數理統計與數據分析第三版習題 第3章 第54-69題

題目54

令$X$,$Y$和$Z$是獨立的$N(0,{\sigma}^2)$.隨機變量$\Theta,\Phi和R$是$(X,Y,Z)$的球形座標：
$\begin{aligned} x&=rsin\phi cos\theta\\ y&=rsin\phi sin\theta\\ z&=rcos\phi\\ 0\leq \phi&\leq \pi, 0\leq \theta\leq 2\pi \end{aligned}$
計算$\Theta,\Phi和R$的聯合密度和邊際密度.(提示:$dxdydz=r^2sin\phi drd\theta d\phi$)

解題思路

根據球形座標的轉換提示，我們直接求聯合密度：
$\begin{aligned} f_{R,\Phi,\Theta}(r,\phi,\theta)&=f_{X,Y,Z}(rsin\phi cos\theta,rsin\phi sin\theta,rcos\phi) r^2sin\phi\\ &=\frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(rsin\phi cos\theta)^2}{2\sigma^2}}\cdot \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(,rsin\phi sin\theta)^2}{2\sigma^2}} \cdot \frac{1}{\sqrt{2\pi}\sigma}e^{-\frac{(rcos\phi)^2}{2\sigma^2}} r^2 \sin\phi\\ &={(2\pi)}^{-\frac32} \cdot \sigma^{-3}\cdot e^{-\frac{r^2sin^2\phi cos^2\theta+r^2\sin^2\phi \sin^2\theta+r^2\cos^2\phi}{2\sigma^2}}r^2 \sin\phi\\ &=\sin\phi \cdot \frac1{2\pi}\cdot \frac1{\sqrt{2\pi}}\cdot \frac1{\sigma^3}\cdot r^2 \cdot e^{-\frac{r^2}{2\sigma^2}} \end{aligned}$

求邊際密度
$\begin{aligned} f_r(r)&=\int_0^\pi\int_0^{2\pi}f_{R,\Phi,\Theta}(r,\phi,\theta)d\theta d\phi\\ &=\frac1{2\pi}\cdot \frac1{\sqrt{2\pi}}\cdot \frac1{\sigma^3}\int_0^\pi\int_0^{2\pi}\sin\phi\cdot r^2\cdot e^{-\frac{r^2}{2\sigma^2}}d\theta d\phi\\ &=\frac1{2\pi}\cdot \frac1{\sqrt{2\pi}}\cdot \frac1{\sigma^3}\int_0^\pi\sin\phi d\phi \cdot\int_0^{2\pi}d\theta \cdot r^2e^{-\frac{r^2}{2\sigma^2}}\\ &=\frac1{2\pi}\cdot \frac1{\sqrt{2\pi}}\cdot \frac1{\sigma^3} \cdot 2 \cdot 2\pi \cdot r^2e^{-\frac{r^2}{2\sigma^2}}\\ &=\frac{\sqrt2}{\sqrt\pi \sigma^3}r^2e^{-\frac{r^2}{2\sigma^2}} \end{aligned}$
$\begin{aligned} f_\Phi(\phi)&=\int_0^{\infty}\int_0^{2\pi}f_{R,\Phi,\Theta}(r,\phi,\theta)d\theta dr\\ &=\frac1{2\pi}\cdot \frac1{\sqrt{2\pi}}\cdot \frac1{\sigma^3}\int_0^\infty\int_0^{2\pi}\sin\phi\cdot r^2\cdot e^{-\frac{r^2}{2\sigma^2}}d\theta dr\\ &=\frac1{2\pi}\cdot \frac1{\sqrt{2\pi}}\cdot \frac1{\sigma^3}\cdot \sin\phi \cdot \int_0^{2\pi}d\theta \int_0^\infty r^2e^{-\frac{r^2}{2\sigma^2}} dr\\ &=\frac1{2\pi}\cdot \frac1{\sqrt{2\pi}}\cdot \frac1{\sigma^3} \cdot \sin\phi \cdot 2\pi \cdot \frac{\sqrt\pi}{\sqrt2}\sigma^3\\ &=\frac{\sin\phi}{2} \end{aligned}$
$\begin{aligned} f_\Theta(\theta)&=\int_0^{\infty}\int_0^{\pi}f_{R,\Phi,\Theta}(r,\phi,\theta)d\phi dr\\ &=\frac1{2\pi}\cdot \frac1{\sqrt{2\pi}}\cdot \frac1{\sigma^3}\int_0^{\pi}\sin\phi d\phi \cdot \int_0^\infty r^2\cdot e^{-\frac{r^2}{2\sigma^2}}dr\\ &=\frac1{2\pi}\cdot \frac1{\sqrt{2\pi}}\cdot \frac1{\sigma^3}\cdot 2 \cdot \frac{\sqrt\pi}{\sqrt2}\sigma^3\\ &=\frac1{2\pi} \end{aligned}$
$f_{R,\Phi,\Theta}(r,\phi,\theta)=f_r(r)f_\Phi(\phi)f_\Theta(\theta)$三個變量轉換爲球形座標後也是獨立的。

