Timeline for Distance between trunctated random walk and its normal form
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 26, 2022 at 11:49 | comment | added | nodis6 | Yes, sorry it was my mistake i read inequality wrong. I'm deleting my last comment. | |
| May 1, 2022 at 14:14 | comment | added | nodis6 | Okay, I'll make a note of it. Thank you very, very much. | |
| May 1, 2022 at 14:07 | comment | added | Iosif Pinelis | @nodis6 : Let $f_v(x):=x^2\,1(|x|>v)$ and let $\mu$ denote the probability distribution of $X_1$, so that $E[ X_1^2; |X_1| > v]=\int_{\mathbb R}f_v(x)\,\mu(dx)$ and $\int_{\mathbb R}g(x)\,\mu(dx)=E[X_1^2]<\infty$, where $g(x):=x^2$. We have $f\le g$ and $f_v(x)\to0$ for each real $x$ as $v\to\infty$. So, by the dominated convergence theorem, $E[ X_1^2; |X_1| > v]=\int_{\mathbb R}f_v(x)\,\mu(dx)\to0$ as $v\to\infty$. | |
| May 1, 2022 at 13:51 | comment | added | nodis6 | I'm so sory but i don't know one thing to understand entire proof. How doeas the the dominated convergence theorem is used here? | |
| May 1, 2022 at 13:02 | vote | accept | nodis6 | ||
| May 1, 2022 at 12:57 | vote | accept | nodis6 | ||
| May 1, 2022 at 12:57 | |||||
| May 1, 2022 at 12:54 | comment | added | Iosif Pinelis | (i) The brackets are optional. By a standard convention, $EY^2:=E[Y^2]$. (ii) We have $\sup_{n\ge1}\sqrt{E[d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right)^2]} =\sup_{n\ge1}\|d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right) \|_2$ because, by definition, $\|d( \mathscr{S}_n^{(v) }, \mathscr{S}_n ) \|_2=\sqrt{E[d( \mathscr{S}_n^{(v) }, \mathscr{S}_n )^2]}$ or, generally, $\|Y\|_2:=\sqrt{EY^2}$. | |
| May 1, 2022 at 12:50 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
added 2 characters in body
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| May 1, 2022 at 9:25 | comment | added | nodis6 | There should be $\sup_{n\ge1}\sqrt{E[d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right)^2]}$ right? But i have another stupid question. Why the $\sup_{n\ge1}\sqrt{E[d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right)^2]} = \sup_{n\ge1}||{d\left( \mathscr{S}_n^{(v) }, \mathscr{S}_n \right)}||_2$ ? | |
| May 1, 2022 at 1:57 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |