概率统计D 02.01 随机变量及其分布

$\color{blue}{第二章 随机变量及其分布}$

$\color{blue}{\S 2.1 随机变量及其分布函数}$

$\color{blue}{一、随机变量}$

用数量来表示试验的基本事件.

$定义1.设试验E的基本空间为\Omega, \Omega = \lbrace \omega \rbrace,如果对试验E的\\ 每一个基本事件\omega,规定一个实数记作X(\omega)与之对应,这样\\ 得到一个定义在基本空间\Omega上的单值实函数X = X(\omega),称\\ 变量X = X(\omega)为随机变量.$
$随机变量常用字母X、Y、Z等表示.或用\xi、\eta等表示.$

$注:\\ 1^°随机变量X = X(\omega)是基本事件的函数,具体问题里具体规定.$
$2^° 对于不同的基本事件,X的取值亦要不同.$
$3^° 每一个基本事件都可用随机变量的取值来表示,\\ 如X(\omega_k) = x_k,则\omega_k = \lbrace X = x_k \rbrace.$
$4^° 当x_k \neq x_r时,事件\lbrace X = x_k \rbrace 与 \lbrace X = x_r \rbrace 互不相容.$
$5^° \lbrace X \leq x \rbrace 表示X取小于等于 x 的每一个值所对应的基本事件的和事件.$

$\color{blue}{二、随机变量的分布函数}$

$定义2.设X是一个随机变量,对任意实数x,令\\ F(x) = P\lbrace X \leq x \rbrace, (-\infty, +\infty) \\ 称F(x)为随机变量X的分布函数.$
$分布函数是定义在(-\infty, +\infty)上的函数,具有如下性质:\\ ① 0 \leq F(x) \leq 1 且F(-\infty) = 0, F(+\infty) = 1. \\ ② F(x) 是单调不减函数.\\ ③ F(x)是右连续的,即 F(x^{+}) = F(x). \\ ④ 对于任意 a < b, 有\\ P\lbrace a < X \leq b \rbrace = F(b) - F(a). \\ P\lbrace a < X < b \rbrace = F(b) - F(a) - P\lbrace X = b \rbrace. \\ P\lbrace a \leq X \leq b \rbrace = F(b) - F(a) + P\lbrace X = a \rbrace.$

$\color{blue}{\S 2.2 离散型随机变量及其概率分布}$

$\color{blue}{一、离散型随机变量及其分布律}$

$定义3.设E是一个试验,X为E中的随机变量,如果X只取有限个数值\\ 或可数无穷多个数值,则称X为离散型随机变量.$

$定义4.分布律:P\lbrace X = x_k \rbrace = p_k,k = 1, 2, \cdots, 即$

$$\begin{array}{l|c c} X & x_1 & x_2 & \cdots & x_k & \cdots \\ \hline P&P_1&P_2 & \cdots & P_k & \cdots \\ \end{array}$$

例如抛硬币的试验：$\begin{array}{l|c|c} X & 1 & 0 \\ \hline P& 1/2 & 1/2 \\ \end{array}$

掷骰子的试验：$\begin{array}{l|c c} X & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\ \end{array}$

$分布律的性质: \\ ① p_k \geq 0, k = 1, 2, \cdots; \\ ②\sum \limits _ {k = 1} ^ {\infty} p_k = 1.$

$例1.某人射击命中率为p, 不断地独立射击目标,直到命中为止,\\ 求发射子弹数X的分布律(概率分布).$
$解:X可取值为1,2,\cdots, k, \cdots, \lbrace X = k \rbrace表示事件前k-1次不中,\\ 第k次击中,则p_k = P\lbrace X = k \rbrace = (1-p)^{k-1} p,因此:$
$\begin{array}{l|c c} X & 1 & 2 & 3 & \cdots & k & \cdots \\ \hline P & p & (1-p)p & (1-p)^2p & \cdots & (1-p)^{k-1}p & \cdots \\ \end{array}$

例2.设$\begin{array}{l|c c} X & 0 & 1 \\ \hline P & 1/2 & 1/2 \\ \end{array}$ $,求X为分布函数F(x).$
$解: F(x) = \left \{ \begin{array}{l} 0, \qquad x < 0 \\ 1/2, \quad 0 \leq x < 1 \\ 1, \qquad x \geq 1 \end{array} \right.$

