Section01_数学期望

定义

离散型

• 设 $\mathbb{P}\{X=x_{i}\} = p_{i},\quad i=1,2,\cdots$，若 $\sum_{i}\limits^{\infty}x_{i}p_{i}$ 绝对收敛（指 $\sum\limits_{i}^{\infty}\vert x_{i} \vert p_{i}$ 收敛），则称 $\sum\limits_{i}^{\infty}x_{i}p_{i}$ 为随机变量的数学期望，记为 $EX$ 或 $E(X)$

常见离散型分布期望推导

1. $X\sim \left( \begin{matrix} 0 & 1 \\ 1-p & p \end{matrix} \right)$，求 $EX$ $$\begin{array}{ll} \because & EX = \sum\limits_{i}^{\infty}x_{i}p_{i} \\ \therefore & EX = 0\times (1-p) + 1\times p = p \\ \end{array}$$
2. $X\sim P(\lambda)$，求 $EX$ $$\begin{array}{ll} \because & \mathbb{P}\{X=k\} = \frac{\lambda^{k}}{k!}e^{-\lambda}, \quad k = 0,1,2,\cdots \\ & EX = \sum\limits_{i}^{\infty}x_{i}p_{i} \\ \therefore & EX = \sum\limits_{k=0}^{\infty}k\cdot\frac{\lambda^{k}}{k!}e^{-\lambda} = \sum\limits_{k=1}^{\infty}k\cdot \frac{\lambda^{k}}{k!}e^{-\lambda} \\ & = e^{-\lambda}\sum\limits_{k=1}^{\infty}\frac{\lambda^{k}}{(k-1)!} \\ & = \lambda e^{-\lambda} \sum\limits_{k=1}^{\infty}\frac{\lambda^{k-1}}{(k-1)!} \\ & = \lambda e^{-\lambda}e^{\lambda} \\ & = \lambda \end{array}$$
3. $X\sim \text{几何分布}$，求 $EX$ $$\begin{array}{ll} \because & \mathbb{P}\{X=k\} = (1-p)^{k-1}p,\quad k = 1,2,3,\cdots \\ & EX = \sum\limits_{i}^{\infty}x_{i}p_{i} \\ \therefore & EX = \sum\limits_{k=1}^{\infty} k(1-p)^{k-1}p \\ & = p\sum\limits_{k=1}^{\infty} k (1-p)^{k-1} \\ & \xlongequal{1-p = x}p \sum\limits_{k=1}^{\infty}kx^{k-1} = p\sum\limits_{k=1}^{\infty} (x^{k})' \\ & = p \bigg(\sum\limits_{k=1}^{\infty}x^{k}\bigg)' = p(\frac{x}{1-x})' = p (-1 + \frac{1}{1-x})' \\ & = p \frac{1}{(1-x)^{2}} = p \frac{1}{p^{2}} = \frac{1}{p} \end{array}$$

连续型

• $X\sim f_{X}(x)$，若 $\displaystyle \int_{-\infty}^{+\infty}\vert x \vert f_{X}(x)\cdot dx$ 收敛，则 $\displaystyle EX = \int_{-\infty}^{+\infty}xf_{X}(x)\cdot dx$ 为随机变量的数学期望

