概率统计D 03.02 多维随机变量及其分布 B

$\color{blue}{\S 3.4 随机变量的相互独立性}$

$定义:设(X, Y)是二维随机变量,若X与Y的联合分布等于\\ 边缘分布的乘积,则称X与Y相互独立.$

$若(X, Y)的分布函数为F(x, y),关于X、Y的边缘分布函数为\\ F_X(x)、F_Y(y),则X与Y相互独立的充分必要条件是\\ F(x, y) = F_X(x) \cdot F_Y(y).$

$对于二维离散型随机变量(X, Y),X与Y相互独立的充分必要条件\\ 是联合分布律等于边缘分布律的乘积,即\\ P\lbrace X = x_i, Y = y_j \rbrace = P\lbrace X = x_i \rbrace P\lbrace Y = y_j \rbrace\begin{pmatrix}i = 1, 2, \cdots \\ j = 1, 2, \cdots \end{pmatrix}$

$对于二维连续型随机变量(X, Y),X与Y相互独立的充分必要条件\\ 是联合概率密度等于边缘概率密度的乘积,即\\ f(x, y) = f_X(x) f_Y(y) \begin{pmatrix}-\infty < x < + \infty \\ -\infty < y < +\infty \end{pmatrix}$

$定理:设随机变量X与Y相互独立,则\\ (1) 对任意常数a, b, c, d, 事件\lbrace a < X < b \rbrace 与 \lbrace c < Y < d \rbrace 相互独立; \\ (2) 对任意常数a, b, c, d, 随机变量aX + b与cY + d相互独立; \\ (3) X^2 与 Y^2相互独立; \\ (4) 对任意连续函数h,g,随机变量h(X)与g(Y)相互独立.$

$例1.第二节例1中的随机变量X与Y不相互独立,\\\\因为p_{11} = \dfrac{1}{6}, 而p_{1 \cdot} = \dfrac{1}{6}, p_{\cdot 1} = \dfrac{1}{2}, p_{11} \neq p_{1\cdot} p_{\cdot 1}$
$例1.设(X, Y)的联合分布律为$
$\begin{array}{c c | c c } & Y & 1 & 2 \\ X & & & & & & \\ \hline 1 & & 1/6 & 0 \\ 2 & & 0 & 1/6 \\ 3 & & 1/6 & 0 \\ 4 & & 0 & 1/6 \\ 5 & & 1/6 & 0 \\ 6 & & 0 & 1/6 \\ \end{array}$

$例2.设二维随机变量(X, Y)的概率密度为$
$f(x, y) = \left \{ \begin{array}{l} xe^{-(x + y)}, x > 0, y > 0 , \\ 0 , 其它 \end{array} \right.$
$问X与Y是否相互独立?$
$解:\\ f_X(x) = \int_{-\infty}^{+\infty} f(x, y) dy \\ = \int _0^{+\infty} f(x, y) dy \\ = \int_0^{+\infty} xe^{-x}e^{-y} dy \\ = xe^{-x}\int_{+\infty}^0 de^{-y} \\ = xe^{-x}[e^{-y}]_{+\infty}^0 \\ = xe^{-x} (x > 0)$
$即f_X(x) = \left \{ \begin{array}{l} xe^{-x}, x > 0, \\ 0 , x \leq 0 \end{array} \right.$
$f_Y(y) = \int_{-\infty}^{+\infty} f(x, y) dx \\ = \int_{0}^{+\infty} f(x, y) dx \\ = \int_0^{+\infty}e^{-y} xe^{-x} dx \\ = e^{-y}\int_0^{+\infty}(-x) de^{-x} \\ = e^{-y}[(-xe^{-x} )_0^{+\infty} - \int_0^{+\infty}e^{-x} d(-x)] \\ = e^{-y}[-(e^{-x})_0^{+\infty}] \\ = e^{-y} (y > 0)$
$即f_Y(y) = \left \{ \begin{array}{l} e^{-y}, y > 0, \\ 0 , y \leq 0 \end{array} \right.$
$对于任意x, y有 f(x,y) = f_X(x) f_Y(y),即X与Y相互独立.$

