175 lines
7.1 KiB
TeX
175 lines
7.1 KiB
TeX
\documentclass[全部作业]{subfiles}
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\input{mysubpreamble}
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\begin{document}
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\setcounter{chapter}{4}
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\setcounter{section}{1}
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\section{练习题\thesection}
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\begin{enumerate}
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\questionandanswerProof[1]{
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设$\{ (X_n,Y_n)\ ;\ n\geqslant 0 \}$是隐马氏链，其中$\{ X_n\ ;\ n\geqslant 0 \}$, $\{ Y_n\ ;\ n\geqslant 0 \}$分别是状态序列和观察序列，证明对任意$n\geqslant 0， m\geqslant 1$，
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$$
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\begin{aligned}
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&P\left( Y_{n+k}=j_{n+k}, 1\leqslant k\leqslant m|X_n=i, Y_l=j_l , 0\leqslant l\leqslant n \right) \\
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=&P\left( Y_{n+k}=j_{n+k}, 1\leqslant k\leqslant m | X_n=i \right) \\
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\end{aligned}
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$$
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}{
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设状态空间为$S$，转移概率矩阵为$(p_{ij})$，观测概率矩阵为$(q_{jk})$，则
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$$
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\begin{aligned}
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&P\left( Y_{n+k}=j_{n+k}, 1\leqslant k\leqslant m|X_n=i, Y_l=j_l , 0\leqslant l\leqslant n \right) \\
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=&\sum_{x \in S} p_{i, x}^{(k)} q_{x, j_{n+k}} \\
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=&P\left( Y_{n+k}=j_{n+k}, 1\leqslant k\leqslant m | X_n=i \right) \\
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\end{aligned}
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$$
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}
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\questionandanswerSolution[2]{
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假设例4.2.1中产品的质量分为1（合格）和0（不合格）。在机器状态为1时，产品质量为0,1的概率分别为0.05,0.95，而在机器状态为2时，产品质量为0,1的概率分别为0.5,0.5。假设在产品检测的间隔时间内机器状态转移概率为
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$$
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p_{1,1} = 0.9, \ p_{1,2}=0.1,\ p_{2,1} =0, p_{2,2}=1
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$$
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若初始机器状态为1，求第4次检测才首次检测到不合格产品的概率并在此条件下计算第5次抽到合格品的概率。
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}{
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设状态序列为$X$，观测序列为$Y$，则$Z=(X,Y)$构成隐马尔科夫过程，初始状态$X_0=1$。由题意可得
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$$
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\begin{tabular}{c|cc}
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P & 1 & 2 \\
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\hline
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1 & 0.9 & 0.1 \\
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2 & 0 & 1 \\
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\end{tabular}
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\qquad
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\begin{tabular}{c|cc}
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Q & 0 & 1 \\
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\hline
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1 & 0.05 & 0.95 \\
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2 & 0.5 & 0.5 \\
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\end{tabular}
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\qquad
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\begin{tabular}{c|cc}
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\pi & 1 & 2 \\
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\hline
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P & 0.9 & 0.1 \\
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\end{tabular}
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$$
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第4次检测才首次检测到不合格产品即观测序列为1110。由于$p_{2,1}=0$，所以可以直接不考虑状态序列的2后跟着1的情况。所以计算的概率如下表所示：
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$$
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\begin{tabular}{cc}
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\toprule
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\diagbox{X}{Y} & 1110 \\
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\midrule
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1111 & 0.9\times 0.95\times 0.9\times 0.95\times 0.9\times 0.95\times 0.9\times 0.05 = 0.028126186875 \\
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1112 & 0.9\times 0.95\times 0.9\times 0.95\times 0.9\times 0.95\times 0.1\times 0.5 = 0.03125131875\\
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1122 & 0.9\times 0.95\times 0.9\times 0.95\times 0.1\times 0.5\times 1\times 0.5 = 0.018275625\\
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1222 & 0.9\times 0.95\times 0.1\times 0.5\times 1\times 0.5\times 1\times 0.5 = 0.0106875\\
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2222 & 0.1\times 0.5\times 1\times 0.5\times 1\times 0.5\times 1\times 0.5 = 0.00625\\
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\midrule
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\sum & 0.028126186875+0.03125131875+0.018275625+0.0106875+0.00625 = 0.094590630625 \\
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\bottomrule
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\end{tabular}
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$$
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所以第4次检测才首次检测到不合格产品的概率为$\mathbf{0.094590630625}$。
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在此条件下计算第5次抽到合格品的概率，则根据马氏链的性质，先计算
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$$
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P(X_4=1) = 0.028126186875
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$$
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$$
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P(X_4=2) = 0.03125131875+0.018275625+0.0106875+0.00625 = 0.06646444375
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$$
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则从第4次开始计算，初始分布变为
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$$
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\begin{tabular}{c|cc}
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\pi & 1 & 2 \\
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\hline
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P & 0.028126186875 & 0.06646444375 \\
