随机过程Hw2

随机过程Hw2
1、设${\xi }_1,{\xi }_2,\cdots ,{\xi }_n$独立同分布，${\xi }_i$~$U(0,1)$,对$0\le t\le 1$定义
$X(t)=\frac{1}{n}{\sum }_{i=1}^n{1}_{{\xi }_i\le t}$
求EX(t)，cov(X(s),X(t)).
solution
(1)
$$\begin{array}{rl}E(x)& =E(\frac{1}{n}\sum _{i=1}^n{1}_{{\xi }_i\le t})\\ & =\frac{1}{n}\sum _{i=1}^{n}P({\xi }_i\le t)=t\end{array}$$
$$\begin{array}{rl}cov(X(s),X(t))& =EX(S)X(t)-EX(S)EX(t)\\ & =E(\frac{1}{n^2}\sum _{i=1}^n\sum _{j=1}^n{1}_{{\xi }_i\le s}{1}_{{\xi }_j\le t})-st\\ & =\frac{(n^2-n)st-n(s\wedge t)}{n^2}-st\\ & =\frac{s\wedge t-st}{n}\end{array}$$
2、令{Zn;n∈Z}\{Z_n;n\in Z\}{Zn​;n∈Z}是两两不相关随机变量序列，$E{Z}_n=0,Var{Z}_n=1$,令
$$X_n=\sum _{i=0}^r{\alpha }_i{Z}_{n-i},n\in Z$$
这里$r\ge 1$,${\alpha }_0,{\alpha }_1,{\alpha }_2,\cdots ,{\alpha }_r$为常数.
求$E{X}_n和Cov(X_n,X_m)$.
Solution
$$E{X}_n=0$$
$$\begin{array}{rl}Cov(X_n,X_m)& =E{X}_n{X}_m-E{X}_{n}E{X}_m\\ & =E(\sum _{i=0}^r{\alpha }_i{Z}_{n-i}\sum _{j=0}^r{\alpha }_j{Z}_{n-j})(不妨设n\le m)\\ & =\{\begin{array}{ll}0& |m-n|>r\\ {\sum }_{k=0}^{r-|m-n|}{\alpha }_k{\alpha }_{k+|m-n|}& |m-n|\le r\end{array}\end{array}$$

3、设$X(t)=At+(1-|A|)B,t\ge 0$,这里A和B独立同分布，$P(A=0)=P(A=1)=P(A=-1)=\frac{1}{3}$
(1)写出{$X(t)$}的所有样本函数
(2)计算$P(X(1)=1),P(X(2)=1),P(X(1)=1,X(2)=1)$.
Solution
(1)
$$X(t)=\{\begin{array}{ll}0& A=0,B=0,\\ 1& A=0,B=1,\\ -1& A=0,B=-1,\\ t& A=1,\\ -t& A--1.\end{array}$$
(2)
$$\begin{gathered} P(X(1)=1)=\frac{4}{9} \\ P(X(2)=1)=\frac{1}{9} \\ P(X(1)=1,X(2)=1)=\frac{1}{9} \end{gathered}$$
4、设$Z(t)=AXt+1-A,t\ge 0$,这里A和X相互独立，$P(A=0)=P(A=1)=\frac{1}{2}$,X~ N(1,1).
(1)计算$P(Z(1)<1),P(Z(2)<2),P(Z(1)<1,Z(2)<2);$
(2)计算${\mu }_Z(t),R_Z(s,t)$
Solution
$$\begin{gathered} (1)P(Z(1)<1)=\frac{1}{4} \\ P(Z(2)<2)=\frac{3}{4} \\ P(Z(1)<1,Z(2)<2)=\frac{1}{4} \end{gathered}$$
(2)
$$\begin{array}{rl}{R}_Z(s,t)& =E(Z(s)Z(t))\\ & =E[A^2{X}^{2}st+A(1-A)X(s+t)+(1-A{)}^2]\\ & =stE{A}^{2}E{X}^2+(s+t)E[A(1-A)]EX+E(1-A{)}^2\\ & =st+\frac{1}{2}\end{array}$$

5、独立重复投掷一颗均匀的骰子，用$Z_n$表示前n次中掷出6点的次数
(1)计算$P(Z_2=1,Z_5=3,Z_7=5)$；
(2)求$P(Z_{18000}>2900)$的近似值
(3)若掷骰子一直到刚好出现20次6点为止，问需掷多于180次的概率近似为多少.
Solution
(1)
$$\begin{array}{rl}& P(Z_2=1,Z_5=3,Z_7=5)\\ & =P(Z_7=5|Z_5=3,Z_2=1)P(Z_5=3|Z_2=1)P(Z_2=1)\\ & =P(Z_2=2)P(Z_3=2)P(Z_2=1)\\ & =\frac{25}{46656}\end{array}$$
(2)
$$\begin{array}{rl}& E{Z}_{18000}=3000,Var{Z}_{18000}=2500\\ & P(Z_{18000}>2900)\\ & =P(\frac{Z_{18000}-3000}{\sqrt{2500}}>\frac{2900-3000}{\sqrt{2500}})\\ & \approx 1-\Phi (-2)\\ & \approx 0.9772\end{array}$$
(3)
将问题转化为，投掷180次后点数等于6的次数小于等于19，类似第二题由中心极限定理可得：
$$\begin{array}{rl}& P(Z_{180}>19)\\ & =P(\frac{Z_{180}-30}{\sqrt{25}}\le \frac{19-30}{\sqrt{25}})\\ & \approx \Phi (-2.2)\\ & \approx 0.0139\end{array}$$