例 6.3

(1)

E(X)=\frac{\theta-1}2

\hat{\theta}=2\overline X+1

(2)

E(X)=m\theta

\hat{\theta}=\frac m{\overline X}

(3)

\begin{align}
E(X)&=\Sigma_{x=1}^{\infin}x(x-1)\theta^2(1-\theta)^{x-2}\
&=\theta^2\Sigma_{x=1}^{\infin}\frac{\partial^2}{\partial \theta}(1-\theta)^x\
&=\theta^2\frac{\partial^2}{\partial \theta}\left(\frac1{\theta}-1\right)\
&=\frac2{\theta}
\end{align}

\hat{\theta}=\frac2{\overline X}

(4)

E(X)=-\Sigma_{x=1}^{\infin}\frac{\theta^x}{ln(1-\theta)}=-\frac{\theta}{(1-\theta)ln(1-\theta)}

ln(1-\hat\theta)\overline X=1-\frac 1{(1-\hat\theta)}

(5)

E(X)=\theta

\hat\theta=\overline X

例 6.4

(1)

E(X)=\frac2{\theta^2}\int_0^\theta(\theta-x)xdx=\frac\theta3

\hat\theta=3\overline X

(2)

E(X)=\int_0^1(\theta+1)x^{\theta+1}dx=\frac{\theta+1}{\theta+2}%=1-\frac1{\theta+2}

\hat\theta=\frac1{1-\overline X}-2

(3)

E(X)=\int_0^1\sqrt\theta x^{\sqrt\theta} dx=\frac{\sqrt{\theta}}{\sqrt{\theta+1}}

\hat\theta=(\frac1{1-\overline X}-1)^2

(4)

f(x;\theta)=\theta\frac{c^\theta}{x^{\theta+1}} （pdf 写错了）

E(X)=\theta c^\theta\int_c^\infin x^{-\theta}dx=\frac{\theta c}{(\theta-1)}

\hat\theta=\frac1{\frac{\overline X}c-1}+1

(5)

E(X)=\int_0^\theta\frac{6x^2(\theta-x)}{\theta^3}dx=\frac\theta2

\hat\theta=2\overline X

(6)

E(X)=\int_0^\infin(\frac\theta x)^2e^{-\frac\theta x}dx=\theta

\hat\theta=\overline X

例 6.5

\begin{align}
E(X)
&=\int_0^\infin\frac{2\theta}{\sqrt\pi}\frac{x^2}{\theta^2}e^{-\frac{x^2}{\theta^2}}d(\frac{x^2}{\theta^2})\
&=\int_0^\infin\frac{2\theta}{\sqrt\pi}xe^{-x}dx\
&=\frac{2\theta}{\sqrt\pi}
\end{align}

\hat\theta=\frac{\sqrt\pi}2\overline X

(2)

例 6.6

设 Y = e^X，其中 X\sim N(\mu, \sigma^2)，求 E(Y) 和 Var(Y) 的矩估计

例 6.7

\hat\sigma=\overline X^2+S^2

例 6.25

(1)

L(\theta)=\Pi_{i=1}^{n}\frac1\theta=(\frac1\theta)^n

ln L(\theta)=-n ln\theta

又 \theta\geq max{X_1,X_2,\ldots,X_n}

\therefore \hat\theta=max{X_1,X_2,\ldots,X_n}

(2)

\begin{align}
lnL(\theta)&=\Sigma_{i=1}^n\left[ln(C_m^{x_i})+x_iln\theta+(m-x_i)ln(1-\theta)\right]\
&=\Sigma_{i=1}^nln(C_m^{x_i})+n\overline Xln\theta+n(m-\overline X)ln(1-\theta)
\end{align}

\frac d{d\theta}lnL(\theta)=n[\frac{\overline X}{\theta}-\frac{m-\overline X}{1-\theta}]

\hat\theta=\frac{\overline X}m

(3)

L(\theta)=\Pi(x-1)\cdot\theta^{2n}(1-\theta)^{n\overline X-2n}

\frac d{d\theta}lnL(\theta)=\frac{2n}\theta-\frac{n\overline X-2n}{1-\theta}

\hat\theta=\frac2{\overline X}

(4)

L(\theta)=\frac{\theta^{n\overline X}}{\Pi x\cdot(ln(1-\theta))^n}

lnL(\theta)=n\overline Xln\theta-\Sigma lnx-nlnln(1-\theta)

\frac d{d\theta}lnL(\theta)=\frac{n\overline X}\theta+\frac{n}{(1-\theta)ln(1-\theta)}

(5)

L(\theta)=\theta^{n\overline X}e^{-\theta\Sigma(\frac1{x!})}

\frac d{d\theta}lnL(\theta)=n\overline Xln\theta-\Sigma_{i=1}^n(\frac1{x_i!})