例3.$\begin{array}{l|c c} X & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline P & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} & \frac{1}{6} \\ \end{array}$,$求分布函数F(x).$
$解: F(x) = \left \{ \begin{array}{l} 0, \qquad x < 1 \\ 1/6, \quad 1 \leq x < 2 \\ 2/6, \quad 2 \leq x < 3 \\ 3/6, \quad 3 \leq x < 4 \\ 4/6, \quad 4 \leq x < 5 \\ 5/6, \quad 5 \leq x < 6 \\ 1, \qquad x \geq 6 \end{array} \right.$

$\color{blue}{二、几种常用的离散型随机变量及其分概率分布}$

$\color{blue}{1. (0-1)分布}$

$若随机变量X只取0与1两个值,其概率分布为\\ P\lbrace X = 0 \rbrace = 1- p, P\lbrace X = 1 \rbrace = p, 0 < p < 1 \\ 或写成 P \lbrace X = x \rbrace = p^x(1 - p)^{1 - x}, x = 0, 1, 则称\\ X服从参数为p的(0-1)分布或两点分布.$
$\begin{array}{l|c c} X & 0 & 1 \\ \hline P & 1-p & p \\ \end{array}$
$分布函数F(x) = \left \{ \begin{array}{l} 0, \qquad \quad x < 0 \\ 1 - p, \quad 0 \leq x < 1, \\ 1, \qquad \quad x \geq 1 \end{array} \right.$

$\color{blue}{2. 二项分布}$

$如果随机变量X的取值为0,1,2, \cdots, n, 其分布为\\ P\lbrace X = k \rbrace = \dbinom{n}{k} p^k(1 - p)^{n -k},k = 0, 1, 2, \cdots, n; 0 < p < 1. \\ 则称X服从参数为n,p的二项分布,证作X \sim b(n, p)(或B(n, p)).$
$当n = 1时,二项分布B(1, p)就是(0-1)分布.$
$在n重伯努利概型中,事件A发生的次数X就服从B(n, p), p = P(A).$

$\color{blue}{3. 泊松分布}$

$如果随机变量X可能取值为0, 1, 2, \cdots, k, \cdots, 并且\\ P\lbrace X = k \rbrace = \dfrac{\lambda^k}{k!}e^{-\lambda}, k = 0, 1, 2, \cdots \\ 其中 \lambda > 0为常数,则称X服从参数为\lambda的泊松分布,记作X \sim \pi(\lambda).$
$显然\\ p_k = P\lbrace X = k \rbrace = \dfrac{\lambda^k}{k!}e^{-\lambda} > 0 \quad k = 0, 1, 2, \cdots \\ \sum \limits _ {k=0} ^ {\infty} p_k = \sum \limits _ {k = 0} ^ {\infty} \dfrac{\lambda^k}{k!}e^{-\lambda} = e^{-\lambda} \cdot e^\lambda = 1.$

$泊松定理:设\lambda > 0为一常数,n 为任意正整数, np_n = \lambda,\\ 则对于任一固定的非负整数k,有 \\ \lim \limits _ {n \rightarrow \infty} \dbinom{n}{k} p_n^k(1 - p)^{n-k} = \dfrac{\lambda^k e^{-\lambda}}{k!}.$

$证:由p_n = \dfrac{\lambda}{n},有\\ \dbinom{n}{k} p_n^k(1 - p)^{n-k} = \dfrac{n(n-1)\cdots (n-k+1)}{k!}(\dfrac{\lambda}{n})^k(1 - \dfrac{\lambda}{n})^{n-k} \\ = \dfrac{\lambda^k}{k!}\Big[1 \cdot (1 - \dfrac{1}{n}) \cdots (1 - \dfrac{k -1}{n}) \Big](1 - \dfrac{\lambda}{n})^n(1 - \dfrac{\lambda}{n})^{-k} \\ \rightarrow \dfrac{\lambda^k}{k!} \cdot 1 \cdot e^{-\lambda} \cdot1 = \dfrac{\lambda^k}{k!}e^{-\lambda}, (n \rightarrow \infty)$
$由泊松定理可知,当n很大,p很小时有近似公式(\lambda = np): \\ \dbinom{n}{k} p^k(1 - p)^{n-k} \approx \dfrac{\lambda^k}{k!}e^{-\lambda} \\ 即二项分布近似泊松分布,而泊松分布有表可查.$

