Timeline for Asymptotic rate for the expected value of the square root of sample average
Current License: CC BY-SA 4.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 21, 2019 at 22:26 | vote | accept | Florian Tramèr | ||
| Mar 21, 2019 at 11:18 | answer | added | Iosif Pinelis | timeline score: 2 | |
| Mar 21, 2019 at 7:59 | comment | added | user35593 | You could Taylor expand one order more, i.e. $\sqrt{x}=1+(x-1)/2-(x-1)^2/8+O(((x-1)^3)$. Then it would be left to show that $E[O((S_n-1)^3)]=o(n^{-1})$. | |
| Mar 21, 2019 at 7:53 | comment | added | Florian Tramèr | So you're essentially switching the big-O and the expectation, i.e., $E[O((S_n -1)^2)] = O(E[(S_n -1)^2]) = O(1/n)$. Is this always valid, or do we need some extra assumptions for this step? | |
| Mar 21, 2019 at 7:42 | comment | added | user35593 | sry $\sigma^2$ of course | |
| Mar 21, 2019 at 7:40 | comment | added | user35593 | $E((S_n-1)^2)=Var(S_n)=Var(\sum_i X_i)/n^2=n\sigma/n^2=\sigma/n$ | |
| Mar 21, 2019 at 7:37 | comment | added | Florian Tramèr | How does the asymptotic growth of the second term follow? E.g., why wouldn't this be $O(1/\sqrt{n})$ or $O(1/\log{n})$ or anything else? | |
| Mar 21, 2019 at 7:33 | comment | added | user35593 | Yes, I could not correct it. expectation of second order term gives $O(1/n)$. | |
| Mar 21, 2019 at 7:26 | comment | added | Florian Tramèr | The second order term in the Taylor expansion should be $O((S_n - 1)^2)$ of course. | |
| Mar 21, 2019 at 7:18 | comment | added | Florian Tramèr | Right, this seems to yield something similar to the expression in terms of the variance I have above. Taking expectations on the Taylor expansion, I'd get $E[\sqrt{Sn}] = 1 + E[O(S_n - 1)]$. I'm not sure what to make of that second term. | |
| Mar 21, 2019 at 7:07 | comment | added | user35593 | Taylor expansion of square root at 1 yields $\sqrt{S_n}=1+(S_n-1)/2+O((S_n-1))$. | |
| Mar 21, 2019 at 3:18 | history | asked | Florian Tramèr | CC BY-SA 4.0 |