7.2 习题课讲解

解 设$Z:=X+Y$，依“概率不变原则”可得$$\begin{array}{}\text{(1)}& \mathrm{d}P=f_{X,Z}(x,z)\mathrm{d}\mu (x,z)=f_{X,Y}(x,y)\mathrm{d}\mu (x,y)\end{array}$$ 因此$$\begin{array}{}\text{(2)}& f_{X,Z}(x,z)=f_{X,Y}(x,y)|det\frac{\partial (x,y)}{\partial (x,z)}|=f_X(x)f_Y(z-x)\end{array}$$ 因此$$\begin{array}{}\text{(3)}& f_Z(z)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{f}_X(x)f_Y(z-x)\mathrm{d}x=(f_X\ast f_Y)(z)\end{array}$$ 其中$f_X\ast f_Y$表示$f_X$和$f_Y$的卷积。同理可得$$\begin{array}{}\text{(4)}& f_{X-Y}(z)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{f}_X(x)f_Y(x-z)\mathrm{d}x\end{array}$$ $◻$

解 设两个点的坐标分别为$(X_1,Y_1)$和$(X_2,Y_2)$，依题意可得$$\begin{array}{}\text{(5)}& \bar{d}={\int }_{[0,1{]}^4}\sqrt{(x_1-x_2{)}^2+(y_1-y_2{)}^2}\mathrm{d}{x}_1\mathrm{d}{y}_1\mathrm{d}{x}_2\mathrm{d}{y}_2\end{array}$$ 这是一个很难计算的四重积分。为此我们需要换一种思路。

设$X:=X_1-X_2$、$Y=Y_1-Y_2$。以$X$为例（$Y$同理），由于$X_1,X_2$为$[0,1]$均匀分布，则$X$为三角分布，其概率分布为$$\begin{array}{}\text{(6)}& f_X(x)={\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{f}_{X_1}(x_1)f_{X_2}(x_1-x)\mathrm{d}{x}_1=\{\begin{array}{ll}1+x,& x\in [-1,0]\\ 1-x,& x\in [0,1]\end{array}\end{array}$$ 再设$U=|X|$，则$U$的分布列为$$\begin{array}{}\text{(7)}& f_U(u)=f_X(u)+f_X(-u)=\{\begin{array}{ll}2(1-u),& u\in [0,1]\\ 0,& \text{otherwise}\end{array}\end{array}$$ 同理可设$V=|Y|$，则$V$的分布列为$$\begin{array}{}\text{(8)}& f_V(v)=\{\begin{array}{ll}2(1-v),& v\in [0,1]\\ 0,& \text{otherwise}\end{array}\end{array}$$ 因此距离的期望为$$\begin{array}{}\text{(9)}& \begin{array}{rl}\bar{d}& ={\int }_{[0,1{]}^2}\sqrt{u^2+v^2}\mathrm{d}P(u,v)={\int }_{[0,1{]}^2}\sqrt{u^2+v^2}{f}_U(u)f_V(v)\mathrm{d}u\mathrm{d}v\\ & =8{\int }_0^1\mathrm{d}u{\int }_0^{1-u}\sqrt{u^2+v^2}(1-u)(1-v)\mathrm{d}v\end{array}\end{array}$$ 利用极坐标换元$(u,v)=(r\cos \theta ,r\sin \theta )$可得$$\begin{array}{}\text{(10)}& \begin{array}{rl}\bar{d}& =8{\int }_0^{\pi /4}\mathrm{d}\theta {\int }_0^{\sec \theta }(1-r\cos \theta )(1-r\sin \theta )r^2\mathrm{d}r\\ & =8{\int }_0^{\pi /4}[\frac{{\sec }^3\theta }{3}-\frac{{\sec }^4\theta }{4}\cos \theta -\frac{{\sec }^4\theta }{4}\sin \theta +\frac{{\sec }^5\theta }{5}\sin \theta \cos \theta ]\mathrm{d}\theta \\ & =8[\frac{1}{12}{\int }_0^{\pi /4}{\sec }^3\theta \mathrm{d}\theta -\frac{1}{20}{\int }_{\sqrt{2}/2}^1\frac{\mathrm{d}t}{t^4}]=\frac{2+\sqrt{2}+5\ln (\sqrt{2}+1)}{15}\end{array}\end{array}$$ $◻$

