Timeline for Сoincidence of discrete random variables
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 17, 2018 at 3:19 | comment | added | VictorZurkowski | '@Iosif I see. You are right, sir | |
| Dec 17, 2018 at 2:27 | comment | added | Iosif Pinelis | @VictorZurkowski : So, you have $E_\eta\xi=\eta$ and $E_\xi\eta=\xi$. How do you get $\xi=\eta$ from here? I think the strict version of the conditional Jensen inequality is essential here. | |
| Dec 16, 2018 at 22:29 | comment | added | VictorZurkowski | '@Iosif, $E(\eta) \le E(E(\xi| \eta) ) = E(\xi) \le E( E(\eta|\xi)) = E(\eta) $, therefore the inequalities are equalities. Then $E(\xi|\eta) - \eta$ is a non-negative function whose integral is 0, hence $E(\xi|\eta) = \eta$ a.e.; likewise $ E(\eta|\xi)) - \xi$ is a non-negative function whose integral is 0, so $ E(\eta|\xi)) = \xi$ a.e. | |
| Dec 16, 2018 at 22:16 | comment | added | Iosif Pinelis | @VictorZurkowski : $g(x)\equiv x$ would not quite do. As specified in the answer, $g$ has to be strictly convex -- which is then used to get the strict inequality in (2) (unless $\xi=E_\eta\xi$ a.s. | |
| Dec 16, 2018 at 18:06 | comment | added | VictorZurkowski | As @Iosif mentions, the argument applies with any $g$. One can take $g(x) = x$. | |
| Dec 12, 2018 at 14:36 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Dec 12, 2018 at 11:31 | vote | accept | Lisa | ||
| Dec 12, 2018 at 6:15 | comment | added | Iosif Pinelis | I have now given a very different proof, which holds in complete generality, without assuming that $\xi$ and $\eta$ are discrete. | |
| Dec 12, 2018 at 6:13 | history | edited | Iosif Pinelis | CC BY-SA 4.0 |
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| Dec 11, 2018 at 23:05 | history | answered | Iosif Pinelis | CC BY-SA 4.0 |