also $(M^1,M^2)^\top$ a local martingale?

I don't think this is true. How would you construct such a sequence of stopping times that satisfies the definition above for all components? I'll think about a counterexample, but maybe someone knows better.

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if $\tau_i, T_i$ work for the two processes, why not $\tau_i \wedge T_i$ ? – mike Nov 13 '12 at 14:56
Well, how do you know that the min of the two makes both a uniformly integrable martingale? This is in my view precisely the problem! – user13655 Nov 13 '12 at 15:05
because it is u.i. up to $\tau_1$, so it is automatically u.i. up to $\tau_1 \wedge T_1$, just because the latter is smaller. – mike Nov 13 '12 at 16:12
@ user13655 : A remark although the definition asserts that the stopped processes are u.i. martingales, if the stopped processes are martingales then are u.i. martingales. So often the definition is given with only the property that the stopped processes are martingales. Best regards – TheBridge Nov 13 '12 at 16:31

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