常见连续型分布的期望推导

1. $X\sim U(a,b)$，求 $EX$ $$\begin{array}{ll} \because & f_{X}(x) = \begin{cases} \frac{1}{b-a}, & a < x <b \\ 0, & \text{其他} \end{cases} \\ & \displaystyle EX = \int_{-\infty}^{+\infty}xf_{X}(x)\cdot dx \\ \therefore & \displaystyle EX = \int_{a}^{b}\frac{x}{b-a}\cdot dx \\ & \displaystyle = \frac{1}{b-a}\bigg(\left.\frac{x^{2}}{2}\bigg)\right\vert^{b}_{a} \\ & \displaystyle = \frac{a+b}{2} \end{array}$$
2. $X\sim E(\lambda)$，求 $EX$ $$\begin{array}{ll} \because & f_{X}(x) = \begin{cases} \lambda e^{-\lambda x}, & x>0 \\ 0, & x \le 0 \end{cases} \\ & \displaystyle EX = \int_{-\infty}^{+\infty}xf_{X}(x)\cdot dx \\ \\ & \text{method 1} \\ \therefore & \displaystyle EX = \int_{0}^{\infty} x \lambda e^{-\lambda x} \cdot dx = -\int_{0}^{\infty}x \cdot de^{-\lambda x}\\ & \displaystyle = -[(xe^{-\lambda x})\vert^{\infty}_{0} - \int_{0}^{\infty}e^{-\lambda x}\cdot dx]\\ & \displaystyle = \left.\frac{1}{\lambda}e^{-\lambda x}\right\vert^{\infty}_{0} = \frac{1}{\lambda} \\\\ & \text{\color{#D0104C}{method 2} }\displaystyle \Gamma(z) = \int_{0}^{\infty}x^{z-1}e^{-x}\cdot dx \xlongequal{z\in \mathbb{N}^{+}} (z-1)! \\ & \displaystyle EX = \int_{0}^{\infty}x \lambda e^{-\lambda x}\cdot dx = \frac{1}{\lambda}\int_{0}^{\infty}\lambda xe^{-\lambda x}\cdot d\lambda x \\ & \displaystyle = \frac{1!}{\lambda} = \frac{1}{\lambda} \end{array}$$
   • 注 $X$ 服从均值为 $\theta$ 的指数分布，则：$f_{X}(x) = \begin{cases} \frac{1}{\theta}e^{-\frac{x}{\theta}}, & x>0 \\ 0, & x\le 0 \end{cases}$
3. $X\sim \mathcal{N}(\mu,\sigma^{2})$，求 $EX$ $$\begin{array}{ll} \because & \displaystyle f_{X}(x) = \frac{1}{\sqrt{2\pi} \sigma}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}\cdot dx\\ & \displaystyle EX = \int_{-\infty}^{+\infty}xf_{X}(x)\cdot dx\\ \therefore & \displaystyle EX = \int_{-\infty}^{+\infty}x \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}}\cdot dx \\ & \displaystyle \xlongequal{\frac{x-\mu}{\sigma} = t} \int_{-\infty}^{+\infty} (\sigma t + \mu)\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{t^{2}}{2}} \cdot d(\sigma t + \mu) \\ & \displaystyle = \mu\int_{-\infty}^{+\infty} \frac{1}{\sqrt{2\pi}}e^{-\frac{t^{2}}{2}}\cdot dt + \int_{-\infty}^{+\infty}\frac{\sigma t}{\sqrt{2\pi}}e^{-\frac{t^{2}}{2}}\cdot dt \\ & = \mu \end{array}$$

求$Eg(X)$和$Eg(X,Y)$

求 $Eg(X)$

• $Eg(X) = \begin{cases} \displaystyle \sum_{i}g(x_{i})\mathbb{P}\{X= x_{i}\} \\ \displaystyle \int_{-\infty}^{+\infty}g(x)f_{X}(x)\cdot dx \end{cases}$
  • 如 $X\sim \left( \begin{matrix} -1 & 0 & 1\\ 1/3 & 1/3 & 1/3 \end{matrix} \right)$ 求 $EX^{2}$
    • 用 $X$ 的分布求 $EX^{2} = (-1)^{2}\times \frac{1}{3} + 0^{2}\times \frac{1}{3} + 1^{2}\times \frac{1}{3} = \frac{2}{3}$
    • 用 $X^{2}$ 的分布求 $X^{2}\sim \left( \begin{matrix} 0 & 1 \\ 1/3 & 2/3 \end{matrix} \right)\Rightarrow EX^{2} = 0\times \frac{1}{3} + 1\times \frac{2}{3} = \frac{2}{3}$
  • 如 $X\sim E(\lambda)$，求 $EX^{2}$
    • $\displaystyle EX^{2} = \int_{0}^{\infty}x^{2}\lambda e^{-\lambda x}\cdot dx = \frac{1}{\lambda^{2}}\int_{0}^{\infty}(\lambda x)^{2}e^{-\lambda x}\cdot d\lambda x = 2 \frac{1}{\lambda^{2}}$
  • 注 连续型时，由 $f_{X}(x)$ 求 $Y=X^{2}$ 的密度 $f_{Y}(y)$ 为第二章的一个难点，一般不用 $f_{Y}(y)$ 求 $EY$

