228 lines
8.4 KiB
TeX
228 lines
8.4 KiB
TeX
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\documentclass[全部作业]{subfiles}
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\begin{document}
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\chapter{单元作业4}
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\begin{enumerate}
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\item 若二维离散型随机变量$(X,Y)$有联合分布列如下：
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\begin{center}
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% \renewcommand\arraystretch{1.5}
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\begin{tabular}{c|ccc}
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% \hline
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\diagbox{$X$}{$Y$} & $0$ & $1$ & $2$ \\
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\hline
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$0$ & $\frac{1}{2}$ & $\frac{1}{8}$ & $\frac{1}{4}$ \\
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% \hline
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$1$ & $\frac{1}{16}$ & $\frac{1}{16}$ & $0$ \\
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% \hline
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\end{tabular}
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\end{center}
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求$X$与$Y$的边际分布列。
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\begin{proof}[解]
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\begin{zhongwen}
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$X$和$Y$的边际分布列分别为
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\begin{center}
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\renewcommand\arraystretch{1.5}
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\begin{tabular}{c|cc}
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$X$ & $0$ & $1$ \\
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\hline
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$P$ & \makecell{$\dfrac{7}{8}$} & $\dfrac{1}{8}$ \\
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\end{tabular} \qquad
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\begin{tabular}{c|ccc}
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$Y$ & $0$ & $1$ & $2$ \\
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\hline
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$P$ & $\dfrac{9}{16}$ & $\dfrac{3}{16}$ & $\dfrac{1}{4}$ \\
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\end{tabular}
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\end{center}
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\end{zhongwen}
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\end{proof}
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\item 设随机变量$(X,Y)$的概率密度函数为
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$$
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p(x,y)=\begin{cases}
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3x,\quad & \text{若}0<y<x<1; \\
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0,\quad & \text{其他}. \\
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\end{cases}
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$$
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求$X$与$Y$的边际概率密度函数，并判断$X$与$Y$是否互相独立。
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\begin{proof}[解]
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\begin{zhongwen}
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\begin{window}[0,r,
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{\includexopp[0.3]{4.2.1}},
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{}]
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$$
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\begin{aligned}
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p_{X}(x)&=\int_{-\infty}^{+\infty} p(x,y) \mathrm{d}y=\begin{cases}
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\int_{0}^{x} 3x \mathrm{d}y,\quad & 0\leqslant x\leqslant 1 \\
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0,\quad & x<0 \text{或} x>1 \\
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\end{cases}\\
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&=\begin{cases}
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3x^{2},\quad & 0\leqslant x\leqslant 1 \\
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0,\quad & x<0 或x>1 \\
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\end{cases}
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\end{aligned}
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$$
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$$
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\begin{aligned}
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p_{Y}(y)&=\int_{-\infty}^{+\infty} p(x,y) \mathrm{d}x=\begin{cases}
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\int_{y}^{1} 3x \mathrm{d}x, \quad & 0\leqslant y\leqslant 1 \\
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0,\quad & x<0 或 x>1 \\
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\end{cases}\\
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&=\begin{cases}
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\frac{3}{2}-\frac{3}{2}y^{2},\quad & 0\leqslant y\leqslant 1 \\
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0,\quad & y<0 或 y>1 \\
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\end{cases}
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\end{aligned}
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$$
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\end{window}
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\noindent$p(1,1)=3,p_{X}(1)=3,p_{Y}(1)=0$，于是$p(1,1)\neq p_{X}(1)\cdot p_{Y}(1)$，所以$X$和$Y$互相不独立。
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\end{zhongwen}
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\end{proof}
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\item 设随机变量$X$与$Y$相互独立，且$X$服从均匀分布$U(0,1)$，$Y$服从指数分布$\operatorname{Exp}(1)$。求
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\begin{enumerate}
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\item $(X,Y)$的联合概率密度函数$p(x,y)$；
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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p_{X}(x)=\begin{cases}
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1,\quad & x\in [0,1] \\
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0,\quad & x<0 或 x>1 \\
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\end{cases}\quad p_{Y}(y)=\begin{cases}
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e^{-y},\quad & y\geqslant 0 \\
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0,\quad & y< 0 \\
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\end{cases}
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$$
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$$
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p(x,y)=\begin{cases}
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e^{-y},\quad & x\in [0,1], y\geqslant 0 \\
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0,\quad & 其他 \\
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\end{cases}
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$$
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\end{zhongwen}
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\end{proof}
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\item 概率$P(X+Y\leqslant 1)$；
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\begin{proof}[解]
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\begin{zhongwen}
