Timeline for Commuting supremum and expectation
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 3, 2019 at 5:21 | history | edited | zhoraster | CC BY-SA 4.0 |
added 63 characters in body
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| Dec 8, 2010 at 8:19 | vote | accept | Johannes | ||
| Dec 5, 2010 at 13:07 | vote | accept | Johannes | ||
| Dec 5, 2010 at 13:07 | |||||
| Dec 4, 2010 at 17:21 | comment | added | Johannes | I changed the original question and included your answer. Do you agree with the assertion? | |
| Dec 4, 2010 at 14:45 | comment | added | zhoraster | Yes, you are right. Don't see the reason why it shouldn't work for a non-Markov process (though I didn't check thoroughly). | |
| Dec 4, 2010 at 10:01 | comment | added | Johannes | I would say, that the Markov property of $S$ would further imply that it sufficies to take the supremum over $\left\lbrace A|\sigma(A)\subseteq\sigma(S_t)\right\rbrace$ | |
| Dec 4, 2010 at 9:58 | comment | added | Johannes | This looks like it would also work for non-markovian processes $S$ and more general $f$, with arbitrary dependence on the path of $S$. Do you agree that your argument can also be used to show: $E\left[\sup\limits_{a\in U}E\left[f(a,\lbrace S\rbrace)\Bigr|\mathcal F_t\right]\right]=\sup\limits_{A,\;\sigma(A)\subseteq{\mathcal F_t},\;A(\omega)\in U}E\left[f(A,\lbrace S\rbrace)\right]$ | |
| Dec 3, 2010 at 19:19 | history | answered | zhoraster | CC BY-SA 2.5 |