Consider two $n-$dimensional random vectors $u$ and $v$ uniformly distributed on the sphere. Define $X\_n :=u\cdot v$. Note that as $n\to \infty$, $\sqrt{n}X\_n \to N(0,1)$ as $n\to \infty$. Fix $\epsilon>0$ (very small).


Can we show that for every $\delta>0$, there exist constants $\alpha>0, \beta>0$ so that
$$
\lim\_{n\to \infty}P\left(\frac{X\_n^{-2}-1}{\epsilon^{-2}-1}<\alpha n^\beta\right)\ge 1-\delta.
$$


Or can we revise this upper bound for $\frac{X\_n^{-2}-1}{\epsilon^{-2}-1}$ depending on $n$.


Since we know that the order $X\_n=O\_p(n^{-1/2})$, then the order of $\frac{X\_n^{-2}-1}{\epsilon^{-2}-1}$ is about $O\_p(n)$. But I am stuck on how get the strict upper bound. (Maybe this question would be helpful: <https://math.stackexchange.com/questions/4593238/can-we-find-c1-so-that-px-le-frac-epsilonc-ge-1-delta?>)




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Let $Y\sim N(0,1)$ (hence we can write $X\_n=n^{-1/2}Y$. Note that
$$\begin{align\*}
\lim\_{n\to \infty}P\left(\frac{X\_n^{-2}-1}{\epsilon^{-2}-1}<\alpha n^\beta\right)&=P\left(n\frac{Y^{-2}-1}{\epsilon^{-2}-1}<\alpha n^\beta\right)\\&=P\left(nY^{-2}<\alpha(\epsilon^{-2}-1)n^{\beta}+1\right)\\&=P\left(Y^2>\frac{n}{\alpha(\epsilon^{-2}-1)n^{\beta}+1}\right)\end{align\*}
$$


I am not sure if we can apply the concentration result of the Gaussian variable to find proper $\alpha, \beta>0$ so that this probability larger than $1-\delta$.



Can we show that for every $\delta>0$, there exist constants $\alpha>0, \beta>0$ so that the following inequality holds with high probability?
$\newcommand\al\alpha\newcommand\be\beta\newcommand\ep\epsilon\newcommand\de\delta$Let $\al=\beta=1$. Let $Z\_n:=\sqrt n\,X\_n$, so that $Z\_n\to Z\sim N(0,1)$ in distribution. Then for real $\ep>0$
$$
P\Big(\frac{X\_n^{-2}-1}{\ep^{-2}-1}<\al n^\be\Big)
=P\Big(|Z\_n|>\sqrt{\frac n{1+(\ep^{-2}-1)n}}\,\Big) \\
\to P\Big(|Z|>\sqrt{\frac1{\ep^{-2}-1}}\,\Big)
\ge 1-\de
$$
if $\ep>0$ is small enough, depending on $\de$. $\quad\Box$




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To address a comment by the OP, alternatively we can take any real $\be>1$ and then
$$
P\Big(\frac{X\_n^{-2}-1}{\ep^{-2}-1}<\al n^\be\Big)
=P\Big(|Z\_n|>\sqrt{\frac n{1+(\ep^{-2}-1)n^\be}}\,\Big) \\
\to P(|Z|>0)=1\ge 1-\de.
$$


