Timeline for Change of time or change of measure
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| Jun 15, 2016 at 7:13 | history | edited | The Bridge | CC BY-SA 3.0 |
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| Jan 4, 2011 at 13:05 | comment | added | SBF | Sure, I understand - it's a very nice correction, thanks. I mean that I need to find change of measure which makes the same change of distributions, i.e. $Q_X\to Q_Y$ such that $$ dX_t = \sigma dw_t\to dY_t = \sqrt{a}\sigma dw_t. $$ I ask this question as a new question mathoverflow.net/questions/51103/… so I will be happy if can forward our discussion to the new question. | |
| Jan 4, 2011 at 12:29 | comment | added | The Bridge | Well what I meant was that $Y_t$ is a $\mathcal{F}_{at}$-measurable random variable as $X_t$ is $\mathcal{F}_{t}$-measurable (where $\mathcal{F}_t$ is the usual augmentation of the Brownian filtration generated by $(W_t)_{t>=0}$). So, I don't see how in a strong sense a change of measure can address the problem, but maybe there are others ways to treat the problem (for example in a weak sense of equality in law but I just can't see it). Regards | |
| Jan 4, 2011 at 10:13 | comment | added | Tim van Beek | Does a deterministic time change change the filtration of Brownian motion? I think only a stochastic time change by a stopping time does that. | |
| Jan 4, 2011 at 10:09 | comment | added | SBF | I think, my question is not correct, I will reask it. | |
| Jan 4, 2011 at 9:54 | history | answered | The Bridge | CC BY-SA 2.5 |