Timeline for Independence of $X+Y$ and $X-Y$
Current License: CC BY-SA 4.0
15 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 22, 2018 at 13:36 | vote | accept | S.Rana | ||
| Oct 22, 2018 at 11:28 | vote | accept | S.Rana | ||
| Oct 22, 2018 at 13:36 | |||||
| Oct 22, 2018 at 11:28 | vote | accept | S.Rana | ||
| Oct 22, 2018 at 11:28 | |||||
| Oct 21, 2018 at 21:00 | history | tweeted | twitter.com/StackStats/status/1054115319705489411 | ||
| Oct 21, 2018 at 17:04 | answer | added | user158565 | timeline score: 10 | |
| Oct 21, 2018 at 17:03 | comment | added | whuber♦ | I suspect "contrivance" may have been a typographical error for "covariance." It is relevant to the question about different dice, because they will have different variances, whence the covariance of $X-Y$ and $X+Y$ will be nonzero, which is prima facie evidence of lack of independence. It won't settle the question for identical dice, though, because a zero covariance does not imply independence in this setting. | |
| Oct 21, 2018 at 11:11 | answer | added | Ben | timeline score: 15 | |
| Oct 21, 2018 at 11:07 | comment | added | NofP | If your dice are known, e.g. with standard numbering from 1 to n, then (X+Y) and (X-Y) are not independent. A simple way of thinking about disproving independence is that you only need to show that there exists at least one outcome of X+Y such that X-Y is known with absolute certainty. if X+Y = 2, then (X-Y) is known and it has to be 0. | |
| Oct 21, 2018 at 10:55 | history | edited | kjetil b halvorsen♦ | CC BY-SA 4.0 |
deleted 11 characters in body; edited tags
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| Oct 21, 2018 at 8:13 | comment | added | S.Rana | @t.f I'm not aware of the term 'contrivance' yet. Yes, they have same no of sides. | |
| Oct 21, 2018 at 8:12 | comment | added | S.Rana | @keiv.fly Do we calculate X-Y and X+Y probabilities for different cases and then use P((X-Y)(X+Y))=P(X-Y)P(X+Y)? Isn't there a shorter, more formal method to do the same? | |
| Oct 21, 2018 at 7:59 | comment | added | keiv.fly | Independence is P(XY)=P(X)P(Y). You can calculate all probabilities for 36 outcomes and show that the equation holds. | |
| Oct 21, 2018 at 7:47 | comment | added | Kozolovska | The contrivance is Var(X) - Var(Y). Also, do the dies have the same number of sides? | |
| Oct 21, 2018 at 7:25 | history | edited | user158565 | CC BY-SA 4.0 |
added 48 characters in body
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| Oct 21, 2018 at 7:09 | history | asked | S.Rana | CC BY-SA 4.0 |