Exercise 4.1: Continuity of Functionals

chapter 4

(a)

Let $\gamma$ be a functional that is continuous in $\Vert \cdot {\Vert }_{\mathrm{\infty }}$ at $F$. By the Glivenko–Cantelli theorem, $\Vert {\hat{F}}_n-F{\Vert }_{\mathrm{\infty }}\to 0$ almost surely, so $|\gamma (F_n)-\gamma (F)|\to 0$ almost surely, hence also in probability.

(b)

(i)

The mean functional is not continuous. Consider $F={\mathbb{1}}_{[0,\mathrm{\infty })}$ and $$\begin{array}{}\text{(1)}& F_n=(1-\frac{1}{n}){\mathbb{1}}_{[0,\mathrm{\infty })}+\frac{1}{n}{\mathbb{1}}_{[n,\mathrm{\infty })}.\end{array}$$ Then $F$ has mean $0$, $F_n$ has mean $1$, and $\Vert F-F_n{\Vert }_{\mathrm{\infty }}\to 0$.

(ii)

The Cramér–von Mises functional is continuous. For this, see that $$\begin{array}{}\text{(2)}& (F-F_0{)}^2-(G-F_0{)}^2=(F+G)(F-G)-2F_0(F-G)\le 4\Vert F-G{\Vert }_{\mathrm{\infty }}\to 0.\end{array}$$

(iii)

The quantile function is also continuous, but only if we assume that $F(x)>\alpha$ for all $x>Q_{\alpha }(F)$. Note that $$\begin{array}{}\text{(3)}& x<Q_{\alpha }(F)⟺F(x)<\alpha ,x\ge Q_{\alpha }(F)⟺F(x)\ge \alpha .\end{array}$$ Let $\epsilon >0$. Then $$\begin{array}{}\text{(4)}& |x-Q_{\alpha }(F)|\le \epsilon ⟸x-\epsilon <Q_{\alpha }(F)\le x+\epsilon ⟺F(x-\epsilon )<\alpha \le F(x+\epsilon ).\end{array}$$ Thus, we desire to show that $$\begin{array}{}\text{(5)}& F(Q_{\alpha }(F_n)-\epsilon )<\alpha \le F(Q_{\alpha }(F_n)+\epsilon ).\end{array}$$ Alternatively, we could show that $$\begin{array}{}\text{(6)}& F_n(Q_{\alpha }(F)-\epsilon )<\alpha \le F_n(Q_{\alpha }(F)+\epsilon ).\end{array}$$ This turns out to be easier.

First, we will show that $F_n(Q_{\alpha }(F)+\epsilon )\ge \alpha$. Note that $$\begin{array}{}\text{(7)}& F_n(Q_{\alpha }(F)+\epsilon )\ge F(Q_{\alpha }(F)+\epsilon )-\Vert F-F_n\Vert .\end{array}$$ Since, by assumption, $F(Q_{\alpha }(F)+\epsilon )>\alpha$ (the strict inequality is important here), we can make $n$ large enough to find $F_n(Q_{\alpha }(F)+\epsilon )>\alpha$. The other inequality $F_n(Q_{\alpha }(F)-\epsilon )<\alpha$ follows similarly.