Chapter 7

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