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 $1-\delta$.


1 Answer 1


$\newcommand\al\alpha\newcommand\be\beta\newcommand\ep\epsilon\newcommand\de\delta$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 $$ P\Big(\frac{X_n^{-2}-1}{\ep^{-2}-1}<\al n^\be\Big) =P\Big(|Z_n|>\sqrt{\frac n{1+(\ep^{-2}-1)n^\be}}\,\Big) \\ \to P(|Z|>0)=1\ge 1-\de. $$

  • 1
    $\begingroup$ @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. $\endgroup$ Commented Feb 27, 2023 at 16:58

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