$例3.设X与Y独立,f_X(x) = \left \{ \begin{array}{l} 1, 0 \leq x \leq 1, \\ 0 , 其它 \end{array} \right. f_Y(y) = \left \{ \begin{array}{l} e^{-y}, y > 0, \\ y \leq 0, \end{array} \right.$
$求P\lbrace 0 < X < \dfrac{1}{2}, 0 < Y < 2 \rbrace.$
$解:\\ P\lbrace 0 < X < \dfrac{1}{2}, 0 < Y < 2 \rbrace \\ = P\lbrace 0 < X < \dfrac{1}{2} \rbrace P\lbrace 0 < Y < 2 \rbrace \\ = \int_0^{\frac{1}{2}} f_X(x) dx \cdot \int_0^2 f_Y(y) dy \\ = \int_0^{\frac{1}{2}} 1 dx \int_0^2 e^{-y} dy \\ = \dfrac{1}{2}(1 - e^{-2})$

$例4.设(X, Y) \sim N( \mu_1, \mu_2, \sigma_ 1^2, \sigma_ 2^2, \rho),\\ 证明X与Y相互独立的充分必要条件是\rho = 0.$
$证明: f(x, y) = \dfrac{1}{2 \pi \sigma_1 \sigma_2 \sqrt{1 - \rho^2}}e^{-\frac{1}{2(1-\rho^2)}[\frac{(x -\mu_1)^2}{\sigma_1^2} - 2\rho \frac{(x - \mu_1)(y - \mu_2)}{\sigma_1 \sigma_2} + \frac{(y - \mu_2)^2}{\sigma_2^2}]} \\ ① 充分性,假设\rho = 0,则有 \\ f(x, y) = \dfrac{1}{2\pi \sigma_1 \sigma_2} e^{-\frac{(x - \mu_1)^2}{2\sigma_1^2} - \frac{(y-\mu_2)^2}{2\sigma_2^2} }, \\ 而f_X(x) = \dfrac{1}{\sqrt{2 \pi } \sigma_1} e^{-\frac{(x - \mu_1)^2}{2\sigma_1^2}}, f_Y(y) = \dfrac{1}{\sqrt{2 \pi} \sigma_2} e^{-\frac{(y - \mu_2)^2}{2\sigma_2^2}}, \\ 于是f(x, y) = f_X(x)f_Y(y),即X与Y相互独立. \\ ②必要性:假设X与Y相互独立,即对任意x, y有f(x, y) = f_X(x) f_Y(y), 即\\ \dfrac{1}{2 \pi \sigma_1 \sigma_2 \sqrt{1 - \rho^2}}e^{-\frac{1}{2(1-\rho^2)}[\frac{(x -\mu_1)^2}{\sigma_1^2} - 2\rho \frac{(x - \mu_1)(y - \mu_2)}{\sigma_1 \sigma_2} + \frac{(y - \mu_2)^2}{\sigma_2^2}]} = \dfrac{1}{\sqrt{2 \pi } \sigma_1} e^{-\frac{(x - \mu_1)^2}{2\sigma_1^2}} \dfrac{1}{\sqrt{2 \pi} \sigma_2} e^{-\frac{(y - \mu_2)^2}{2\sigma_2^2}} \\ 特别的,令x = \mu_1, y = \mu_2，则得\\ \dfrac{1}{2 \pi \sigma_1 \sigma_2 \sqrt{1 - \rho^2}} = \dfrac{1}{2\pi \sigma_1 \sigma_2} \\ 从而有 \rho = 0$

$\color {blue}{\S 3.5 两个随机变量的函数的分布}$

$定义:设(X, Y)是二维随机变量, z = f(x, y)是二元函数, \\ 若当(X, Y)取值(x, y)时,随机变量Z取值为z = f(x, y), \\ 则称Z是X,Y的函数,记作Z = f(X,Y).$