解 记$S_n=\sum _{i=1}^n{X}_i$，依据题意可得$$\begin{array}{}\text{(12)}& \mathrm{P}(S_n\le c)=|D|,D=\{(x_1,\cdots ,x_n)\mid x_1+\cdots +x_n\le c\}\cap [0,1{]}^n\end{array}$$ 则$N$的分布列为$$\begin{array}{}\text{(13)}& \mathrm{P}(N=n)=\{\begin{array}{ll}0,& n\le \lfloor c\rfloor \\ \mathrm{P}(S_{n-1}\le c)-\mathrm{P}(S_n\le c),& n>\lfloor c\rfloor \end{array}\end{array}$$

我们首先证明：$$\begin{array}{}\text{(14)}& |D_n|=\frac{c^n}{n!},D_n=\{(x_1,\cdots ,x_n)\mid x_1+\cdots +x_n\le c\}\cap [0,+\mathrm{\infty }{)}^n\end{array}$$ 设$1\le m\le n$，记$D_m=\{(x_1,\cdots ,x_m)\mid x_1+\cdots +x_m\le c\}\cap [0,+\mathrm{\infty }{)}^n$，定义$$\begin{array}{}\text{(15)}& I(n,m):={\int }_{D_m}\frac{{(c-\sum _{i=1}^m{x}_i)}^{n-m}}{(n-m)!}\mathrm{d}{x}_1\cdots \mathrm{d}{x}_m\end{array}$$ 则$|D_n|=I(n,n)$。记$A=c-\sum _{i=1}^{m-1}{x}_i$，注意到$$\begin{array}{}\text{(16)}& \begin{array}{rl}I(n,m)& ={\int }_{D_{m-1}}\mathrm{d}{x}_1\cdots \mathrm{d}{x}_{m-1}{\int }_0^A\frac{(A-x_m{)}^{n-m}}{(n-m)!}\mathrm{d}{x}_m\\ & ={\int }_{D_{m-1}}\frac{A^{n-m+1}}{(n-m+1)!}\mathrm{d}{x}_1\cdots \mathrm{d}{x}_{m-1}=I(n,m-1)\end{array}\end{array}$$ 因此$$\begin{array}{}\text{(17)}& |D_n|={\int }_{D_n}\mathrm{d}{x}_1\cdots \mathrm{d}{x}_n=I(n,n)=I(n,0)=\frac{c^n}{n!}\end{array}$$

当$c\le 1$时，$D=D_n$，此时有$\mathrm{P}(S_n\le c)=\frac{c^n}{n!}$。当$c>1$时，定义$$\begin{array}{}\text{(18)}& E_i:=\{(x_1,\cdots ,x_n)|x_i>1,x_1,\cdots ,x_{i-1},x_{i+1},\cdots ,x_n\ge 0,\sum _{i=1}^n{x}_i\le c\}\end{array}$$ 显然有$E_i\cap D=\varnothing$，且$$\begin{array}{}\text{(19)}& D=D_n\setminus \bigcup _{i=1}^n{E}_i\end{array}$$ 根据容斥原理可得$$\begin{array}{}\text{(20)}& \begin{array}{rl}\left|\bigcup _{i=1}^n{E}_i\right|& =\sum _{i=1}^n\left|E_i\right|-\sum _{1\le i<j\le n}|E_i\cap E_j|+\cdots +(-1{)}^{n-1}|E_1\cap \cdots \cap E_n|\\ & =\sum _{k=1}^{min\{\lfloor c\rfloor ,n\}}(-1{)}^{k-1}\sum _{1\le i_1<\cdots <i_k\le n}|E_{i_1}\cap \cdots \cap E_{i_k}|\\ & =\sum _{k=1}^{min\{\lfloor c\rfloor ,n\}}(-1{)}^{k-1}\sum _{1\le i_1<\cdots <i_k\le n}\frac{(c-k{)}^n}{n!}=\sum _{k=1}^{min\{\lfloor c\rfloor ,n\}}(-1{)}^{k-1}(\genfrac{}{}{0ex}{}{n}{k})\frac{(c-k{)}^n}{n!}\end{array}\end{array}$$ 其中$$\begin{array}{}\text{(21)}& \begin{array}{rl}\bigcup _{j=1}^k{E}_{i_j}& =\{(x_1,\cdots ,x_n)|x_{i_1},\cdots ,x_{i_k}>1,x_1,\cdots ,x_n\ge 0,\sum _{i=1}^n{x}_i\le c\}\\ & =\{(x_1^{\prime },\cdots ,x_n^{\prime })|x_{i_1}^{\prime },\cdots ,x_{i_k}^{\prime }>0,x_1^{\prime },\cdots ,x_n^{\prime }\ge 0,\sum _{i=1}^n{x}_i^{\prime }\le c-k\}\\ & \Rightarrow \left|\bigcup _{j=1}^k{E}_{i_j}\right|=\frac{(c-k{)}^n}{n!}\end{array}\end{array}$$ 因此$$\begin{array}{}\text{(22)}& \mathrm{P}(S_n\le c)=|D|=|D_n|-\left|\bigcup _{i=1}^n{E}_i\right|=\frac{1}{n!}\sum _{k=0}^{min\{\lfloor c\rfloor ,n\}}(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})(c-k{)}^n\end{array}$$

