【概率论基础进阶】多维随机变量及其分布-两个随机变量函数Z=g(X,Y)的分布

$X,Y$均为离散型随机变量

与一维离散型类似（画表，加和）

$X,Y$均为连续型随机变量

可用公式
$$F_Z(z)=P\left\{Z\le z\right\}=P\left\{g(X,Y)\le z\right\}=\underset{g(x,y)\le z}{\iint }f(x,y)dxdy$$

例1：$Z=X+Y$的分布
设$(X,Y)$的概率密度为$f(x,y)$，则$Z=X+Y$的分布函数
$$\begin{array}{rl}{F}_Z(z)& =P\left\{X+Y\le z\right\}\\ & =\underset{x+y\le z}{\iint }f(x,y)dxdy\\ & ={\int }_{-\infty }^{+\infty }dx{\int }_{-\infty }^{z-x}f(x,y)dy或{\int }_{-\infty }^{+\infty }dy{\int }_{-\infty }^{z-y}f(x,y)dx\end{array}$$
由此可以得到$Z=X+Y$的概率密度为
$$f_Z(z)={\int }_{-\infty }^{+\infty }f(x,z-x)dx={\int }_{-\infty }^{+\infty }f(z-y,y)dy$$

这里说实话我不知道为啥$(F_Z(z){)}^{\prime }=f_Z(z)$，后续尽量补上

特别是当$X$和$Y$相互独立时，$f(x,y)=f_X(x)\cdot f_Y(y)$，则
$$f_Z(z)={\int }_{-\infty }^{+\infty }{f}_X(x)f_Y(z-x)dx={\int }_{-\infty }^{+\infty }{f}_X(z-y)f_Y(y)dy$$
这个公式称为卷积公式，记为
$$f_Z=f_X\ast f_Y$$

$X$为离散型随机变量，$Y$为连续型随机变量

一般对离散型随机变量$X$的各种可能用全概率公式把它们展开

$Y$为连续型，$Z=g(X,Y)$，则
$$\begin{array}{rl}{F}_Z(z)& =P\left\{Z\le z\right\}\\ & =P\left\{g(X,Y)\le z\right\}\\ & =\sum _{i}P\left\{X=x_i\right\}P\left\{g(X,Y)\le z|X=x_i\right\}\\ & 即P(x_i)=p_i\\ & =\sum _i{p}_{i}P\left\{g(X,Y)\le z|X=x_i\right\}\end{array}$$

$Z=\max \left\{X,Y\right\}$的分布

由$\max \left\{X,Y\right\}$不大于$z$等价于$X$和$Y$都不大于$z$，即
$$\begin{array}{rl}{F}_Z(z)& =P\left\{Z\le z\right\}\\ & =P\left\{X\le z,Y\le z\right\}\\ & =P\left\{X\le z\right\}P\left\{Y\le z\right\}\\ & =F_X(z)F_Y(z)\end{array}$$

$Z=\min \left\{X,Y\right\}$的分布

$$\begin{array}{rl}{F}_Z(z)& =P\left\{Z\le z\right\}\\ & =1-P\left\{Z>z\right\}\\ & =1-P\left\{X>z,Y>z\right\}\\ & =1-P\left\{X>z\right\}P\left\{Y>z\right\}\\ & =1-(1-F_X(z))(1-F_Y(z))\\ & =F_X(z)+F_Y(z)-F_X(z)F_Y(z)\\ & 也可\\ F_Z(z)& =P\left\{Z\le z\right\}\\ & =P\left\{\min \left\{X,Y\right\}\le z\right\}\\ & =P\left\{X\le z\cup Y\le z\right\}\\ & =P\left\{X\le z\right\}+P\left\{Y\le z\right\}-P\left\{X\le z,Y\le z\right\}\\ & =1-(1-F_X(z))(1-F_Y(z))\\ & =F_X(z)+F_Y(z)-F_X(z)F_Y(z)\end{array}$$

以上结果可以推广至$n$个相互独立的随机变量

例2：设随机变量$X$与$Y$相互独立，$X\sim E({\lambda }_1),Y\sim E({\lambda }_2),{\lambda }_1,{\lambda }_2>0$，令$Z=\min \left\{X,Y\right\}$，求$Z$的概率密度函数$f_Z(z)$

由公式
$$F_Z(x)=F_X(z)+F_Y(z)-F_X(z)F_Y(z)$$
当$z>0$时
$$\begin{array}{rl}{f}_Z(z)=F_Z^{\prime }(x)& =f_X(z)+f_Y(z)-f_X(z)F_Y(z)-F_X(z)f_Y(z)\\ & =({\lambda }_1+{\lambda }_2)e^{-({\lambda }_1+{\lambda }_2)z}\end{array}$$
当$z\le 0$时
$$f_Z(z)=0$$
故
$$Z\sim E({\lambda }_1+{\lambda }_2)$$

