# Empirical CDF for Gaussian Distribution

Let $$X_1, ..., X_n \stackrel{iid}{\sim} N(0,1)$$, define $$\hat F_n(x) \equiv \frac{1}{n} \sum_{i=1}^n \mathbf{1}(X_i \leq x)$$ and suppose $$a_n$$ is a sequence of (deterministic) numbers such that

$$a_ne^{\frac{a_n^2}{2}} = n \quad \quad (1)$$

I want to

(i) show that $$\hat F_n(a_n) \xrightarrow{P} 1$$ and

(ii) determine the limiting distribution of $$n(1- \hat F_n(a_n))$$.

I don't really have an idea of how to do this properly, but I've tried the following:

For (i): $$a_n \rightarrow \infty$$, for if it is bounded or goes to $$-\infty$$ we can't have the LHS in (1) equal to $$n$$. Now let $$\epsilon > 0$$ and choose $$M \in \mathbb{R}$$ such that $$1-\Phi(M) < \epsilon$$ where $$\Phi(x) \equiv \int_{-\infty}^x \frac{e^{\frac{-x^2}{2}}}{\sqrt{2 \pi}}dx$$. There must exist an $$N$$ such that $$n \ge N \implies a_n > M$$, in which case $$1 \geq \hat F_n(a_n) \geq \hat F_n(M) \xrightarrow{a.s.} \Phi(M) > 1- \epsilon$$

where the almost sure convergence is by Glivenko-Cantelli and then $$\hat F_n(a_n) \xrightarrow{a.s.} 1$$ which is even stronger than what I need. Does this work?

For the second one, we have to find the limiting distribution of $$Y_n:=\sum_{i=1}^n\mathbf 1\{X_i\gt a_n\}$$. Using independence, the characteristic function of $$Y_n$$ is $$\varphi_n(t)=\left(1+\left(e^{it}-1\right)\mathbb P\{X_1>a_n\}\right)^n.$$ Using the fact that $$a_n\to+\infty$$ and the lower and upper tail for Gaussian distribution, we find that $$n\mathbb P\{X_1>a_n\}\to 1/\sqrt{2\pi}$$. We can conclude using the fact that $$(1+x_n/n)^n\to e^x$$ if $$x_n\to x\in (0,+\infty)$$.
• Wouldn't the Gaussian tail bounds imply that $nP(X_1 > a_n) \rightarrow \frac{1}{\sqrt{2\pi}}$? Apr 18, 2020 at 17:29
• Yes.${}{}{}{}{}$ Apr 18, 2020 at 17:31