当$c\ge n$时，显然有$$\begin{array}{}\text{(23)}& \mathrm{P}(S_n\le c)=\frac{1}{n!}\sum _{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})(c-k{)}^n=1,c\ge n\end{array}$$ 上式为一个关于$c$的多项式，其在$c\ge n$时恒为$1$，故有$$\begin{array}{}\text{(24)}& \sum _{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})(c-k{)}^n\equiv n!,c\in \mathbb{R}\end{array}$$ 因此$N$的数学期望为$$\begin{array}{}\text{(25)}& \begin{array}{rl}\mathbb{E}[N]& =\sum _{n=\lfloor c\rfloor +1}^{+\mathrm{\infty }}n\mathrm{P}(N=n)=\sum _{n=\lfloor c\rfloor +1}^{+\mathrm{\infty }}n[\mathrm{P}(S_{n-1}\le c)-\mathrm{P}(S_n\le c)]\\ & =\sum _{n=\lfloor c\rfloor +1}^{+\mathrm{\infty }}[(n-1)\mathrm{P}(S_{n-1}\le c)-n\mathrm{P}(S_n\le c)]+\sum _{n=\lfloor c\rfloor +1}^{+\mathrm{\infty }}\mathrm{P}(S_{n-1}\le c)\\ & =\lfloor c\rfloor +1+\sum _{n=\lfloor c\rfloor +1}^{+\mathrm{\infty }}\frac{1}{n!}\sum _{k=0}^{\lfloor c\rfloor }(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})(c-k{)}^n\\ & =\lfloor c\rfloor +1+\sum _{k=0}^{\lfloor c\rfloor }\frac{(-1{)}^k}{k!}(c-k{)}^k[{\mathrm{e}}^{c-k}-\sum _{m=0}^{\lfloor c\rfloor -k}\frac{(c-k{)}^m}{m!}]\\ & =\sum _{k=0}^{\lfloor c\rfloor }\frac{(-1{)}^k}{k!}(c-k{)}^k{\mathrm{e}}^{c-k}+\lfloor c\rfloor +1-\sum _{k=0}^{\lfloor c\rfloor }\sum _{m=0}^{\lfloor c\rfloor -k}\frac{(-1{)}^k(c-k{)}^{k+m}}{k!m!}\end{array}\end{array}$$ 借助$(\text{24})$式，可对$n$使用数学归纳法证明$$\begin{array}{}\text{(26)}& \sum _{k=0}^n\sum _{m=0}^{n-k}\frac{(-1{)}^k(c-k{)}^{k+m}}{k!m!}=n+1,n\in {\mathbb{N}}^{\ast },c\in \mathbb{R},c\ge n\end{array}$$ 因此$$\begin{array}{}\text{(27)}& \mathbb{E}[N]=\sum _{k=0}^{\lfloor c\rfloor }\frac{(-1{)}^k}{k!}(c-k{)}^k{\mathrm{e}}^{c-k}\end{array}$$ $◻$

