156 lines
8.5 KiB
TeX
156 lines
8.5 KiB
TeX
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\documentclass[全部作业]{subfiles}
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\setlength{\textheight}{210em}
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\setlength{\paperheight}{210em}
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\begin{document}
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\chapter{单元作业6}
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\begin{enumerate}
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\item 证明随机变量$X$的特征函数是实值函数当且仅当$X$与$-X$同分布。
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\begin{proof}
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\begin{zhongwen}
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设$X$的特征函数为$f_{X}(t)$，则
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$$
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\begin{aligned}
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X的特征函数是实值函数 \iff& f_{X}(t)=\overline{f_{X}(t)} \\
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\iff& f_{X}(t)=f_{X}(-t) \\
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\iff& Ee^{itX}=Ee^{i(-t)X} \\
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\iff& Ee^{itX}=Ee^{it(-X)} \\
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\iff& f_{X}(t)=f_{-X}(t) \\
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\iff& X与-X同分布 \\
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 设随机变量$X$的特征函数为$f(t)=\left( \frac{2-it}{2} \right) ^{-2}$，求$EX$和$\operatorname{Var}X$。
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\begin{proof}[解]
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\begin{zhongwen}
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% 根据唯一性定理的推论，可知$X$的概率密度函数为
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% $$
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% p(x)=\frac{1}{2\pi }\int_{-\infty}^{+\infty} e^{-itx}\left( \frac{2-it}{2} \right) ^{-2} \mathrm{d}t
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% $$
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根据特征函数的性质，可知
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$$
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iEX = \left. f'(t) \right|_{t=0}=\left. \left( -\frac{i}{2}\cdot (-2)\left( \frac{2-it}{2} \right) ^{-3} \right) \right|_{t=0}=i\cdot 1^{-3}=i
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$$
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$$
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-EX^{2}=i^{2}EX^{2}=\left. f''(t) \right|_{t=0}=\left.\left( i\cdot \left( -\frac{i}{2} \right) \cdot (-3)\left( \frac{2-it}{2} \right) ^{-4} \right) \right|_{t=0}=\left( -\frac{3}{2} \right) \cdot 1^{-4}=-\frac{3}{2}
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$$
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所以有
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$$
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EX=1, \qquad EX^{2}=\frac{3}{2}, \qquad \operatorname{Var}X=EX^{2}-(EX)^{2}=\frac{3}{2}-1^{2}=\frac{1}{2}
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$$
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\end{zhongwen}
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\end{proof}
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\item 设$X_i$独立同分布，且$X_i\sim \operatorname{Exp}(\lambda ),i=1,2, \ldots ,n$。试用特征函数的方法证明：
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$$
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Y_n=\sum_{i=1}^{n}X_i\sim \operatorname{Ga}(n,\lambda )
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$$
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\begin{proof}
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\begin{zhongwen}
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根据题意可知$X_i$的概率密度函数为$p(x)=\begin{cases}
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\lambda e^{-\lambda x},\quad & x>0 \\
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0,\quad & x\leqslant 0 \\
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\end{cases}$。根据特征函数的定义，计算$X_i$的特征函数$f(t)$：
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$$
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\begin{aligned}
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f(t)&=Ee^{itX_{i}}=\int_{-\infty}^{+\infty} e^{itx}p(x) \mathrm{d}x=\int_{0}^{+\infty} e^{itx}\lambda e^{-\lambda x} \mathrm{d}x=\frac{\lambda}{it-\lambda }\int_{0}^{+\infty} e^{(it-\lambda )x} \mathrm{d}(it-\lambda )x \\
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&=\frac{\lambda}{it-\lambda }\cdot \left. e^{(it-\lambda )x} \right|_{0}^{+\infty}=\frac{\lambda }{it-\lambda } \cdot \left( \lim_{x \to +\infty}e^{itx-\lambda x}-1 \right)=\frac{\lambda}{it-\lambda }\cdot \left( \lim_{x \to +\infty}e^{-\lambda x}e^{itx} -1 \right) \\
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\end{aligned}
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$$
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对于$\displaystyle \lim_{x \to +\infty}e^{-\lambda x}e^{itx}$，设$a=e^{-\lambda x}e^{itx}$，将其看作复数的模长辐角表示法，则$a$的模长为$e^{-\lambda x}$，辐角为$tx$。由于$\lambda >0$，所以当$x \to +\infty$时，$a$的模长$e^{-\lambda x} \to 0$（辐角$tx \to +\infty$，但不重要），因此$\displaystyle \lim_{x \to +\infty}e^{-\lambda x}e^{itx}=\lim_{x \to +\infty}a=0$。
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于是$\displaystyle f(t)=\frac{\lambda}{it-\lambda }\cdot (0-1)=\frac{\lambda }{\lambda -it}$。
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由于$X_i$相互独立，所以$Y_n$的特征函数为：
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$$
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g(t)=\left[ f(t) \right] ^{n}=\left( \frac{\lambda }{\lambda -it} \right) ^{n}
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$$
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再计算$Y\sim \operatorname{Ga}(n,\lambda )$的特征函数：
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$$
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\begin{aligned}
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h(t)&=Ee^{itY}=\int_{-\infty}^{+\infty} e^{itx}p(x) \mathrm{d}x=\int_{0}^{+\infty} e^{itx} \frac{\lambda ^{n}}{\Gamma (n)}x^{n-1}e^{-\lambda x} \mathrm{d}x \\
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&=\frac{\lambda ^{n}}{(\lambda -it)^{n}}\int_{0}^{+\infty} \frac{(\lambda -it)^{n}}{\Gamma (n)}x^{n-1}e^{(it-\lambda )x} \mathrm{d}x \\
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\end{aligned}
