题目要求：

给出解决代码如下：

import numpy as np


def miu_calc(pii, pi, qi, yj):
    up_b = pii * pi ** yj * (1 - pi) ** (1 - yj)
    up_c = (1 - pii) * qi ** yj * (1 - qi) ** (1 - yj)
    return up_b / (up_b + up_c)


yj = np.array([1, 1, 0, 1, 0, 0, 1, 0, 1, 1])

pi0 = 0.46
p0 = 0.55
q0 = 0.67

miu = miu_calc(pi0, p0, q0, yj)

pi_n = miu.mean()
p_n = (miu * yj).sum() / miu.sum()
q_n = ((1 - miu) * yj).sum() / (1 - miu).sum()

for ii in range(100):
    miu = miu_calc(pi_n, p_n, q_n, yj)

    pi_n = miu.mean()
    p_n = (miu * yj).sum() / miu.sum()
    q_n = ((1 - miu) * yj).sum() / (1 - miu).sum()

print(pi_n, p_n, q_n)

 算法解析：

def miu_calc(pii, pi, qi, yj):
    up_b = pii * pi ** yj * (1 - pi) ** (1 - yj)
    up_c = (1 - pii) * qi ** yj * (1 - qi) ** (1 - yj)
    return up_b / (up_b + up_c)

 

for ii in range(100):
    miu = miu_calc(pi_n, p_n, q_n, yj)

    pi_n = miu.mean()
    p_n = (miu * yj).sum() / miu.sum()
    q_n = ((1 - miu) * yj).sum() / (1 - miu).sum()

 最终得到结果：

代码来源及参考文章：统计学习方法 EM算法三硬币问题python代码 - 知乎