Timeline for Is it true that $F_U(U(\bar\omega)) = F_X(z)$ implies $z=X(\bar\omega)$?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Jan 31, 2014 at 21:23 | vote | accept | Martin Van der Linden | ||
| Sep 26, 2013 at 18:50 | comment | added | mpiktas | Yeah it is complicated at first, since there are multiple ways to compare random variables. But when you finally understand lots of things become much clearer and it is so much fun :) | |
| Sep 26, 2013 at 18:16 | history | edited | Martin Van der Linden | CC BY-SA 3.0 |
added 71 characters in body
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| Sep 26, 2013 at 18:09 | comment | added | Martin Van der Linden | Good to know the convention, I edit! Thanks for all you comment, really helpfull at clarifying the difference between equality of r.v. and equal distribution of r.v. | |
| Sep 26, 2013 at 17:33 | comment | added | mpiktas | Final nitpicking, usually the notation is $X\sim F$ where X is a random variable, $F$ is the distribution function and $\sim$ means distributed as. To note that two random variables have the same distribution notation $X=^dZ$ is used. The notation though is a minor matter, the understanding is much more significant, keep up the good work! | |
| Sep 26, 2013 at 16:49 | comment | added | Martin Van der Linden | Thanks a lot for you comment. I edited my answer given my understanding of what you wrote. I hope this works now... | |
| Sep 26, 2013 at 16:48 | history | edited | Martin Van der Linden | CC BY-SA 3.0 |
edit following comment
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| Sep 26, 2013 at 16:22 | comment | added | mpiktas | My comment above answers clarifies your doubt about the last step. The equality of distribution functions implies equality in distribution, but not almost sure equality. Sorry for writing the previous comment without reading the whole answer. | |
| Sep 26, 2013 at 16:19 | comment | added | mpiktas | In fact your statement can be simplified to this one: if $F_X(Z)$ is distributed as $U(0,1)$ then $Z$ must be distributed as $F_X$, where $F_X$ is a distribution function of random variable $X$. I would not write $X(\omega)=Z(\omega)$ since this means that $P(X=Z)=1$, which is too strong, since if we take two independent copies of $X$, $X_1$ and $X_2$ then $F_X(X_1)\sim U(0,1)$ and $F_X(X_2)\sim U(0,1)$, but $P(X_1=X_2)\neq 1$ and in general can be made to be 0 for symmetrical distributions. | |
| Sep 26, 2013 at 15:49 | history | edited | Martin Van der Linden | CC BY-SA 3.0 |
expressing some doubts
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| Sep 26, 2013 at 15:34 | history | answered | Martin Van der Linden | CC BY-SA 3.0 |