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 \(x\in(-\infty,\infty)\).

證明：分佈函式序列的點態收斂到一個連續的分佈函式，則可以推出一致收斂。

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

證明：連續分佈函式序列的點態收斂可以推出相應的分位數隨機變數序列的依概率收斂。

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.