\hat\theta=e^{{\Sigma_{i=1}^n(\frac1{x_i!})}/{\Sigma_{i=1}^nx_i}}

例 6.26

(1)

\frac d{d\theta}lnL(\theta)=\Sigma(\frac1{\theta-x})-\frac{2n}\theta=\Sigma(\frac{2x-\theta}{\theta(\theta-x)})

(2)

\frac d{d\theta}lnL(\theta)=\frac n{\theta+1}+\Sigma lnx

\hat\theta=\frac n{\Sigma_{i=1}^nlnx_i}-1

(3)

同上一题

\sqrt{\hat\theta}-1=\frac n{\Sigma_{i=1}^nlnx_i}-1

\hat\theta=(\frac n{\Sigma_{i=1}^nlnx_i})^2

(4)

f(x;\theta)=\theta\frac{c^\theta}{x^{\theta+1}} （pdf 写错了）

L(\theta)=\theta^nc^{n\theta}\cdot\frac{1}{(\Pi x)^{\theta+1}}

\frac d{d\theta}lnL(\theta)=\frac n\theta+nlnc-\Sigma(lnx)

\hat\theta=\frac n{\Sigma_{i=1}^nln(\frac xc)}

(5)

\frac d{d\theta}lnL(\theta)=\Sigma_{i=0}^n\frac1{\theta-x_i}-\frac{3n}\theta

(6)

\frac d{d\theta}lnL(\theta)=\frac{2n}\theta-\Sigma\frac1x

\hat\theta=\frac{2n}{\Sigma_{i=1}^n\frac1{x_i}}

例 4.42

X_n \xrightarrow{P} X\Rightarrow \forall\epsilon>0, lim_{n\rightarrow\infin}P(|X_n-X|>\epsilon)=0

Y_n \xrightarrow{P} Y\Rightarrow \forall\epsilon>0, lim_{n\rightarrow\infin}P(|Y_n-Y|>\epsilon)=0

\therefore\forall\epsilon_0>0,\forall n,P(|X_n-X+Y_n-Y|>\epsilon_0)<P(|X_n-X|+|Y_n-Y|<\epsilon_0)

又对 \epsilon=\frac{\epsilon_0}2, lim_{n\rightarrow\infin}P(|X_n-X|>\epsilon)=0, lim_{n\rightarrow\infin}P(|Y_n-Y|>\epsilon)=0

\therefore lim_{n\rightarrow\infin}P(|X_n-X|+|Y_n-Y|<\epsilon_0)=0\Rightarrow lim_{n\rightarrow\infin}P(|X_n-X+Y_n-Y|>\epsilon_0)=0

即 X_n+Y_n\xrightarrow{P}X+Y

例 4.43

设 X_n\sim Ge(\frac{\lambda}n) 为一随机变量序列，\lambda>0 为常数，定义 Y_n=\frac1nX_n, 求证 Y_n 依分布收敛于 Y\sim Exp(\lambda).

F_{X_n}(k)=1-(1-\frac{\lambda}n)^{\lfloor k\rfloor}

F_{Y_n}(k)=1-(1-\frac{\lambda}n)^{\lfloor nk\rfloor}

lim_{n\rightarrow\infin}F_{Y_n}(k)=e^{-\lambda k}

故 Y_n\xrightarrow{\mathcal L}Y

例 4.45

A\sim B(1,0.2)

E(A)=0.2, Var(A)=0.2\times0.8=0.16

记 A 发生的次数为 S. 由中心极限定理可知 S\sim N(100,80).

故 P(80\leq S\leq120)=P(|S-100|<20)=1-\frac{80}{20^2}=0.8

例 4.46

设 X_1,X_2,X_3,\ldots 为一列独立同分布的随机变量，满足 E(X_i^k) = \alpha_k, k = 1,2,3,4, 利用中心极限定理说明 \Sigma_{i=1}^nX_i^2 的渐近分布是什么.

Var(X_i^2)=\alpha_4-\alpha_2^2

由中心极限定理，\Sigma_{i=1}^nX_i^2\sim N(n\alpha_1,n(\alpha_4-\alpha_2^2))

例 4.48

(1)

记正常工作的部件数量为 X.

则 X\sim B(100,0.9)

P(X\geq85)\approx1-\Phi(\frac53)=0.9522(2)

记正常工作的部件数量为X.

则X\sim B(n,0.9)%N(0.9n, 0.09n)P(X\geq0.8n)=1-\Phi(\frac{0.1n}{0.3\sqrt n})>0.95n\geq24.35即n\geq 25$.

