Timeline for A question on Ito integral
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 10, 2015 at 0:52 | answer | added | Sergio Almada | timeline score: 2 | |
| Dec 29, 2014 at 7:50 | comment | added | ofer zeitouni | This is an Orenstein-Uhlenbeck process with high damping (solution to $dX_t=-\beta X_t dt+dW_t$, starting at 0). Of course it goes to $0$ as $\beta\to\infty$.... | |
| Dec 28, 2014 at 22:18 | comment | added | Matthias Ludewig | Is this a homework problem? | |
| Dec 28, 2014 at 21:16 | comment | added | P.Windridge | @Nate Eldredge - that was my initial reaction too. However the integrand on the RHS is not bounded by an integrable function as $\beta \to \infty$ ($\beta e^{-\beta t} \ge \beta(1 - \beta t) \ge \beta/2$ for $t \le 1/2\beta$, in particular any (uniform in $\beta$) upper bound $g$ satisfies $g(x) \ge 1/4x$). On the other hand Ito isometry yields $$ E\left[ \left( e^{-\beta t} \int_0^t e^{\beta s} dW_s \right)^2 \right] = \frac{1}{2\beta}(1 - e^{-2\beta t} ) \to 0 $$ as $\beta \to \infty$. Also I did a quick simulation in R. I think the OP's statement is correct. | |
| Dec 28, 2014 at 19:37 | comment | added | Nate Eldredge | If I have done it correctly, stochastic integration by parts gives $$e^{-\beta t} \int_0^t e^{\beta t}\,dW_s = W_t - \int_0^t W_s \beta e^{-\beta(t-s)}\,ds$$ which seems to show this is not true, as the second term on the right side goes to 0 as $\beta \to \infty$. | |
| Dec 28, 2014 at 4:22 | history | asked | yangmengqh | CC BY-SA 3.0 |