Timeline for On the expectation of a path integral involving Brownian motion up to a random time
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 28, 2013 at 15:38 | vote | accept | epsilon | ||
| Aug 28, 2013 at 15:38 | vote | accept | epsilon | ||
| Aug 28, 2013 at 15:38 | |||||
| Aug 27, 2013 at 18:45 | comment | added | ofer zeitouni | As I wrote above: solve first in a bounded interval $[b,R]$ with both boundary conditions $0$. This gives you an $R$ dependent solution, and now take $R\to\infty$. This should give you the answer. (There is a technical point that indeed convergence occurs, but it is not too hard to prove, because of the fact that when starting at $x$, the probability to hit $+R$ decays exponentially as $R\to\infty$ while the gain in the value is only polynomial.) | |
| Aug 27, 2013 at 16:33 | comment | added | epsilon | Professor Zeitouni, thanks a lot for your answer. I solved the ODE with two free parameters: $u(x)=A \exp(2\mu x/\sigma^2)+x^2/2\mu + \sigma^2 x/2\mu^2 + B$. The boundary condition $u(b)=0$ gives me one equation for the parameters $A,B$; but the other condition $u(\infty)=\infty$ does not give me an explicit equation. In fact, for any $A\ge 0$, I can find a $B$ so that both boundary conditions are satisfied. In other words, the solution to the ODE is not unique? Maybe I am still missing something. | |
| Aug 27, 2013 at 15:45 | history | edited | ofer zeitouni | CC BY-SA 3.0 |
edited body
|
| Aug 27, 2013 at 7:53 | history | answered | ofer zeitouni | CC BY-SA 3.0 |