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令$Z=X+Y$，则随机变量$Z$的概率密度函数为
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$$
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p_{Z}(z)=\int_{-\infty}^{+\infty} p_{X}(x)p_{Y}(z-x) \mathrm{d}x=\begin{cases}
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\int_{0}^{1} e^{x-z} \mathrm{d}x ,\quad & z>1\\
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\int_{0}^{z} e^{x-z} \mathrm{d}x,\quad & 0\leqslant z\leqslant 1 \\
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0,\quad & z<0 \\
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\end{cases}=\begin{cases}
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e^{1-z}-e^{-z},\quad & z>1 \\
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1-e^{-z},\quad & 0\leqslant z\leqslant 1 \\
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0,\quad & z<0 \\
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\end{cases}
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$$
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所以
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$$
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P(X+Y\leqslant 1)=\int_{-\infty}^{1} p_{Z}(z) \mathrm{d}z=\int_{0}^{1} (1-e^{-z}) \mathrm{d}z=1+(\frac{1}{e}-1)=\frac{1}{e}
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$$
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\end{zhongwen}
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\end{proof}
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\item 概率$P(X\leqslant Y)$。
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\begin{proof}[解]
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\begin{zhongwen}
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令随机变量$Z=X-Y$，则$Z$的概率密度函数为
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$$
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p_{Z}(z)=\int_{-\infty}^{+\infty} p(x,x-z) \mathrm{d}x=\begin{cases}
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0,\quad & z>1 \\
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\int_{z}^{1} e^{z-x} \mathrm{d}x,\quad & 0\leqslant z\leqslant 1 \\
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\int_{0}^{1} e^{z-x} \mathrm{d}x,\quad & z<0 \\
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\end{cases}=\begin{cases}
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0,\quad & z>1 \\
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1-e^{z-1},\quad & 0\leqslant z\leqslant 1 \\
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e^{z}-e^{z-1},\quad & z<0 \\
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\end{cases}
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$$
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所以
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$$
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P(X\leqslant Y)=P(X-Y\leqslant 0)=\int_{-\infty}^{0} (e^{z}-e^{z-1}) \mathrm{d}z=1-\frac{1}{e}
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$$
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\end{zhongwen}
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\end{proof}
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\end{enumerate}
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\item 设随机变量$(X,Y)$的概率密度函数为
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$$
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p(x,y)=\begin{cases}
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Ae^{-(3x+2y)},\quad & x>0,y>0; \\
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0,\quad & \text{otherwise}. \\
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\end{cases}
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$$
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求：
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\begin{enumerate}
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\item 常数$A$的值；
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&\int_{-\infty}^{+\infty} \int_{-\infty}^{+\infty} p(x,y) \mathrm{d}x \mathrm{d}y=\int_{0}^{+\infty} \int_{0}^{+\infty} Ae^{-(3x+2y)} \mathrm{d}x \mathrm{d}y=\int_{0}^{+\infty} \left. -\frac{1}{3}Ae^{-(3x+2y)} \right|_{0}^{+\infty} \mathrm{d}y\\
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=&\int_{0}^{+\infty} \frac{1}{3}Ae^{-2y} \mathrm{d}y=\left. -\frac{1}{6}Ae^{-2y} \right|_{0}^{+\infty}=\frac{1}{6}A=1
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\end{aligned}
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$$
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所以常数$A$的值为$6$。
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\end{zhongwen}
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\end{proof}
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\item 分布函数$F(x,y)$；
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\begin{proof}[解]
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\begin{zhongwen}
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由题意可知，当$x\leqslant 0$ 或$y\leqslant 0$时，$F(x,y)=0$。
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当$x>0,y>0$时，$
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\displaystyle F(x,y)
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=\int_{0}^{y} \int_{0}^{x} 6e^{-(3u+2v)} \mathrm{d}u \mathrm{d}v
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=\int_{0}^{y} \left. -2e^{-(3u+2v)} \right|_{0}^{x} \mathrm{d}v
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=\int_{0}^{y} \left(2e^{-2v}-2e^{-(3x+2v)}\right) \mathrm{d}v
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=\left. -e^{-2v} \right|_{0}^{y}+\left. e^{-(3x+2v)} \right|_{0}^{y}
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=-e^{-2y}+1+e^{-(3x+2y)}-e^{-3x}
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=e^{-(3x+2y)}-e^{-3x}-e^{-2y}+1
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$
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所以
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$$
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\begin{aligned}
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F(x,y)=\begin{cases}
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e^{-(3x+2y)}-e^{-3x}-e^{-2y}+1,\quad & x>0,y>0 \\
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0,\quad & \text{otherwise} \\
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\end{cases}
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\item 概率$P(-2<X\leqslant 2,-3<Y\leqslant 3)$。
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\begin{proof}[解]
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\begin{zhongwen}
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$$
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\begin{aligned}
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&P(-2<X\leqslant 2,-3<Y\leqslant 3)=P(X\leqslant 2,Y\leqslant 3)=F(2,3)\\
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=&e^{-(3\times 2+2\times 3)}-e^{-3\times 2}-e^{-2\times 3}+1 \\
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=&1 +e^{-12} - 2e^{-6}
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\end{aligned}
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$$
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\end{zhongwen}
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\end{proof}
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\end{enumerate}
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\end{enumerate}
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\end{document}
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