数字特征典型例题

10 minIntermediate2026/6/14

数字特征部分的典型例题精选,涵盖期望、方差、协方差、相关系数等核心知识点。

1. 数学期望

例题1

XX 的密度函数为 f(x)=12exf(x) = \dfrac{1}{2}e^{-|x|},求 E(X)E(X)E(X2)E(X^2)

E(X)=+x12exdx=0E(X) = \int_{-\infty}^{+\infty} x \cdot \frac{1}{2}e^{-|x|} \, dx = 0(奇函数在对称区间上积分

E(X2)=+x212exdx=0+x2exdx=Γ(3)=2E(X^2) = \int_{-\infty}^{+\infty} x^2 \cdot \frac{1}{2}e^{-|x|} \, dx = \int_0^{+\infty} x^2 e^{-x} \, dx = \Gamma(3) = 2

例题2

XU(0,1)X \sim U(0, 1),求 E(Xn)E(X^n)

E(Xn)=01xndx=1n+1E(X^n) = \int_0^1 x^n \, dx = \frac{1}{n+1}

例题3

某人有 nn 把钥匙,其中只有一把能开门。他随机取一把试开,不能开门则除去,求试开次数 XX 的期望。

XX 的分布律为 P(X=k)=1nP(X = k) = \dfrac{1}{n}k=1,2,,nk = 1, 2, \cdots, n)。

E(X)=k=1nk1n=n+12E(X) = \sum_{k=1}^n k \cdot \frac{1}{n} = \frac{n+1}{2}

2. 方差

例题4

XX 的密度为 f(x)={3x2,0<x<10,其他f(x) = \begin{cases} 3x^2, & 0 < x < 1 \\ 0, & \text{其他} \end{cases},求 D(X)D(X)

E(X)=01x3x2dx=34E(X) = \int_0^1 x \cdot 3x^2 \, dx = \frac{3}{4}

E(X2)=01x23x2dx=35E(X^2) = \int_0^1 x^2 \cdot 3x^2 \, dx = \frac{3}{5}

D(X)=E(X2)[E(X)]2=35916=484580=380D(X) = E(X^2) - [E(X)]^2 = \frac{3}{5} - \frac{9}{16} = \frac{48 - 45}{80} = \frac{3}{80}

例题5

XB(10,0.4)X \sim B(10, 0.4),求 E(3X+2)E(3X + 2)D(3X+2)D(3X + 2)

E(X)=10×0.4=4,D(X)=10×0.4×0.6=2.4E(X) = 10 \times 0.4 = 4, \quad D(X) = 10 \times 0.4 \times 0.6 = 2.4

E(3X+2)=3E(X)+2=14E(3X + 2) = 3E(X) + 2 = 14

D(3X+2)=9D(X)=21.6D(3X + 2) = 9D(X) = 21.6

例题6

设随机变量 XX 的期望 E(X)=10E(X) = 10D(X)=4D(X) = 4,求 E[(X+1)2]E[(X+1)^2]

E[(X+1)2]=E[X2+2X+1]=E(X2)+2E(X)+1E[(X+1)^2] = E[X^2 + 2X + 1] = E(X^2) + 2E(X) + 1

=[D(X)+E(X)2]+2E(X)+1=(4+100)+20+1=125= [D(X) + E(X)^2] + 2E(X) + 1 = (4 + 100) + 20 + 1 = 125

3. 协方差与相关系数

例题7

(X,Y)(X, Y) 的联合密度为

f(x,y)={1,0<x<1,0<y<10,其他f(x, y) = \begin{cases} 1, & 0 < x < 1, 0 < y < 1 \\ 0, & \text{其他} \end{cases}

Cov(X,Y)\text{Cov}(X, Y)ρXY\rho_{XY}

XXYY 独立(非零区域为矩形,联合密度可分解),故 Cov(X,Y)=0\text{Cov}(X, Y) = 0ρXY=0\rho_{XY} = 0

例题8

(X,Y)(X, Y) 的联合密度为

f(x,y)={2,0<y<x<10,其他f(x, y) = \begin{cases} 2, & 0 < y < x < 1 \\ 0, & \text{其他} \end{cases}

Cov(X,Y)\text{Cov}(X, Y)ρXY\rho_{XY}

E(X)=23,E(Y)=13,E(XY)=14E(X) = \frac{2}{3}, \quad E(Y) = \frac{1}{3}, \quad E(XY) = \frac{1}{4}

Cov(X,Y)=1429=136\text{Cov}(X, Y) = \frac{1}{4} - \frac{2}{9} = \frac{1}{36}

E(X2)=010x2x2dydx=12,D(X)=1249=118E(X^2) = \int_0^1 \int_0^x 2x^2 \, dy \, dx = \frac{1}{2}, \quad D(X) = \frac{1}{2} - \frac{4}{9} = \frac{1}{18}

E(Y2)=01y12y2dxdy=16,D(Y)=1619=118E(Y^2) = \int_0^1 \int_y^1 2y^2 \, dx \, dy = \frac{1}{6}, \quad D(Y) = \frac{1}{6} - \frac{1}{9} = \frac{1}{18}

