Following this question:[Can we find such $k$ so that the following inequality holds?](https://mathoverflow.net/questions/436036/can-we-find-such-k-so-that-the-following-inequality-holds/436040#436040).


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
> $$ 
> \lim\_{n\to \infty}\mathbb{P}\left(\frac{X\_1^2+X\_2^2+\dots+X\_k^2}{X\_1^2}>\frac{1}{\epsilon^2}\right)>1- \delta ?
> $$
> 
> 
> 



Can we find the following $k$ so that the following inequality holds for asymptotic normal?
$\newcommand\ep\epsilon\newcommand{\R}{\mathbb R}$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](https://mathoverflow.net/a/436051/36721),
\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](https://mathoverflow.net/a/436068/36721).


