Section05_由(X,Y)的分布，求Z=g(X,Y)的分布

$(X,Y)$ 为二维离散型

计算方法

• $\mathbb{P}\{X=x_{i},Y = y_{j}\}= p_{ij},\quad i,j=1,2,\cdots$，则 $\mathbb{P}\{Z=g(x_{i},y_{j})\} =p_{ij}$，若 $g(x_{i},y_{j})$ 有相同值，应合并

例题

1. 随机变量 $X,Y$ 的分布律如下所示，求$Z=X+Y, Z=\max\{X,Y\}$ 的分布律 $$\begin{array}{c|cc} & Y = 0 & Y=1 \\ \hline X=0 & 1/8 & 1/8 \\ X=1 & 1/8 & 5/8 \\ \end{array}$$
   • 解 $$\begin{array}{ll} & Z = X+Y \sim \left( \begin{matrix} 0 & 1 & 2 \\ \frac{1}{8} & \frac{1}{4} & \frac{5}{8} \end{matrix} \right) \\ & Z = \max\{X,Y\}\sim \left( \begin{matrix} 0 & 1 \\ \frac{1}{8} & \frac{7}{8} \end{matrix} \right) \end{array}$$

$(X,Y)$ 为二维连续型

三种情况

• $(X,Y)\sim f(x,y)$，则 $Z=g(X,Y)$ 可能为
  1. 离散型 (求分布律)
  2. 连续型 (先求$F_{Z}(z)$，再求$f_{Z}(z)$)
  3. 混合型 (仅求$F_{Z}(z)$)

计算方法

分布函数法

$$\begin{split} F_{Z}(z) & = \mathbb{P}\{Z\le z\} = \mathbb{P}\{g(X,Y)\le z\} \\ & = \iint\limits_{g(x,y)\le z}f(x,y)\cdot d\sigma \Rightarrow f_{Z}(z) \end{split}$$

• ==Leibniz integral rule==

  Want to know more? $$\color{#D0104C} \begin{split} &\frac{d}{dz}\int_{\alpha(z)}^{\beta(z)} g(x, z)\cdot dx = \\ &\int_{\alpha(z)}^{\beta(z)}g'_{z}(x,z)\cdot dx + [g(\beta(z),z)\beta'(z) + g(\alpha(z),z)\alpha'(z)] \end{split}$$

$U = \max\{X,Y\}$ 与 $V = \min\{X,Y\}$ 的分布

$$\begin{split} & F_{U}(x) = \mathbb{P}\{U\le x\} = \mathbb{P}\{\max(X,Y)\le x\} = \mathbb{P}\{X\le x, Y\le x\} \\ & \xlongequal{\text{若}X,Y\text{独立}} \mathbb{P}\{X\le x\} \mathbb{P}\{Y\le x\} = F_{X}(x)F_{Y}(x) \\ & \xlongequal{\text{若}X,Y\text{同分布}} \mathbb{P}\{X\le x\} \mathbb{P}\{X\le x\} = F_{X}^{2}(x)\\ \\\\ & F_{V}(x) = \mathbb{P}\{V\le x\} = \mathbb{P}\{\min(X,Y)\le x\} = 1 - \mathbb{P}\{X > x, Y > x\} \\ & \xlongequal{\text{若}X,Y\text{独立}} 1 - [1- \mathbb{P}\{X\le x\}][1- \mathbb{P}\{Y\le x\}] = 1 - [1-F_{X}(x)][1- F_{Y}(x)] \\ & \xlongequal{\text{若}X,Y\text{同分布}} 1 - [1-F_{X}(x)]^{2} \end{split}$$

例题

1. 设连续型随机变量 $(X,Y)$ 的概率密度为 $f(x,y) = \begin{cases} 1, & 0 \le x\le 1, 0\le y\le 1 \\ 0, & \text{其他} \end{cases}$，求
   1. $Z=X+Y$ 的概率密度 $f_{Z}(z)$
   2. $Z = \vert X-Y \vert$ 的概率密度 $f_{Z}(z)$

      

$$\begin{array}{ll} & F_{Z}(z) = \mathbb{P}\{Z\le z\} = \mathbb{P}\{X+Y\le z\},\quad z\in \mathbb{R} \\ \because & X\in (0,1), Y\in(0,1) \\ \therefore & Z = X+Y \in (0,2)\\ \therefore & z < 0 \text{时, } F_{Z}(z) = 0\\ & 0 \le z < 1 \text{时, } \displaystyle F_{Z}(z) = \iint\limits_{x+y <z}f(x,y) \cdot d\sigma \\ & \displaystyle = \int_{0}^{z}dx \int_{0}^{z-x}1\cdot dy \\ & 1 \le z < 2 \text{时, } \displaystyle F_{Z}(z) = \iint\limits_{x+y < z}f(x,y) \cdot d\sigma \\ & \displaystyle = 1 - \int_{z-1}^{1}dx \int_{z-x}^{1}1\cdot dy\\ \therefore & f_{Z}(z) = \begin{cases} z, & 0< z < 1 \\ 2-z,& 1 < z < 2\\ 0, & \text{其他} \end{cases} \\ \end{array}$$

