Timeline for Is every (finite) moving average (weakly) stationary?
Current License: CC BY-SA 4.0
5 events
| when toggle format | what | by | license | comment | |
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| Dec 26, 2019 at 16:49 | comment | added | mlofton | Thanks Richard. Since Re-instate Monica constructed such a beautiful answer, I'll just leave as is. Merry Christmas. | |
| Dec 25, 2019 at 18:29 | comment | added | Richard Hardy | Regarding a possible correction, I would try to make the statement as precise as possible, but it is your call. Merry Christmas! :) | |
| Dec 25, 2019 at 16:08 | comment | added | mlofton | @Richard Hardy: You are correct and Reinstate Monica gave a way better answer anyway so I'll leave as is unless you think it should be modified. I didn't include it because I think of that stationarity of MA as meaning non-explosive but you ( and the reinstate monica ) are absolutely right that variance needs to be finite and independent of time also. Thanks for correction. | |
| Dec 24, 2019 at 8:49 | comment | added | Richard Hardy | This is because any linear combination of zero mean variables still has an expectation of zero. This statement is imprecise because stationarity is not determined by mean alone. Weak stationarity considers second moments while strong stationarity considers entire distributions. | |
| Dec 24, 2019 at 4:32 | history | answered | mlofton | CC BY-SA 4.0 |