例 4.49

记每次取款额度为 X, 总额度为 S

则 E(X)=5.5, Var(X)=8.25

\therefore S\sim N(1100,1650)

S_{0.95}=1100+1.645\sqrt{1650}=1166.8 （百元）

例 4.52

盈利 200 万元即赔款 S 少于 1000 万元（10000k 元）。

记车辆的赔偿额度为 A, 赔偿次数为 B，总额为 C

则 A\sim U(1000,5000)

B\sim P(2)

C=\Sigma_{i=1}^BA_i

E(C)=E(A)E(B)=6000

Var(C)=E(B)Var(A)+E^2(A)Var(B)=20666.7

\therefore S\sim N(14400 000, 49600 000 000)%var=222711

P(S>10000 000)\approx 1

例 4.15

f(x)=\frac1{\pi} (|x|<\frac{\pi}2)

E(sinX)=\int_{-\frac{\pi}2}^{\frac{\pi}2}sinxf(x)dx=0

E(cosX)=\frac{2}{\pi}

E(XcosX)=0

例 4.16

(1)

E(sgn^2(X))=1

E(sgn(X))=-0.5

Var(sgn(X))=E(sgn(X))^2-E(sgn^2(X))=0.75

E(Xsgn(X))=E(|X|)=\frac2{2\pi}\int_0^{\infin}e^{-\frac{x^2}2}xdx=\frac1{\pi}

例 4.20

看不懂什么是《相同的小球》，暂且理解为不同的小球。

P(X_a=b)=C_m^b\cdot p_a^b\cdot(1-p_a)^{m-b}

p’_2=\frac{p_2}{1-p_1}

P(X_2=x|X_1=k)=C_{m-k}^x p’_2{}^x(1-p’_2)^{m-k-x}

E(X_2|X_1=k)=\Sigma_0^{m-k}x\cdot C_{m-k}^x p’_2{}^x(1-p’_2)^{m-k-x}\=(m-k)\cdot\frac{p_2}{1-p_1}

Var(X_2|X_1=k)=(m-k)\cdot\frac{p_2}{1-p_1}\cdot\frac{1-p_1-p_2}{1-p_1}

E(X_1+X_2)=m(p_1+p_2)

Var(\Sigma_{i=1}^kX_i)=m(\Sigma_{i=1}^kp_i)(1-\Sigma_{i=1}^kp_i)

例 4.26

E(X_1+X_2+X_3)=0

E((X_1+X_2+X_3)^2)=E(X_1^2+X_2^2+X_3^2)+2E(X_1X_2)+2E(X_2X_3)+2E(X_3X_1)=1

Var(X_1+X_2+X_3)=1

法二：

0\leq x\leq\pi,0\leq y\leq2\pi

\begin{align}
E(&(cosx+sinxcosy+sinxsiny)^2)\
=E(&(cosx+\sqrt2sinxsin(y+\frac{\pi}4))^2)\
=E(&1+\sqrt2cosxsinxsin(y+\frac{\pi}4)\
&+sin^2x(2sin^2(y+\frac{\pi}4)-1))\
=1&+E(sin^2x)(2E(sin^2(y+\frac{\pi}4))-1)\
=1&
\end{align}

Var(X_1+X_2+X_3)=1

例 4.32

Cov(\alpha X+\beta Y,\alpha X-\beta Y)=\alpha^2Var(X)-\beta^2Var(Y)=(\alpha^2-\beta^2)\sigma^2

当 |\alpha|=|\beta| 时两变量独立。

例 4.33

存在，\alpha=\beta=1

由于上题并不要求 Cov(X,Y)=0, 故结论在本题仍然成立。

类似的题在先前的作业中也有过。

例 4.36

Var(Z)=Cov(Z,Z)=(\pi^2+(1-\pi^2)+2\pi(1-\pi)\rho_{X,Y})\sigma^2<\sigma^2

由对称性，\pi=0.5 时风险最小。

例 4.37

\begin{align}
&Cov(X_1,E(X_2|X_1))\
=&E([X_1-E(X_1)][E(X_2|X_1)-E_{X_1}(E(X_2|X_1))])\
=&\int_{X_1}[X_1-E(X_1)][E(X_2|X_1)-E(X_2)]f_{X_1}(x)dx\
=&\int_{X_1}[X_1-E(X_1)][E(X_2)-E(X_2)]f_{X_1}(x)dx\
=&Cov(X_1,X_2)
\end{align}

例 4.39

E(S|N=n)=\frac np

Var(S|N=n)=\frac{n(1-p)}{p^2}

P(N=m)=C_n^mq^m(1-q)^{n-m}

E(S)=\Sigma_{i=0}^nC_n^iq^i(1-q)^{n-i}\frac ip=\frac{E(N)}p=\frac{nq}p

\begin{align}
Var(S)&=E((S^2-E(S))^2)\
&=E(S^2)-E(S)^2\
&=\Sigma_{m=0}^n[P(N=m)E(S^2|N=m)]-E(S)^2\
&=\Sigma_{m=0}^n[P(N=m)(E(S|N=m)^2+Var(S|N=m))]-E(S)^2\
&=\Sigma_{m=0}^n[C_n^mq^m(1-q)^{n-m}\frac{m^2+m-mp}{p^2}]-\frac{n^2q^2}{p^2}\
&=\frac1{p^2}(E(N^2)+(1-p)E(N))-\frac{n^2q^2}{p^2}\
&=\frac1{p^2}(n^2q^2+nq-nq^2+(1-p)nq)-\frac{n^2q^2}{p^2}\
&=\frac {nq}{p^2}(2-p-q)
\end{align}