$$\begin{array}{ll} & F_{Z}(z) = \mathbb{P}\{Z\le z\} = \mathbb{P}\{\vert X- Y \vert \le z\} \\ & = \mathbb{P}\{-z \le X-Y \le z\} \quad z\in \mathbb{R} \\ \because & X \in (0,1), Y \in (0,1) \\ \therefore & Z = \vert X-Y \vert \in (0,1) \\ \therefore & z < 0 \text{时，} F_{Z}(z) = 0 \\ & \displaystyle 0<z<1 \text{时, } F_{Z}(z) = \iint\limits_{\vert X+Y \vert < z} f(x,y)\cdot d\sigma \\ & = \displaystyle 1 - \iint\limits_{x - y > z} f(x,y)\cdot d\sigma - \iint\limits_{y-x > z} f(x,y)\cdot d\sigma \\ & \displaystyle = 1 - \int_{z}^{1}dx \int_{0}^{x-z}1\cdot dy - \int_{z}^{1}dy \int_{0}^{y- z}1\cdot dx \\ & \displaystyle = 1 - 2\int_{z}^{1}dx \int_{0}^{x-z}1\cdot dy \\ & z > 1 \text{时, } F_{Z}(z) =1 \\ \therefore & f_{Z}(z) = \begin{cases} 2(1-z), & 0 < z < 1 \\ 0, & \text{其他} \end{cases} \\ \end{array}$$

1. 设随机变量 $X,Y$ 相互独立，且 $X\sim \mathcal{N}(0,1), Y\sim \mathcal{N}(0,1)$，求 $Z = X+Y$ 的密度函数 $f_{Z}(z)$

   

$$\begin{array}{ll} & \mathbb{P}\{Z\le z\} = \mathbb{P}\{X+Y\le z\}\quad z\in \mathbb{R} \\ \because & X\sim \mathcal{N}(0,1), Y\sim \mathcal{N}(0,1) \\ \therefore & X\in (-\infty,+\infty), Y\in(-\infty,+\infty) \\ \therefore & Z = X+Y \in (-\infty,+\infty) \\ \because & X,Y \text{独立} \\ \therefore & f(x,y) = f_{X}(x)f_{Y}(y) = \frac{1}{2\pi}e^{-\frac{x^{2}+y^{2}}{2}} \\ \because & \displaystyle F_{Z}(z) = \iint\limits_{x+y \le z}f(x,y)\cdot d\sigma \\ \therefore & \displaystyle F_{Z}(z) = \int_{-\infty}^{+\infty} dx \int_{-\infty}^{z - x}\frac{1}{2\pi}e^{-\frac{x^{2}+y^{2}}{2}}\cdot dy\\ \therefore & \displaystyle f_{Z}(z) = \int_{-\infty}^{+\infty} \frac{1}{2\pi}e^{- \frac{x^{2} + (z-x)^{2}}{2}} \cdot dx \\ & \displaystyle = \int_{-\infty}^{+\infty}\frac{1}{2\pi}e^{-\frac{2x^{2} - 2zx + z^{2}}{2}}\cdot dx = e^{-\frac{\frac{1}{2}z^{2}}{2}} \int_{-\infty}^{+\infty}\frac{1}{2\pi}e^{\frac{2(x-\frac{1}{2}z)^{2}}{2}} \\ & \displaystyle = \frac{1}{\sqrt{\pi} 2}e^{-\frac{z^{2}}{4}} \int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi} \sqrt{\frac{1}{2}}}e^{-\frac{(x-\frac{1}{2}z)^{2}}{2\times \frac{1}{2}}} \\ & \displaystyle = \frac{1}{\sqrt{2\pi}\sqrt{2}} e^{-\frac{z^{2}}{2\times \sqrt{2}^{2}}}\\ \therefore & Z\sim \mathcal{N}(0, 2) \\ \end{array}$$

• ==注== 一般地 $X\sim \mathcal{N}(\mu_{1},\sigma_{1}^{2}),Y\sim \mathcal{N}(\mu_{2},\sigma_{2}^{2})$，$X,Y$ 独立，则 $Z=aX+bY+c\sim \mathcal{N}(a\mu_{1}+b\mu_{2}+c, a^{2}\sigma_{1}^{2} + b^{2}\sigma_{2}^{2})$
• 设 $X,Y$ 相互独立，且 $X\sim E(1), Y\sim E(2)$，求 $Z = \min\{X,Y\}$ 的概率密度 $f_{Z}(x)$ $$\begin{array}{ll} \because & X\sim E(1), Y \sim E(2) \\ \therefore & f_{X}(x) = \begin{cases} e^{-x}, & x > 0 \\ 0 , & x \le 0 \end{cases} \\ & f_{Y}(y) = \begin{cases} 2e^{-2y}, & y > 0 \\ 0 , & y \le 0 \end{cases} \\ \because & F_{Z}(x) = \mathbb{P}\{Z \le x\} = \mathbb{P}\{\min(X,Y)\le x\} \\ & = 1 - \mathbb{P}\{X>x, Y>x\} \\ \because & X,Y \text{独立} \\ \therefore & F_{Z}(z) = 1 - [1-F_{X}(x)][1-F_{Y}(x)] \\ & = 1 - e^{-x}e^{-2x} = 1- e^{-3x} \\ \therefore & f_{Z}(x) = \begin{cases} 3 e^{-3x}, & x > 0 \\ 0, & x \le 0 \end{cases} \\ \therefore & Z\sim E(3) \\ \end{array}$$
• ==注== 一般地 $X\sim E(\lambda_{1}), Y\sim E(\lambda_{2})$，且 $X,Y$ 独立，则 $Z=\min\{X,Y\} \sim E(\lambda_{1}+\lambda_{2})$