# Convergence of the empirical distribution function

Let $\alpha\in\mathbb R^d$, with $\alpha\neq 0$. Take a sequence of iid random variables $X^{1},\dots,X^{n}$ with values in $\mathbb R^d$ and denote $R$ the cdf of $\alpha X^1$ (where the product is a scalar product). Now consider $\alpha^n=g(X^{1},\dots,X^{n})$ for some function $g$, and suppose we know that $\alpha^n\to\alpha$ almost surely as $n\to\infty$. Define
$\hat R_n(t)=\frac{1}{n}\displaystyle\sum_{i=1}^n \mathbf 1_{\left\{\alpha^nX^i\leq t\right\}},$
which is the empirical distribution of $\alpha^nX^{1},\dots, \alpha^nX^{n}$.
Can we say that $\hat R_n(t)\to R(t)$ as $n\to\infty$?

• What kind of convergence are you looking for? Point-wise or uniform in $t$? – Santiago Jan 2 '14 at 16:58
• I'm interested in anything that can be said about this convergence, of course the stronger result would be better! – splinter123 Jan 3 '14 at 11:54

The answer is yes for $d=1$. By denoting $F$ the cdf of $X^1$, you have that $\hat{F}_n(t)$ converges uniformly to $F(t)$ by the Glivenko–Cantelli theorem. You want to show that $\hat{F}_n(t/\alpha_n)$ converges uniformly to $F(t/\alpha)$. You can check this on open sets $(-\infty,t)$. You get $$\left|\int_{-\infty}^{t/\alpha_n} d\hat{P}_n-\int_{-\infty}^{t/\alpha} dP\right|\le \left|\int_{-\infty}^{t/\alpha} d\hat{P}_n-dP\right|+\left|\int_{-\infty}^{t/\alpha_n-t/\alpha} d\hat{P}_n-dP\right|+\left|\int_{-\infty}^{t/\alpha_n-t/\alpha} dP\right|$$
Now use uniform convergence of $\hat{F}_n$ to $F$ to make the first two terms as small as you want. Then you can use uniform convergence of $t/\alpha_n\rightarrow t/\alpha$ to make the third term small (assuming that $\alpha\neq 0)$.