Timeline for Backshift Operator: Is it well-defined?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Feb 24, 2020 at 16:42 | history | edited | whuber♦ | CC BY-SA 4.0 |
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| Feb 24, 2020 at 16:14 | comment | added | whuber♦ | Yes, because $BX$ is automatically measurable with respect to the sigma algebra at time $t-1.$ In fact, this is called a "filtration" because anything measurable at time $t$ is measurable at all later times. | |
| Feb 24, 2020 at 15:11 | vote | accept | ObnoxiousFrog | ||
| Feb 24, 2020 at 15:09 | comment | added | ObnoxiousFrog | Thank you for your response! This is what I have been looking for. So essentially, the backshift operator exists because of the definition of stochastic process. To be specific, $BX$ is measurable with respect to the filtration at time $t-1$ AND $t$. Would I be correct in saying this? | |
| Feb 24, 2020 at 14:44 | history | answered | whuber♦ | CC BY-SA 4.0 |