求常见分布的 $EX^{2}$

1. $X\sim \left( \begin{matrix} 0 & 1 \\ 1-p & p \end{matrix} \right)$，求 $EX^{2}$ $$\begin{array}{ll} & \displaystyle EX^{2} = \sum_{i}^{\infty}x_{i}^{2}p_{i} = 0^{2}\times (1-p) + 1^{2}\times p = p \end{array}$$
2. $X\sim P(\lambda)$，求 $EX^{2}$ $$\begin{array}{ll} & \displaystyle EX^{2} = \sum_{i}^{\infty}x_{i}^{2}p_{i} = \sum_{k=0}^{\infty}k^{2}\frac{\lambda^{k}}{k!}e^{-\lambda} = e^{-\lambda}\sum_{k=1}^{\infty}k^{2}\frac{\lambda^{k}}{k!} \\ & \displaystyle = e^{-\lambda}\sum_{k=1}^{\infty}\bigg[k(k-1)\frac{\lambda^{k}}{k!} + k \frac{\lambda^{k}}{k!} \bigg] \\ & \displaystyle = \lambda^{2}e^{-\lambda} \sum_{k=2}^{\infty}\frac{\lambda^{k-2}}{(k-2)!} + \lambda e^{-\lambda} \sum_{k=1}^{\infty} \frac{\lambda^{k-1}}{(k-1)!} \\ & = \lambda^{2} + \lambda \end{array}$$
3. $X\sim \text{几何分布}$，求 $EX^{2}$ $$\begin{array}{ll} & \displaystyle EX^{2} = \sum_{i}^{\infty}x_{i}^{2}p_{i} = \sum_{k=1}^{\infty}k^{2}(1-p)^{k-1}p \\ & \displaystyle = p\sum_{k=1}^{\infty}[k(k+1)(1-p)^{k-1} - k (1-p)^{k-1}] \\ & \displaystyle \xlongequal{x = 1-p} p\bigg[ \sum_{k=1}^{\infty} (x^{k+1})'' - \sum_{k=1}^{\infty}(x^{k})'\bigg] \\ & \displaystyle = p\bigg[\bigg(\sum_{k=1}^{\infty}x^{k+1}\bigg)'' - \bigg(\sum_{k=1}^{\infty}x^{k}\bigg)'\bigg] \\ & \displaystyle = p\bigg[\bigg(\frac{x^{2}}{1-x}\bigg)'' - \bigg(\frac{x}{1-x}\bigg)'\bigg] \\ & \displaystyle = p\bigg[\frac{2}{(1-x)^{3}} - \frac{1}{(1-x)^{2}}\bigg]\\ & \displaystyle = \frac{2-p}{p^{2}} \end{array}$$
4. $X\sim U(a,b)$，求 $EX^{2}$ $$\begin{array}{ll} & \displaystyle EX^{2} = \int_{-\infty}^{+\infty}x^{2}f_{X}(x)\cdot dx = \frac{1}{b-a} \int_{a}^{b}x^{2}\cdot dx \\ & \displaystyle = \frac{1}{b-a} \times \left.\frac{x^{3}}{3}\right\vert_{a}^{b} = \frac{b^{3}-a^{3}}{3(b-a)} \\ & \displaystyle = \frac{(b-a)(a^{2} + b^{2} +ab)}{3(b-a)} = \frac{a^{2} + b^{2} + ab}{3} \end{array}$$
5. $X\sim E(\lambda)$，求 $EX^{2}$ $$\begin{array}{ll} & \displaystyle EX^{2} = \int_{-\infty}^{+\infty}x^{2}f_{X}(x)\cdot dx = \int_{0}^{\infty} x^{2}\lambda e^{-\lambda x}\cdot dx \\ & \displaystyle = \frac{1}{\lambda^{2}}\int_{0}^{\infty}(\lambda x)^{2}e^{-\lambda x}\cdot d \lambda x = \frac{2}{\lambda^{2}} \end{array}$$
6. $X\sim \mathcal{N}(\mu,\sigma^{2})$，求 $EX^{2}$ $$\begin{array}{ll} & \displaystyle EX^{2} = \int_{-\infty}^{+\infty}x^{2}f_{X}(x)\cdot dx = \int_{-\infty}^{+\infty} x^{2}\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(x-\mu)^{2}}{2\sigma^{2}}} \\ & \displaystyle \xlongequal{t = \frac{x-\mu}{\sigma}} \int_{-\infty}^{+\infty} (\sigma t - \mu)^{2}\frac{1}{\sqrt{2\pi}}e^{- \frac{t^{2}}{2}}\cdot dt \\ & \displaystyle = \mu^{2}\int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{t^{2}}{2}}\cdot dt + \sigma^{2} \int_{-\infty}^{+\infty}\frac{t^{2}}{\sqrt{2\pi}}e^{-\frac{t^{2}}{2}}\cdot dt- 2\sigma\mu \int_{-\infty}^{+\infty}\frac{t}{\sqrt{2\pi}}e^{-\frac{t^{2}}{2}}\cdot dt \\ & \displaystyle = \mu^{2} - \sigma^{2}\int_{-\infty}^{+\infty}\frac{t}{\sqrt{2\pi}} \cdot d e^{-\frac{t^{2}}{2}} \\ & \displaystyle = \mu^{2} - \sigma^{2}\bigg(te^{-\frac{t^{2}}{2}}\vert^{+\infty}_{-\infty} - \int_{-\infty}^{+\infty}\frac{1}{\sqrt{2\pi}}e^{-\frac{t^{2}}{2}}\cdot dt\bigg) \\ & = \mu^{2} + \sigma^{2} \end{array}$$

求 $Eg(X,Y)$

• $Eg(X,Y) = \begin{cases} \displaystyle \sum_{i}^{}\sum_{j}^{}g(x_{i},y_{j}) \mathbb{P}\{X=x_{i}, Y=y_{j}\} \\ \displaystyle \int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}g(x,y)f(x,y)\cdot dxdy \end{cases}$