$\color{blue}{一、Z = X +Y的分布}$

$例1.设X,Y相互独立,X \sim \pi(\lambda_1),Y \sim \pi(\lambda_2),求Z = X + Y的分布.$
$解:Z = X + Y的可取值为0, 1, 2, \cdots, 对任意正整数k,有\\ P\lbrace Z = k \rbrace = P\lbrace X +Y = k \rbrace \\ = P(\sum \limits _{i = 0}^k \lbrace X = i, Y = k-i \rbrace ) \\ = \sum \limits _{i = 0}^k P \lbrace X = i, Y = k - i \rbrace \\ = \sum \limits _{i = 0}^k P\lbrace X = i \rbrace P \lbrace Y = k - i \rbrace \\ = \sum \limits _{i = 0}^k \dfrac{\lambda_1^ie^{-\lambda_1}}{i!} \dfrac{\lambda_2^{k-i}e^{-\lambda_2}}{(k-i)!} \\ = e^{-(\lambda_1 + \lambda_2)} \sum \limits _{i=0}^k \dfrac{k!}{i!(k-i)!}\lambda_1^i \lambda_2^{k-i} \cdot \dfrac{1}{k!} \\ = \dfrac{(\lambda_1 + \lambda_2)^k}{k!} e^{-(\lambda_1 + \lambda_2)}, \\ 即 Z = X + Y \sim \pi(\lambda_1 + \lambda_2).$

$设二维连续型随机变量(X, Y)的概率密度为f(x, y),\\ 则Z= X+ Y的分布函数为F_Z(z) = P\lbrace Z \leq z \rbrace \\ = \iint \limits _{x + y \leq z} f(x, y) dxdy \\ = \int_{-\infty}^{+\infty} [\int_{-\infty}^{z - x} f(x, y) dy] dx \\ 由于 \int_{-\infty}^{z - x} f(x, y) dy \overset{y = u - x}{=} \int_{-\infty}^{z} f(x, u-x) du, \\ 所以 F_Z(z) = \int_{-\infty}^{+\infty}[\int_{-\infty}^z f(x, u -x) du] dx \\ \overset{假设}{=} \int_{-\infty}^{z} [\int_{-\infty}^{+\infty} f(x, u-x) dx] du, \\ 从而得到Z = X + Y的概率密度为\\ f_Z(z) = \int_{-\infty}^{+\infty} f(x, z-x)dx (-\infty < z < +\infty) \\ 同理可得\\ f_Z(z) = \int_{-\infty}^{+\infty} f(z - y, y) dy (-\infty < z < +\infty)$

$当X与Y相互独立时,有\\ f_Z(z) = \int_{-\infty}^{+\infty} f_X(z - y)f_Y(y) dy , \\ 或\\ f_Z(z) = \int_{-\infty}^{+\infty} f_X(x)f_Y(z - x) dx.$

$例2.设X,Y相互独立,均服从标准正态分布,求Z = X + Y的概率密度.$
$解:f_X(x) = \dfrac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}, -\infty < x < +\infty, \\ f_Y(y) = \dfrac{1}{\sqrt{2 \pi}}e^{-\frac{y^2}{2}}, -\infty < y < +\infty, \\ f_Z(z) = f_X * f_Y = \int_{-\infty}^{+\infty} f_X(x) f_Y(z - x) dx \\ = \int_{-\infty}^{+\infty} \dfrac{1}{\sqrt{2 \pi}}e^{-\frac{x^2}{2}}\dfrac{1}{\sqrt{2 \pi}}e^{-\frac{(z -x)^2}{2}} dx \\ = \dfrac{1}{2 \pi}e^{-\frac{z^2}{4}}\int_{-\infty}^{+\infty} e^{-(x - \frac{z}{2})^2} dx \\ = \dfrac{1}{2 \pi}e^{-\frac{z^2}{4}}\int_{-\infty}^{+\infty} e^{-t^2} dt \\ = \dfrac{1}{2 \pi} e^{-\frac{z^2}{4}} \sqrt{\pi} \\ = \dfrac{1}{\sqrt {2 \pi} \sqrt{2}} e^{-\frac{z^2}{2(\sqrt{2})^2}} \\ 即Z = X + Y \sim N(0, (\sqrt{2})^2).$

