192 lines
8.5 KiB
TeX
192 lines
8.5 KiB
TeX
\documentclass[全部作业]{subfiles}
|
||
\input{mysubpreamble}
|
||
\begin{document}
|
||
\setcounter{chapter}{2}
|
||
\section{《随机过程》第四周作业}
|
||
\setcounter{subsection}{1}
|
||
\subsection{练习题\thechapter.\thesubsection}
|
||
\begin{enumerate}
|
||
\questionandanswer[]{
|
||
已知$W$是初值为$0$的随机游动，步长分布为
|
||
$$
|
||
P(X_n=-2)=P(X_n=-1)=P(X_n=1)=\frac{1}{3}
|
||
$$
|
||
}{}
|
||
\begin{enumerate}
|
||
\questionandanswerSolution[]{
|
||
求概率$P(W_3=-2)$，
|
||
}{
|
||
设$m,n,k$分别为选择$-2,-1,1$的次数，那么
|
||
$$
|
||
\begin{cases}
|
||
m+n+k=3 \\
|
||
-2m-n+k=-2 \\
|
||
\end{cases}
|
||
$$
|
||
由此可得$3m+2n=5$。由于$m,n,k$均为非负整数，因此
|
||
$$
|
||
(m,n,k)=(1,1,1)
|
||
$$
|
||
所以
|
||
$$
|
||
P(W_3=-2)=p_3(0,-2)=C_{3}^{1}C_{2}^{1} \left( \frac{1}{3} \right) ^{3}=\frac{2}{9}
|
||
$$
|
||
}
|
||
\questionandanswerSolution[]{
|
||
求概率$P(W_3=-2,W_7=0)$。
|
||
}{
|
||
$$
|
||
P(W_3=-2,W_7=0)=p_3(0,-2)p_4(-2,0)=p_3(0,-2)p_4(0,2)
|
||
$$
|
||
继续同上方法
|
||
$$
|
||
\begin{cases}
|
||
m+n+k=4 \\
|
||
-2m-n+k=2 \\
|
||
\end{cases} \Longrightarrow (m,n,k)=(0,1,3)
|
||
$$
|
||
$$
|
||
p_4(0,2)=\mathrm{C}_{4}^{1}\left( \frac{1}{3} \right) ^{4} = \frac{4}{81}
|
||
$$
|
||
所以
|
||
$$
|
||
P(W_3=-2,W_7=0)=\frac{2}{9}\times \frac{4}{81} = \frac{8}{729}
|
||
$$
|
||
}
|
||
\questionandanswerSolution[]{
|
||
求概率$P(W_1>0,W_2>0, \cdots ,W_5>0,W_6=2)$。
|
||
}{
|
||
% $$
|
||
% \frac{\mathrm{d}}{\mathrm{d}x}\frac{\mathrm{d}}{\mathrm{d}x}\frac{\mathrm{d}}{\mathrm{d}x}\frac{\mathrm{d}}{\mathrm{d}x}\frac{\mathrm{d}}{\mathrm{d}x}(x^{3}+x+1)^{5} = 360360 x^{10} + 772200 x^{8} + 475200 x^{7} + 554400 x^{6} + 604800 x^{5} + 302400 x^{4} + 201600 x^{3} + 88200 x^{2} + 21600 x + 3720
|
||
% $$
|
||
根据反射原理，
|
||
$$
|
||
\begin{aligned}
|
||
&P(W_1>0,W_2>0, \cdots ,W_5>0,W_6=2) \\
|
||
=&P(W_1=1)P(W_2>0, \cdots ,W_5>0,W_6=2) \\
|
||
=&P(W_1=1)P(W_2>0, \cdots ,W_5>0,W_6=2) \\
|
||
=&P(W_1=1)p_5(-1,2) = P(W_1=1)p_5(0,3) \\
|
||
\end{aligned}
|
||
$$
|
||
继续同上方法
|
||
$$
|
||
\begin{cases}
|
||
m+n+k=5 \\
|
||
-2m-n+k=3 \\
|
||
\end{cases} \Longrightarrow (m,n,k)=(0,1,4)
|
||
$$
|
||
$$
|
||
p_5(0,3)=\mathrm{C}_{5}^{1}\left( \frac{1}{3} \right) ^{5} = \frac{5}{243}
|
||
$$
|
||
所以
|
||
$$
|
||
\text{原式}=P(W_1=1)p_5(0,3)=\frac{1}{3}\times \frac{5}{243} = \frac{5}{729}
