統計學習方法 第十章習題答案

10.1

題目：

給定盒子和球組成的隱馬爾可夫模型$\lambda = (A,B,\pi)$，其中，$A =\begin{bmatrix} 0.5 & 0.2 & 0.3\\ 0.3 & 0.5 & 0.2\\ 0.2 & 0.3 & 0.5 \end{bmatrix}, B=\begin{bmatrix} 0.5 & 0.5\\ 0.4 & 0.6\\ 0.7 & 0.3 \end{bmatrix}, \pi=(0.2, 0.4, 0.4)^T$
設$T = 4, O = (紅，白， 紅， 白)$，試用後向算法計算$P(O|\lambda)$

解答

按照算法10.3
（1）計算初值
$\beta_4(i) = 1 \quad i=1,2,3$
（2）遞推計算
$\beta_3(1)=\sum^3_{j=1}a_{1j}b_j(O_4)\beta_4(j)=0.25+0.12+0.09=0.46$
$\beta_3(2)=\sum^3_{j=1}a_{2j}b_j(O_4)\beta_4(j)=0.15+0.3+0.06=0.51$
$\beta_3(3)=\sum^3_{j=1}a_{3j}b_j(O_4)\beta_4(j)=0.1+0.18+0.15=0.43$

$\beta_2(1)=\sum^3_{j=1}a_{1j}b_j(O_3)\beta_3(j)=0.25*0.46+0.08*0.51+0.21*0.43=0.2461$
$\beta_2(2)=\sum^3_{j=1}a_{2j}b_j(O_3)\beta_3(j)=0.15*0.46+0.2*0.51+0.14*0.43=0.2312$
$\beta_2(3)=\sum^3_{j=1}a_{3j}b_j(O_3)\beta_3(j)=0.1*0.46+0.12*0.51+0.35*0.43=0.2577$

$\beta_1(1)=\sum^3_{j=1}a_{1j}b_j(O_2)\beta_2(j)=0.25*0.2461+0.12*0.2312+0.09*0.2577 = 0.112462$
$\beta_1(2)=\sum^3_{j=1}a_{2j}b_j(O_2)\beta_2(j)=0.15*0.2461+0.3*0.2312+0.06*0.2577 = 0.121737$
$\beta_1(3)=\sum^3_{j=1}a_{3j}b_j(O_2)\beta_2(j)=0.1*0.2461+0.18*0.2312+0.15*0.2577 = 0.104881$
（3）終止
$P(O|\lambda) = \sum^3_{i=1}\pi_{i}b_i(O_1)\beta_1(i) = 0.2*0.5*0.112462+0.4*0.4*0.121737+0.4*0.7*0.104881=0.0600908$

10.2

題目

考慮盒子和球組成的隱馬爾可夫模型$\lambda = (A,B,\pi)$，其中，$A =\begin{bmatrix} 0.5 & 0.1 & 0.4\\ 0.3 & 0.5 & 0.2\\ 0.2 & 0.2 & 0.6 \end{bmatrix}, B=\begin{bmatrix} 0.5 & 0.5\\ 0.4 & 0.6\\ 0.7 & 0.3 \end{bmatrix}, \pi=(0.2, 0.3, 0.5)^T$
設$T = 8, O = (紅，白，紅，紅，白，紅，白，白)$，試用前向後向算法計算$P(i_4=q_3|O,\lambda)$

解答

這題就偷個小懶，讓python幫忙解決叭(╹▽╹)
代碼：

import numpy as np

def fore_algorithm(A, B, p_i, o, T, N):
    # 設初值，alpha_1(i) = pt_(i)b_i(o(i))
    alpha = np.zeros((T, N))
    for i in range(N):
        h = o[0]
        alpha[0][i] = p_i[i] * B[i][h]
    #遞推
    for t in range(T-1):
        h = o[t+1]
        for i in range(N):
            a = 0
            for j in range(N):
                a += (alpha[t][j] * A[j][i])
            alpha[t+1][i] = a * B[i][h]
    #終止
    P = 0
    for i in range(N):
        P += alpha[T-1][i]
    return P, alpha

def back_algorithm(A, B, p_i, o, T, N):
    #設置初值，beta_t(i)=1
    beta = np.ones((T, N))
    #遞推
    for t in range(T-1):
        t = T - t - 2
        h = o[t + 1]
        h = int(h)