計算上邊三個邊際密度時，有一個關鍵積分公式：
$\int_0^\infty x^{2n}e^{-\frac{x^2}{\alpha^2}}dx=\sqrt\pi\frac{(2n)!}{n!}(\frac{\alpha}{2})^{2n+1}$

題目55

從單位元內部按如下規則生成一點：半徑R是[0,1]上的均勻隨機變量，角度$\Theta是[0,2\pi]$上我均勻隨機變量，並且與R獨立

a.計算$X=R\cos\theta,Y=R\sin\theta$的隨便密度
b.計算$X和Y$的邊際密度
c.密度是圓盤上的均勻分佈嗎？如果不是修正該方法是密度是均勻分佈

解題思路

a.
$\begin{aligned} &R\sim U[0,1]\\ &\Theta\sim U[0,2\pi]\\ &(X=R\cos\Theta,Y=R\sin\Theta)\Rightarrow (R=\sqrt{X^2+Y^2},\Theta=\tan^{-1}\frac{Y}{X})\\ &|J|=\begin{Vmatrix} \frac{\partial r}{\partial x} & \frac{\partial r}{\partial y} \\ \frac{\partial \theta}{\partial x} & \frac{\partial \theta}{\partial y} \end{Vmatrix} =\frac{1}{\sqrt{x^2+y^2}}\\ f_{XY}(x,y)&=f_{R\Theta}(\sqrt{x^2+y^2},\tan^{-1}\frac{y}{x})\cdot\frac{1}{\sqrt{x^2+y^2}}\\ &=\frac1{2\pi}\cdot\frac{1}{\sqrt{x^2+y^2}} \qquad x^2+y^2\leq 1 \end{aligned}$
b.計算$X和Y$的邊際密度
$\begin{aligned} f_X(x)&=\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}f_{XY}(x,y)dy\\ &=\int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\frac1{2\pi}\cdot\frac{1}{\sqrt{x^2+y^2}}dy\\ &=\frac1\pi\int_{0}^{\sqrt{1-x^2}}\frac{1}{\sqrt{x^2+y^2}}dy\\ &=\frac1\pi[\ln(y+\sqrt{x^2+y^2})] \Bigg |_0^{\sqrt{1-x^2}}\\ &=\frac1\pi\ln(\frac{\sqrt{1-x^2}+1}{|x|}) \quad -1\leq x \leq 1\\ 同樣道理：\\ f_Y(y)=&\frac1\pi\ln(\frac{\sqrt{1-y^2}+1}{|y|}) \quad -1\leq y \leq 1\\ \end{aligned}$
c.密度是圓盤上的均勻分佈嗎？如果不是修正該方法是密度是均勻分佈
$\begin{aligned} f_R(r)=1\Rightarrow F_R(r)=r \end{aligned}$
如果要在圓盤上均勻分佈，則應該按面積進行均勻分佈如果整個圓是1 ，則：
$F_R(r)=\frac{2\pi r^2}{2\pi}=r^2 很明顯 r!=r^2$,不是均勻分佈

