Timeline for When is the limit of Martingales a Martingale?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Jan 22, 2016 at 18:26 | history | edited | Ben | CC BY-SA 3.0 |
fixed title
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| Jan 22, 2016 at 18:24 | vote | accept | Ben | ||
| Jan 22, 2016 at 15:47 | answer | added | Nate Eldredge | timeline score: 6 | |
| Jan 22, 2016 at 14:21 | history | edited | Ben | CC BY-SA 3.0 |
fixed spacing
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| Dec 13, 2011 at 3:18 | comment | added | Ben | @George. Yes, of course. I got confused about the definition of a martingale. Your calculation shows that for fixed $t,s$, $E[X(t)|F_s] = X(s)$ a.s., and I thought I wanted something like the statement, a.s. for every fixed $t,s$, $E[X(t)|F_s] = X(s)$ which in retrospect doesn't make sense since the equality is only defined up to measure zero sets. Thanks. | |
| Dec 13, 2011 at 2:55 | comment | added | George Lowther | But, the fact that $X_n(t)$ converges to $X(t)$ in $L^1$ for each time $t$ is enough. For any $0\le s < t\le1$ and $A\in F_s$, you have$$\mathbb{E}[1_A(X(t)-X(s))]=\lim_n\mathbb{E}[1_A(X_n(t)-X_n(s))]=0.$$ | |
| Dec 13, 2011 at 2:53 | comment | added | George Lowther | @Gerald: Yes, but he didn't say anything about convergence of X(T) for stopping times T. | |
| Dec 13, 2011 at 1:51 | comment | added | Gerald Edgar | Isn't saying $X(t)$ is a martingale for $(F_t)$ equivalent to saying $E[X(T)] = E[X(0)]$ for all stopping times $T$? If that is so, then won't your uniform convergence preserve this? | |
| Dec 13, 2011 at 1:35 | history | asked | Ben | CC BY-SA 3.0 |