        for i in range(N):
            beta[t][i] = 0
            for j in range(N):
                beta[t][i] += A[i][j] * B[j][h] * beta[t+1][j]
    #終止
    P = 0
    for i in range(N):
        h = o[0]
        h = int(h)
        P += p_i[i] * B[i][h] * beta[0][i]
    return P, beta

if __name__ == "__main__":
    T = 8
    N = 3
    A = [[0.5, 0.1, 0.4], [0.3, 0.5, 0.2], [0.2, 0.2, 0.6]]
    B = [[0.5, 0.5], [0.4, 0.6], [0.7, 0.3]]
    pi = [0.2, 0.3, 0.5]
    O = ['紅', '白', '紅', '紅', '白', '紅', '白', '白']
    o = np.zeros(T, np.int)
    for i in range(T):
        if O[i] == '白':
            o[i] = 1
        else:
            o[i] = 0
    PF, alpha = fore_algorithm(A, B, pi, o, T, N)
    PB, beta = back_algorithm(A, B, pi, o, T, N)
    print("PF:", PF, "PB:", PB)
    #P(i_4=q_3|O,\lambda) = alpah_4(3)* beta_4(3)
    P = alpha[4-1][3-1] * beta[4-1][3-1]
    print("前向後向概率計算可得 P(i4=q3|O,lambda)=", P / PF)

運行結果：
PF爲前向算法得到的$P(O|\lambda)$
PB爲後向算法得到的$P(O|\lambda)$
（改一下main函數中的輸出值，可以驗算一下第一題的結果）

10.3

題目

在習題10.1中，試用維特比算法求最優路徑

解答

（1）初始化
$\delta_1(1) = \pi_1*b_1(o_1)=0.2*0.5=0.1$
$\delta_1(2) = \pi_2*b_2(o_1)=0.4*0.4=0.16$
$\delta_1(3) = \pi_3*b_3(o_1)=0.4*0.7=0.28$

（2）遞推
$\delta_t(i) = max_{1\leq j\leq N}[\delta_{t-1}(j)a_{ji}]b_i(o_t)$

t = 2
$\delta_2(1) = max\{0.1*0.5, 0.16*0.3, 0.28*0.2\}*0.5 = 0.028; \quad\psi_2(1) = 3$
$\delta_2(2) = max\{0.1*0.2, 0.16*0.5, 0.28*0.3\}*0.6 = 0.0504;\quad\psi_2(2) = 3$
$\delta_2(3) = max\{0.1*0.3, 0.16*0.2, 0.28*0.5\}*0.3 = 0.042;\quad\psi_2(3) = 3$

t=3
$\delta_3(1) = max\{0.028*0.5, 0.0504*0.3, 0.042*0.2\}*0.5 = 0.00756; \quad\psi_3(1) = 2$
$\delta_3(2) = max\{0.028*0.2, 0.0504*0.5, 0.042*0.3\}*0.4 = 0.01008; \quad\psi_3(2) = 2$
$\delta_3(3) = max\{0.028*0.3, 0.0504*0.2, 0.042*0.5\}*0.7 = 0.0147; \quad\psi_3(3) = 3$

t=4
$\delta_4(1) = max\{0.00756*0.5, 0.01008*0.3, 0.0147*0.2\}*0.5 = 0.00189; \quad\psi_4(1) = 1$
$\delta_4(2) = max\{0.00756*0.2, 0.01008*0.5, 0.0147*0.3\}*0.6 = 0.003024;\quad\psi_4(2) = 2$
$\delta_4(3) = max\{0.00756*0.3, 0.01008*0.2, 0.0147*0.5\}*0.3 = 0.002205;\quad\psi_4(3) = 3$

（3）終止
$P^*=max_{1\leq i\leq 3}\delta_4(i) = 0.003024$
$i_4^*=argmax_i[\delta_4(i)] = 2$

（4）最優路徑回溯
$i_3^*=\psi_4(i_4^*) = 2$
$i_2^*=\psi_3(i_3^*) = 2$
$i_1^*=\psi_2(i_2^*) = 3$

$I^*=(i_1^*,i_2^*,i_3^*,i_4^*)=(3,2,2,2)$
用代碼驗算一下~~

import numpy as np

def viterbi(A, B, pai, o, T, N):
    a = []
    i_star = np.zeros(T, np.int)
    # 初始化
    delta = np.zeros((T, N))
    psi = np.zeros((T, N), np.int)

    for i in range(N):
        delta[0][i] = pai[i] * B[i][o[0]]
        psi[0][i] = 0