$例4.一部电话交换台每分钟接到的呼叫次数\\ 服从参数为4的泊松分布,求:\\ (1) 每分钟恰有6次呼叫的概率; \\ (2) 每分钟的呼叫次数大于5的概率.$
$解:以X表示每分钟呼叫的次数,则X \sim \pi(4). \\ P\lbrace X = k \rbrace = \dfrac{4^k}{k!}e^{-4}, k = 0, 1, 2, \cdots \\ (1) P\lbrace X = 6 \rbrace = \dfrac{4^6}{6!}e^{-4} = 0.1042 \\ (2) P\lbrace X > 5 \rbrace = 1 - P\lbrace X \leq 5 \rbrace \\ = 1 - [P\lbrace X = 0 \rbrace + P\lbrace X = 1 \rbrace + \cdots + P\lbrace X = 5 \rbrace] \\ = 1 - (0.0183 + 0.0733 + 0.1465 + 0.1954 + 0.1954 + 0.1563) \\ = 0.2148$

$例5.某人进行射击,每次命中率为0.02,独立射击400次,\\ 求至少命中两次的概率.$
$解: X \sim B(400, 0.02) \\ P\lbrace X = k \rbrace = \dbinom{400}{k} 0.02^k(1-0.02)^{400 - k}, k = 0, 1, 2 \cdots, 400 \\ \approx \dfrac{\lambda^k}{k!}e^{-\lambda} = \dfrac{8^k}{k!}e^{-8}, \lambda = np = 8 \\ 所求概率为\\ p = P\lbrace X \geq 2 \rbrace = 1 - p\lbrace X = 0 \rbrace - P\lbrace X = 1 \rbrace \\ \approx 1 - e^{-8} - 8e^{-8} \\ = 0.997$

$\color{blue}{\S 2.3连续型随机变量及其概率密度}$

$\color{blue}{一、连续型随机变量的概率密度及其性质}$

$定义4.设随机变量X的分布函数为F(x),如果存在一个非负函数f(x),\\ 使对于任意实数x,有\\ F(x) = \int_{-\infty}^x f(x) dx, -\infty < x < \infty \\ 则称X为连续型随机变量,其中f(x)称为X的概率密度.$
$性质:\\ ① f(x) \geq 0 \\ ② \int_{-\infty} ^ {+\infty} f(x) dx = 1 \\ ③ F(x)连续\\ ④ P\lbrace X = a \rbrace = 0 \\ ⑤ P\lbrace x_1 < X < x_2 \rbrace = P\lbrace x_1 \leq X \leq x_2 \rbrace \\ = F(x_2) - F(x_1) = \int_ {x_1} ^ {x_2}f(x) dx \\ ⑥ f(x) = F^{\prime}(x)$

$例1.设X的概率密度为f(x) = \left \{ \begin{array}{l} ke^{-3x}, x > 0 \\ 0 , x \leq 0 \end{array} \right.$
$(1) 求k; (2)求P\lbrace X > 0.1 \rbrace; (3) F(x).$
$解: \\ (1) 由\int_{-\infty}^{+\infty} f(x) dx = 1,有k = 3. \\ (2) P\lbrace X > 0.1 \rbrace = \int_{0.1}^{+\infty} f(x) dx = 0.7408$
$(3) F(x) = \left \{ \begin{array}{l} 1 - e^{-3x}, x > 0, \\ 0, x \leq 0. \end{array} \right.$

$例2.设连续型随机变量X的概率密度为f(x) = \left \{ \begin{array}{l} A(8x - 3x^2), 0 < x < 2, \\ 0, 其他 \end{array} \right.$
$(1)求A, (2) 求P\lbrace X > 1 \rbrace; (3)分布函数F(x).$
$解：\\ (1) \int_{-\infty} ^{+\infty} f(x) dx = A[4x^2 - x^3]_0^2 = 8A = 1 \\ A = \dfrac{1}{8} \\ (2) P\lbrace X > 1 \rbrace = \int_1^{\infty} f(x) dx = \int_1^2 f(x) dx = \dfrac{1}{8}[4x^2 - x^3]_1^2 = \dfrac{5}{8}$
$(3) F(x) = \left \{ \begin{array}{l} 1, x \geq 2 \\ \dfrac{4x^2 - x^3}{8}, 0 \leq x < 2 \\ 0, x < 0 \end{array} \right.$

