Timeline for Show that $\mathbb{E}[(S+\xi)^2]\rightarrow 0$ as $n\rightarrow\infty$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Feb 2, 2021 at 17:42 | comment | added | Parseval | @rubikscube09- what do you mean by "remove the last term"? I can't just remove it because it will change the value of the expression. EDIT: Ok I think I understand what you mean now. | |
| Feb 2, 2021 at 17:39 | vote | accept | Parseval | ||
| Feb 2, 2021 at 17:34 | comment | added | rubikscube09 | I assume he did the following: Remove the last term on the left hand sum, the first term on the right hand sum. Then you get two sums from $j =0$ to $n-2$. Combine those sums, and then add back in the terms you popped off separately. | |
| Feb 2, 2021 at 17:22 | comment | added | Parseval | I missed the $j$ factor above that multiplies the $W'$s. What happens to them? Things are clear till the 4th line, I don't see how you go from 4th to 5th. | |
| Feb 2, 2021 at 17:07 | comment | added | Parseval | Ah, alright, so you did not actually discretize $TW(T)$ but only approximated the integral with the Riemann sum. But how did you combine the last two sums, when they differ in indexes? I assume that what you've done is this: $$\frac{T}{n}\left(\sum_{j=0}^{n-1}W\left(\frac{(j+1)T}{n}\right)-\sum_{j=-1}^{n-2}W\left(\frac{jT}{n}\right)\right)=\frac{T}{n}\sum_{j=0}^{n-1}\left(W\left(\frac{(j+1)T}{n}\right)-W\left(\frac{jT}{n}\right)\right)=\frac{T}{n}W(T)$$ | |
| Feb 2, 2021 at 16:27 | history | answered | Gordon | CC BY-SA 4.0 |