Timeline for Convergence of probabilities imply convergence of joint probability
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Mar 22 at 12:02 | comment | added | Grigori | Indeed, @IosifPinelis, thank you for saying that :) I think simple triangle inequality will solve everything here. | |
| Mar 22 at 2:21 | review | Close votes | |||
| Apr 6 at 3:09 | |||||
| Mar 21 at 16:53 | comment | added | Iosif Pinelis | Now this is a simple fact about sequences of numbers. | |
| Mar 21 at 15:26 | comment | added | Grigori | @tsnao, fair point. I assume now that these RV are defined on the same space. I also added the "dependence vanishing condition". Would it help somehow? | |
| Mar 21 at 15:25 | history | edited | Grigori | CC BY-SA 4.0 |
added 260 characters in body
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| Mar 21 at 15:25 | comment | added | tsnao | It's just that (1) and (2) make sense even if $X$s and $Y$s are defined on different probability spaces. Last formula requires them to be defined on the same probability space. And there is infinite number of ways of coupling two laws. | |
| Mar 21 at 15:21 | comment | added | Grigori | @ChristianRemling, thank you, I misunderstood you. I know something about the dependence of $X_n$ and $Y_n$, I will add it to the question. | |
| Mar 21 at 14:59 | comment | added | Christian Remling | @Grigori: The difference is zero identically, $X_n$, $\tilde{X_n}$ have the same distribution. (There is no $n$ dependence in my example, I'm just writing this to be in line with your notation.) | |
| Mar 21 at 14:46 | comment | added | Grigori | @ChristianRemling, but in this case there won't be convergence of $|P(X_n \in A) - P(\tilde{X}_n \in A)| \to 0$. | |
| Mar 21 at 14:34 | comment | added | Christian Remling | This is hopeless without information on the joint distribution of $X_n,Y_n$. For example, suppose that $P(X_n=1)=P(X_n=0)=1/2$, $Y_n=X_n$, and the tilde variables are independent with the same distribution. Then $P(X_n=Y_n=1)=1/2$ while $P(\tilde{X_n}=\tilde{Y_n}=1)=1/4$. | |
| Mar 21 at 13:35 | comment | added | Grigori | @tsnao, thank you for the comment, I corrected the question, there was a typo: $X_n$ and $Y_n$ could be dependent, but $\tilde{X}_n$ and $\tilde{Y}_n$ are independent for any $n$. | |
| Mar 21 at 13:34 | history | edited | Grigori | CC BY-SA 4.0 |
edited body
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| Mar 21 at 13:32 | comment | added | tsnao | It all depends on how $X_n$ and $Y_n$ depend on each other... | |
| Mar 21 at 13:10 | history | asked | Grigori | CC BY-SA 4.0 |