Timeline for Transfer independance from $\mathbb{N}$ to $\mathbb{N}^2$
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 18, 2011 at 22:07 | vote | accept | kaleidoscop | ||
| Aug 18, 2011 at 20:33 | answer | added | Noah Stein | timeline score: 5 | |
| Aug 18, 2011 at 18:45 | comment | added | kaleidoscop | Ok well it does not seem clear so let me rephrase: If for every (any) $m,n$ $\psi(X_n,X_m)$ is a.s. equal to a constant, then I am not interested. | |
| Aug 18, 2011 at 18:22 | comment | added | Robert Israel | So what you really mean is that $\psi$ should not be constant almost everywhere. | |
| Aug 18, 2011 at 15:53 | comment | added | kaleidoscop | Ok then let's just say that $\psi$ should not be constant...Concerning your example i assume you are working by default with uniform variables, but then what happens with $y=0$ should not matter because it happens almost never, so then we're back to a constant function equal to $0$. | |
| Aug 18, 2011 at 15:31 | comment | added | Emil Jeřábek |
Then it is poorly expressed, indeed. What does “arrival space” mean? Is it the same as the range of the function? And why is this condition stated only in the description of the generalization, not in the original question about $\psi\colon[0,1]^2\to[0,1]$? Anyway, what about $$\psi(x,y)=\begin{cases}x,&\text{if }y=0,\\0,&\text{otherwise}\end{cases}$$
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| Aug 18, 2011 at 14:02 | comment | added | kaleidoscop | @Emil: No because the arrival space has to take two values (maybe it is poorly expressed, but it means the constant function does not work). | |
| Aug 18, 2011 at 13:53 | comment | added | Emil Jeřábek | Any constant function $\psi$ works. You presumably do not want that. | |
| Aug 18, 2011 at 13:29 | history | asked | kaleidoscop | CC BY-SA 3.0 |