$F_R(r)=r^2\Rightarrow f_R(r)=2r$

$f_{R\Theta}(r,\theta)=2r\cdot\frac1{2\pi}=\frac r\pi$

題目56

如果$X$和$Y$是獨立的隨機變量，計算點$（X，Y）$的極座標R和$\Theta$的聯合分佈，R和$\Theta$獨立嗎？

解題思路

未找到答案

題目57

假設$Y_1$和$Y_2$服從二元正態分佈，具有參數$\mu_{Y_1}=\mu_{Y_2}=0,\sigma_{Y_1}^2=1,\sigma_{Y_2}^2=2,且\rho=1/\sqrt{2}$.線性變換 $x_1=a_{11}y_1+a_{12}y_2,x_2=a_{21}y_1+a_{22}y_2$ 使得$x_1$和$x_2$是獨立的標準的正態分佈？

解題思路

幾個關鍵公式
1.二元正態分佈密度函數
$f(y_1,y_2)=\frac{1}{2\pi\sigma_{Y_1}\sigma_{y_2}\sqrt{1-\rho^2}}exp{\left [ -\frac{1}{2(1-\rho^2)}\left( \frac{(y_1-\mu_{Y_1})^2}{\sigma_{y_1}^2}-\frac{2\rho(y_1-\mu_{Y_1})(y_2-\mu_{Y_2})}{\sigma_{Y_1}\sigma_{Y_2}}+ \frac{(y_2-\mu_{Y_2})^2}{\sigma_{y_2}^2}\right)\right]}$
2.設$(Y_1,Y_2)\sim N(\mu_{Y_1},\mu_{Y_2},\sigma_{Y_1}^2,\sigma_{Y_2}^2,\rho)$則$Y_1$與$Y_2$的線性組合仍然服從正態分佈,且$aY_1+bY_2 \sim N(a\mu_{Y_1}+b\mu_{Y_2},a^2\sigma_{Y_1}^2+b^2\sigma_{Y_2}^2+2ab\rho\sigma_{Y_1}\sigma_{Y_2})$

