Timeline for When are events in tail $\sigma$-algebra the limsup of some sequence of events?
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Oct 6, 2019 at 7:22 | vote | accept | yellowello | ||
| Oct 6, 2019 at 0:32 | comment | added | Bjørn Kjos-Hanssen | @yellowello I think you're right but you may want to write it more carefully (than I did) | |
| Oct 5, 2019 at 18:34 | comment | added | Bjørn Kjos-Hanssen | @RW Thanks I think it may be fixed now | |
| Oct 5, 2019 at 18:33 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 4.0 |
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| Oct 5, 2019 at 15:11 | comment | added | yellowello | I assume that each $X_n$ denotes the value of the $n^{\text{th}}$ number in the binary sequence. This is my attempt to show that $B$ is not the limsup of any $A_n$: Suppose there are infinitely many $A_n$ such that $1 \in A_n$. Then since $\limsup{A_n}$ contains sequences that have infinitely many 1s, this choice is not possible. This means that for sufficiently large $N$, $A_n = \{0\}$ for $n \geq N$. This however excludes all sequences that have finitely many 1s but contains 1s after the $N^\text{th}$ digit. Therefore, no choice of $A_n$ is possible. Am I correct? | |
| Oct 5, 2019 at 14:47 | comment | added | R W | Which space are you talking about? First you say that all $\mathcal F_n$ are the powerset of $\{0,1\}$, meaning that the state space is just the two-point set $\{0,1\}$. However, in the next sentence you are talking about subsets of the space of binary sequences. | |
| Oct 5, 2019 at 7:45 | history | edited | Bjørn Kjos-Hanssen | CC BY-SA 4.0 |
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| Oct 5, 2019 at 7:39 | history | answered | Bjørn Kjos-Hanssen | CC BY-SA 4.0 |