    # 遞推
    for t in range(T-1):
        t = t + 1
        for i in range(N):
            for j in range(N):
                a.append(delta[t-1][j] * A[j][i])
            delta[t][i] = np.max(a) * B[i][o[t]]
            psi[t][i] = np.argmax(a, axis=0)
            a = []
    psi = psi + 1

    # 終止
    P_star = np.max(delta[T-1])
    i_star[T-1] = np.argmax(delta[T-1], axis=0) +1

    # 最優路徑回溯
    for t in range(T-1):
        t = T - t - 2
        a = t + 1
        b = i_star[t+1]-1
        i_star[t] = psi[a][b]

    return i_star, P_star

if __name__ == "__main__":
    T = 4
    N = 3
    O = ['紅', '白', '紅', '白']
    A = [[0.5, 0.2, 0.3],
         [0.3, 0.5, 0.2],
         [0.2, 0.3, 0.5]]
    B = [[0.5, 0.5],
         [0.4, 0.6],
         [0.7, 0.3]]
    p_i = [0.2, 0.4, 0.4]
    o = np.zeros(T, np.int)
    for i in range(T):
        if O[i] == '白':
            o[i] = 1
        else:
            o[i] = 0
    I, P = viterbi(A, B, p_i, o, T, N)
    print('最優路徑： ')
    print(I)
    print('最優路徑的概率：')
    print(P)

運行結果：

10.4

題目

試用前向概率和後向概率推導

$P(O|\lambda) = \sum^N_{i=1}\sum^N_{j=1}\alpha_t(i)a_{ij}b_j(o_{t+1})\beta_{t+1}(j), t=1,2,...,T-1$
這個式子是在書中的公式(10.21)，書上是用前向概率和後向概率文字推導得出，$\alpha_t(i)a_{ij}b_j(o_{t+1})\beta_{t+1}(j)$等價於$p(o_1,o_2,...,o_T,i_t=q_i,i_{t+1}=q_j|\lambda)$，兩次求和則$q_i,q_j$取到所有可能的情況。
詳細的數學證明可以參考：https://blog.csdn.net/weixin_35479108/article/details/87907429
這裏補充一個數學歸納法的證明
（1）t = 1
$P(O|\lambda)\\ = \sum^N_{i=1}\sum^N_{i=1}\alpha_1(i)a_{ij}b_j(o_2)\beta_{2}(j)\\=\sum^N_{i=1}\alpha_1(i)\sum^N_{i=1}a_{ij}b_j(o_2)\beta_{2}(j)\\=\sum^N_{i=1}\alpha_1(i)\beta_1(i)\\=\sum^N_{i=1}\pi_ib_i(o_1)\beta_1(i)$
這個即爲公式（10.21），後向算法的定義式，滿足條件
（2）假設t=k時等式成立（t = 1,2,…T-2)
即$P(O|\lambda) = \sum^N_{i=1}\sum^N_{j=1}\alpha_k(i)a_{ij}b_j(o_{k+1})\beta_{k+1}(j)$
那麼對於t=k+1，有
$\sum^N_{i=1}\sum^N_{j=1}\alpha_{k+1}(i)a_{ij}b_j(o_{k+2})\beta_{k+2}(j)\\=\sum^N_{i=1}\alpha_{k+1}(i)\sum^N_{j=1}a_{ij}b_j(o_{k+2})\beta_{k+2}(j)\\=\sum^N_{i=1}\alpha_{k+1}(i)\beta_{k+1}(i)\\=\sum^N_{i=1}\beta_{k+1}(i)[\sum_{j=1}^N\alpha_k(j)a_{ij}]b_i(o_{k+1})\\=\sum^N_{i=1}\sum^N_{j=1}\alpha_k(i)a_{ij}b_j(o_{k+1})\beta_{k+1}(j)\\=P(O|\lambda)$

得證

解答

10.5

題目

比較維特比算法中變量$\delta$的計算和前向算法中變量$\alpha$的計算的主要區別.

解答

計算變量$\alpha$的時候直接對上個的結果進行數值計算，而計算變量$\delta$需要在上個結果計算的基礎上選擇最大值。