随机变量的依分布收敛

我们来指出淡收敛在分布函数和随机变量上的体现.

定义1(淡收敛) 设\(\{F_n\}\)和\(F\)是分布函数, 对应的概率测度为\(\{\mu_n\}\)和\(\mu\), 若\(\mu_n\xrightarrow{v}\mu\), 则称\(F_n\)于\(F\), 记作\(F_n\xrightarrow{v}F\).

推论设分布函数\(F\)的连续点所构成的集合为\(C\), 则\(F_n\xrightarrow{v}F\)当且仅当 \[ F_n(x)\to F(x),\quad\forall x\in C. \]

证明一方面, 设\(F_n\xrightarrow{v}F\), 则对任意的\(\mu\)的连续性区间\((a, b]\), 都有 \[ \mu_n(a, b]=F_n(b)-F_n(a)\to\mu(a, b]=F(b)-F(a). \] 令\(a=x\), \(b=-\infty\), 则对任意的\(x\in C\), 都有\(F_n(x)\to F(x)\).

另外一方面, 由\(F(x)\)是分布函数知\(C\)在\(\mathbb{R}\)中稠密, 若对任意的\(x\in C\), 都有\(F_n(x)\to F(x)\), 则对任意的\(a, b\in C\), 都有 \[ \lim_{n\to\infty}\mu_n(a, b]=\lim_{n\to\infty}(F_n(b)-F_n(a))=F(b)-F(a)=\mu(a, b], \] 这便说明了\(\mu_n\xrightarrow{v}\mu\), 从而\(F_n\xrightarrow{v}F\).

定义2(依分布收敛) 设\(\{X_n\}\)和\(X\)是随机变量, 对应的分布函数为\(\{F_n\}\)和\(F\), 若\(F_n\xrightarrow{v}F\), 则称\(X_n\)于\(X\), 记作\(X_n\xrightarrow{d}X\).

推论设\(F_n, F\)是随机变量\(X_n, X\)的分布函数, \(F\)的连续点所构成的集合为\(C\), 则\(X_n\xrightarrow{d}X\)当且仅当 \[ F_n(x)\to F(x),\quad\forall x\in C. \]

证明应用上述结论即可.

命题1(蕴含关系) 若\(X_n\xrightarrow{p}X\), 则\(X_n\xrightarrow{d}X\).

证明设\(F_n, F\)是随机变量\(X_n, X\)的分布函数. 一方面, 对任意的\(x\in\mathbb{R}\)及\(\varepsilon>0\), 注意到 \[ \begin{aligned} \{X\le x\}&=\{X\le x, X_n\le x+\varepsilon\}\cup\{X\le x, X_n>x+\varepsilon\}\\ &\subset\{X_n\le x+\varepsilon\}\cup\{|X_n-X|>\varepsilon\}, \end{aligned} \] 因此 \[ F(x)\le F_n(x+\varepsilon)+\mathbb{P}(|X_n-X|>\varepsilon). \] 由\(X_n\xrightarrow{p}X\)知\(\mathbb{P}(|X_n-X|>\varepsilon)\to 0\), 令\(n\to\infty\)得 \[ F(x)\le\liminf_{n\to\infty}F_n(x+\varepsilon). \] 另外一方面, 有 \[ \begin{aligned} \{X>x\}&=\{X>x, X_n>x-\varepsilon\}\cup\{X>x, X_n\le x-\varepsilon\}\\ &\subset\{X_n>x-\varepsilon\}\cup\{|X_n-X|>\varepsilon\}, \end{aligned} \] 因此 \[ \begin{aligned} &1-F(x)\le 1-F_n(x-\varepsilon)+\mathbb{P}(|X_n-X|>\varepsilon)\\ &\implies F(x)\geq F_n(x-\varepsilon)-\mathbb{P}(|X_n-X|>\varepsilon). \end{aligned} \] 令\(n\to\infty\)得 \[ F(x)\geq\limsup_{n\to\infty}F_n(x-\varepsilon). \] 综上有 \[ \limsup_{n\to\infty}F_n(x-\varepsilon)\le F(x)\le\liminf_{n\to\infty}F_n(x+\varepsilon), \] 若\(x\)是\(F\)的连续点, 令\(\varepsilon\to 0\), 则 \[ F(x)=\lim_{n\to\infty}F_n(x)=\limsup_{n\to\infty}F_n(x)=\liminf_{n\to\infty}F_n(x), \] 这便说明了\(X_n\xrightarrow{d}X\).

命题2(蕴含关系) 设\(c\in\mathbb{R}\), 则\(X_n\xrightarrow{p}c\)当且仅当\(X_n\xrightarrow{d}c\).

证明只需证明当\(X_n\xrightarrow{d}c\)时有\(X_n\xrightarrow{p}c\). 记\(c\)的分布函数 \[ F(x)=\begin{cases} 0, & x<c, \\ 1, & x\geq c, \end{cases} \] 连续点所构成的集合为\(\mathbb{R}\setminus\{c\}\). 设\(F_n\)是\(X_n\)的分布函数, 则 \[ F_n(x)=\mathbb{P}(X_n\le x)\to F(x),\quad\forall x\in\mathbb{R}\setminus\{c\}. \] 对任意的\(\varepsilon>0\), 计算得 \[ \begin{aligned} \mathbb{P}(|X_n-c|>\varepsilon)&=\mathbb{P}(X_n<c-\varepsilon)+\mathbb{P}(X_n>c+\varepsilon)\\ &\le F_n(c-\varepsilon)+1-F_n(c+\varepsilon), \end{aligned} \] 令\(n\to\infty\)可得\(\mathbb{P}(|X_n-c|>\varepsilon)\to 0\), 这便说明了\(X_n\xrightarrow{p}c\).

