Timeline for Limit of the stochastic process at time 0
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 23, 2012 at 1:41 | comment | added | Kamil | right, there is something missing: $dY_t=\frac{S_t−Y_t}{t}dt$, this simply comes from writing the SDE for the process defined in the original question. So it does have a solution: it is an average of the path over time interval $[0,t]$. But not clear how that affects the solution of the pde at $t=0$, at least to me. | |
| Nov 22, 2012 at 20:46 | comment | added | The Bridge | @kamil : something is missing in your sde for $Y$, moreover once modification done can you show that the sde has a solution ? Best regards | |
| Nov 22, 2012 at 16:14 | comment | added | Kamil | related to time $0$ question: If I assume $dS_t = adt+bdB_t$ and from above $dY_t = \frac{1}{t}(S_t-Y_t)$ then I can use Ito formula for $u(t,S_t,Y_t)$. I would like that to be a martingale and thus require $u_t+0.5b^2u_{ss}+au_s+\frac{1}{t}(s-y)u_y=0$. The last term is not defined at $t=0$ but from above answer as $t$ approaches $0$ I have $\frac{1}{t}(s-y)=\frac{0}{0}$. Is there anything can be said about the last term in the pde? | |
| Nov 21, 2012 at 13:30 | vote | accept | Kamil | ||
| Nov 21, 2012 at 13:30 | comment | added | Kamil | yes, thanks the Bridge and Guillaume, I think the answer now is complete as I was looking for some omega mentioning. Thus, I have the convergence for every single omega and therefore convergence in a.s. sense. | |
| Nov 20, 2012 at 19:24 | comment | added | The Bridge | @ Guillaume : 1up for honesty ;-) | |
| Nov 20, 2012 at 18:36 | comment | added | Guillaume | actually The Bridge had already answered in the same way in the comments. | |
| Nov 20, 2012 at 18:30 | history | answered | Guillaume | CC BY-SA 3.0 |