Timeline for How $E(X/X+Y)=E(Y/X+Y)$ when $X,Y$ are i.i.d's [closed]
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
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| Feb 22, 2020 at 21:15 | comment | added | Nascimento de Cos | @StubbornAtom If it is conditional expectation(2nd case as you have pointed) E(X|T=X+Y) = E(Y|T) (by exchangeability of i.i.d s). Is this argument suffice. But the way without using exchangeability, how to prove this? kindly explain | |
| Feb 22, 2020 at 20:51 | comment | added | StubbornAtom | I am no sir, but you linked to a lecture about conditional expectations and added the relevant tag as well. That tells me it is the second one. | |
| Feb 22, 2020 at 20:20 | comment | added | Nascimento de Cos | @StubbornAtom sir i have gone through the link u sent. Thank you | |
| Feb 22, 2020 at 20:20 | comment | added | Nascimento de Cos | @StubbornAtom sir in that youtube link professor is talking about 1st one right? pls correct me if i am wrong. I am a beginner. Started with harvard lectures. How the proof varies in case of 2nd one. Can you pls elaborate? | |
| Feb 22, 2020 at 20:11 | history | closed | whuber♦ | Needs details or clarity | |
| Feb 22, 2020 at 10:52 | comment | added | StubbornAtom | See stats.stackexchange.com/a/374997/119261. | |
| Feb 22, 2020 at 10:43 | comment | added | StubbornAtom | Did you mean $E\left[\frac{X}{X+Y}\right]$ or $E[X\mid X+Y]$? Looks like the second one but still you accepted an answer that uses the first one. | |
| Feb 22, 2020 at 8:36 | answer | added | gunes | timeline score: 0 | |
| Feb 22, 2020 at 8:12 | answer | added | Stéphane Laurent | timeline score: 2 | |
| Feb 22, 2020 at 8:07 | vote | accept | Nascimento de Cos | ||
| Feb 22, 2020 at 8:02 | answer | added | quester | timeline score: 2 | |
| Feb 22, 2020 at 7:35 | review | Suggested edits | |||
| Feb 22, 2020 at 20:11 | |||||
| Feb 22, 2020 at 6:17 | history | asked | Nascimento de Cos | CC BY-SA 4.0 |