|
||
$$
|
||
}
|
||
\end{enumerate}
|
||
\questionandanswerSolution[5]{
|
||
分别记甲乙两种药物对某种疾病的治愈率为$p_1,p_2$。为了比较它们的治愈率
|
||
大小, 安排了一系列如下的临床对比试验: 每次试验同时治疗两个病人,一
|
||
个接受甲药治疗，另一个接受乙药治疗。观察每次治疗效果。假设每次试验
|
||
是独立的; 病人对药物的疗效无本质影响。试用恰当的整数值随机游动模型
|
||
对这样一系列的试验中治愈病人数差异的变化建模。 对任意$x\in \mathbb{Z}$,试写该
|
||
模型的一步转移概率$p(x)$。
|
||
}{
|
||
设随机变量$X_0=1$，随机变量序列$\{ X_n\ ;\ n\geqslant 1 \}$表示每次试验增加的治愈病人数，而随机过程$W=\{ \sum_{k=0}^{n} X_k\ ;\ n\geqslant 0 \}$表示这样一系列的试验中总治愈病人数，则$\{ X_n\ ;\ n\geqslant 0 \}$相互独立且只能取整数值，且对$\forall n\geqslant 1$，$X_n$的分布相同，从而$W$是整数值的随机游动。则该模型的一步转移概率为
|
||
$$
|
||
p(x)=\begin{cases}
|
||
(1-p_1)(1-p_2),\quad & x=0 \\
|
||
p_1(1-p_2)+p_2(1-p_1),\quad & x=1 \\
|
||
p_1p_2,\quad & x=2 \\
|
||
\end{cases}\quad =\begin{cases}
|
||
1-p_1-p_2+p_1p_2,\quad & x=0 \\
|
||
p_1+p_2-2p_1p_2,\quad & x=1 \\
|
||
p_1p_2,\quad & x=2 \\
|
||
\end{cases}
|
||
$$
|
||
}
|
||
% $$
|
||
% 1-\sum_{i=1}^{n} \frac{1}{2i-1}\binom{2i}{i}\left( \frac{1}{2} \right) ^{2i} - \binom{2n}{n}\left( \frac{1}{2} \right) ^{2n} = 0 = = - \frac{\Gamma(n + \frac{1}{2})}{\sqrt{\pi} \Gamma(n + 1)} + 1
|
||
% $$
|
||
\questionandanswer[6]{
|
||
对初值为$0$的简单对称随机游动$W$,
|
||
}{}
|
||
\begin{enumerate}
|
||
\questionandanswerProof[]{
|
||
对任意$n\geqslant 1$，证明$P(W_1\neq 0, W_2\neq 0, \cdots ,W_{2n}\neq 0)=P(W_{2n}=0)$。
|
||
}{
|
||
$$
|
||
\text{左边}=P(\tau_0>2n), \quad \text{右边}=\mathrm{C}_{2n}^{n}\left( \frac{1}{2} \right) ^{n}\left( \frac{1}{2} \right) ^{n}
|
||
$$
|
||
$$
|
||
\begin{aligned}
|
||
P(\tau_0>2n)=1-P(\tau_0\leqslant 2n)=1-\sum_{m=1}^{2n} P(\tau_0=m)
|
||
\end{aligned}
|
||
$$
|
||
由于$P(\tau_0 \text{为奇数})=0$，所以
|
||
$$
|
||
\begin{aligned}
|
||
P(\tau_0>2n)=1-\sum_{i=1}^{n} P(\tau_0=2i)=1-\sum_{i=1}^{n} \frac{1}{2i-1}\mathrm{C}_{2i}^{i} \left( \frac{1}{2} \right) ^{2i} =\frac{\Gamma(n + \frac{1}{2})}{\sqrt{\pi} \Gamma(n + 1)}
|
||
\end{aligned}
|
||
$$
|
||
使用Python计算可得
|
||
$$
|
||
\frac{\Gamma(n + \frac{1}{2})}{\sqrt{\pi} \Gamma(n + 1)} - \binom{2n}{n}\left( \frac{1}{2} \right) ^{2n} = 0
|
||
$$
|
||
即
|
||
$$
|
||
P(\tau_0>2n)=\frac{\Gamma(n + \frac{1}{2})}{\sqrt{\pi} \Gamma(n + 1)} = \binom{2n}{n}\left( \frac{1}{2} \right) ^{2n} = \mathrm{C}_{2n}^{n}\left( \frac{1}{2} \right) ^{n}\left( \frac{1}{2} \right) ^{n}
|
||
$$
|
||
因此
|
||
$$
|