$\color{blue}{二、几种常见的连续型随机变量及其概率密度}$

$\color{blue}{1.均匀分布}$

$若连续型随机变量X具有概率密度f(x) = \left \{ \begin{array}{l} \dfrac{1}{b - a}, a < x < b, \\ 0, 其他. \end{array} \right.$
$则称X在区间(a, b)上服从均匀分布.其分布函数为$
$F(x) = \left \{ \begin{array}{l} 0, x \leq a, \\ \dfrac{x - a}{b - a}, a < x < b, \\ 1 , x \geq b. \end{array} \right.$

$对于任意x_1, x_2 \in (a, b),若x_1 < x_2, 则有$
$P\lbrace x_1 < X < x_2 \rbrace = \int_{x_1}^{x_2} f(x) dx = \int_{x_1}^{x_2} \dfrac{1}{b - a} dx = \dfrac{x_2 - x_1}{b - a}.$
$这说明X取值落在(a, b)内任一子区间(x_1, x_2)内的概率,\\ 只依赖于子区间的长度x_2 - x_1,而与子区间位置无关.$

$例3.设连续型随机变量X的分布函数为F(x) = \left \{ \begin{array}{l} 0, x < -3, \\ \dfrac{x + 3}{12}, -3 \leq x < 9, \\ 1 , x \geq 9. \end{array} \right.$
$(1)求密度f(x); \\ (2) 若-3 < a < a + l < 9, 求P\lbrace a < X < a + l \rbrace.$
$解:$
$(1) f(x) = F^{\prime}(x) = \left \{ \begin{array}{l} \dfrac{1}{12}, -3 \leq x < 9, \\ 0 , 其他. \end{array} \right.$
$(2) P\lbrace a < X < a + l \rbrace = \int_{a}^{a+l} \dfrac{1}{12}dx = \dfrac{l}{12}$

$\color{blue}{2.指数分布}$

$若连续型随机变量X的概率密度为f(x) = \left \{ \begin{array}{l} \lambda e^{-\lambda x}, x > 0 \\ 0, 其他 \end{array} \right.$
$其中\lambda > 0为常数,则称X服从参数为\lambda的指数分布.其分布函数为:$
$F(x) = \left \{ \begin{array}{l} 1 - e^{-\lambda x}, x > 0, \\ 0, x \leq 0. \end{array} \right.$

$例4.已知某种电子管的寿命X(单位小时)服从指数分布,其概率密度为$
$f(x) = \left \{ \begin{array}{c} \dfrac{1}{1000}e^{-\frac{x}{1000}}, x > 0, \\ 0, x \leq 0, \end{array} \right.$
$求这种电子管能使用1000小时以上的概率.$
$解: P\lbrace X > 1000 \rbrace = \int_{1000}^{\infty} f(x) dx \\ = \int_{1000}^{\infty} \dfrac{1}{1000}e^{-\frac{x}{1000}} dx \\ = -[e^{-\frac{x}{1000}}]_{1000}^{\infty} \\ = 0.368$

$\color{blue}{\S 2.4正态分布}$

$\color{blue}{一、正态分布的定义与性质}$

$若连续型随机变量X的概率密度为:\\ f(x) = \dfrac{1}{\sqrt{2\pi} \sigma } e^{-\frac{(x-\mu)^2}{2 \sigma^2}}, -\infty < x < +\infty \\ 其中\mu, \sigma (\sigma > 0)为常数,则称X服从参数为\mu, \sigma^2的正太分布,\\ 记作X \sim N(\mu, \sigma^2). 其分布函数为:\\ F(x) = \dfrac{1}{\sqrt{2\pi} \sigma } \int_{-\infty}^{x} e^{-\frac{(t-\mu)^2}{2 \sigma^2 }} dt, -\infty < x < +\infty$

