# Can we show that for every $\delta>0$, there exist constants $\alpha>0, \beta>0$ so that the following inequality holds with high probability?

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 $$1-\delta$$.

$$\newcommand\al\alpha\newcommand\be\beta\newcommand\ep\epsilon\newcommand\de\delta$$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 $$\de$$. $$\quad\Box$$
To address a comment by the OP, alternatively we can take any real $$\be>1$$ and then $$P\Big(\frac{X_n^{-2}-1}{\ep^{-2}-1}<\al n^\be\Big) =P\Big(|Z_n|>\sqrt{\frac n{1+(\ep^{-2}-1)n^\be}}\,\Big) \\ \to P(|Z|>0)=1\ge 1-\de.$$
• @Hermi : Your comment has now been addressed. I thought you would want $\beta$ to be as small as possible. Then one must choose $\beta=1$, and $\epsilon$ will have to depend on $\delta$. However, if any $\beta>0$ is OK with you, then we can take any $\beta>1$, and then any $\epsilon>0$ will do. Please let me know if you are now fully satisfied with the answer. Commented Feb 27, 2023 at 16:58