Question 2
\(\{F_n(x),n\geq1\}\) is a sequence of c.d.f.'s and \(F_n(x)\rightarrow F(x)\) for each \(x\in(-\infty,\infty)\), where both \(F_n(x)\) and \(F(x)\) are continuous and strict increasing c.d.f.'s. Assume \(\xi\sim U(0,1)\), show that \(F_n^{-1}(\xi)\stackrel{P}{\rightarrow}F^{-1}(\xi)\).
翻译：连续分布函数序列的点态收敛可以推出相应的分位数随机变量序列的依概率收敛。
Proof 2
We want to show that for any \(\epsilon>0,\delta>0\), \(\exist N(\epsilon,\delta)>0\), s.t. \(\forall n>N\), \(P(|F_n^{-1}(\xi)-F^{-1}(\xi)|\leq\delta)\geq1-\epsilon\), then by the arbitrariness of \(\epsilon\) and \(\delta\), we have \(F_n^{-1}(\xi)\stackrel{P}{\rightarrow}F^{-1}(\xi)\), as \(n\rightarrow\infty\).
\(\forall \epsilon>0,\delta>0\), first we can find \(M(\delta)>0\) large enough, s.t. \(F(M)-F(-M)>1-\epsilon/2\), and then for fixed \(M\), we can find \(k(\delta)>0,m(\delta)>0\), s.t. \(1/k<\delta,m/k>M\), for fixed \(m\), we can further find \(h(\epsilon,\delta)>0\), s.t. \(h<\epsilon/[4(2m+1)]\).
Since \(\sup_x|F_n(x)-F(x)|\rightarrow0\), as \(n\rightarrow\infty\), uniformly for all \(x\in(-\infty,\infty)\), \(\exist N(\epsilon,\delta)>0\), s.t. \(\forall n>N\),
\[|F_n(x)-F(x)|<h,\quad \forall x\in(-\infty,\infty).
\]
Then
\[\begin{align*}
P(|F_n^{-1}(\xi)-F^{-1}(\xi)|\leq\delta)&\geq P\left(|F_n^{-1}(\xi)-F^{-1}(\xi)|\leq\frac{1}{k}\right)\\
&\geq P\left\{\sum_{i=-m}^m\left[\left(|F_n^{-1}(\xi)-\frac{i}{k}|\leq\frac{1}{2k}\right)\cap\left(|F^{-1}(\xi)-\frac{i}{k}|\leq\frac{1}{2k}\right)\right]\right\}\\&=\sum_{i=-m}^mP\left\{\left[F_n\left(\frac{i}{k}-\frac{1}{2k}\right)\leq\xi\leq F_n\left(\frac{i}{k}+\frac{1}{2k}\right)\right]\cap\left[F\left(\frac{i}{k}-\frac{1}{2k}\right)\leq\xi\leq F\left(\frac{i}{k}+\frac{1}{2k}\right)\right]\right\}\\&=\sum_{i=-m}^mP\left(\max\left(F_n\left(\frac{i}{k}-\frac{1}{2k}\right),F\left(\frac{i}{k}-\frac{1}{2k}\right)\right)\leq\xi\leq \min\left(F_n\left(\frac{i}{k}+\frac{1}{2k}\right),F\left(\frac{i}{k}+\frac{1}{2k}\right)\right)\right)\\&\geq\sum_{i=-m}^mP\left(F\left(\frac{i}{k}-\frac{1}{2k}\right)+h\leq\xi\leq F\left(\frac{i}{k}+\frac{1}{2k}\right)-h\right)\\&=\sum_{i=-m}^m\left[F\left(\frac{i}{k}+\frac{1}{2k}\right)-F\left(\frac{i}{k}-\frac{1}{2k}\right)-2h\right]\\&=F\left(\frac{m}{k}+\frac{1}{2k}\right)-F\left(-\frac{m}{k}-\frac{1}{2k}\right)-2(2m+1)h\\&>1-\epsilon/2-\epsilon/2=1-\epsilon,\quad \forall n>N,
\end{align*}
\]
which completes the proof.