$性质:f(x)关于直线x = \mu 对称,并在 x = \mu 处取最大值\dfrac{1}{\sqrt{2 \pi} \sigma}.$

$实际问题中,许多随机变量都服从正太分布,如测量长度的误差,\\ 机器包装货物的重量等.$

$\color{blue}{二、标准正态分布}$

$设X \sim N(\mu, \sigma^2),如果\mu = 0, \sigma = 1,则X服从标准正态分布.\\ 记作X \sim N(0, 1).X的概率密度为\\ \varphi(x) = \dfrac{1}{\sqrt{2 \pi}}e^{-\frac{x^2}{2}}, -\infty < x < +\infty \\ 分布函数为\\ \Phi(x) = \int_{-\infty} ^ {x} \dfrac{1}{\sqrt{2 \pi}}e^{-\frac{t^2}{2}} dt, -\infty < x < +\infty$

$\Big( \int_0^{+\infty} e^{-x^2} dx = \dfrac{\sqrt{\pi}}{2}, \int_{-\infty} ^{+\infty} e^{-x^2} dx = \sqrt{\pi} \Big)$

$性质:若X \sim N(0, 1), 则\Phi(x) = P\lbrace X \leq x \rbrace 有表可查.$
$P\lbrace X > x_1 \rbrace = 1 - P\lbrace X \leq x_1 \rbrace = 1 - \Phi(x_1)$
$P\lbrace x_1 < X < x_2 \rbrace = P\lbrace x_1 \leq X \leq x_2 \rbrace = \Phi(x_2) - \Phi(x_1)$
$\Phi(-x) = 1 - \Phi(x)$

$例5.设 X \sim N(0, 1).\\ 求P\lbrace |X| \leq 1 \rbrace,P\lbrace |X| \leq 3 \rbrace,P\lbrace |X| > 1.96 \rbrace$
$解:P\lbrace |X| \leq 1 \rbrace = P\lbrace -1 \leq X \leq 1 \rbrace = \Phi(1) - \Phi(-1) = 2\Phi(1) - 1 = 0.6826, \\ P\lbrace |X| \leq 3 \rbrace = 2\Phi(3) - 1 = 0.9974, \\ P\lbrace |X| > 1.96 \rbrace = 1 - P\lbrace |X| \leq 1.96 \rbrace = 2[1 - \Phi(1.96)] = 0.05.$

$\color{blue}{三、一般正态分布与标准正态分布的关系}$

$设X \sim N(\mu, \sigma^2)分布函数为F(x),则\\ F(x) = \dfrac{1}{\sqrt{2\pi} \sigma}\int_{-\infty}^{+\infty} e^{-\frac{(t - \mu)^2}{2 \sigma^2}} dt \overset{u = \frac{t - u}{\sigma}}{=} \dfrac{1}{\sqrt{2 \pi}}\int_{-\infty}^{\frac{x-\mu}{\sigma}} e^{-\frac{u^2}{2}} du = \Phi(\dfrac{x - \mu}{\sigma}) \\ 即F(x) = \Phi(\dfrac{x - \mu}{\sigma}) \quad Y = \dfrac{X - \mu}{\sigma} \sim N(0, 1)$

