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?)

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 $1-\delta$.