Timeline for Does the martingale property holds after changing filtration?
Current License: CC BY-SA 3.0
12 events
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| Mar 11, 2016 at 20:55 | comment | added | The Bridge | @CodeGofl : Giving second thought about your problem this is not a filtration shrinkage as I first thought, neither it is a classical enlargement of filtration as it is more an enlargement of the probability basis. I still think it might be true but it is a bit more involved that I thought. Best regards. | |
| Mar 2, 2016 at 14:24 | comment | added | CodeGolf | I shall read this paragraph. Thx a lot! | |
| Mar 2, 2016 at 12:37 | comment | added | The Bridge | Hi if you can show that the filtration $\mathcal{F^B}$ can be seen a proper shrinkage of filtration (not totally clear but I think this is the case by using some simple trick like seeing $\Omega$ in one-to -one correspondence with $\Omega \times [0,t]$ for a fixed $t$), then Protter shows in his book (Lemma page 368 Chap VI) that in this case your process has to be a martingale (i.e. the optional projection of the original process on the shrank filtration, that means that you have to show that Z is this optional projection), hope this helps. Best regards | |
| Feb 25, 2016 at 8:07 | comment | added | CodeGolf | Yes, $\{\lambda\in\Lambda: T(\lambda)\le s\}=\Omega\times [0,s]$ | |
| Feb 25, 2016 at 7:47 | comment | added | The Bridge | @ Codegolf : Ok so as T is the canonical projection it means, unless mistaken, that $\{\lambda\in\Lambda: T(\lambda)\le s\}=[0,s]$ right ? Best regards. | |
| Feb 24, 2016 at 18:04 | comment | added | CodeGolf | $\{T\le s\}$ stands for the subset $\{\lambda\in\Lambda: T(\lambda)\le s\}$. | |
| Feb 24, 2016 at 9:20 | comment | added | The Bridge | @ CodeGolf : Here my interpretation problem is on the set $\{T\le s\} $ which constitutes the "basis" for the sigma algebra $\mathcal{F}^T_t$ over the space $[0,t]$. What is the meaning of $\{T\le s\} $ in this context for a fixed $s leq t$? Best regards | |
| Feb 24, 2016 at 8:59 | comment | added | CodeGolf | We may write equally the filtration this way: Set $\mathcal F_t^T:=\sigma(\{T\le s\} \mbox{ for all } s\le t)$, then $\mathcal F_t=\mathcal F^B_t \vee \mathcal F^T_t$. | |
| Feb 23, 2016 at 13:17 | comment | added | The Bridge | @ CodeGolf : Hi your definition of the canonical filtration $(\mathcal{F}_t)_{t\ge 0}$ is not completely clear to me, could you elaborate on this point (for example exhibiting a simple set at fixed $t$ from $\mathcal{F}_t$). Best regards | |
| S Feb 17, 2016 at 11:37 | history | suggested | Jean Duchon | CC BY-SA 3.0 |
spelling (plus unnecessary but possible changes to meet the threshold...)
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| Feb 17, 2016 at 11:28 | review | Suggested edits | |||
| S Feb 17, 2016 at 11:37 | |||||
| Feb 16, 2016 at 14:03 | history | asked | CodeGolf | CC BY-SA 3.0 |