Timeline for Asymptotic scaling of mean and variance for non-central chi distribution
Current License: CC BY-SA 4.0
17 events
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| Aug 30, 2023 at 17:06 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 30, 2023 at 15:25 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 30, 2023 at 15:00 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 30, 2023 at 14:21 | comment | added | Iosif Pinelis | @user1172131 : I did not say that no CLT holds here ever. What I said was that $Y^2$ never has a Gaussian distribution. Also, even when a distribution (say $P$) is close to a Gaussian distribution (say $G$), it does not necessarily follow that any moments of $P$ are close to those of $G$. | |
| Aug 30, 2023 at 14:17 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 30, 2023 at 14:15 | comment | added | Iosif Pinelis | @CarloBeenakker : You wrote: "in any case, the numerics leaves little doubt that the scaling I derived is correct, doesn't it?" No, of course this is not true. First here, the issue is not a numerical heuristics but a formal proof. Second, as I wrote in my previous comment (and to which you have not responded), the asymptotic will very much depend on how the noncentrality parameter (that depends on the $\mu_i/\sigma_i$'s) varies with $k$. So, you may not in general assume that $\mu_i=\mu$ and $\sigma_i=\sigma$ for all $i$. | |
| Aug 30, 2023 at 14:06 | comment | added | user1172131 | @IosifPinelis I see the non triviality if the $X_i$ are not identically distributed but assuming the identically distributed hypothesis as the special case treated in the answer, then we have a sum of i.i.d. random variables with finite variance; then CLT must follows. then the question become how to extend the conclusion to the more general case of independent non-i.d. random variables Am I missing something? | |
| Aug 30, 2023 at 14:05 | comment | added | user1172131 | @IosifPinelis why $Y^2$ shouldn't follow the CLT? after all is the sum of fast decaying independent random variables, no? | |
| Aug 30, 2023 at 14:05 | comment | added | Iosif Pinelis | How is "no appreciable weight" defined, mathematically, and do you prove the numerous so far unproved claims in your answer, when you don't even have clear definitions? | |
| Aug 30, 2023 at 14:03 | comment | added | user1172131 | @CarloBeenakker Thank you for the detailed answer I have only two doubts: 1) Could you report explicitly the passages for the $V$ derivation? 2) Also is not clear to me how to pass to the second equality of the second equation ($Var[Y]$), could you report the intermediate steps also there? | |
| Aug 30, 2023 at 13:59 | comment | added | Carlo Beenakker | for large $k$ the mean of $Y^2$ is larger than the standard deviation by order $\sqrt k$, so the tails of the Gaussian have no appreciable weight on the negative axis; in any case, the numerics leaves little doubt that the scaling I derived is correct, doesn't it? | |
| Aug 30, 2023 at 13:56 | comment | added | Iosif Pinelis | It is also false that "The generalisation to arbitrary $\mu_i,\sigma_i$ is straightforward." The asymptotic will very much depend on how the noncentrality parameter (that depends on the $\mu_i/\sigma_i$'s) varies with $k$. | |
| Aug 30, 2023 at 13:47 | comment | added | Iosif Pinelis | It is obviously false that for any $k$, large or not, "the [positive random -- I.P.] variable $Y^2$ has a Gaussian distribution". | |
| Aug 30, 2023 at 11:50 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 30, 2023 at 11:42 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 30, 2023 at 11:24 | history | edited | Carlo Beenakker | CC BY-SA 4.0 |
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| Aug 30, 2023 at 11:12 | history | answered | Carlo Beenakker | CC BY-SA 4.0 |