# 随机变量的数字特征练习题

\begin{align*} \newcommand{\dif}{\mathop{}\!\mathrm{d}} \newcommand{\p}{\partial} \newcommand{\Cov}{\text{Cov}} \end{align*}

## 题型

### 利用定义与性质求期望和方差

• $E(aX+bY+c)=aE(X)+bE(Y)+c$
• 若 $X,Y$ 相互独立，则 $D(aX+bY+c)=a^2 D(X)+b^2D(Y)$

$$\because X,Y 独立\\ \therefore (X-Y) \sim N(0, 1)\\$$

\begin{align} E(|U|)&=\int_{-\infty}^{+\infty} |u| \cdot \frac{1}{\sqrt{2\pi}} e^{-\frac{u^2}{2}} \dif u\\ &=\frac{2}{\sqrt{2\pi}} \int_0^{+\infty} u\cdot e^{-\frac{u^2}{2}} \dif u\\ &= \frac{2}{\sqrt{2\pi}} \int_0^{+\infty} (\frac{u^2}{2})^0 e^{-\frac{u^2}{2}} \dif (\frac{u^2}{2})\\ &= \frac{2}{\sqrt{2\pi}} \end{align}
$$D(|U|)=E(|U|^2)-[E(|U|)]^2\\ 而 E(|U|^2)=E(U^2)=D(U)+[E(U)]^2=1\\ D(|U|)=1-\frac{2}{\pi}$$

$$f(x)=\frac{1}{\sqrt{2\pi} \cdot \sqrt{\frac{1}{2}} } \cdot e^{-\frac{(x-1)^2}{2\left(\sqrt{1/2}\right)^2}}\\ X\sim N(1,\frac{1}{2})\\ EX=1, DX=\frac{1}{2}$$

$$1=EX^2-3EX+2\\ \because EX^2 = DX+(EX)^2=\lambda + \lambda^2\\ \therefore \lambda + \lambda^2 - 3 \lambda + 2 =1\\ \lambda^2-2\lambda +1=0\Rightarrow \lambda=1$$

$$p=\int_\frac{\pi}{3}^\pi \frac{1}{2}\cos \frac{x}{2} \dif x = \frac{1}{2} \\ \therefore Y \sim B(4,\frac{1}{2})\\ EY = 2 \; DY=1\\ EY^2 = DY+(EY)^2=5$$