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\end{tabular}
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$$
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所以可以继续计算第4和第5步的概率（由于已经确定了$Y_4=0$，所以不需要乘$q_{1,0}$或$q_{2,0}$）：
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$$
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\begin{tabular}{cc}
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\toprule
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\diagbox{X}{Y} & 01 \\
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\midrule
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11 & 0.028126186875\times 0.9\times 0.95 = 0.024047889778125 \\
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12 & 0.028126186875\times 0.1\times 0.5 = 0.00140630934375 \\
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22 & 0.06646444375\times 1\times 0.5 = 0.033232221875 \\
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\midrule
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\sum & 0.024047889778125 + 0.00140630934375 + 0.033232221875 = 0.058686420996875 \\
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\bottomrule
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\end{tabular}
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$$
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所以在此条件下第5次抽到合格品的概率为$\mathbf{0.058686420996875}$。
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}
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\questionandanswerSolution[4]{
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在例4.2.5设定下，若前3天价格波动恰为$Y_0=1, Y_1=0, Y_2=-1$，求$X_2$的分布，并预测$X_3$最可能的取值。
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}{
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已知$Z_n=(X_n,Y_n)$为隐马氏链，
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$$
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\begin{tabular}{c|cc}
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P & -1 & 1 \\
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\hline
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-1 & 0.75 & 0.25 \\
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1 & 0.2 & 0.8 \\
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\end{tabular}
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\qquad
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\begin{tabular}{c|ccc}
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Q & -1 & 0 & 1 \\
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\hline
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-1 & 0.6 & 0.2 & 0.2 \\
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1 & 0.2 & 0.3 & 0.5 \\
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\end{tabular}
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\qquad
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\begin{tabular}{c|cc}
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\pi & -1 & 1 \\
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\hline
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P & 0.2 & 0.1 \\
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\end{tabular}
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$$
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将$X_0,X_1,X_2$的取值分情况讨论，得下表：
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$$
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\begin{tabular}{c|c}
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(a,b,c) & P(X_0=a,Y_0=1,X_1=b,Y_1=0,X_2=c,Y_2=-1) \\
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\hline
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(1,1,1) & 0.8\times 0.5\times 0.8\times 0.3\times 0.8\times 0.2 = 0.01536 \\
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(1,1,-1) & 0.8\times 0.5\times 0.8\times 0.3\times 0.2\times 0.6 = 0.01152 \\
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(1,-1,1) & 0.8\times 0.5\times 0.2\times 0.2\times 0.25\times 0.2 = 0.0008 \\
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(1,-1,-1) & 0.8\times 0.5\times 0.2\times 0.2\times 0.75\times 0.6 = 0.0072 \\
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(-1,1,1) & 0.2\times 0.2\times 0.25\times 0.3\times 0.8\times 0.2 = 0.00048 \\
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(-1,1,-1) & 0.2\times 0.2\times 0.25\times 0.3\times 0.2\times 0.6 = 0.00036 \\
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(-1,-1,1) & 0.2\times 0.2\times 0.75\times 0.2\times 0.25\times 0.2 = 0.0003 \\
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(-1,-1,-1) & 0.2\times 0.2\times 0.75\times 0.2\times 0.75\times 0.6 = 0.0027 \\
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\end{tabular}
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$$
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将$X_2$的情况提取出来可以计算得
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$$
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P(Y_0=1,Y_1=0,X_2=1,Y_2=-1)=0.01536+0.0008+0.00048+0.0003 = 0.01694
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$$
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$$
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P(Y_0=1,Y_1=0,X_2=-1,Y_2=-1)=0.01152+0.0072+0.00036+0.0027 = 0.02178
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$$
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所以
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$$
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\begin{aligned}
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&P(\mathbf{X_2=1}|Y_0=-1,Y_1=0,Y_2=-1) = \frac{P(Y_0=1,Y_1=0,X_2=1,Y_2=-1)}{P(Y_0=-1,Y_1=0,Y_2=-1)} \\
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&=\frac{0.01694}{0.01694+0.02178} = \mathbf{0.4375} \\
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\end{aligned}
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$$
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所以$P(\mathbf{X_2=-1}|Y_0=1,Y_1=0,Y_2=-1)=1-0.4375 = \mathbf{0.5625}$。\\
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即为$X_2$的分布。
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令$P'(\cdot )=P(\cdot |Y_0=1,Y_1=0,Y_2=-1)$，则由于$X_n$是马氏链，可得
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$$
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P'(X_3=1)=P'(X_2=1)p_{1,1}+P'(X_2=-1)p_{-1,1} = 0.4375\times 0.8+0.5625\times 0.25 = 0.490625
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$$
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$$
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P'(X_3=-1)=P'(X_2=1)p_{1,-1}P'(X_2=-1)p_{-1,-1}=0.4375\times 0.2+0.5625\times 0.75 = 0.509375
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$$
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所以$X_3$最可能的取值为$\mathbf{-1}$。
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}
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\end{enumerate}
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\end{document} |