Timeline for Can we order random variables in a measurable way in a general setup?
Current License: CC BY-SA 4.0
13 events
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| May 21, 2019 at 16:02 | comment | added | 0xbadf00d | Thank you very much for your detailed answer! Just one side note: I'm interested in the scenario, where $(E,d)$ is a metric space and we consider the distance to a fixed element $x\in E$. In order to obtain a total order, we would first need to define the equivalence relation $y\sim_xz:\Leftrightarrow d(x,y)=d(x,z)$, $y,z\in E$, and then consider the factor space $E/\sim_x$ on which we can define a total order $\le$ by $y\le z:\Leftrightarrow d(x,y)\le d(x,z)$, $y,z\in E$. Since we need to go over to $E/\sim_x$, I'm unsure whether we still can apply your answer. | |
| May 20, 2019 at 4:58 | vote | accept | 0xbadf00d | ||
| May 19, 2019 at 18:01 | comment | added | Iosif Pinelis | @0xbadf00d : I have added a detailed formal description of your desired permutation. | |
| May 19, 2019 at 17:59 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| May 19, 2019 at 10:44 | comment | added | 0xbadf00d | (assuming $[n]$ is equipped with the $\sigma$-algebra $2^{[n]}$) (c) Does your definition of $X_{n:j}$ satisfy the desired index condition (i.e. does it always yield the smallest index variable if two or more of them are equal)? That doesn't seem to be the case. | |
| May 19, 2019 at 10:40 | comment | added | 0xbadf00d | Let me try to understand this: (a) I get that your definition of $X_{n:j}(\omega)$ yields the desired ordering. I guess the measurable dependence on $\omega$ follows from the observation that it is the finite composition of measurable functions, right? (b) As described in my comment below the question, it's clear to me that there is a $\pi:\Omega\to S_n$ such that $X_{\pi(\omega)(1)}(\omega)\le\cdots\le X_{\pi(\omega)(n)}(\omega)$ for all $\omega$. Now, as $X_{\pi(\omega)(j)}=X_{1:j}(\omega)$, this is measurable in $\omega$ by (a), right? But is $\pi$ itself measurable? | |
| May 19, 2019 at 4:42 | comment | added | Iosif Pinelis | Previous comment continued: That is, we will have $X_{n:j}(\omega)=X_{\pi(\omega)(j)}(\omega)$ for all $j\in[n]$ and all $\omega\in\Omega$. So, the situation will be exactly of the same kind as for the usual order statistics (en.wikipedia.org/wiki/Order_statistic) of real-valued random variables. | |
| May 19, 2019 at 4:41 | comment | added | Iosif Pinelis | @AnthonyQuas : I did not claim that $(X_{n:j})_1^n$ will be a permutation of $(X_i)_1^n$, or even a random permutation of $(X_i)_1^n$. Of course, in general that will not be the case. However, again in general, the construction given in my answer seems to be the only reasonable one. On the other hand, after the OP's editing the question to assume now that the order is total, of course $(X_{n:j})_1^n$ given by the same construction will indeed be a (clearly measurable) random permutation $\pi\colon\Omega\to S_n$ of $(X_i)_1^n$, where $S_n$ is the set of all permutations of the set $[n]$. | |
| May 18, 2019 at 0:55 | comment | added | Anthony Quas | In the case you describe, there is no reason to think that the random variables $X_{n,j}$ should be a permutation of the $(X_j)$. | |
| May 17, 2019 at 18:46 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| May 17, 2019 at 18:30 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| May 17, 2019 at 18:05 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| May 17, 2019 at 17:59 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |