Timeline for Standard Deviation of an Exponentially-weighted Mean
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 15, 2023 at 16:34 | comment | added | Hannes Landeholm | +1 Note that the paper contains another constant "a" which is equal to 1 - alpha. You have to look very carefully to see if it's a or alpha. | |
| Feb 3, 2022 at 7:02 | comment | added | Rok Povsic | The paper is available at archive.org here. For future reference, the paper's title is "Incremental calculation of weighted mean and variance" written by Tony Finch, Feb 2009 (in case this link also gets broken). | |
| Jan 8, 2022 at 12:25 | comment | added | Dylan Kerler | +1 is there an updated link for the Section 9? What's the intuition behind the formula being shaped like that instead of how @AlbertNetymk suggested. | |
| Aug 19, 2020 at 15:18 | comment | added | Albert Netymk |
The Section 9 link is broken, and the shape of the formula looks odd to me. I was expecting sth like (1-a)*sth + a*sth, instead of (1-a)*(...).
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| Aug 17, 2014 at 3:42 | vote | accept | Mariska | ||
| Aug 15, 2014 at 7:30 | comment | added | Roman Shapovalov | Using $\alpha = 0.98$ you also get the mean = 4.98, which is equally useless. :) Using such coefficient, you put almost all weight on the last measurement. More realistic values of $\alpha$ are close to zero, in that case they account for long-range average. For your example, try $\alpha = 0.2$, but in practice you will probably need to average more measurements, so the values around $\alpha = 0.01$ are more realistic. | |
| Aug 15, 2014 at 2:12 | comment | added | Mariska | using the above formula and the list [1,2,3,4,5], I got SD = 0.144, whereas the normal Sample SD is 1.58. There is a factor of 10x between the two different SD's. Is this normal? | |
| Aug 14, 2014 at 19:09 | history | answered | Roman Shapovalov | CC BY-SA 3.0 |