$例3.设X与Y相互独立,且都在区间(0, 1)上服从均匀分布, \\ 求Z = X + Y的概率密度.$
$解:f_X(x) = \left \{ \begin{array}{l} 1, 0 < x < 1, \\ 0, 其它 \end{array} \right. f_Y(y) = \left \{ \begin{array}{l} 1, 0 < y < 1, \\0 , 其它. \end{array} \right.$
$由卷积公式有\\ f_Z(z) = \int_{-\infty}^{+\infty} f_X(z-y)f_Y(y) dy \\ =\int_0^1 f_X(z - y)dy \\ \overset{t = z-y}{=} \int_{z-1}^{z} f_X(t) dt$
$(1)当z < 0时, z - 1 < 0, 此时f_Z(z) = 0.$
$(2) 当0 \leq z < 1时, z - 1 < 0, 此时f_Z(z) = \int_0^z 1 dt = z.$
$(3) 当1 \leq z < 2时, 0 \leq z - 1 < 1,此时f_Z(z) = \int_{z-1}^1 1 dt = 2 - z.$
$(4) 当 z \geq 2时, z-1 \geq 1,此时f_Z(z) = 0.$
$因此 f_Z(z) = \left \{ \begin{array}{l} z, 0 \leq z < 1, \\ 2 - z, 1 \leq z < 2, \\ 0, z \geq 2 或 z < 0. \end{array} \right.$

$例4.设随机变量X和Y相互独立,并且都服从正态分布\\ N(0, \sigma^2)(\sigma > 0),求Z = \sqrt{x^2 + y^2}的分布.$
$解:f_X(x) = \dfrac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{x^2}{2\sigma^2}}, -\infty < x < +\infty \\ f_Y(y) = \dfrac{1}{\sqrt{2 \pi} \sigma} e^{-\frac{y^2}{2\sigma^2}}, -\infty < y < +\infty \\ f(x, y) = f_X(x)f_Y(y) = \dfrac{1}{2\pi \sigma}e^{-\frac{x^2 + y^2}{2\sigma^2}} \\ F_Z(z) = P \lbrace Z \leq z \rbrace = P\lbrace \sqrt {x^2 + y^2 } \leq z \rbrace \\ = \iint \limits _{\sqrt{x^2 + y^2} \leq z} f(x, y) dxdy \\ = \iint \limits _{\sqrt{x^2 + y^2} \leq z} \dfrac{1}{2 \pi \sigma^2} e^{-\frac{x^2 + y^2}{2\sigma^2}} dxdy \\ = \dfrac{1}{2 \pi} \int_0^{2 \pi} d\theta \int_0^{z} \dfrac{1}{\sigma^2}e^{-\frac{r^2}{2\sigma^2}}rdr \\ = 1 - e^{-\frac{z^2}{2\sigma^2}} \\ 当z \leq 0时,F_Z(z) = P\lbrace \sqrt{X^2 + Y^2} \leq z \rbrace = P(\phi) = 0.$
$综上 F_Z(z) = \left \{ \begin{array}{l} 1 - e^{-\frac{z^2}{2\sigma^2}}, z \geq 0, \\ 0, z < 0 \end{array} \right.$
$从而得到Z的密度为$
$f_Z(z) = \left \{ \begin{array}{l} \dfrac{z}{\sigma^2} e^{-\frac{z^2}{2\sigma^2}}, z \geq 0, \\ 0, z < 0 \end{array} \right.$