例 4.40

E_t(N)=Var_t(N)=\lambda t

E(N(T))=E(\lambda t)=\lambda a

Cov(T,N(t))=E(TN)-E(T)E(N)

E(TN(t))=E(T\cdot E_t(N))=E(T\cdot\lambda T)

\therefore Cov(T,N(t))=\lambda E(T^2)-E(T)\lambda a=\lambda Var(T)=\lambda b

(2)

Var(N)=E(Var_t(N))+Var(E_t(N))=\lambda a+\lambda b

例 4.3

E(X) = 0.5\times0+0.5\times4=2

例 4.4

\begin{align}
E(X)&=\int_0^{\infin}xf(x)dx=\int_0^{\infin}\frac{x^2}{\sigma^2}e^{-\frac{x^2}{2\sigma^2}}dx\
&=\sigma\int_0^{\infin}x^2e^{-\frac{x^2}2}dx\
&=-\sigma\int_0^{\infin}xd(e^{-\frac{x^2}2})\
&=\sqrt2\sigma\int_0^{\infin}e^{-x^2}dx\
&=\sqrt{\frac{\pi}2}\sigma
\end{align}

\begin{align}
Var(X)&=E(X^2)-E(X)^2\
&=\int_0^{\infin}\frac{x^3}{\sigma^2}e^{-\frac{x^2}{2\sigma^2}}dx
-\frac{\pi\sigma^2}2\
&=4\sigma^2\int_0^{\infin}xe^{-x}dx-\frac{\pi\sigma^2}2\
&=(4-\frac{\pi}2)\sigma^2
\end{align}

2. $$
   E(X)=\int_1^0x\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1}d(1-x)\
   =\frac1{\beta}\int_1^0\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha}d(1-x)^{\beta}
   $$

$$
\begin{align}
E(1-X)&=\int_1^0(1-x)\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha-1}(1-x)^{\beta-1}d(1-x)\
&=\frac1{\alpha}\int_0^1\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}(1-x)^{\beta}dx^{\alpha}
\end{align}
$$

故

$-\beta E(X)+\alpha E(1-x)=
\int_0^1\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}d((1-x)^{\beta}x^{\alpha})=0$

又显然 $E(X)+E(1-X)=1$

故 $E(X)=\frac{\alpha}{\alpha+\beta}$.

$$
\begin{align}
E(X^2)&=\int_1^0x\frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)}x^{\alpha}(1-x)^{\beta-1}d(1-x)\
&=\int_1^0x\frac{\frac1{\alpha+\beta}\Gamma(\alpha+\beta+1)}{\frac1{\alpha}\Gamma(\alpha+1)\Gamma(\beta)}x^{\alpha}(1-x)^{\beta-1}d(1-x)\
&=\frac{\alpha}{\alpha+\beta}\cdot\frac{\alpha+1}{\alpha+\beta+1}
\end{align}
$$

$$
\begin{align}
Var(X)&=E(X^2)-E(X)^2\
&=\frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}
\end{align}
$$

3. $$
   f(x)dx=e^{-(\frac x{\lambda})^k}d(\frac x{\lambda})^k
   $$

$$
\begin{align}
E(X)&=\int_0^{\infin}xf(x)dx\
&=\lambda\int_0^{\infin}xd(-e^{-x^k})\
&=\lambda\int_0^{\infin}e^{-x^k}dx\
&=\lambda\int_0^{\infin}e^{-t}d(t^{\frac1k})\
&=\lambda\frac1k\int_0^{\infin}t^{\frac {1-k}k}e^{-t}dt\
&=\lambda\frac1k\Gamma(\frac{1-k}k)\
&(???)
\end{align}
$$

例 4.7

记空盒子个数为 X.

