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Sep 5, 2023 at 17:39 comment added Carlo Beenakker see also mathoverflow.net/q/454009/11260
Sep 4, 2023 at 15:12 vote accept user1172131
Sep 1, 2023 at 14:31 comment added Iosif Pinelis @user1172131 : Do you have a further response to the answers below?
Aug 30, 2023 at 22:04 answer added Iosif Pinelis timeline score: 1
Aug 30, 2023 at 21:15 comment added user1172131 @IosifPinelis yes, added in the post now
Aug 30, 2023 at 21:14 history edited user1172131 CC BY-SA 4.0
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Aug 30, 2023 at 21:12 comment added Iosif Pinelis @user1172131 : Are the $X_i$'s independent?
Aug 30, 2023 at 21:10 comment added user1172131 @IosifPinelis Ok, let's focus on the Gaussian starting case; I will ask the extension in a independent post
Aug 30, 2023 at 21:06 history edited user1172131 CC BY-SA 4.0
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Aug 30, 2023 at 20:34 history edited Michael Hardy CC BY-SA 4.0
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Aug 30, 2023 at 18:11 history became hot network question
Aug 30, 2023 at 17:57 comment added Iosif Pinelis (i) Are the $X_i$'s independent? (ii) What is $P_X$? (iii) Is $P_X(X_i=x)$ your way to denote the value of the pdf of $X_i$ at $x$? (iv) Is $o(x^{-n})$ uniform in $i$? (v) What is $o(x^{-n})$ for $x<0$? (vi) Under such general conditions that you seem to want here, there will probably no definite asymptotics. (vii) According to a MathOverflow policy, there should be only one question in one post. (viii) For all these reasons, I strongly suggest you roll back the changes and only add the condition that the $X_i$'s are independent.
Aug 30, 2023 at 17:33 comment added user1172131 Thank you for the comment @CarloBeenakker; I added the hypothesis of fast decaying tails
Aug 30, 2023 at 17:32 history edited user1172131 CC BY-SA 4.0
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Aug 30, 2023 at 17:08 comment added Carlo Beenakker for your more general case you would also need to know the fourth moment of $X_i$ (knowledge of the first two moments is not sufficient if the distribution of $X_i$ is not a Gaussian); moreover, you would want the limit $\lim_{k\rightarrow\infty}k^{-1}\sum_{i=1}^k(\mu_i/\sigma_i)^2$ to exist, I don't think it is sufficient to have the two limits $\mu,\sigma^2$.
Aug 30, 2023 at 14:48 history edited user1172131 CC BY-SA 4.0
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Aug 30, 2023 at 14:42 comment added Iosif Pinelis Yes, then we can say something specific. If you include this additional hypothesis in your post, I may be able to provide a detailed answer (except that you should write something like $\lim_{k\to\infty}\lambda/\sqrt k=L$ instead of $\lim_{k\to\infty}\lambda=L\sqrt k$, because $\lim_{k\to\infty}x_k$ cannot depend on the dummy variable $k$.)
Aug 30, 2023 at 14:29 comment added user1172131 @IosifPinelis Thank you for the comment; just to be sure that we are on the same page namewise, by noncentrality parameter you mean $\lambda = \sqrt{\sum_{i=1}^k \left( \frac{\mu_i}{\sigma_i} \right)^2}$, right? Could we conclude something if we add the hypothesis that $\lim_{k \rightarrow \infty} \lambda = \sqrt{k} L$, with L being some finit constant, i.e. $0<L<\infty$?
Aug 30, 2023 at 13:57 comment added Iosif Pinelis This asymptotic will very much depend on how the noncentrality parameter varies with $k$.
Aug 30, 2023 at 11:12 answer added Carlo Beenakker timeline score: 3
Aug 30, 2023 at 10:08 history asked user1172131 CC BY-SA 4.0