步驟1.求出$(X_1,X_2)$的聯合密度函數
$J(y_1,y_2)=\begin{Vmatrix} \frac{\partial x_1}{\partial y_1} & \frac{\partial x_1}{\partial y_2} \\ \frac{\partial x_2}{\partial y_1} & \frac{\partial x_2}{\partial y_2} \end{Vmatrix} =\begin{Vmatrix} a_{11} & a_{12}\\ a_{21} & a_{22} \end{Vmatrix}=a_{11}a_{22}-a_{12}a_{21}$
$\begin{aligned} y_1&=\frac{a_{12}x_2-a_{22}x_1}{a_{12}a_{21}-a_{22}a_{11}}\\ y_2&=\frac{a_{11}x_2-a_{21}x_1}{a_{22}a_{11}-a_{21}a_{12}}\\ f_{X_1X_2}(x_1,x_2)&=J^{-1}(y_1,y_2)\cdot f_{Y_1Y_2}(\frac{a_{12}x_2-a_{22}x_1}{a_{12}a_{21}-a_{22}a_{11}},\frac{a_{11}x_2-a_{21}x_1}{a_{22}a_{11}-a_{21}a_{12}})\\ 把題目的中參數代&入:\\ &=\frac{1}{a_{11}a_{22}-a_{12}a_{21}}\cdot\frac{1}{2\pi}exp{\left [-\left( \frac{(y_1)^2}{1}-\frac{2\frac{1}{\sqrt2} y_1 y_2}{1\sqrt2}+ \frac{(y_2)^2}{2}\right)\right]}\\ &=\frac{1}{a_{11}a_{22}-a_{12}a_{21}}\cdot\frac{1}{2\pi}exp{\left [-\left( (\frac{a_{12}x_2-a_{22}x_1}{a_{12}a_{21}-a_{22}a_{11}})^2- (\frac{a_{12}x_2-a_{22}x_1}{a_{12}a_{21}-a_{22}a_{11}})(\frac{a_{11}x_2-a_{21}x_1}{a_{22}a_{11}-a_{21}a_{12}})+ \frac{(\frac{a_{11}x_2-a_{21}x_1}{a_{22}a_{11}-a_{21}a_{12}})^2}{2}\right)\right]}\\ \end{aligned}$
步驟2.求$f_{X_1}(x_1)和f_{X_2}(x_2)$
$\begin{aligned} f_{X_1}&=\frac1{\sqrt{2\pi}\sqrt{a_{11}^2\sigma_{Y_1}^2+a_{12}^2\sigma_{Y_2}^2+2a_{11}a_{12}\rho\sigma_{Y_1}\sigma_{Y_2}}}exp\left[ -\frac{x_1-(a_{11}\mu_{Y_1}+a_{12}\mu_{Y_2})}{2(a_{11}^2\sigma_{Y_1}^2+a_{12}^2\sigma_{Y_2}^2+2a_{11}a_{12}\rho\sigma_{Y_1}\sigma_{Y_2})}\right]\\ &=\frac1{\sqrt{2\pi}\sqrt{a_{11}^2+2a_{12}^2+2a_{11}a_{12}}}exp\left[ -\frac{x_1}{2(a_{11}^2+a_{12}^2\cdot2+2a_{11}a_{12})}\right]\\ f_{X_2}&=\frac1{\sqrt{2\pi}\sqrt{a_{21}^2\sigma_{Y_1}^2+a_{22}^2\sigma_{Y_2}^2+2a_{21}a_{22}\rho\sigma_{Y_1}\sigma_{Y_2}}}exp\left[ -\frac{x_2-(a_{21}\mu_{Y_1}+a_{22}\mu_{Y_2})}{2(a_{21}^2\sigma_{Y_1}^2+a_{22}^2\sigma_{Y_2}^2+2a_{21}a_{22}\rho\sigma_{Y_1}\sigma_{Y_2})}\right]\\ &=\frac1{\sqrt{2\pi}\sqrt{a_{21}^2+2a_{22}^2+2a_{21}a_{22}}}exp\left[ -\frac{x_2}{2(a_{21}^2+a_{22}^2\cdot2+2a_{21}a_{22})}\right]\\ f_{X1}f_{X_2}&=\frac1{2\pi\sqrt{a_{11}^2+2a_{12}^2+2a_{11}a_{12}}\sqrt{a_{21}^2+2a_{22}^2+2a_{21}a_{22}}}exp\left[ -\frac{x_1}{2(a_{11}^2+a_{12}^2\cdot2+2a_{11}a_{12})}-\frac{x_2}{2(a_{21}^2+a_{22}^2\cdot2+2a_{21}a_{22})}\right] \end{aligned}$
根據題意要求，即標準的獨立的二元正太分佈，則標準差爲1，$\rho=0$,整理後得到如下等式：
$\begin{aligned} a_{11}a_{22}-a_{12}a_{21}=1\\ a_{11}^2+2a_{12}^2+2a_{11}a_{12}=1\\a_{21}^2+2a_{22}^2+2a_{21}a_{22}=1\\ 2a_{12}a_{22}+a_{12}a_{21}+a_{11}a_{22}+a_{11}a_{21}=0 \end{aligned}$
$a_{11}=1 ,a_{12}=0 ,a_{21}=-1,a_{22}=1$
當然這並不是的解。解可有很多，滿足上邊4個等式即可