P_n(X\geq k)=\frac1{n^n}\cdot

例 4.9

E(W_1)=\frac a{a+b}

E(W_2)=E(W_1)+\frac {a+E(W_1)}{a+b+1}=\frac{a+b+2}{a+b+1}E(W_1)+\frac a{a+b+1}

E(W_2)+a=\frac{a+b+2}{a+b+1}(E(W_1)+a)

E(W_n)+a=\frac{a+b+2}{a+b+1}(E(W_{n-1})+a)

\Rightarrow E(W_n)=(\frac{a+b+2}{a+b+1})^{n-1}\cdot\frac{a+a^2+ab}{a+b}-a

例 4.13

f(x)=2(x-1),1<x<2

E(Y)=\int_1^2yf(x)dx=2e=5.4366

E(Z)=\int_1^2zf(x)dx=2-2ln2=0.6137

例 4.18

f_{Y|X}(y|x)=\frac1x,\space x=1,2

f(y)=\sum_xf_{Y|X}(y|x)P(X=x)=\begin{cases}
\frac34, &0<y<1\
\frac14, &1<y<2
\end{cases}

(2)

E(Y)=\int_0^2yf(y)dy=\frac34

例 4.19

由例 3.47 得 X+Y 与 X-Y 线性无关。

Var(X-Y)=Var(X)+Var(Y)-2Cov(X,Y)=0.5

E(X-Y)=0

f_Z(z)=2\frac{e^{-z^2}}{\sqrt{\pi}} (???)

E(Z)=\frac1{\sqrt{\pi}}

(2)

Var(X+Y)=Var(X)+Var(Y)+2Cov(X,Y)=1.5

U=|X+Y|+|X-Y|,V=|X+Y|-|X-Y|

例 4.21

(1)

\begin{align}
P(X=k|X+Y=m)
&=\frac{\frac{\lambda^k\mu^{m-k}}{k!(m-k)!}}{\sum_{i=0}^m\frac{\lambda^i\mu^{m-i}}{i!(m-i)!}}\
&=\frac{C_m^k\lambda^k\mu^{m-k}}{\sum_{i=0}^mC_m^i\lambda^i\mu^{m-i}}\
&=\frac{C_m^k\lambda^k\mu^{m-k}}{(\mu+\lambda)^m}
\end{align}

\begin{align}
E(X|X+Y=m)&=\sum_{k=0}^mkP(X=k|X+Y=m)\
&=\frac{\mu^m}{(\mu+\lambda)^m}\sum_{k=0}^mkC_m^k\left(\frac{\lambda}{\mu}\right)^k
\end{align}

\begin{align}
\sum_{k=0}^mkC_m^k\left(\frac{\lambda}{\mu}\right)^k
&=\left.\left[\sum_{k=0}^mC_m^k\left(x\right)^{k+1}\right]^{\prime}\right|_{x=\frac{\lambda}{\mu}}\
&=\left.\left[x(x+1)^m\right]^{\prime}\right|_{x=\frac{\lambda}{\mu}}\
&=\left.\left[(mx+x+1)(x+1)^{m-1}\right]\right|_{x=\frac{\lambda}{\mu}}\
&=\frac{(\lambda+\mu)^{m-1}((m+1)\lambda+\mu)}{\mu^m}
\end{align}

\therefore E(X|X+Y=m)=\frac{(m+1)\lambda+\mu}{\lambda+\mu}

(2)

Cov(X+Y, X-Y)=Var(X)-Var(Y)=0

故 X+Y,X-Y 相互独立。

\begin{align}
P(X=k|X+Y=m)&=P(X-Y=2k-m)\
&=\frac{C_n^kC_n^{m-k}p^m(1-p)^{2n-m}}
{p^m(1-p)^{2n-m}\sum_{x=0}^mC_n^xC_n^{m-x}}\
&=\frac{C_n^kC_n^{m-k}}
{\sum_{x=0}^mC_n^xC_n^{m-x}}\
\end{align}

E(X|X+Y=m)=\frac{E(X-Y)+m}2=\frac m2

例 4.23

E(\omega X+(1-\omega)Y)=\omega\mu+(1-\omega)\cdot2\mu=(2-\omega)\cdot\mu

\begin{align}
Var(\omega X+(1-\omega)Y)&=
\omega^2Var(X)+(1-\omega)^2Var(Y)+2\omega(1-\omega)Cov(X,Y)\
&=\omega^2\sigma^2+3(1-\omega)^2\sigma^2+2\omega(1-\omega)\cdot0.5\times\sqrt3\sigma^2\
&=\sigma^2\left(4\omega^2-6\omega+3+\sqrt3\omega-\sqrt3\omega^2\right)\
&=\left[(4-\sqrt3)\omega^2-(6-\sqrt3)\omega+3\right]\sigma^2
\end{align}

R(\omega)^2=\frac{(2-\omega)^2}{(4-\sqrt3)\omega^2-(6-\sqrt3)\omega+3}\frac{\mu^2}{\sigma^2}

\frac{\mathrm{d}}{\mathrm{d\omega}}lnR(\omega)^2=-\frac2{2-\omega}-\frac{2(4-\sqrt3)\omega-6+\sqrt3}{(4-\sqrt3)\omega^2-(6-\sqrt3)\omega+3}=0

