Timeline for Can we construct close martingales if their terminal marginal laws are close?
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14 events
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| Dec 22, 2023 at 20:04 | comment | added | Fawen90 | Yes. That's exactly the problem. To get the continuity, we need in general the filtration is Brownian, otherwise I don't know how to estimate $\mathbb E[|Y_t-Y_s|^p]$ for $p>0$. My feeling is to approximate first $M$ by a discrete martingale and add as many (independent) Brownian motions as possible to obtain a Brownian filtration | |
| Dec 22, 2023 at 15:24 | comment | added | Nate River | I added the argument for continuity at a fixed $T$ in the post. | |
| Dec 22, 2023 at 15:23 | history | edited | Nate River | CC BY-SA 4.0 |
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| Dec 22, 2023 at 15:17 | comment | added | Nate River | Hm, there is a problem - with my argument we can get that for every $t \in [0, 1]$, $Y$ is almost surely continuous at $t$, however it is not clear to me how to extend this to almost sure continuity at all times $t$. | |
| Dec 22, 2023 at 15:14 | comment | added | Fawen90 | Do you mind specifying this argument in the post? I don't see how it is applied here, see e.g. almostsuremath.com/2009/12/20/martingale-convergence | |
| Dec 22, 2023 at 15:12 | history | edited | Nate River | CC BY-SA 4.0 |
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| Dec 22, 2023 at 15:08 | comment | added | Nate River | @Fawen90 Wait, isn't $Y$ continuous by the almost sure martingale convergence theorems? Apply the forward almost sure convergence theorem to the $L^2$ bounded martingale $Y_t$ between $t = 0$ and $T$ for each $T$ in $[0, 1]$ to get left continuity, and then similarly apply the backward convergence theorem to get right continuity. | |
| Dec 22, 2023 at 15:05 | comment | added | Fawen90 | My feeling is to first "discretise" $M$ and then introduce a large number of independent Brownian motions. This may yield the desired nice filtration, while I still don't know how to achieve it | |
| Dec 22, 2023 at 15:03 | comment | added | Nate River | Ah, I missed the continuity requirement on $Y$ as well. Hmm.. | |
| Dec 22, 2023 at 15:02 | comment | added | Fawen90 | Thanks for your answer. Indeed, this is my first thought while the delicate part is related to the filtration $\mathcal F_t$. Note that, $X$ is a general martingale and its natural filtration may not be "nice" enough to ensure that $Y$ is continuous, even to ensure the right-continuity, some conditions for $\mathcal F_t$ are needed | |
| Dec 22, 2023 at 15:01 | history | edited | Nate River | CC BY-SA 4.0 |
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| Dec 22, 2023 at 14:52 | history | edited | Nate River | CC BY-SA 4.0 |
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| Dec 22, 2023 at 14:41 | history | edited | Nate River | CC BY-SA 4.0 |
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| Dec 22, 2023 at 14:34 | history | answered | Nate River | CC BY-SA 4.0 |