Timeline for Infinite-time, Path-Dependent Expected Value of an Orstein-Uhlenbeck process
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 3, 2018 at 11:51 | comment | added | Nawaf Bou-Rabee | Hinit: use limit rules. | |
| Jan 3, 2018 at 4:11 | comment | added | Jackie Lu | Actually $\frac{1}{\int_{0}^{\infty}f(X_s)ds}=0$ almost surely does not imply $\lim_{t\rightarrow\infty}\mathbb{E}\left\{\left(\frac{1}{\int_{0}^{t}f(X_s)ds}\right)^2\right\}=0$, but what about $\mathbb{E}\left\{\left(\frac{1}{\int_{0}^{\infty}f(X_s)ds}\right)^2\right\}$? | |
| Jan 3, 2018 at 3:44 | comment | added | Jackie Lu | Sorry but can you explain the last step a bit more? (3) implies that $\frac{1}{\int_{0}^{\infty}f(X_s)ds}=0$ almost surely, and that does not necessarily mean $\mathbb{E}\left\{\left(\frac{1}{\int_{0}^{\infty}f(X_s)ds}\right)^2\right\}=0$, right? | |
| Jan 2, 2018 at 19:47 | history | undeleted | Nawaf Bou-Rabee | ||
| Jan 2, 2018 at 19:45 | history | edited | Nawaf Bou-Rabee | CC BY-SA 3.0 |
deleted 305 characters in body
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| Jan 2, 2018 at 19:28 | history | deleted | Nawaf Bou-Rabee | via Vote | |
| Jan 2, 2018 at 17:33 | history | edited | Nawaf Bou-Rabee | CC BY-SA 3.0 |
removed an unnecessary expectation
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| Jan 2, 2018 at 17:18 | history | answered | Nawaf Bou-Rabee | CC BY-SA 3.0 |