題目58

如果$X_1$和$X_2$的聯合分佈是二元正態的，則$Y_1=a_1X_1+b_1,Y_2=a_2X_2+b_2$的聯合分佈也是二元正態的

解題思路

根據線性變換的雅可比行列式進行轉換
$J(x_1,x_2)=\begin{Vmatrix} \frac{\partial y_1}{\partial x_1} & \frac{\partial y_1}{\partial x_2} \\ \frac{\partial y_2}{\partial x_1} & \frac{\partial y_2}{\partial x_2} \end{Vmatrix} =\begin{Vmatrix} a_{1} & 0\\ 0 & a_{2} \end{Vmatrix}=a_1a_2$
$x_1=\frac{y_1-b_1}{a_1},x_2=\frac{y_2-b_2}{a_2}$
$f_{X_1X_2}(x_1,x_2)=\frac{1}{2\pi\sigma_{X_1}\sigma_{X_2}\sqrt{1-\rho^2}}exp{\left [ -\frac{1}{2(1-\rho^2)}\left( \frac{(x_1-\mu_{X_1})^2}{\sigma_{X_1}^2}-\frac{2\rho(x_1-\mu_{X_1})(x_2-\mu_{X_2})}{\sigma_{X_1}\sigma_{X_2}}+ \frac{(x_2-\mu_{X_2})^2}{\sigma_{X_2}^2}\right)\right]}$
$\begin{aligned} f_{Y_1Y_2}(y_1,y_2)&=f_{X_1X_2}(\frac{y_1-b_1}{a_1},\frac{y_2-b_2}{a_2})\cdot|J|^{-1}\\ &=\frac{1}{a_1a_2}\frac{1}{2\pi\sigma_{X_1}\sigma_{X_2}\sqrt{1-\rho^2}}\exp{\left [ -\frac{1}{2(1-\rho^2)}\left( \frac{(\frac{y_1-b_1}{a_1}-\mu_{X_1})^2}{\sigma_{X_1}^2}-\frac{2\rho(\frac{y_1-b_1}{a_1}-\mu_{X_1})(\frac{y_2-b_2}{a_2}-\mu_{X_2})}{\sigma_{X_1}\sigma_{X_2}}+ \frac{(\frac{y_2-b_2}{a_2}-\mu_{X_2})^2}{\sigma_{X_2}^2}\right)\right]}\\ &=\frac{1}{2\pi a_1\sigma_{X_1}a_2\sigma_{X_2}\sqrt{1-\rho^2}}\exp\left[ -\frac{1}{2(1-\rho^2)}\left(\frac{(y_1-(b_1+a_1\mu_{X_1}))^2}{(a_1\sigma_{X_1})^2}-\frac{2\rho(y_1-(b_1+a_1\mu_{X_1}))(y_2-(b_2+a_2\mu_{X_2}))}{a_1\sigma_{X_1}a_2\sigma_{X_2}}+\frac{(y_2+(b_2+a_2\mu_{X_2}))^2}{(a_2\sigma_{X_2})^2}\right)\right] \end{aligned}$
令,$\mu_{Y_1}=b_1+a_1\mu_{X_1},\mu_{Y_2}=b_2+a_2\mu_{X_2},\sigma_{Y_1}=a_1\sigma_{X_1},\sigma_{Y_2}=a_2\sigma_{X_2}$則:
$(Y_1,Y_2)\sim N(\mu_{Y_1},\mu_{Y_2},\sigma_{X_2}^2,\sigma_{X_2}^2,\rho)$

題目59

如果$X_1$和$X_2$是獨立的標準正態隨機變量，證明：$Y_1$和$Y_2$的聯合分佈是二元正態的
$\begin{aligned} Y_1=a_{11}X_1+a_{12}X_2+b_1\\ Y_2=a_{21}X_1+a_{22}X_2+b_2 \end{aligned}$

解題思路

令:
$\begin{aligned} Z_1=a_{11}X_1+a_{12}X_2\\ Z_2=a_{21}X_1+a_{22}X_2\\ \end{aligned}$
則有:
$\begin{aligned} Y_1=Z_1+b_1\\ Y_2=Z_2+b_2 \end{aligned}$
如果X_1和X_2是標準正態隨機變量則他們的聯合分佈是二元正態的即：
$(X_1,X_2)\sim N(0,0,1,1,0)$
二元正態隨機變量的線性組合依然是二元正態的即：
$(Z_1,Z_2)$是二元正態的
再根據58題的結果即可以證明$(Y_1,Y_2)$也是二元正態的(線性變換)