也可用
$$\begin{array}{rl}{F}_Z(z)& =1-\left\{P\left\{X>z\right\}P\left\{Y>z\right\}\right\}\\ & =\{\begin{array}{rlr}& 1-e^{{\lambda }_{1}x}{e}^{-{\lambda }_{2}x}=1-e^{-({\lambda }_1+{\lambda }_2)x}& z>0\\ & 0& z\le 0\end{array}\\ f_Z(z)& =F_Z^{\prime }(z)=\{\begin{array}{rlr}& ({\lambda }_1+{\lambda }_2)e^{-({\lambda }_1+{\lambda }_2)x}& z>0\\ & 0& z\le 0\end{array}\end{array}$$
故
$$Z\sim E({\lambda }_1+{\lambda }_2)$$

例3：设二维随机变量$(X,Y)$的概率密度为
$$f(x,y)=\{\begin{array}{rlr}& 2-x-y& 0<x<1,0<y<1\\ & 0& 其他\end{array}$$

• 求$P\left\{X>2Y\right\}$
• 求$Z=X+Y$的概率密度$f_Z(z)$

$$\begin{array}{rl}P\left\{X>2Y\right\}& =\underset{x>2y}{\iint }f(x,y)dxddy\\ & =\underset{D}{\iint }(2-x-y)dxdy\\ & ={\int }_0^{1}dx{\int }_0^{\frac{1}{2}x}(2-x-y)dy\\ & =\frac{7}{24}\end{array}$$
有
$$f_Z(z)={\int }_{-\infty }^{+\infty }f(x,z-x)dx$$
先考虑被积函数$f(x,z-x)$中第一个自变量$x$的变化范围，只有当$0<x<1$时，$f(x,z-x)$才不等于$0$，因此被积函数上下限最大范围为$(0,1)$
再考虑被积函数$f(x,z-x)$中第二个自变量$z-x$的变化范围，只有当$0<z-x<1$时，$f(x,z-x)$不为$0$，因此需要对$z$分区间讨论
当$z\le 0$时，由于$0<x<1$，故$z-x<0$，所以
$$f_Z(z)=0$$
当$0<z\le 1$时，
$$f_Z(z)={\int }_0^z(2-z)dx=2z-z^2$$
当$1<z\le 2$
$$f_Z(z)={\int }_{z-1}^1(2-z)dx=4-4z+z^2$$
当$2<z$时，由于$0<x<1$，故$z-x>0$，所以
$$f_Z(z)=0$$
因此
$$f_Z(z)=\{\begin{array}{rlr}& 2z-z^2& 0<z\le 1\\ & 4-4z+z^2& 1<z\le 2\\ & 0& 其他\end{array}$$

这里直接用最基本的方法比卷积公式简单

$$F_Z(z)=P\left\{Z\le z\right\}=P\left\{X+Y\le z\right\}=\underset{x+y\le z}{\iint }f(x,y)dxdy$$
当$z\le 0$时，
$$F_Z(z)=0$$
当$0<z\le 1$时，
$$\begin{array}{rl}{F}_Z(z)& =\underset{D_1}{\iint }f(x,y)dxdy\\ & ={\int }_0^{x}dx{\int }_0^{z-x}(2-x-y)dy\\ & =z^2-\frac{1}{3}{z}^3\end{array}$$
当$1<z\le 2$时
$$\begin{array}{rl}{F}_Z(z)& =1-\underset{x+y>z}{\iint }f(x,y)dxdy\\ & =1-\underset{D_2}{\iint }f(x,y)dxdy\\ & =1-{\int }_{z-1}^{1}dx{\int }_{z-x}^1(2-x-y)dy\\ & =\frac{1}{3}{z}^3-2z^2+4z-\frac{5}{3}\end{array}$$
当$2<z$时，
$$F_Z(z)=1$$
所以
$$f_Z(z)=F_Z^{\prime }(z)=\{\begin{array}{rlr}& 2z-z^2& 0<z\le 1\\ & 4-4z+z^2& 1<z\le 2\\ & 0& 其他\end{array}$$

例4：设二维随机变量$(X,Y)$服从正态分布$N(1,0;1,1;0)$，则$P\left\{XY-Y<0\right\}=()$

由于$\rho =0$，因此$X$与$Y$相互独立，且$X\sim N(1,1),Y\sim N(0,1)$，也就有$(X-1)\sim N(0,1)$与$Y$相互独立
根据正态分布密度的对称性，有
$$P\left\{X-1<0\right\}=P\left\{X-1>0\right\}=P\left\{Y<0\right\}=P\left\{Y>0\right\}=\frac{1}{2}$$
因此
$$\begin{array}{rl}P\left\{XY-Y<0\right\}& =P\left\{(X-1)Y<0\right\}\\ & =P\left\{X-1<0,Y>0\right\}+P\left\{X-1>0,Y<0\right\}\\ & =P\left\{X-1<0\right\}P\left\{Y>0\right\}+P\left\{X-1>0\right\}P\left\{Y<0\right\}\\ & =\frac{1}{2}\end{array}$$

CSDN话题挑战赛第2期
参赛话题：学习笔记