Section04_已知X分布，求X=g(X)的分布

$X$ 为离散型

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

例题

1. 已知随机变量$X$的分布律 $X\sim \left( \begin{matrix} -1 & 0 & 1 & 2 \\ 0.1 & 0.2 & 0.3 &0.4 \end{matrix} \right)$，求 $X^{2}$ 和 $\max\{X,1\}$ 的分布律 $$\begin{array}{ll} X^{2} \sim \left( \begin{matrix} 0 & 1 & 4 \\ 0.2& 0.4 & 0.4 \end{matrix} \right) \\ \max \{X, 1\} \sim \left( \begin{matrix} 1 & 2 \\ 0.6 & 0.4 \end{matrix} \right) \end{array}$$

$X$ 为连续型 $X\sim f_{X}(x)$

• $y= g(X) \text{可能为} \begin{cases} \text{离散型} & \text{求分布律} \\ \text{连续型} & \text{先求}f_{y}(y)\text{再求}f_{y}(y) \\ \text{混合型} & \text{只求}f_{y}(y) \end{cases}$
• 分布函数法 $y = g(X)$ 的分布函数 $$f_{y}(y) = \mathbb{P}\{Y \le y\} = \mathbb{P}\{g(X) \le y\} = \int_{g(x)\le y}^{} f_{X}(t)\cdot dt$$
  • 关键在于寻找 $g(x)\le y$ 的区间
  • 若 $Y$ 为连续型，则 $\displaystyle f_{Y}(y) = F'_{Y}(y)$，可
    • 先积分后求导
    • 直接用变先积分求导
      • $\displaystyle \frac{d}{dx}\int_{a}^{x}f(t)\cdot dt = f(x)$
      • $\displaystyle \frac{d}{dx}\int_{a}^{\varphi(x)}f(t)\cdot dt = \varphi'(x)f[\varphi(x)]$
      • $\displaystyle \frac{d}{dx}\int_{\psi (x)}^{\varphi(x)}f(t)\cdot dt = \varphi'(x)f[\varphi(x)] - \psi'(x)f[\psi(x)]$
• 解题步骤
  1. 讨论 $Y$ 的取值，分段讨论
  2. 找出 $g(x)\le y$ 对应的 $x$ 的区间
  3. 先积后导或采用变限积分的求导

例题

1. $X\sim U(-1,2), y = \text{sgn }x = \begin{cases} -1 & x<0 \\ 0, & x=0 \\ 1, & x>0 \\ \end{cases}$，则 $Y$ 的分布律为 $$Y\sim \left( \begin{matrix} -1 & 0 & 1 \\ \frac{1}{3} & 0 & \frac{2}{3} \end{matrix} \right)$$
2. 设随机变量 $X$ 的分布函数为 $F_{X}(x)$，则 $Y = 2X+1$ 的分布函数 $F_{Y}(y)$ 为 $$\begin{array}{ll} & F_{Y}(y) = \mathbb{P}\{Y \le y\} = \mathbb{P}\{2X + 1\le y\} = \mathbb{P}\{X \le \frac{y-1}{2}\} \\ \therefore & F_{Y}(y) = F_{X}(\frac{y-1}{2}) \\ \end{array}$$
3. 设 $X\sim \mathcal{N}(\mu,\sigma^{2})$，证明 $Y = \frac{X-\mu}{\sigma}\sim \mathcal{N}(0,1)$ $$\begin{array}{ll} \because & X\sim \mathcal{N}(\mu,\sigma^{2}) \\ \therefore & X\sim f(x) = \frac{1}{\sqrt{2\pi}\sigma} e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}, X \text{的取值}(-\infty,+\infty)\\ \because & Y = \frac{X-\mu}{\sigma} \\ \therefore & Y \text{的取值范围} (-\infty,+\infty) \\ \because & F_{Y}(y) = \mathbb{P}\{Y \le y\} = \mathbb{P}\{\frac{X-\mu}{\sigma}\le y\} = \mathbb{P}\{X\le y\sigma+\mu\} \\ \therefore &\displaystyle F_{Y}(y) = F_{X}(y\sigma + \mu) = \int_{-\infty}^{y\sigma + \mu} \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(y\sigma)^{2}}{2\sigma}}\\ \because & f_{Y}(y) = F'_{Y}(y) \\ \therefore & f_{Y}(y) = \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{y^{2}}{2}}\cdot \sigma = \frac{1}{\sqrt{2\pi}}e^{-\frac{y^{2}}{2}} \\ \therefore & Y \sim \mathcal{N}(0,1) \\ \end{array}$$
4. 设随机变量 $X$ 的概率密度为 $f(x)= \begin{cases} \frac{3}{2}x^{2}, & -1 < x < 1 \\ 0, & \text{其他} \end{cases}$ 求随机变量 $Y = X^{2} + 1$ 的概率密度函数 $f(y)$ $$\begin{array}{ll} \because & X \in (-1, 1) \\ \therefore & Y = X^{2} + 1\in [1,2) \\\\ & F(y) = \mathbb{P}\{Y\le y\} = \mathbb{P}\{X^{2} + 1 \le y\} = \mathbb{P}\{-\sqrt{y-1} \le X\le \sqrt{y-1}\} \\ \therefore & F(y) = \displaystyle \int_{x^{2}\le y}^{} f(x)\cdot dx = \int_{-\sqrt{y-1}}^{\sqrt{y-1}}f(x)\cdot dx \\ \therefore &\displaystyle \text{When y}\in [1,2), F(y) = \int_{-\sqrt{y-1}}^{\sqrt{y-1}}\frac{3}{2}x^{2}\cdot dx \\ \therefore & f(y) = F'(y) = (\sqrt{y-1})'\frac{3}{2}(y-1) - (-\sqrt{y-1})' \frac{3}{2}(y-1) = (\sqrt{y-1})'3(y-1) \\ & = \frac{3}{2}\sqrt{y-1} \\ \therefore & f(y) = \begin{cases} \frac{3}{2}\sqrt{y-1}, & 1\le y < 2 \\ 0,& \text{others} \end{cases} \\ \end{array}$$
5. $X\sim E(\lambda), Y = \min\{X,2\}$，求 $F_{Y}(y)$，并请回答 $Y$ 是否为连续变量？ $$\begin{array}{ll} \because & X\sim E(\lambda) \\ \therefore & f_{X}(x) = \begin{cases} \lambda e^{-\lambda x}, & x > 0 \\ 0, & x\le 0 \end{cases}, X \text{的取值范围}(0,+\infty) \\ \because & Y = \min\{X,2\} \\ \therefore & Y \text{的取值范围}(0,2) \\ \therefore & Y < 0, F(y) = 0\\ & Y \ge 2, F(y) = 1 \\ & 0 \le Y < 2: \\ & \displaystyle F(y) = \mathbb{P}\{Y \le y\} = \mathbb{P}\{X\le y\} = F_{X}(y) \\ & \displaystyle = \int_{0}^{y}\lambda e^{-\lambda x} = 1 - e^{-\lambda y} \\ \therefore & F(y) = \begin{cases} 0, & y < 0 \\ 1- e^{-\lambda y}, & 0 \le y < 2 \\ 1, & y \ge 2 \\ \end{cases} \\ \because & f(2) \ne f(2-0) \\ \therefore & Y \text{不为连续型随机变量，无密度} \\ \end{array}$$