Exercise 4.1: Continuity of Functionals
(a)
Let \(\gamma\) be a functional that is continuous in \(\norm{\vardot}_\infty\) at \(F\). By the Glivenko–Cantelli theorem, \(\norm{\hat F_n - F}_\infty \to 0\) almost surely, so \(\abs{\gamma(F_n) - \gamma(F)} \to 0\) almost surely, hence also in probability.
(b)
(i)
The mean functional is not continuous. Consider \(F = \ind_{[0, \infty)}\) and \begin{equation} F_n = \parens{1 - \tfrac{1}{n}} \ind_{[0, \infty)} + \tfrac{1}{n} \ind_{[n, \infty)}. \end{equation} Then \(F\) has mean \(0\), \(F_n\) has mean \(1\), and \(\norm{F - F_n}_\infty \to 0\).
(ii)
The Cramér–von Mises functional is continuous. For this, see that \begin{equation} (F - F_0)^2 - (G - F_0)^2 = (F + G)(F - G) - 2F_0(F - G) \le 4 \norm{F - G}_\infty \to 0. \end{equation}
(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{equation} x < Q_\alpha(F) \iff F(x) < \alpha, \qquad x \ge Q_\alpha(F) \iff F(x) \ge \alpha. \end{equation} Let $\e > 0$. Then \begin{equation} \abs{x - Q_\alpha(F)} \le \e \impliedby x - \e < Q_\alpha(F) \le x + \e \iff F(x - \e) < \alpha \le F(x + \e). \end{equation} Thus, we desire to show that \begin{equation} F(Q_\alpha(F_n) - \e) < \alpha \le F(Q_\alpha(F_n) + \e). \end{equation} Alternatively, we could show that \begin{equation} F_n(Q_\alpha(F) - \e) < \alpha \le F_n(Q_\alpha(F) + \e). \end{equation} This turns out to be easier.
First, we will show that $F_n(Q_\alpha(F) + \e) \ge \alpha$. Note that \begin{equation} F_n(Q_\alpha(F) + \e) \ge F(Q_\alpha(F) + \e) - \norm{F - F_n}. \end{equation} Since, by assumption, $F(Q_\alpha(F) + \e) > \alpha$ (the strict inequality is important here), we can make $n$ large enough to find $F_n(Q_\alpha(F) + \e) > \alpha$. The other inequality $F_n(Q_\alpha(F) - \e) < \alpha$ follows similarly.