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  1. Let $M=(M_t)$ be a continuous local martingale, i.e. it exists a sequences of stopping times $(\tau_n)$ converging to $+\infty$ $P-$a.s. such that $M^{\tau_n}=(M_{\tau_n\wedge t})$ is a martingale. Now I was able to show that that every continuous adapted process $X$ is locally bounded, i.e. there exists a sequence of stopping times $\sigma_n$ such that $(X_{\sigma_n\wedge t})$ is bounded by a constant depending on $n$. Applying this result to the continuous local martingale $M$, I want to show that $\rho_n:=\tau_n\wedge\sigma_n$ is a stopping time (clear), it converges to $+\infty$ $P-$a.s. such that the continuous local martingale $M$ becomes a bounded martingale, i.e. $(M_{\rho_n\wedge t})$ is a bounded martingale. Boundedness is clear, but why is it again a martingale for this new sequences?

  2. Let $M$ be a continuous local martingale, $L^2_{loc}(M)$ be the space of all predictable process (equivalence clases) such that there exists a sequence of stopping times $\tau_n$ converging to $+\infty$ $P-$a.s. such that for all $n$ we have $E[\int_0^{\tau_n}H_s^2d\langle M\rangle_s]<\infty$, where $\langle M\rangle $ is the bracket process of $M$. I want to show that $L^2_{loc}=\{\mbox{all pred. process }H;\int_0^tH_s^2d\langle M \rangle_s<\infty \mbox{ }P-a.s. \mbox{ for all }t\ge 0\}$. For the inclusion "$\supset$" I have the following approach: define $\tau_n:=\inf\{t\ge 0;\int_0^t H^2_s d\langle M\rangle_s > n\}$. Is this really a stopping time? If $\{\tau_n\le t\}=\{\int_0 ^t H_s^2 d\langle M\rangle_s > n\}$ then it's clear, since $\int_0^t\cdots$ is $\mathcal{F}_t$ measurable. Are these to sets really equal? Assuming they are, we have $\tau_n\to +\infty$ $P-$a.s. since $\int_0 ^t H_s^2 d\langle M\rangle_s < \infty$ for all $t\ge 0$ and with the usual convention that Infimum over an empty set is defined as $+\infty$ (in the stopping time sense). About the other inclusion I have no idea.

  3. Why is every locally bounded predictable process $H$ in $L^2_{loc}(M)$ for every continuous local martingale $M$

Thank you for your help!

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    Regarding 1., you know that $X=M^{\tau_n}$ is a martingale and, for $\tau=\rho_n$, you ask a reason why $X^\tau=(X_{\tau\wedge t})_t$ is a martingale?2012-07-15
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    @ did: by the definition of local martingale, I know there exists (maybe) just one sequence of stopping time $(\tau_n)$ such that $M^{\tau_n}$ is a martingale. Now my question is simple, why is this true for the new sequence of stopping times $\rho_n:=\tau_n\wedge \sigma_n$2012-07-15
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    You did not read my comment. If you had, you would have seen that there is no *local* martingale in the picture anymore, only bona fide martingales. (Unrelated: adding a space between @ and one's name effectively prevents the comment to be signaled to the user (I saw your comment by chance).)2012-07-15
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    @did: Thx for pointing out about the space between @ and the name. I didn't know that. Maybe I do not understand your comment. I know that there exists a localizing sequence. Why do I know that another sequence of stopping times (converging to $\infty$ P-a.s.) still works, i.e. is also a localizing sequence? In fact, can I choose any sequence of stopping times converging to $\infty$ P-a.s. sucht that the stopped process is a martinagle?2012-07-16
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    Because $\rho_n\leqslant\tau_n$, hence $M^{\rho_n}$ is obtained by stopping the (true) martingale $M^{\tau_n}$ and $M^{\rho_n}$ is (trivially) also a (true) martingale. By the way, without the condition $\rho_n\leqslant\tau_n$, my first comment is wrong hence, simply *checking* this comment would have led you to the solution.2012-07-16
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    *This question has not received enough attention*... Did you get something from the comments above?2012-07-17
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    @did: of course, but your comment was (just) about my first question and I'm very thankful for that! (sorry if this sounds impolite). However, question 2 and 3 are also of particular interest for me.2012-07-17
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    @ hulik : about 3/, don't you think that is simply because $$ is a finite variation process. So for any such process H, you have $Z^{\tau_n}_t=\int_0^{t\wedge \tau_n} H^2_s d_s_{t\wedge \tau_n}$ is finite if $\tau_n$ is a sequence of stopping time such that the $\tau_n$ stopped process $H^{\tau_n}$ is (predictible) bounded by $K_n$.2012-07-17
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    @TheBridge : Thanks for your comment! Maybe you can also answer question 2) and post it, so that you can get the bounty2012-07-18
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    @ hulik : regarding 2/ I don't think those sets are equal your inclusion is correct and indeed your $\tau_n$ are valid stopping times, but the reverse implication seems to good to be true. When it's about local martingales even $L^2$ local martingales you always have to be very cautious with such tempting results. I don't have at hand a counterexample though, but I 'll try to find one if I have the time. Best Regards2012-07-20
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    @TheBridge : Thanks for your comment. If the sets are not equal, how can I prove the statement? The statement is given in my lecture notes, so it should be ture.2012-07-20
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    @ hulik : If it is a result then it might be true, I'll try to look at it with this in mind. Best Regards.2012-07-20
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    @ hulik : It is indeed true look theorem 2 here :http://almostsure.wordpress.com/2010/04/01/continuous-local-martingales/2012-07-20
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    @TheBridge : I have some trouble seeing the connection. I don't think, that Theorem 2 is my question, isn't it?In my post, do you think that tha sets $\{\tau_n\le t\}$ and $\{\int_0^t M_s^2 d\langle M\rangle_s > n\}$ are equal? I am not able to prove this. How would you prove the other inclusion? I'm very thankful for your help!2012-07-20
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    @ hulik : If you show that the set $L^2_loc$ as you defined it is the set of process for which stochastic integral makes sense ( this is the classical way of showing things) then as the other set you defined is the same as in th 2 and as th 2 says that this second set is exactely the one for which stochastic integral makes sense you have the result. Best regards2012-07-20

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