$例6.设X \sim N(1, 4),求\\ (1)P\lbrace X > 0 \rbrace; (2) P\lbrace -1 < X < 2 \rbrace; \\ (3) P\lbrace |X - 2| < 1 \rbrace; (4) P\lbrace |X - 1| \geq 1 \rbrace.$
$解: \mu = 1, \sigma = 2 \\ P\lbrace X > 0 \rbrace \\ = 1 - P\lbrace X \leq 0 \rbrace \\ = 1 - F(0) \\ = 1 - \Phi(\dfrac{0 - 1}{2}) \\ = 1 - \Phi(-0.5) \\ = 1 - [1 - \Phi(0.5)] \\ = \Phi(0.5) \\ = 0.6915 \\ (2) P\lbrace -1 < X < 2 \rbrace \\ = F(2) - F(-1) \\ = \Phi(\dfrac{2 - 1}{2}) - \Phi(\dfrac{-1 - 1}{2}) \\ = \Phi(\dfrac{1}{2}) - \Phi(-1) \\ = \Phi(0.5) - 1 + \Phi(1) \\ = 0.6915 - 1 + 0.8413 \\ = 0.5328 \\ (3) P\lbrace |X - 2| < 1 \rbrace \\ = P\lbrace 1 \leq X \leq 3 \rbrace \\ = F(3) - F(1) \\ = \Phi(\dfrac{3 - 1}{2}) - \Phi(\dfrac{1 - 1}{2}) \\ = \Phi(1) - \Phi(0) \\ = 0.8413 - 0.5 \\ = 0.3413 \\ (4) P\lbrace |X - 1| \geq 1 \rbrace \\ = 1 - P\lbrace |X - 1| < 1 \rbrace \\ = 1 - P\lbrace 0 < X < x \rbrace \\ = 1 - F(2) + F(0) \\ = 1 - \Phi(0.5) + \Phi(-0.5) \\ = 2[1 - \Phi(0.5)] \\ = 2(1 - 0.6915) \\ = 0.6170$

$例7.设X \sim N(\mu, \sigma^2),求 P\lbrace |X - \mu| < 3 \sigma \rbrace$
$解: P\lbrace |X - \mu| < 3 \sigma \rbrace \\ = P\lbrace \mu - 3 \sigma < X < \mu + 3 \sigma \rbrace \\ = F(\mu + 3 \sigma) - F(\mu - 3 \sigma) \\ = \Phi(3) - \Phi(-3) \\ = 2\Phi(3) - 1 \\ = 0.9974$

$P\lbrace |X - \mu| < k \sigma \rbrace = 2\Phi(k) - 1$

$\color{blue}{四、标准正态分布的上\alpha分位点}$

$设X \sim N(0, 1),对于给定的\alpha( 0 < \alpha < 1).若点 \mu _{\alpha} 满足$
$P\lbrace X > \mu _{\alpha} \rbrace = \alpha,则称点 \mu_{\alpha} 为标准正态分布的上\alpha分位点.$
$-\mu_{\alpha} = \mu_{1 - \alpha}, \Phi(\mu_{\alpha}) = 1 - \alpha .$
$当\alpha = 0.025时,由\Phi(\mu_{\alpha}) = 1 - \alpha = 0.975, \\ 查表得\mu_{0.025} = 1.96$

$\color{blue}{\S 2.5 随机变量的函数分布}$

$定义.设有函数y = f(x), X与Y是两个随机变量,\\ 如果当随机变量X取x时,随机变量Y取值为y = f(x),\\ 则称随机变量Y是随机变量X的函数,记作Y = f(X).$
$本节举例说明由随机变量X的分布,求Y = f(X)的分布方法.$

$例1.设离散型随机变量X的分布律为$
$\begin{array}{l | c c} X & -1 & 0 & 1 & 2 & 3 \\ \hline P & \dfrac{1}{10} & \dfrac{2}{10} & \dfrac{3}{10} & \dfrac{3}{10} & \dfrac{1}{10} \end{array}$
$求Y_1 = -2X和Y_2 = X^2的分布律.$
$解:$
$\begin{array}{c | c c} X & -1 & 0 & 1 & 2 & 3 \\ \hline Y_1 = -2X & 2 & 0 & -2 & -4 & -6 \\ \hline Y_2 = X^2 & 1 & 0 & 1 & 4 & 9 \\ \hline P & \dfrac{1}{10} & \dfrac{2}{10} & \dfrac{3}{10} & \dfrac{3}{10} & \dfrac{1}{10} \end{array}$

$\begin{array}{c | c c} Y_1 & 2 & 0 & -2 & -4 & -6 \\ \hline P & \dfrac{1}{10} & \dfrac{2}{10} & \dfrac{3}{10} & \dfrac{3}{10} & \dfrac{1}{10} \end{array}$

$\begin{array}{c | c c} Y_2& 0 & 1 & 4 & 9 \\ \hline P & \dfrac{2}{10} & \dfrac{4}{10} & \dfrac{3}{10} & \dfrac{1}{10} \end{array}$

