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Following this question:Can we find such $k$ so that the following inequality holds?.

Consider a sequence of independent $n-$dimensional random vectors $u, v_1, v_2,\dots, v_k$ uniformly distributed on the unit sphere on $\mathbb{R}^n$. Define $X_i=\sqrt{n}(u\cdot v_i)$ for $i=1,\dots, k$. Note that by central limit theorem, we have $X_i\to N(0,1)$ as $n\to \infty$ for $i=1,\dots, k$. Can we still prove this similarly inequality?

Fix $\epsilon\in (0,1)$, can prove that for any $\delta>0$ there exists $1\le k=k(\epsilon, \delta)$ (some number) so that $$ \lim_{n\to \infty}\mathbb{P}\left(\frac{X_1^2+X_2^2+\dots+X_k^2}{X_1^2}>\frac{1}{\epsilon^2}\right)>1- \delta ? $$

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    $\begingroup$ Does not my proof still work? Now $X_i$'s are not independent, but what we actually need is that the distribution of $(X_1,\ldots,X_n)$ is symmetric. $\endgroup$
    – Fedor Petrov
    Commented Dec 7, 2022 at 18:36
  • $\begingroup$ Does it seem that $X_i$ are still iid? $\endgroup$
    – Hermi
    Commented Dec 7, 2022 at 18:38
  • $\begingroup$ Well, if $u$ is chosen on the sphere, they indeed are iid (for any fixed $u$ the common distribution is the same) $\endgroup$
    – Fedor Petrov
    Commented Dec 7, 2022 at 21:14

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$\newcommand\ep\epsilon\newcommand{\R}{\mathbb R}$By the spherical symmetry, conditionally on $u$, the $X_i$'s are iid random variables (r.v.'s) each with a conditional distribution not depending on $u$. So, even unconditionally, the $X_i$'s are iid r.v.'s each equal $X:=\sqrt n\,e_1\cdot v$ in distribution, where $v:=v_1$ and $e_1$ is the first vector in the standard basis of $\R^n$.

Next, the random vector $v$ equals $\frac{(G_1,\dots,G_n)}{\sqrt{G_1^2+\dots+G_n^2}}$ in distribution, where $(G_1,\dots,G_n)$ is a standard Gaussian random vector in $\R^n$. So, $X^2$ equals $\frac{G_1^2}{(G_1^2+\dots+G_n^2)/n}$ in distribution. So, by the law of large numbers, $X^2$ converges to $G^2$ in distribution as $n\to\infty$, where $G:=G_1$.

Because the $X_i$'s are independent copies of $X$, it follows that \begin{equation} \frac{X_1^2+\dots+X_k^2}{X_1^2}\to\frac{G_1^2+\dots+G_k^2}{G_1^2} \end{equation} in distribution as $n\to\infty$.

So, by the previous answer, \begin{equation} L:=\lim_{n\to\infty}P\Big(\frac{X_1^2+\dots+X_k^2}{X_1^2}<C\Big) \\ =P\Big(\frac{G_1^2+\dots+G_k^2}{G_1^2}<C\Big) \le 2\exp-\frac{k-1}{2(1+\sqrt C)^2}, \end{equation} where $C:=1/\ep^2>1$.

Similarly one can get a better upper bound on $L$ by using the other previous answer.

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  • $\begingroup$ Thank you! But why did you assume that $u=e_1$? I feel that without this hypothesis, we are still being led by $\frac{X_1^2+\dots+X_k^2}{X_1^2}\to\frac{G_1^2+\dots+G_k^2}{G_1^2}$ in distribution. $\endgroup$
    – Hermi
    Commented Dec 8, 2022 at 4:38
  • $\begingroup$ @Hermi : I never assumed or said that the random vector $u$ is the same as the true, non-random vector $e_1$. As I wrote, by the spherical symmetry, the conditional distribution of $(X_1,\dots,X_k)$ given $u$ does not depend on the value of $u$. So, we can choose any particular value of $u$ on the unit sphere; $e_1$ is one of those values. (It is not a good idea to denote random variables/elements by lower-case letters; lower-case letters should be used for values of random variables/elements.) $\endgroup$
    – Iosif Pinelis
    Commented Dec 8, 2022 at 16:12

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