Question 1

\(\{F_n(x),n\geq1\}\) is a sequence of c.d.f.'s and \(F_n(x)\rightarrow F(x)\) for each \(x\in(-\infty,\infty)\), where \(F(x)\) is a continuous c.d.f.. Prove that \(\sup\limits_{x\in\{-\infty,\infty\}}|F_n(x)-F(x)|\rightarrow0\), as \(n\rightarrow\infty\), i.e. \(\{F_n(x)\}\) converge to \(F(x)\) uniformly for all \(x\in(-\infty,\infty)\).

证明：分布函数序列的点态收敛到一个连续的分布函数，则可以推出一致收敛。

Proof 1

For any \(\epsilon>0\), we first can always find a large enough positive constant \(M(\epsilon)>0\), such that \(1-F(x)<\epsilon/2,\forall x>M\) and \(F(x)<\epsilon/2,\forall x<-M\)，since \(F(x):(-\infty,\infty)\rightarrow[0,1]\) is a continuous c.d.f..

We know \(F(x)\) is uniformly continuous in the close interval \([-M,M]\), so we can find \(k(\epsilon)\in \mathbb{N}^+\) large enough (then \(M/(k-1)\) is small enough for \(\epsilon\)), and let \(x_i=-M+\frac{2M}{k-1}(i-1),i=1,\ldots,k\) be points of division of \([-M,M]\), such that \(F(x_{i+1})-F(x_i)<\epsilon/2,1\leq i<k\). In addition, we denote \(x_0:=-\infty\) and \(x_{k+1}:=\infty\), then

\[F(x_{i+1})-F(x_i)<\epsilon/2,\quad 0\leq i< k+1. \]

For each \(x_i\), since \(F_n(x_i){\rightarrow}F(x_i),n\rightarrow\infty\), we have \(\exist N(\epsilon)>0,\forall n>N\),

\[|F_n(x_i)-F(x_i)|<\epsilon/2,\quad 0\leq i\leq k+1. \]

Now \(\forall x\in(-\infty,\infty)\), \(\exist i\in\{0,\ldots,k\}\), s.t. \(x\in(x_i,x_{i+1}]\), we have

\[F(x_i)-\epsilon/2<F_n(x_i)\leq F_n(x)\leq F(x_{i+1})<F(x_{i+1})+\epsilon/2, \]

where the middle two inequalities are from the non-decreasing property of c.d.f. \(F_n(x)\). And then, continuously by this property, we have

\[F_n(x)-F(x)<F(x_{i+1})+\epsilon/2-F(x)\leq F(x_{i+1})-F(x_i)+\epsilon/2<\epsilon;\\ F_n(x)-F(x)>F(x_i)-\epsilon/2-F(x)\geq F(x_i)-F(x_{i+1})-\epsilon/2>-\epsilon. \]

In conclusion, \(\forall\epsilon>0,\exist N(\epsilon)>0\), s.t. \(\forall n>N,\forall x\in(-\infty,\infty),|F_n(x)-F(x)|<\epsilon\), which completes the proof.

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 \(F_n^{-1}(\xi)\stackrel{P}{\rightarrow}F^{-1}(\xi)\).

证明：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.