Timeline for Expectation of time integral of Wiener process
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| May 1, 2022 at 8:20 | comment | added | user20672 | Unclear why you're able to conclude that the Wiener integral on the right is normally distributed with these properties | |
| Sep 8, 2017 at 1:29 | comment | added | David Addison | This is the canonical answer according to: math.nyu.edu/faculty/goodman/teaching/StochCalc2004/notes/… | |
| Oct 18, 2010 at 15:22 | comment | added | MarkV | Slightly technical, but not bad: If h is a function with bounded variation on [a,b] and g is continuous on [a,b], then the Riemann-stieltjes integral \int_a^b g(x) dh(x) can be defined by approximating with step functions. Then, applying integration by parts, you can define the integral \int h(x) dg(x) as h(x)g(x)|_a^b - \int g(x) dh(x). Since W is almost surely continuous, and (T-t) has bounded variation, you apply this fact omega by omega to g(t) = W(t,\omega). See pages 12 and 13 in Kuo's book Introduction to Stochastic Integration for more detail. | |
| Oct 15, 2010 at 4:29 | comment | added | Cosmonut | This is really cool! I had worked out the variance of \Int_0_T {W_s ds} the hard way by partitioning [0, T] into n parts, calculating the variance of the sum and then letting n go to infinity. I got the correct answer, (thanks for verifying that :)) but was once again worried about limit interchange issues. Although, I'm guessing there must be some technical isses involved in justifying the integration by parts formula and equating the time integral to a stochastic integral. (I guess you are implicitly using Ito's lemma ?) | |
| Oct 14, 2010 at 17:23 | history | edited | MarkV | CC BY-SA 2.5 |
added 1 characters in body
|
| Oct 14, 2010 at 14:16 | history | answered | MarkV | CC BY-SA 2.5 |