\omega=\frac{42}{73}-\frac{2\sqrt3}{73}\approx0.528

R_{max}=\frac49(7-2\sqrt3)\approx1.5715

例 4.25

f(X)=2e^{-2x}

F(X)=P(X<x)=1-e^{-2x}

F(min{X_1,X_2})=F_{X_1}(x)F_{X_2}(x)=(1-e^{-2x})^2

E(min{X_1,X_2})=\int_0^{\infin}(1-F(min{X_1,X_2}))dx=

F(max{X_1,X_2})=1-(1-F_{X_1}(x))(1-F_{X_2}(x))=1-e^{-4x}

E(max{X_1,X_2})=\int_0^{\infin}(1-F(max{X_1,X_2}))dx=\frac14

4.24 更新：注释了部分会导致渲染错误的公式，请自行脑补。

---

例 3.20

f(x,y)=\frac14 (0\leq x,y\leq2)

$$
f(x,z)=(f(x,x-z)+f(x,x+z))|\frac{\partial(x,y)}{\partial(x,z)}|=
\begin{cases}
\frac12 &(z\leq x,z\leq 2-x)\\
\frac14 &(z\leq x\cap z> 2-x \cup z>x\cap z\leq2-x)\\
0 &(z> x,z> 2-x)
\end{cases}
$$

$$
f_Z(z)=\int_{0}^{2}f(x,z)dx=\begin{cases}
(2-2z)\cdot\frac12+2z\cdot\frac14 &0\leq z\leq1\\
(2z-2)\cdot0+(4-2z)\cdot\frac14 &1< z\leq2
\end{cases}
$$

\therefore f_Z(z)=1-\frac z2 (0\leq z\leq2)

例 3.21

f(x,y)=1 (x<y<2-x,0<x<1)

f_X(x)=\int_0^{min{x,2-x}}f(x,y)dy=min{x,2-x}

f_Y(y)=\int_y^{2-y}f(x,y)dx=2-2y

f_{X|Y}(x|y)=\frac1{2-2y} (y<x<2-y,0\leq y<1)

例 3.29

XY>0 时：

f_{XY}(t)=p\cdot N(\mu,\sigma^2)=p\frac{e^{-\frac{(x-\mu)^2}{2\sigma^2}}}{\sigma\sqrt{2\pi}}

XY=0 时：

P(XY=0)=1-p

例 3.35

(2)

f(x,y)=A\cdot e^{-3x}\cdot e^{-4y}

故 X,Y 独立

(1)

\iint_{x,y>0}f(x,y)dxdy=A\cdot\int_0^{\infin}e^{-3x}dx\cdot\int_0^{\infin}e^{-4y}dy=1\Rightarrow A=12

(3)

f_X(x)=3e^{-3x}

f_Y(y)=4e^{-4y}

f_Z(z)=\int_0^zf_X(x)f_Y(z-x)dx=12e^{-4z}(e^z-1)

\int_0^z3e^{-3x}4e^{-4z+4x}dx=12e^{-4z}\int_0^ze^xdx=12e^{-4z}(e^z-1)

(4)

\begin{align}
P(X>0.5|X+Y=1)&=
\frac{\int_{0.5}^1f_X(x)f_Y(1-x)dx}{\int_0^1f_X(x)f_Y(1-x)dx}\
&=\frac{\int_{0.5}^13e^{-3x}4e^{-4+4x}dx}{\int_0^13e^{-3x}4e^{-4+4x}dx}\
&=\frac{e-e^{\frac12}}{e-1}
\end{align}

例 3.39

f(x,y)=\frac12 (|x|-1<y<1-|x|,-1<x<1)

f_X(x)=\frac12\cdot2(1-|x|)=1-|x|

f_Y(y)=\frac12\cdot2(1-|y|)=1-|y|

(2)

不相互独立，因为 f(x,y)\neq f_X(x)f_Y(y)

例 3.40

(1)

f_X(x)=3x\cdot x=3x^2 (0<x<1)

f_Y(y)=\int_y^13xdx=\frac32(1-y^2) (0<y<1)

(2)

不相互独立，因为 f(x,y)\neq f_X(x)f_Y(y)

(3)

\begin{align}
P(X+Y\leq1)
&=\int_0^1dx\cdot 3x\cdot min{x,1-x}\
&=\int_0^{0.5}3x^2dx
+\int_{0.5}^13x(1-x)dx\
&=\frac38
\end{align}

例 3.41

(1)

F_X(x)=lim_{y->\infin}F(x,y)=1-(x+1)e^{-x}

F_y(y)=lim_{x->\infin}F(x,y)=\frac y{1+y}

(2)

$f(x,y)=\frac{\part F(x,y)}{\part x\part y}=xe^{-x}\cdot\frac1{(1+y)^2}$

f_X(x)=xe^{-x}

f_Y(y)=\frac1{(1+y)^2}

(3)

F(x,y)=F_X(x)F_Y(y)