||
\text{左边}=\text{}右边
|
||
$$
|
||
}
|
||
\questionandanswerSolution[]{
|
||
令$F_1(s)=\sum_{k=1}^{\infty} P(\tau_1=k|W_0=0)s^{k}$，其中$\left\vert s \right\vert <1$，求$F_1(s)$。
|
||
}{
|
||
% $$
|
||
% \sum_{k=1}^{\infty} \binom{k}{\frac{k-1}{2}}\left( \frac{1}{2} \right) ^{2k} s^{k} = \frac{\sum_{k=1}^{\infty} \frac{2^{- k} s^{k} \Gamma(\frac{k}{2} + 1)}{\Gamma(\frac{k + 3}{2})}}{\sqrt{\pi}}
|
||
% $$
|
||
$$
|
||
\begin{aligned}
|
||
F_1(s)&=\sum_{k=1}^{\infty} P_0(\tau_1=k)s^{k}=\sum_{k=1}^{\infty} P(\tau_1=k,W_k=1)s^{k} \\
|
||
&=\sum_{k=1}^{\infty} P(\tau_1=k|W_k=1)P(W_k=1)s^{k} \\
|
||
\end{aligned}
|
||
$$
|
||
根据对称原理，从$0$经过$k$步到$1$中间不碰$1$，相当于从$0$经过$k$步到$1$中间不碰$0$。
|
||
$$
|
||
\begin{aligned}
|
||
\text{上式}&=\sum_{k=1}^{\infty} P(\tau_0>k|W_k=1)P(W_k=1)s^{k} \\
|
||
&=\sum_{k=1}^{\infty}\frac{1}{k}\mathrm{C}_{k}^{\frac{k-1}{2}} \left( \frac{1}{2} \right) ^{2k} s^{k} \\
|
||
&=\sum_{k=1}^{\infty}\frac{1}{k}\left( \frac{1}{2} \right) ^{2k} \frac{k!s^{k}}{\frac{k+1}{2}! \frac{k-1}{2}!}\\
|
||
\end{aligned}
|
||
$$
|
||
而$\sqrt{1-x}$在$x=0$点的幂级数展开为
|
||
$$
|
||
\sqrt{1-x}=\sum_{n=1}^{\infty} (-1)^{n}\left( \frac{1}{2} \right) ^{n}n!
|
||
$$
|
||
所以
|
||
$$
|
||
F_1(s) = \frac{\sum_{k=1}^{\infty} \frac{2^{- k} s^{k} \Gamma(\frac{k}{2})}{\Gamma(\frac{k + 3}{2})}}{2 \sqrt{\pi}}
|
||
$$
|
||
}
|
||
\end{enumerate}
|
||
\questionandanswerProof[7]{
|
||
对初值为$0$的简单随机游动$W$，证明：对任意正整数$N$以及与$N$同奇偶的整数$a\in [-N,N]$，
|
||
$$
|
||
P(W_k<k,k=1, \cdots ,N-1|W_N=a)=\frac{N-a}{2N}
|
||
$$
|
||
}{
|
||
设$W$是$(q,p)$简单随机游动
|
||
$$
|
||
\begin{aligned}
|
||
&P(W_k<k,k=1, \cdots ,N-1|W_N=a)=\frac{P(W_k<k,k=1, \cdots ,N-1, W_N=a)}{P(W_N=a)} \\
|
||
=&\frac{q \cdot P(W_N=a|W_1=-1)}{P(W_N=a|W_0=0)} = \frac{q\cdot P(W_{N-1}=a+1|W_0=0)}{P(W_N=a|W_0=0)} \\
|
||
=&\frac{q\cdot \mathrm{C}_{N-1}^{\frac{N+a}{2}}p^{\frac{N+a}{2}}q^{\frac{N-a}{2}-1}}{\mathrm{C}_{N}^{\frac{N+a}{2}}p^{\frac{N+a}{2}}q^{\frac{N-a}{2}}} = \frac{\mathrm{C}_{N-1}^{\frac{N+a}{2}}}{\mathrm{C}_{N}^{\frac{N+a}{2}}} = \frac{\frac{(N-1)!}{(\frac{N+a}{2})!(\frac{N-a}{2}-1)!}}{\frac{N!}{(\frac{N+a}{2})!(\frac{N-a}{2})!}}=\frac{\frac{1}{1}}{\frac{N}{\frac{N-a}{2}}}=\frac{\frac{N-a}{2}}{N} \\
|
||
=&\frac{N-a}{2N} \\
|
||
\end{aligned}
|
||
$$
|
||
}
|
||
\end{enumerate}
|
||
\end{document} |