定理3(Slutsky定理) 设\(X_n\xrightarrow{d}X\), \(Y_n\xrightarrow{p}c\).

\(X_n+Y_n\xrightarrow{d}X+c\);
\(Y_nX_n\xrightarrow{d}cX\).

证明 (1) 设\(F_n, G_n, F\)是\(X_n+Y_n, X_n+c, X+c\)的分布函数, 并且根据\(X_n\xrightarrow{d}X\), 知\(X_n+c\xrightarrow{d}X+c\), 因此\(G_n\xrightarrow{v}F\). 设\(x\)是\(F\)的连续点, 对任意的\(\varepsilon>0\), 一方面, 注意到 \[ \begin{aligned} \{X_n+Y_n>x+\varepsilon\}&=\{X_n+Y_n>x+\varepsilon, |Y_n-c|\le\varepsilon\}\cup\{X_n+Y_n>x+\varepsilon, |Y_n-c|>\varepsilon\}\\ &\subset\{X_n+c>x\}\cup\{|Y_n-c|>\varepsilon\}, \end{aligned} \] 因此 \[ \begin{aligned} &1-F_n(x+\varepsilon)\le 1-G_n(x)+\mathbb{P}(|Y_n-c|>\varepsilon)\\ &\implies G_n(x)\le F_n(x+\varepsilon)+\mathbb{P}(|Y_n-c|>\varepsilon), \end{aligned} \] 令\(n\to\infty\)可得 \[ F(x)=\lim_{n\to\infty}G_n(x)\le\liminf_{n\to\infty}F_n(x+\varepsilon); \] 另外一方面, 注意到 \[ \begin{aligned} \{X_n+Y_n\le x-\varepsilon\}&=\{X_n+Y_n\le x-\varepsilon, |Y_n-c|\le\varepsilon\}\cup\{X_n+Y_n\le x-\varepsilon, |Y_n-c|>\varepsilon\}\\ &\subset\{X_n+c\le x\}\cup\{|Y_n-c|>\varepsilon\}, \end{aligned} \] 因此 \[ G_n(x)\geq F_n(x-\varepsilon)-\mathbb{P}(|Y_n-c|>\varepsilon), \] 令\(n\to\infty\)可得 \[ F(x)=\lim_{n\to\infty}G_n(x)\geq\limsup_{n\to\infty}F_n(x-\varepsilon). \] 考虑到\(x\)是\(F\)的连续点, 令\(\varepsilon\to 0\), 则 \[ F(x)=\lim_{n\to\infty}F_n(x)=\liminf_{n\to\infty}F_n(x)=\limsup_{n\to\infty}F_n(x), \] 这便说明了\(X_n+Y_n\xrightarrow{d}X+c\).

设\(F_n, G_n, F\)是\(Y_nX_n, cX_n, cX\)的分布函数, 并且根据\(X_n\xrightarrow{d}X\), 知\(cX_n\xrightarrow{d}cX\), 因此\(G_n\xrightarrow{v}F\). 设\(x\)是\(F\)的连续点, 对任意的\(\varepsilon>0\), 一方面, 注意到 \[ \begin{aligned} \left\{Y_nX_n>x+\dfrac{x}{c}\cdot\varepsilon\right\}&=\left\{Y_nX_n>x+\dfrac{x}{c}\cdot\varepsilon, |Y_n-c|\le \varepsilon\right\}\\ &\cup\left\{Y_nX_n>x+\dfrac{x}{c}\cdot\varepsilon, |Y_n-c|>\varepsilon\right\}\\ &\subset\{cX_n>x\}\cup\{|Y_n-c|>\varepsilon\}, \end{aligned} \] 因此 \[ \begin{aligned} &1-F_n\left(x+\dfrac{x}{c}\cdot\varepsilon\right)\le 1-G_n(x)+\mathbb{P}(|Y_n-c|>\varepsilon)\\ &\implies G_n(x)\le F_n\left(x+\dfrac{x}{c}\cdot\varepsilon\right)+\mathbb{P}(|Y_n-c|>\varepsilon) \end{aligned} \] 令\(n\to\infty\)可得 \[ F(x)=\lim_{n\to\infty}G_n(x)\le\liminf_{n\to\infty}F_n\left(x+\dfrac{x}{c}\cdot\varepsilon\right); \] 另外一方面, 注意到 \[ \begin{aligned} \left\{Y_nX_n\le x-\dfrac{x}{c}\cdot\varepsilon\right\}&=\left\{Y_nX_n\le x-\dfrac{x}{c}\cdot\varepsilon, |Y_n-c|\le \varepsilon\right\}\\ &\cup\left\{Y_nX_n\le x-\dfrac{x}{c}\cdot\varepsilon, |Y_n-c|>\varepsilon\right\}\\ &\subset\{cX_n\le x\}\cup\{|Y_n-c|>\varepsilon\}, \end{aligned} \] 因此 \[ G_n(x)\geq F_n\left(x-\dfrac{x}{c}\cdot\varepsilon\right)-\mathbb{P}(|Y_n-c|>\varepsilon), \] 令\(n\to\infty\)可得 \[ F(x)=\lim_{n\to\infty}G_n(x)\geq\limsup_{n\to\infty}F_n\left(x-\dfrac{x}{c}\cdot\varepsilon\right). \] 考虑到\(x\)是\(F\)的连续点, 令\(\varepsilon\to 0\), 则 \[ F(x)=\lim_{n\to\infty}F_n(x)=\liminf_{n\to\infty}F_n(x)=\limsup_{n\to\infty}F_n(x), \] 这便说明了\(Y_nX_n\xrightarrow{d}cX\).