題目60

利用上一題的結果，描述構造由獨立僞隨機均勻變量生成具有二元正態分佈的僞隨機變量的方法

解題思路

step1.採用Box Muller方法由兩個均勻分佈$(u,v)$，得到兩個獨立的正態分佈：
$z_1=\sqrt{-2logu}cos2\pi v\\ z_2=\sqrt{-2logu}sin2\pi$
step2.利用59題的結果將得到的兩個正態分佈樣本進行變量生成的$(y_1,y_2)$即滿足二元正態分佈。

題目61

令$X$和$Y$是具有聯合分佈的連續隨機變量，求$U=a+bX$和$V=c+dY$的聯合分佈密度的表達式

解:

令(X,Y)的聯合密度爲$f_{XY}(x,y)$
利用雅可比行列式$f_{UV}(u,v)=\frac{1}{J(x,y)}f_{XY}(U^{-1},V^{-1})$

$J(x,y)=\begin{Vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{Vmatrix} =\begin{Vmatrix} b & 0\\ 0 & d \end{Vmatrix}=bd$
所以答案：
$f_{UV}(u,v)=\frac{1}{bd}f_{XY}(\frac{u-a}{b},\frac{v-c}{d})$

題目62

令$X$和$Y$是獨立標準正態隨機變量，求$P(X^2+Y^2\leq 1)$

解:

轉化爲極座標：
$\begin{aligned} f_{R\Theta}(r,\theta)&=rf_{XY}(r\cos\theta,r\sin\theta)\\ &=\frac{1}{2\pi}re^{-\frac{r^2}{2}} \end{aligned}$

根據條件$X^2+Y^2\leq 1$,即極座標下$r\leq1$

$\begin{aligned} P(X^2+Y^2\leq 1)&=P(r\leq1)\\ &=\int_0^{2\pi}\int_0^1\frac{1}{2\pi}re^{-\frac{r^2}{2}}drd\theta\\ &=\int_0^{2\pi}d\theta\int_0^1\frac{1}{2\pi}re^{-\frac{r^2}{2}}dr\\ &=2\pi\cdot \frac1{2\pi}\int_0^1re^{-\frac{r^2}{2}}dr\\ &=\int_0^1e^{-\frac{r^2}{2}}d\frac{r^2}{2}\\ 令u=\frac{r^2}{2}\\ &=\int_0^\frac12e^{-u}du\\ &=1-e^{-\frac12} \end{aligned}$

題目63

令$X$和$Y$是具有聯合分佈的連續隨機變量
a.討論$X+Y$和$X-Y$的聯合密度表達式
b.討論$XY$和$X/Y$的聯合密度表達式
c.在$X$和$Y$獨立的情況下,特殊表示a和b中的表達式

解:

a.令
$\begin{aligned} U=X+Y,V=X-Y\\ X=\frac{U+V}{2}, Y=\frac{U-V}{2}\\ J(x,y)=\begin{Vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{Vmatrix} =\begin{Vmatrix} 1 & 1\\ 1 & -1 \end{Vmatrix}=2\\ f_{UV}=\frac12f_{XY}(\frac{u+v}{2},\frac{u-v}2) \end{aligned}$
b.令
$\begin{aligned} U=XY,V=X/Y\\ X=(UV)^{\frac12} Y=(\frac{U}V)^{\frac12} J(x,y)=\begin{Vmatrix} \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{Vmatrix} =\begin{Vmatrix} y & x\\ -\frac1{x^2} &\frac1x \end{Vmatrix}=2\frac{x}y\\ f_{UV}=2|\frac xy|f_{XY}((xy)^\frac12,(\frac xy)^\frac12) \end{aligned}$

題目64

計算$X+Y$和$X/Y$的聯合密度，其中$X$和$Y$是參數爲$\lambda$的獨立指數隨機變量，證明$X+Y$和$X/Y$獨立的

解:

