连续型随机变量精讲

含有绝对值 $|x|$，先按正负分段：

$$e^{-|x|} = \begin{cases} e^{x}, & x < 0 \\ e^{-x}, & x \ge 0 \end{cases}$$

归一化积分：

$$\begin{aligned} \int_{-\infty}^{+\infty} f(x)\,\mathrm{d}x &= a\left[\int_{-\infty}^{0} e^{x}\,\mathrm{d}x + \int_{0}^{+\infty} e^{-x}\,\mathrm{d}x\right] \\ &= a\Big[\,e^{x}\Big|_{-\infty}^{0} \;+\; (-e^{-x})\Big|_{0}^{+\infty}\Big] \\ &= a\big[(1-0) + (0-(-1))\big] = a \cdot 2 = 1 \end{aligned}$$

$$\therefore \; a = \frac{1}{2}$$

$F(x) = \int_{-\infty}^{x} f(t)\,\mathrm{d}t$，分段讨论：

当 $x < 0$ 时：

$$F(x) = \int_{-\infty}^{x} \frac{1}{2}e^{t}\,\mathrm{d}t = \frac{1}{2}e^{t}\Big|_{-\infty}^{x} = \frac{1}{2}e^{x}$$

当 $x \ge 0$ 时：

$$\begin{aligned} F(x) &= \int_{-\infty}^{0} \frac{1}{2}e^{t}\,\mathrm{d}t + \int_{0}^{x} \frac{1}{2}e^{-t}\,\mathrm{d}t \\ &= \frac{1}{2} + \frac{1}{2}\big[-e^{-t}\big]_{0}^{x} \\ &= \frac{1}{2} + \frac{1}{2}(1 - e^{-x}) = 1 - \frac{1}{2}e^{-x} \end{aligned}$$

$$F(x) = \begin{cases} \dfrac{1}{2}e^{x}, & x < 0 \\[8pt] 1 - \dfrac{1}{2}e^{-x}, & x \ge 0 \end{cases}$$

方法一（分布函数相减）：

$$P(0 < X < 1) = F(1) - F(0) = \left(1 - \frac{1}{2}e^{-1}\right) - \frac{1}{2} = \frac{1}{2} - \frac{1}{2e}$$

方法二（直接积分）：

$$P(0 < X < 1) = \int_{0}^{1} \frac{1}{2}e^{-x}\,\mathrm{d}x = \frac{1}{2}\big[-e^{-x}\big]_{0}^{1} = \frac{1}{2}(1 - e^{-1})$$

仅非零区间 $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ 上积分：

$$\int_{-\pi/2}^{\pi/2} a\cos x\,\mathrm{d}x = a\big[\sin x\big]_{-\pi/2}^{\pi/2} = a(1 - (-1)) = 2a = 1$$

$$\therefore \; a = \frac{1}{2}$$

当 $x < -\dfrac{\pi}{2}$ 时：$F(x) = 0$

当 $-\dfrac{\pi}{2} \le x \le \dfrac{\pi}{2}$ 时：

$$F(x) = \int_{-\pi/2}^{x} \frac{1}{2}\cos t\,\mathrm{d}t = \frac{1}{2}\big[\sin t\big]_{-\pi/2}^{x} = \frac{1}{2}(\sin x + 1)$$

当 $x > \dfrac{\pi}{2}$ 时：$F(x) = 1$

$$F(x) = \begin{cases} 0, & x < -\dfrac{\pi}{2} \\[8pt] \dfrac{1}{2}(\sin x + 1), & -\dfrac{\pi}{2} \le x \le \dfrac{\pi}{2} \\[8pt] 1, & x > \dfrac{\pi}{2} \end{cases}$$

方法一（分布函数相减）：

$$P\!\left(0 < X < \frac{\pi}{4}\right) = F\!\left(\frac{\pi}{4}\right) - F(0) = \frac{1}{2}\!\left(\frac{\sqrt{2}}{2} + 1\right) - \frac{1}{2} = \frac{\sqrt{2}}{4}$$

方法二（直接积分）：

$$P\!\left(0 < X < \frac{\pi}{4}\right) = \int_{0}^{\pi/4} \frac{1}{2}\cos x\,\mathrm{d}x = \frac{1}{2}\big[\sin x\big]_{0}^{\pi/4} = \frac{\sqrt{2}}{4}$$

$$f(x) = \begin{cases} \dfrac{1}{b-a}, & a \le x \le b \\[8pt] 0, & \text{其他} \end{cases}$$

$$F(x) = \begin{cases} 0, & x < a \\[6pt] \dfrac{x-a}{b-a}, & a \le x \le b \\[8pt] 1, & x > b \end{cases}$$

由 $P(X < 2) = \dfrac{1}{3}$，即 $F(2) = \dfrac{2-a}{b-a} = \dfrac{1}{3}$ …… ①

由 $P(X > 5) = \dfrac{1}{4}$，即 $1 - F(5) = 1 - \dfrac{5-a}{b-a} = \dfrac{1}{4}$，

得 $\dfrac{5-a}{b-a} = \dfrac{3}{4}$ …… ②

由 ①：$3(2-a) = b-a \;\Rightarrow\; b-a = 6 - 3a \;\Rightarrow\; b = 6 - 2a$

代入 ②：$\dfrac{5-a}{(6-2a)-a} = \dfrac{5-a}{6-3a} = \dfrac{3}{4}$

交叉相乘：$4(5-a) = 3(6-3a) \;\Rightarrow\; 20 - 4a = 18 - 9a \;\Rightarrow\; 5a = -2 \;\Rightarrow\; a = -0.4$

回代：$b = 6 - 2(-0.4) = 6.8$

$$\therefore \; f(x) = \begin{cases} \dfrac{5}{36}, & -0.4 \le x \le 6.8 \\[8pt] 0, & \text{其他} \end{cases}$$