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$$
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注意到$\displaystyle \frac{(\lambda -it)^{n}}{\Gamma (n)}x^{n-1}e^{(it-\lambda )}$（零延拓后）为$\operatorname{Ga}(n, \lambda -it)$的概率密度函数，根据分布函数的正则性，$\displaystyle \int_{0}^{+\infty} \frac{(\lambda -it)^{n}}{\Gamma (n)}x^{n-1}e^{(it-\lambda )} \mathrm{d}x=1$，所以
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$$
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h(t)=\frac{\lambda ^{n}}{(\lambda -it)^{n}}=\left( \frac{\lambda }{\lambda -it} \right) ^{n}=g(t)
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$$
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根据特征函数的唯一性，$Y_n$与$Y$同分布，即
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$$
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Y_n=\sum_{i=1}^{n}X_i\sim \operatorname{Ga}(n,\lambda )
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$$
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\end{zhongwen}
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\end{proof}
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\item 设$X_n\xrightarrow{P}X,X_n\xrightarrow{P}Y$。证明$P(X=Y)=1$。
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\begin{proof}
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\begin{zhongwen}
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根据依概率收敛的运算性质，有
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$$
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0=X_n-X_n\xrightarrow{P}X-Y
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$$
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即对任意的$\varepsilon>0$，有
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$$
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\lim_{n \to \infty} P(\left\vert 0-(X-Y) \right\vert \geqslant \varepsilon)=\lim_{n \to \infty}P(\left\vert X-Y \right\vert \geqslant \varepsilon)=P(\left\vert X-Y \right\vert \geqslant \varepsilon)=0
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$$
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令$\varepsilon \to 0$，则根据概率的下连续性，可得
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$$
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P(\left\vert X-Y \right\vert >0)=0
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$$
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由于$\left\vert X-Y \right\vert \geqslant 0$，所以
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$$
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P(\left\vert X-Y \right\vert =0)=1-P(\left\vert X-Y \right\vert >0)=1
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$$
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即$P(X=Y)=1$。
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\end{zhongwen}
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\end{proof}
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\item 证明$\displaystyle \lim_{n \to \infty}E\left( \frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert } \right) =0$当且仅当$X_n\xrightarrow{P}0$。
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\begin{proof}
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\begin{zhongwen}
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对任意的$\varepsilon>0$，$1=1_{\{ \left\vert X_n \right\vert \leqslant \varepsilon \}}+1_{\{ \left\vert X_n \right\vert >\varepsilon \}}$。所以有如下不等式：
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$$
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\begin{aligned}
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\frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert }&=\frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert }1_{\{ \left\vert X_n \right\vert \leqslant \varepsilon \}}+\frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert }1_{\{ \left\vert X_n \right\vert >\varepsilon \}} \\
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&\leqslant \left\vert X_n \right\vert 1_{\{ \left\vert X_n \right\vert \leqslant \varepsilon \}}+1_{\{ \left\vert X_n \right\vert >\varepsilon \}}\leqslant \varepsilon+1_{\{ \left\vert X_n \right\vert >\varepsilon \}} \\
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\end{aligned}
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$$
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等式两边同时取期望，则
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$$
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E\left( \frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert } \right) \leqslant E(\varepsilon+1_{\{ \left\vert X_n \right\vert >\varepsilon \}})=\varepsilon+E(1_{\{ \left\vert X_n \right\vert >\varepsilon \}})
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$$
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若$X_n\xrightarrow{P}0$，则$\displaystyle \lim_{n \to \infty}P(\left\vert X_n \right\vert >\varepsilon)=0$，所以$\displaystyle \lim_{n \to \infty}E(1_{\{ \left\vert X_n \right\vert >\varepsilon \}})=0$。
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于是上述不等式两边同时取$n \to 0$时的极限可得
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$$
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\lim_{n \to \infty}E\left( \frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert } \right) \leqslant \varepsilon
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$$
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所以$\displaystyle \lim_{n \to \infty}E\left( \frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert } \right) =0$。
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若$\displaystyle \lim_{n \to \infty}E\left( \frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert } \right) =0$，这里使用和书上不同的方法。根据马尔可夫不等式的一般形式，有
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$$
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P(\left\vert X_n \right\vert >\varepsilon)\leqslant \frac{E\left( \frac{\left\vert X_n \right\vert }{1+\left\vert X_n \right\vert } \right) }{\frac{\varepsilon}{1+\varepsilon}}
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$$
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不等式两边同时取$n \to 0$时的极限可得$\displaystyle \lim_{n \to \infty}P(\left\vert X_n \right\vert >\varepsilon)=0$，即$X_n\xrightarrow{P}0$。
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\end{zhongwen}
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\end{proof}
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\end{enumerate}
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\end{document}
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