令$U=X+Y,V=X/Y \Rightarrow X=\frac{UV}{V+1},Y=\frac{U}{V+1}$

證明是獨立需要證明:$f_U(u)\cdot f_V(v)=f_{UV}(u,v)$

有:
$\begin{aligned} J(x,y)&=\begin{Vmatrix}\\ \frac{\partial u}{\partial x} & \frac{\partial u}{\partial y} \\ \frac{\partial v}{\partial x} & \frac{\partial v}{\partial y} \end{Vmatrix} =\begin{Vmatrix} 1 & 1\\ \frac1y & -\frac x{y^2} \end{Vmatrix}=-\frac{x-y}{y^2}\\ |J(x,y)|^{-1} &=\frac{y^2}{x+y}=\frac{u}{(v+1)^2} \end{aligned}$
$\begin{aligned} f_{UV}(u,v)&=|J(x,y)|^{-1}f_{XY}(\frac{uv}{v+1},\frac{u}{v+1})\\ &=\frac{u}{(v+1)^2} \cdot \lambda e^{-\lambda \frac{uv}{v+1} } \cdot \lambda e^{-\lambda \frac{u}{v+1}}\\ &=\frac{u}{(v+1)^2}\lambda^2e^{-\lambda u} \end{aligned}$
接下來，分別求 $f_U(u)和f_V(v)$
$\begin{aligned} f_U(u)&=\int_{-\infty}^{\infty}f_X(x)f_Y(u-x)dx=\lambda^2ue^{-\lambda u}\\ f_V(v)&=\int_{-\infty}^{\infty}|x|f_X(x)f_Y(xv)dx\\ &=\lambda^2\int_{-\infty}^{\infty}xe^{-\lambda x(1+v)}dx\\ 令a=-\lambda(1+v),則積分下限爲0，上限爲\infty\\ &=\lambda^2\int_0^{\infty}xe^{ax}dx\\ &=\lambda^2\frac1{a^2}(ax-1)e^{ax} \bigg |_0^{\infty}\\ &=\frac1{1+v^2} \end{aligned}$
所以$f_U(u)\cdot f_V(v)=f_{UV}(u,v)$是成立的，U和V是獨立的。

題目65

假定系統元件串聯在一起，且元件的壽命爲參數爲$\lambda_i$獨立指數隨機變量，證明系統壽命服從參數是$\sum\lambda_i$的指數分佈。

解:

參考書中例題
令$V$代表系統整體壽命,根據書中公式有
$\begin{aligned} f_V(v)&=nf(v)[1-F(v)]^{n-1}\\ 其中F(v)=1-e^{-\lambda v}\\ &=n\lambda e^{-\lambda v}(e^{-\lambda v})^{n-1}\\ &=n\lambda e^{-n\lambda v} \end{aligned}$

題目66

下圖系統中的每一個元件壽命都獨立的服從參數爲$\lambda$的指數分佈，計算系統壽命的cdf和密度
圖不畫了，解釋一下，6個元件分成3組，組內兩個元件串連，3組並聯

解:

令串聯部分的壽命分佈爲$F_V(t)$
$F_V(t)=1-[1-e^{-\lambda t}]^2=1-e^{-2\lambda t}$
令整體系統壽命分佈爲$F_U(t)$
$F_U(t)=[F_V(v)]^3=(1-e^{-2\lambda t})^3$
求上式求導
$f(t)=6\lambda e^{-2\lambda t}(1-e^{-2\lambda t})^2$

題目67

卡片含有n個芯片和一個糾錯元件，這樣如果只有一個芯片失效，卡片仍能正常工作；如果有兩個或兩個以上的芯片失效，卡片將不能正常工作。如果每個芯片的壽命服從參數爲 $\lambda$的指數分佈，計算卡片壽命的密度函數。

解:

