Timeline for Exchangeable or iid random variables and linear conditioning
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Nov 26, 2016 at 3:15 | comment | added | Leonid Petrov | @Henry how about just $E(X\mid X+Y, X+Z)$ ? | |
| Nov 25, 2016 at 23:24 | comment | added | Henry | Take the random variables uniformly distrusted on $[1,2)$ so continuous. Then again $X_1 + 2 X_2 + \cdots + 2^n X_n$ will determine the values | |
| Nov 24, 2016 at 12:20 | history | edited | Leonid Petrov | CC BY-SA 3.0 |
added 79 characters in body
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| Nov 24, 2016 at 12:19 | comment | added | Leonid Petrov | @PatDevlin thank you! these are interesting examples, I think I saw some problems like this recently on the arxiv. Let's say RV's are continuous, I'll make an edit | |
| Nov 24, 2016 at 4:56 | comment | added | Pat Devlin | (Even simple looking expressions like $X_1 + 2 X_2 + \cdots + 2^n X_n$ can completely determine the $X_i$ if the variables take values in $\{0, 1\}$) I think it might be interesting to consider the entropy perhaps? [Not that this comment says anything.] | |
| Nov 24, 2016 at 4:49 | comment | added | Pat Devlin | Of course this should depend on the distribution. Say the r.v.s are distributed over some subset of the rationals, and we consider the linear expression $\pi X_1 + \pi^2 X_2 + \cdots + \pi^n X_n$. Then the entire sequence of r.v.s is determined by the value of this sum. Whereas if the variables take the values $\{1, \pi, \pi^2, \ldots \}$ then (much) less can be deduced from this sum. But yea, that's pretty weird at first glance. I don't know any references, but that doesn't say much. | |
| Nov 23, 2016 at 14:36 | history | asked | Leonid Petrov | CC BY-SA 3.0 |