ρXY=1/361/18×1/18=1/361/18=12\rho_{XY} = \frac{1/36}{\sqrt{1/18 \times 1/18}} = \frac{1/36}{1/18} = \frac{1}{2}

例题9

XXYY 的相关系数 ρ=0.5\rho = 0.5D(X)=D(Y)=1D(X) = D(Y) = 1,求 D(XY)D(X - Y)

D(XY)=D(X)+D(Y)2Cov(X,Y)=1+12×0.5=1D(X - Y) = D(X) + D(Y) - 2\text{Cov}(X, Y) = 1 + 1 - 2 \times 0.5 = 1

例题10

XN(0,1)X \sim N(0, 1)Y=X2Y = X^2,求 Cov(X,Y)\text{Cov}(X, Y)

E(X)=0,E(Y)=E(X2)=1E(X) = 0, \quad E(Y) = E(X^2) = 1

E(XY)=E(X3)=0E(XY) = E(X^3) = 0(正态分布的奇数阶矩为零)

Cov(X,Y)=00=0\text{Cov}(X, Y) = 0 - 0 = 0

XXYY 不相关,但 Y=X2Y = X^2,显然不独立。

4. 综合题

例题11

X1,X2,,XnX_1, X_2, \cdots, X_n 独立同分布,E(Xi)=μE(X_i) = \muD(Xi)=σ2D(X_i) = \sigma^2,设 Xˉ=1ni=1nXi\bar{X} = \dfrac{1}{n}\sum_{i=1}^n X_i,求 E(Xˉ)E(\bar{X})D(Xˉ)D(\bar{X})

E(Xˉ)=1ni=1nE(Xi)=μE(\bar{X}) = \frac{1}{n} \sum_{i=1}^n E(X_i) = \mu

D(Xˉ)=1n2i=1nD(Xi)=σ2nD(\bar{X}) = \frac{1}{n^2} \sum_{i=1}^n D(X_i) = \frac{\sigma^2}{n}

例题12

X1,X2,,XnX_1, X_2, \cdots, X_n 独立同分布,E(Xi)=μE(X_i) = \muD(Xi)=σ2D(X_i) = \sigma^2,求 E[i=1n(XiXˉ)2]E\left[\sum_{i=1}^n (X_i - \bar{X})^2\right]

i=1n(XiXˉ)2=i=1nXi2nXˉ2\sum_{i=1}^n (X_i - \bar{X})^2 = \sum_{i=1}^n X_i^2 - n\bar{X}^2

E[i=1nXi2]=n(μ2+σ2)E\left[\sum_{i=1}^n X_i^2\right] = n(\mu^2 + \sigma^2)

E[nXˉ2]=n[D(Xˉ)+E(Xˉ)2]=n(σ2n+μ2)=σ2+nμ2E[n\bar{X}^2] = n[D(\bar{X}) + E(\bar{X})^2] = n\left(\frac{\sigma^2}{n} + \mu^2\right) = \sigma^2 + n\mu^2

E[i=1n(XiXˉ)2]=n(μ2+σ2)σ2nμ2=(n1)σ2E\left[\sum_{i=1}^n (X_i - \bar{X})^2\right] = n(\mu^2 + \sigma^2) - \sigma^2 - n\mu^2 = (n-1)\sigma^2

例题13

XXYY 独立,XN(1,4)X \sim N(1, 4)YN(2,9)Y \sim N(2, 9),求 E(2X3Y+1)E(2X - 3Y + 1)D(2X3Y+1)D(2X - 3Y + 1)

E(2X3Y+1)=2×13×2+1=3E(2X - 3Y + 1) = 2 \times 1 - 3 \times 2 + 1 = -3

D(2X3Y+1)=4×4+9×9=16+81=97D(2X - 3Y + 1) = 4 \times 4 + 9 \times 9 = 16 + 81 = 97

例题14

XU(0,2π)X \sim U(0, 2\pi),求 E(sinX)E(\sin X)E(cosX)E(\cos X)

E(sinX)=12π02πsinxdx=0E(\sin X) = \frac{1}{2\pi}\int_0^{2\pi} \sin x \, dx = 0

E(cosX)=12π02πcosxdx=0E(\cos X) = \frac{1}{2\pi}\int_0^{2\pi} \cos x \, dx = 0

例题15

X1,X2,,X100X_1, X_2, \cdots, X_{100} 独立同分布,XiB(1,0.6)X_i \sim B(1, 0.6),利用切比雪夫不等式估计 P(i=1100Xi70)P\left(\sum_{i=1}^{100} X_i \geq 70\right)

:设 S=i=1100XiS = \sum_{i=1}^{100} X_iE(S)=60E(S) = 60D(S)=24D(S) = 24

P(S70)=P(S6010)P(S6010)D(S)102=24100=0.24P(S \geq 70) = P(S - 60 \geq 10) \leq P(|S - 60| \geq 10) \leq \frac{D(S)}{10^2} = \frac{24}{100} = 0.24