# 六、随机变量的数字特征 ## 1. 数学期望 **定义**: 离散型:设$P\{X = x_k\} = p_k$,若$\sum_{k=1}^{\infty} x_k p_k$绝对收敛,则 $$E(X) = \sum_{k=1}^{\infty} x_k p_k$$ 连续型:若$\int_{-\infty}^{\infty} xf(x)dx$绝对收敛,则 $$E(X) = \int_{-\infty}^{\infty} xf(x)dx$$ **随机变量函数的期望**: 设$Y = g(X)$,g是连续函数 - 离散型:$E(Y) = E[g(X)] = \sum_{k=1}^{\infty} g(x_k)p_k$ - 连续型:$E(Y) = E[g(X)] = \int_{-\infty}^{\infty} g(x)f(x)dx$ **数学期望性质**: 1. 设C是常数,则$E(C) = C$ 2. 设X是随机变量,C是常数,则$E(X + C) = E(X) + C$ 3. 设X是随机变量,C是常数,则$E(CX) = CE(X)$ 4. 设X,Y是两个随机变量,则$E(X \pm Y) = E(X) \pm E(Y)$(可推广到任意有限个) 5. 设X,Y是相互独立的随机变量,则$E(XY) = E(X)E(Y)$(可推广到任意有限个) --- ## 2. 方差与标准差 **定义**:$D(X) = E\{[X - E(X)]^2\}$ **计算公式**:$D(X) = E(X^2) - [E(X)]^2$ **标准差**:$\sigma(X) = \sqrt{D(X)}$ **方差的计算**: 离散型:$D(X) = \sum_{k=1}^{\infty} [x_k - E(X)]^2 p_k$ 连续型:$D(X) = \int_{-\infty}^{\infty} [x - E(X)]^2 f(x)dx$ **方差性质**: 1. 设C是常数,则$D(C) = 0$ 2. 设X是随机变量,C是常数,则$D(CX) = C^2D(X)$,$D(X + C) = D(X)$ 3. 设X,Y是两个随机变量,则 $$D(X \pm Y) = D(X) + D(Y) \pm 2Cov(X,Y)$$ 特别地,若X,Y相互独立,则$D(X \pm Y) = D(X) + D(Y)$ 4. $D(X) = 0$的充要条件是X以概率1取常数$E(X)$,即$P\{X = E(X)\} = 1$ --- ## 3. 协方差 **定义**:$Cov(X,Y) = E\{[X - E(X)][Y - E(Y)]\}$ **计算公式**:$Cov(X,Y) = E(XY) - E(X)E(Y)$ **性质**: 1. $Cov(X,Y) = Cov(Y,X)$(对称性) 2. $Cov(X,C) = 0$(C为常数) 3. $Cov(X,X) = D(X)$ 4. $Cov(aX, bY) = ab \cdot Cov(X,Y)$,a,b是常数 5. $Cov(X_1 + X_2, Y) = Cov(X_1, Y) + Cov(X_2, Y)$(双线性) 6. 若X,Y相互独立,则$Cov(X,Y)=0$ **与方差的关系**: $$D(X + Y) = D(X) + D(Y) + 2Cov(X,Y)$$ --- ## 4. 相关系数 **定义**: $$\rho_{XY} = \frac{Cov(X,Y)}{\sqrt{D(X)}\sqrt{D(Y)}}$$ **性质**: 1. $|\rho_{XY}| \leq 1$ 2. $|\rho_{XY}| = 1$的充要条件是,存在常数a,b使$P\{Y = a + bX\} = 1$(线性关系) 3. 若X,Y相互独立,则$\rho_{XY} = 0$(不相关) 4. **不相关 ≠ 独立**:$\rho_{XY} = 0$只说明X,Y没有线性关系,可能有非线性关系 **不相关的等价条件**(以下四条等价): - $\rho_{XY} = 0$ - $Cov(X,Y) = 0$ - $E(XY) = E(X)E(Y)$ - $D(X + Y) = D(X) + D(Y)$ --- ## 5. 矩 **定义**:设X和Y是随机变量 | 矩的类型 | 定义 | 说明 | |---------|------|------| | **k阶原点矩** | $E(X^k)$,$k = 1,2,...$ | 一阶原点矩就是期望E(X) | | **k阶中心矩** | $E\{[X - E(X)]^k\}$,$k = 2,3,...$ | 二阶中心矩就是方差D(X) | | **k+l阶混合矩** | $E(X^k Y^l)$,$k,l = 1,2,...$ | | | **k+l阶混合中心矩** | $E\{[X-E(X)]^k[Y-E(Y)]^l\}$ | 二阶混合中心矩就是协方差Cov(X,Y) | --- ## 6. 切比雪夫不等式 设$E(X)=\mu$,$D(X)=\sigma^2$存在,则对任意$\varepsilon>0$, $$P\{|X-\mu| \ge \varepsilon\} \le \frac{\sigma^2}{\varepsilon^2}$$ 等价地, $$P\{|X-\mu| < \varepsilon\} \ge 1 - \frac{\sigma^2}{\varepsilon^2}$$ --- ## 7. 数字特征典型例题 **例**:设随机变量$X \sim N(\mu, \sigma^2)$,$Y \sim N(\mu, \sigma^2)$,且设X,Y相互独立,求$Z_1 = \alpha X + \beta Y$和$Z_2 = \alpha X - \beta Y$的相关系数(其中$\alpha, \beta$是不为零的常数)。 **解**:由于$X, Y \sim N(\mu, \sigma^2)$,可得 $$E(X) = E(Y) = \mu, \quad D(X) = D(Y) = \sigma^2$$ $Z_1$和$Z_2$的相关系数: $$\rho_{Z_1Z_2} = \frac{E(Z_1Z_2) - E(Z_1) \cdot E(Z_2)}{\sqrt{D(Z_1)} \cdot \sqrt{D(Z_2)}}$$ 由$E(Z_1) = E(\alpha X + \beta Y) = \alpha E(X) + \beta E(Y) = (\alpha + \beta)\mu$ $E(Z_2) = E(\alpha X - \beta Y) = \alpha E(X) - \beta E(Y) = (\alpha - \beta)\mu$ 又$E(Z_1Z_2) = E[(\alpha X + \beta Y)(\alpha X - \beta Y)] = E(\alpha^2 X^2 - \beta^2 Y^2) = \alpha^2 E(X^2) - \beta^2 E(Y^2)$ $= (\alpha^2 - \beta^2)(\sigma^2 + \mu^2)$ $D(Z_1) = D(\alpha X + \beta Y) = \alpha^2 D(X) + \beta^2 D(Y) = (\alpha^2 + \beta^2)\sigma^2$ $D(Z_2) = D(\alpha X - \beta Y) = \alpha^2 D(X) + \beta^2 D(Y) = (\alpha^2 + \beta^2)\sigma^2$ 于是 $$\rho_{Z_1Z_2} = \frac{(\alpha^2 - \beta^2)(\sigma^2 + \mu^2) - (\alpha + \beta)\mu(\alpha - \beta)\mu}{\sqrt{(\alpha^2 + \beta^2)\sigma^2} \cdot \sqrt{(\alpha^2 + \beta^2)\sigma^2}} = \frac{(\alpha^2 - \beta^2)\sigma^2}{(\alpha^2 + \beta^2)\sigma^2} = \frac{\alpha^2 - \beta^2}{\alpha^2 + \beta^2}$$