例题

1. 设二维随机变量 $(X,Y)$ 的分布律如下所示，分别计算$E(X),E(XY)$ $$\begin{array}{c|cccc} (X,Y) & (0,0) & (0,1) & (1,0) & (1,1) \\ \hline P & 0.2 & 0.1 & 0.4 & 0.3 \end{array}$$ $$\begin{array}{ll} E(X) = 0\times 0.2 + 0\times 0.1 + 1\times 0.4 + 1\times 0.3 = 0.7 \\ E(XY) = 0\times 0.2 + 0\times 0.1 + 0\times 0.4 + 1\times 0.3 = 0.3 \end{array}$$
2. 设随机变量 $X$ 的概率密度为 $f_{X}(x) = \begin{cases} 1, & -\frac{1}{2} < x < \frac{1}{2} \\ 0, & \text{其他} \end{cases}$，设 $Y = \cos x$，试分别计算 $E(X), [E \vert X \vert]^{2}, EX^{2}, E(XY)$ $$\begin{array}{ll} & \displaystyle E(X) = \int_{-\infty}^{+\infty}xf_{X}(x)\cdot dx = \int_{-\frac{1}{2}}^{\frac{1}{2}}x\cdot dx = \left.\frac{x^{2}}{2}\right\vert^{\frac{1}{2}}_{-\frac{1}{2}} = 0\\ & \displaystyle [E \vert X \vert]^{2} = \bigg[\int_{-\infty}^{+\infty}\vert x \vert f_{X}(x)\cdot dx\bigg]^{2} = \bigg[2\int_{0}^{\frac{1}{2}}x\cdot dx\bigg]^{2} = \frac{1}{16} \\ & \displaystyle EX^{2} = \int_{-\frac{1}{2}}^{\frac{1}{2}} x^{2}\cdot dx = \frac{1}{12} \\ & \displaystyle E(XY) = \int_{-\frac{1}{2}}^{\frac{1}{2}}x\cos x\cdot dx = 0 \end{array}$$
   • 注
     1. $E \vert X \vert \ne \vert EX \vert$
     2. $EX^{2} \ne (EX)^{2}$
3. 设二维随机变量 $(X,Y)$ 的概率密度为 $f(x,y) = \begin{cases} \frac{1}{\pi}, & x^{2} + y^{2} \le 1 \\ 0, & \text{其他} \end{cases}$，分别计算 $EY^{2},EXY$ $$\begin{array}{ll} & \displaystyle EY^{2} = \iint\limits_{x^{2}+y^{2}\le 1}y^{2}f(x,y)\cdot d\sigma \\ & \displaystyle \xlongequal{\substack{x = r\cos \theta\\ y = r\sin \theta}} \iint\limits_{r\le 1}r^{2}\sin^{2}\theta f(x,y)\cdot rdrd\theta \\ & \displaystyle = \int_{0}^{1}r^{3} \cdot dr \int_{-\pi}^{\pi}\frac{\sin^{2}\theta}{\pi}\cdot d\theta = 4 \int_{0}^{1}r^{3}\cdot dr \int_{0}^{\frac{\pi}{2}}\frac{\sin^{2}\theta}{\pi} \\ & \displaystyle = \int_{0}^{1}r^{3}\cdot dr = \frac{1}{4} \\\\ & \displaystyle EXY = \iint\limits_{x^{2}+y^{2}\le 1} xy f(x,y)\cdot d\sigma \\ & \displaystyle \xlongequal{\substack{x = r\cos \theta\\ y = r\sin \theta}} \iint\limits_{r\le 1} r^{2}\cos \theta\sin \theta\cdot rdrd\theta \\ & \displaystyle = \int_{0}^{1}r^{3}\cdot dr \int_{-\pi}^{\pi}\cos \theta\sin \theta\cdot d\theta = 0 \end{array}$$
   • 注 本题中 $EXY = EX\cdot EY$，但 $X,Y$ 并不独立；但当 $X,Y$ 独立时，$EXY = EX\cdot EY$

期望的性质

1. $E(c) = c,\quad c \text{为常数}$
2. $E(aX) = aEX,\quad a \text{为常数}$
3. $E(X+Y) = EX + EY; E(aX+bY + c) = aEX + bEY + c$
4. $EXY = EX\cdot EY \Leftrightarrow X,Y \text{不相关(无线性关系)}\substack{\nrightarrow\\\rightarrow} X,Y \text{独立(无任何关系)}$
5. $X\ge c\Rightarrow EX\ge c$
6. 若 $X,Y$ 独立，则 $X^{2}$ 与 $Y^{2}$ 仍独立 $EX^{2}Y^{2} = EX^{2}\cdot EY^{2}$