X,Y 独立

例 3.43

f_X(x)=f_Y(y)=\frac12

f(x,y)\neq f_X(x)f_Y(y), 故 X,Y 不互相独立。

记 U=X2,V=Y^2

$f(u,v)=\frac{1+\sqrt{uv}+1-\sqrt{uv}+1+\sqrt{uv}+1-\sqrt{uv}}4\frac{\part (x,y)}{\part(u,v)}=\frac{1}{4\sqrt{uv}}$ $(0<u,v<1)$

f_U(u)=\frac1{2\sqrt u}

f_V(v)=\frac1{2\sqrt v}

f(u,v)=f_U(u)f_V(v)

故 U,V 独立。

例 3.47

记 m=\frac{x-\mu_1}{\sigma_1},n=\frac{y-\mu_2}{\sigma_2}

f(x,y)=\frac1{2\pi\sigma_1\sigma_2\sqrt{1-\rho^2}}exp{-\frac{m^2-2\rho mn+n^2}{2(1-\rho^2)}}

例 4.2

(1)

\begin{align}
E(X)&=
\Sigma_{n=1}^{\infin}nP(X=n)\
&=\Sigma_{n=1}^{\infin}n(P(X\leq n)-P(X\leq n+1))\
&=\Sigma_{n=1}^{\infin}P(X\leq n)
\end{align}

(2)

\begin{align}
E(X)&=\int_0^{\infin}xdF(x)\
&=\int_0^{\infin}xdF(x)\
&=\int_0^{\infin}d(xF(x))-F(x)dx\
&=-\int_0^{\infin}xd(1-F(x))\
&=\int_0^{\infin}(-d(x(1-F(x)))+(1-F(x))dx)\
&=-(x(1-F(x)))|_0^{\infin}+\int_0^{\infin}(1-F(x))dx\
&=\int_0^{\infin}(1-F(x))dx
\end{align}

(3)

\begin{align}
E(X)&=\int_0^{\infin}xdF(x)\
&=\int_0^{\infin}dF(x)\cdot\int_0^xdy\
&=\int_0^{\infin}dy\cdot\int_y^{\infin}dF(x)\
&=\int_0^{\infin}dy(F(\infin)-F(y))\
&=\int_0^{\infin}(1-F(x))dx
\end{align}

或者：

\begin{align}
E(X)&=\Sigma_xxP(X=x)\
&=\Sigma_x(P(X=x)\int_x^{\infin}dy)\
&=\int_0^{\infin}dxP(X\geq x)\
&=\int_0^{\infin}(1-F(x))dx
\end{align}

例 3.11

f_X(x)=\int_0^xe^{-x}dy=xe^{-x}

f_Y(y)=\int_y^\infin e^{-x}dx=e^{-y}

(1)

f_{Y|X}(y|x)=\frac{f(x,y)}{f_X(x)}=\begin{cases}
\frac1x,&0<y<x \\
0,&\text{其它}
\end{cases}

(2)

\begin{align}
P(X\leq1|Y\leq1)&=\frac{P(X\leq1,Y\leq1)}{P(Y\leq1)}\
&=\frac{P(X\leq1)}{P(Y\leq1)}\
&=\frac{1-\frac2e}{1-\frac1e}\
&=\frac{e-2}{e-1}
\end{align}

例 3.13

f(x,y)=\begin{cases}
Ax^2,& 0<|x|<y<1 \
0,& \text{其它}
\end{cases}

f_Y(y)=\int_{-y}^yAx^2dx=\frac23Ay^3

\int_0^1f_y(y)dy=\frac16A=1\Rightarrow A=6

P(X<0.25|Y=0.5)=\frac{\int_{-0.5}^{0.25}f(x,0.5)dx}{f_Y(0.5)}=\frac{9}{16}

例 3.14

f(x)=\frac1{\sqrt{2\pi}}e^{-\frac{x^2}2}

F(x)=\frac1{\sqrt{2\pi}} \int_{-\infin}^xe^{-\frac{t^2}2}dt

G(z)=\begin{cases}
\frac1{2\sqrt{2\pi}} \int_{-\infin}^ze^{-\frac{t^2}2}dt, &z<0 \
\frac12+\frac1{2\sqrt{2\pi}} \int_{-\infin}^ze^{-\frac{t^2}2}dt, &z\geq0
\end{cases}

例 3.17

(1)

f(x,y)=f_X(x)f_{Y|X}(y|x)=\begin{cases}
\frac{9y^2}x, &0<y<x<1 \
0,&\text{其它}
\end{cases}

(2)

f_Y(y)=\int_y^1f(x,y)dx=-9y^2lny (0<y<1)

例 3.19

f_X(x)=e^{-x}

P(y=i|x=\lambda)=\frac{e^{-\lambda}\lambda^i}{i!}=\frac{e^{-x}x^y}{y!}

\begin{align}
P(y=i)&=\int_{x=0}^\infin\frac{e^{-x}x^y}{y!}\cdot e^{-x}dx\
&=\int_{x=0}^\infin\frac{e^{-x}x^y}{y!2^{y+1}}dx\
&=\frac1{2^{y+1}}
\end{align}