另解 $X_i$的特征函数为$$\begin{array}{}\text{(28)}& {\phi }_{X_i}(t)={\int }_0^1{\mathrm{e}}^{\mathrm{i}tx}\mathrm{d}x=\frac{{\mathrm{e}}^{\mathrm{i}t}-1}{\mathrm{i}t}\end{array}$$ 记$S_n=\sum _{i=1}^n{X}_i$，则$S_n$的特征函数为$$\begin{array}{}\text{(29)}& {\phi }_{S_n}(t)=\prod _{i=1}^n{\phi }_{X_i}(t)={\left(\frac{{\mathrm{e}}^{\mathrm{i}t}-1}{\mathrm{i}t}\right)}^n\end{array}$$ 因此$S_n$的概率密度函数为$$\begin{array}{}\text{(30)}& f_{S_n}(s)=\frac{1}{2\pi }{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}{\left(\frac{{\mathrm{e}}^{\mathrm{i}t}-1}{\mathrm{i}t}\right)}^n{\mathrm{e}}^{-\mathrm{i}st}\mathrm{d}t={\mathcal{F}}^{-1}\left[{\left(\frac{{\mathrm{e}}^{\mathrm{i}t}-1}{\mathrm{i}t}\right)}^n\right](s)\end{array}$$ 注意到$$\begin{array}{}\text{(31)}& \begin{array}{rl}{\mathcal{F}}^{-1}[({\mathrm{e}}^{\mathrm{i}t}-1{)}^n](s)& =\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})(-1{)}^{n-k}{\mathcal{F}}^{-1}\left[{\mathrm{e}}^{\mathrm{i}kt}\right](s)=\sum _{k=0}^n(\genfrac{}{}{0ex}{}{n}{k})(-1{)}^{n-k}\delta (s-k)\\ {\mathcal{F}}^{-1}\left[\frac{1}{\mathrm{i}t}\right]& =\frac{1}{2\pi }\mathrm{PV}{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{{\mathrm{e}}^{-\mathrm{i}st}}{\mathrm{i}t}\mathrm{d}t=-{\int }_{-\mathrm{\infty }}^{+\mathrm{\infty }}\frac{\sin st}{2\pi t}\mathrm{d}t=-\frac{1}{2}\mathrm{sgn}s\end{array}\end{array}$$ 结合Fourier变换的卷积性质$$\begin{array}{}\text{(32)}& \mathcal{F}[f\ast g](t)=\mathcal{F}[f](t)\cdot \mathcal{F}[g](t)\end{array}$$ 利用数学归纳法逐次卷积可得$$\begin{array}{}\text{(33)}& f_{S_n}(s)=\frac{1}{2(n-1)!}\sum _{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})(s-k{)}^{n-1}\mathrm{sgn}(s-k)\end{array}$$ 当$s\ge n$时，显然有$$\begin{array}{}\text{(34)}& f_{S_n}(s)=\frac{1}{2(n-1)!}\sum _{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})(s-k{)}^{n-1}=0,s\ge n\end{array}$$ 上式为一个关于$s$的多项式，其在$s\ge n$时恒为$0$，故有$$\begin{array}{}\text{(35)}& \sum _{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})(s-k{)}^{n-1}\equiv 0,s\in \mathbb{R}\end{array}$$ 借助$(\text{35})$可得$$\begin{array}{}\text{(36)}& \begin{array}{rl}\mathrm{P}(S_n\le c)& ={\int }_0^c{f}_{S_n}(s)\mathrm{d}s=\frac{1}{(n-1)!}\sum _{k=0}^n(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k}){\int }_0^c(s-k{)}^{n-1}[\theta (s-k)-\frac{1}{2}]\mathrm{d}s\\ & =\frac{1}{(n-1)!}\sum _{k=0}^{min\{\lfloor c\rfloor ,n\}}(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k}){\int }_0^c(s-k{)}^{n-1}-\frac{1}{2(n-1)!}{\int }_0^{c}0\mathrm{d}s\\ & =\frac{1}{n!}\sum _{k=0}^{min\{\lfloor c\rfloor ,n\}}(-1{)}^k(\genfrac{}{}{0ex}{}{n}{k})(c-k{)}^n,0\le c\le n\end{array}\end{array}$$ 后续解题步骤与前述解法相同。 $◻$

解 由Green公式可得$$\begin{array}{}\text{(38)}& \begin{array}{rl}I& =-{\int }_D[\frac{\partial }{\partial x}\left(\sqrt{x^2+y^2}\frac{\partial f}{\partial y}\right)-\frac{\partial }{\partial y}\left(\sqrt{x^2+y^2}\frac{\partial f}{\partial x}\right)]\mathrm{d}x\mathrm{d}y\\ & =-{\oint }_{\partial D}\sqrt{x^2+y^2}(\frac{\partial f}{\partial x}\mathrm{d}x+\frac{\partial f}{\partial y}\mathrm{d}y)=-R{\oint }_{\partial D}\mathrm{d}f=0\end{array}\end{array}$$ $◻$