Timeline for The difference of $\sum_{i=1}^{n}X_{i}$ and $\sum_{i=1}^{n}X_{(i)}$
Current License: CC BY-SA 4.0
8 events
| when toggle format | what | by | license | comment | |
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| Aug 25, 2023 at 12:50 | comment | added | Sycorax♦ | If you shuffle a deck of cards, the sum of the numbered cards is the same every time. It’s also the same if you sort the cards. | |
| Aug 25, 2023 at 11:17 | comment | added | Sextus Empiricus | > But I know how to caculate the p.d.f. of $$2\sum\limits_{i=1}^{n}(X_{(i)}-X_{(1)})/\theta$$ Here you actually use the same principle as the one that you are asking for. The sum of ordered exponentially distributed variables $(X_{(i)}-X_{(1)})$ is similar to the sum of unordered exponentially distributed variables. | |
| Aug 25, 2023 at 11:05 | comment | added | Sextus Empiricus | > I don't know how to caculate the p.d.f. of $$2\sum\limits_{i=1}^{n}(X_{i}-X_{(1)})/\theta$$ > But I know how to caculate the p.d.f. of $$2\sum\limits_{i=1}^{n}(X_{(i)}-X_{(1)})/\theta$$This is exactly the trick, use the memoryless property of the exponential distribution, well done. | |
| Aug 25, 2023 at 10:16 | comment | added | StubbornAtom | Original problem discussed at stats.stackexchange.com/q/272385/119261. | |
| Aug 25, 2023 at 3:49 | answer | added | kjetil b halvorsen♦ | timeline score: 4 | |
| Aug 25, 2023 at 3:41 | history | edited | Arya McCarthy |
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| S Aug 25, 2023 at 3:39 | review | First questions | |||
| Aug 25, 2023 at 4:10 | |||||
| S Aug 25, 2023 at 3:39 | history | asked | Inforz | CC BY-SA 4.0 |