$例2.设X的概率密度为f_X(x) = \left \{ \begin{array}{l} \dfrac{x}{8}, 0 < x < 4, \\ 0, 其他. \end{array} \right.$
$求Y = 2X + 8的概率密度.$
$解: 先求Y的分布函数F_Y(y). \\ F_Y(y) = P\lbrace Y \leq y \rbrace \\ = P\lbrace 2X + 8 \leq y \rbrace \\ = P\lbrace X \leq \dfrac{y - 8}{2} \rbrace \\ = F_X(\dfrac{y - 8}{2}) \\ f_Y(y) = \dfrac{d}{dy}F_Y(y) = \dfrac{d}{dy}F_X(\dfrac{y - 8}{2}) \\ = f_X(\dfrac{y-8}{2})\cdot (\dfrac{y - 8}{2})^{\prime} \\ = \dfrac{1}{2}f_X(\dfrac{y-8}{2})$
$=\left \{ \begin{array}{l}\dfrac{1}{2} \dfrac{y-8}{16}, 0 < \dfrac{y-8}{2} < 4 \\ 0, 其他 \end{array} \right.$
$=\left \{ \begin{array}{l} \dfrac{y-8}{32}, 8 < y< 12 \\ 0, 其他 \end{array} \right.$

$例3.已知X \sim N(\mu, \sigma^2),求Y= \dfrac{X - \mu}{\sigma}的概率密度.$
$解:f_X(x) = \dfrac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{(x - \mu)^2}{2\sigma^2}}, -\infty < x < +\infty, \\ F_Y(y) = P\lbrace Y \leq y \rbrace = P\lbrace \dfrac{X - \mu}{\sigma} \leq y \rbrace = P\lbrace X \leq \sigma y + \mu \rbrace \\ = F_x(\sigma + \mu). \\ f_Y(y) = \dfrac{d}{dy}F_Y(y) = f_X(\sigma y + \mu) \cdot (\sigma y + \mu)^{\prime} = \sigma f_X(\sigma y + \mu) \\ = \dfrac{1}{\sqrt{2 \pi}} e^{-\frac{y^2}{2}}, -\infty < y < +\infty. \\ 即 Y = \dfrac{X - \mu}{\sigma} \sim N(0, 1).$

$例4.设X \sim N(0, 1),求Y = X^2的概率密度.$
$解:由于Y = X^2的取值非负,故当 y \leq 0时, \\ F_Y(y) = P\lbrace Y \leq y \rbrace = 0 \quad f_Y(y) = 0. \\ 当y > 0时, \\ F_Y(y) = P\lbrace Y \leq y \rbrace = P\lbrace X^2 \leq y \rbrace = P \lbrace -\sqrt{y} \leq X \leq \sqrt{y} \rbrace \\ = \Phi(\sqrt{y}) - \Phi(-\sqrt{y}) = 2\Phi(\sqrt{y}) - 1 \\ f_Y(y) = \dfrac{d}{dy}F_Y(y) = 2 \varphi(\sqrt{y})(\sqrt{y})^{\prime} = \dfrac{1}{\sqrt{y}} \varphi(\sqrt{y}) \\ = \dfrac{1}{\sqrt{2\pi y}} e^{-\frac{y}{2}}$

$练习:1.设,$ $\begin{array}{l | c c } X & -1 & 0 & 1 & 2 & 3 \\ \hline P & \dfrac{1}{5} & \dfrac{1}{6} & \dfrac{1}{5} & \dfrac{1}{15} & \dfrac{11}{30} \end{array}$ $求Y = 2X^2 + 1的分布律$
$解:$$\begin{array}{l | c c } X & -1 & 0 & 1 & 2 & 3 \\ \hline Y=2X^2 + 1 & 3 & 1 & 3 & 9 & 19 \\ \hline P & \dfrac{1}{5} & \dfrac{1}{6} & \dfrac{1}{5} & \dfrac{1}{15} & \dfrac{11}{30} \end{array}$
$\begin{array}{l | c c } Y & 1 & 3 & 9 & 19 \\ \hline P & \dfrac{1}{6} & \dfrac{2}{5} & \dfrac{1}{15} & \dfrac{11}{30} \end{array}$

$2.设X \sim N(\mu, \sigma^2),求Y = a X + b的密度.$

$3.设X \sim N(0, 1),求Y = e^X的密度.$