例 3.27

f(x,t)=\frac{2e^{-x-\frac{t^2}2}}{\sqrt{2\pi}}

例 3.30

f(x,y)=\begin{cases}
\frac12e^{-\frac y2} &0<x<1,y>0\
0 &\text{其它}
\end{cases}

(2)

题目条件等价于 X^2>Y, 即 X>\sqrt Y

\therefore P(X^2>Y)=\int_0^1(1-\sqrt Y)\cdot\frac12e^{-\frac y2}dy=0.144376

例 3.34

易知 Z_n 满足 {0,1,2,\ldots,K-1} 上的均匀分布。

当 Z_1=z_1,Z_2=z_2 时，z_2-z_1\equiv Y\pmod K

故 Y 取唯一值，同时 X 取唯一值。此时 Z_i (i>2) 为单点分布。故 Z_n 不独立。

接下来考虑 P(Z_b=z_b|Z_a=z_a). 不妨设 b>a. 记 a-b 与 K 的最小公倍数为 c=(a-b,K). 则 P(Z_b=z_b, Z_a=z_a)=\frac c {K^2}. 故 P(Z_b=z_b|Z_a=z_a)=\frac{P(Z_b=z_b,Z_a=z_a)}{P(Z_a=z_a)}=\frac c K. 所以，当且仅当 b-a 与 K 互质时 Z_a 与 Z_b 独立。

例 3.36

(1)

f(x,y)=\frac12 (|x|+|y|\leq1)

f_X(x)=1-|x|,f_Y(y)=1-|y|

(2)

f(\frac23,\frac23)=0\neq f_X(\frac23)f_Y(\frac23)

X,Y 不相互独立

(3)

f_{Y|X}(y|x)=\frac{f(x,y)}{f_X(x)}=\frac{1}{2(1-x)}

例 3.42

f_{x|yz}(x,y,z)=\frac{1-sinxsinysinz}{\int_0^{2\pi}(1-sinxsinysinz)dx}=\frac{1-sinxsinysinz}{2\pi}

例 2.42

记 Y 的分布函数为 G(y), 概率密度为 g(y), X 的概率密度为 f(x)

则 \frac{dy}{dx}=F^\prime(x)=f(x)

g(y)=f(x)\cdot\frac1{\frac{dy}{dx}}=1 (0<y<1)

故 Y 满足 (0,1) 上的均匀分布。

例 2.46

f(x)=\lambda e^{-\lambda x} (x>0)

p(y)=f(x)\frac{dx}{dy}=\begin{cases}
\lambda e^{-\lambda y}&y\geq1 \\
\lambda e^{-\lambda x}\cdot\frac1{-2x}=\frac{\lambda e^{\lambda\sqrt y}}{2\sqrt y}&-1<y<0
\end{cases}

例 2.48

\int_0^3f(x)dx=1 \Rightarrow a=9

g(y)=f(x)\cdot\frac{dx}{dy}=\frac{y^2}9 (1<y<2)

(2)

P(X\leq Y)=P(X\leq 2)=\int_0^2f(x)dx=\frac8{27}

例 3.5

P(X=-1)=0.2+a

P(X+Y=0)=a+b=0.5

P(X=-1,X+Y=0)=a

由于 {X = −1} 和 {X +Y = 0} 相互独立，故

P(X=-1,X+Y=0)=P(X+Y=0)\cdot P(X=-1) \Rightarrow a=0.2

\therefore b=0.3

例 3.8

由于 \rho=0, 故 X,Y 相互独立。

P(XY-Y<0)=P(Y(X-1)<0)=P(Y<0)P(X-1>0)+P(Y>0)P(X-1<0)=\frac12

例 3.10

可见 X,Y 相互独立。

f_1(x)=\int_0^{\frac\pi2}f(x,y)dy=cosx

f_2(y)=\int_0^{\frac\pi2}f(x,y)dx=cosy

F(x,y)=G(x)G(y), 其中

G(t)=\begin{cases}
sint&0<t<\frac\pi2 \\
1&t\geq\frac\pi2 \\
0&t\leq0
\end{cases}

(2)

P(0

例 3.16

(1)

\iint_{x^2+y^2<R^2}f(x,y)dxdy=\int_0^Rc(R-r)\cdot2rdr=c\cdot\frac13R^3=1

\therefore c=\frac3{R^3}

(2)

P(x^2+y^2\leq r^2)=\iint_{x^2+y^2\leq r^2}f(x,y)dxdy=\frac{r^3}{R^3}

例 3.23

易知 X,Y 相互独立

P(U=0,V=0)=\frac14

P(U=0,V=1)=0

P(U=1,V=0)=\frac14

P(U=1,V=1)=\frac12