令
$F_A(t)$系統的壽命分佈
$F_0(t)$所有芯片都沒有失效的概率分佈
$F_1(t)$只有一個芯片失效的概率分佈
$F(t)$單一的一個芯片失效的概率分佈
則有:
$\begin{aligned} F_A(t)&=1-F_0(t)-F(t)\\ &=1-(1-F(t))^n-nF(t)(1-F(t))^{n-1}\\ &=1-(1-(1-e^{-\lambda t}))^n-n(1-e^{-\lambda t})(1-(1-e^{-\lambda t}))^{n-1}\\ &=1-e^{-n\lambda t} -n(1-e^{-\lambda t})e^{-(n-1)\lambda t}\\ &=1-e^{-n\lambda t}-ne^{-(n-1)\lambda t}+ne^{-n\lambda t} \end{aligned}$
對上式求導則爲密度函數
$\begin{aligned} f_A(t)&=F_A^{'}(t)\\ &=-e^{-n\lambda t}\cdot -n\lambda -ne^{-(n-1)\lambda t} \cdot -(n-1)\lambda +ne^{-n\lambda t}\cdot -n\lambda\\ &=n\lambda e^{-n\lambda t}+n(n-1)\lambda e^{-(n-1)\lambda t}-n\cdot n\lambda e^{-n\lambda t}\\ &=n(n-1)\lambda e^{-(n-1)\lambda t}+n(1-n)\lambda e^{-n\lambda t}\\ &=n(n-1)\lambda ( e^{-(n-1)\lambda t}- e^{-n\lambda t}) \end{aligned}$

題目68

令$U_1$、$U_2$和$U_3$是獨立的均勻隨機變量
a.計算$U_{(1)}$、$U_{(2)}$和$U_{(3)}$的聯合密度
b.在一英里的高速公路上，獨立且隨機地建造三個加油站，任意兩個加油站之間的距離不小於$\frac13$英里的概率是多少

解:

a.模擬對最水值和最小值聯合密度的計算方法
令$U=U_{(1)},V=U_{(2)},W=U_{(3)}$
$\begin{aligned} f(u,v,w)&=\frac{n!}{1!1!1!}f(u)f(v)f(w)\\ 代入n=3,f(u)=f(v)=f(w)=1\\ f(u,v,w)&=6,u\leq v\leq w \end{aligned}$
b.在一英里的高速公路上，獨立且隨機地建造三個加油站，任意兩個加油站之間的距離不小於$\frac13$英里的概率是多少
我們可以認爲3個加油站的位置服從$U(0,1)$,令X爲第1個加油站的位置,Y爲第二個加油站的位置Z爲第三個加油站的位置,則X,Y,Z是3個均勻分佈的順序統計量，則$f_{XYZ}(x,y,z)=6(根據a問題的結果)$
任意兩個加油站之間的距離大於$\frac13$則必須有$Y-X>\frac13 並且 Z-Y>\frac13$
$\begin{aligned} P(Y-X>\frac13且 Z-Y>\frac13)&=\int_0^\frac13\int_{x+\frac13}^{\frac23}\int_{y+\frac13}^1f(x,y,z)dzdydx\\ &=\int_0^\frac13\int_{x+\frac13}^{\frac23}4-6ydy\\ &=\int_0^\frac13\frac13-2x-3x^2dx\\ &=\frac1{27} \end{aligned}$
積分上下限是結合$0<x<y<z<1$且$y-x>\frac13且z-t>\frac13$得出

題目69

計算$n$個獨立威布爾隨機變量最小值的密度，每個變量具有密度：
$f(t)=\beta\alpha^{-\beta}t^{-\beta-1}e^{-(t/\alpha)^{\beta}},t \geq 0$

解:

令$V$代表$n$個獨立威布爾變量的最小值隨機變量
根據3.7節中的公式
$f_V(v)=nf(v)[1-F(v)]^{n-1}$
其中：
$f_(v)=\beta\alpha^{-\beta}v^{-\beta-1}e^{-(v/\alpha)^{\beta}}$
$F_(v)=1-e^{-(v/\alpha)^{\beta}}$
代入上式經計算結果如下：
$f_V(v)=n\beta\alpha^{-\beta}v^{\beta-1}e^{-n(v/\alpha)^{\beta}}$