Question

我们有两个硬币，分别具有正面Theta1和Theta2的概率。以下数据为我们提供了经过10次翻转和5次迭代后出现的正面的数量：'x1 = 5，x2 = 9，x3 = 8，x4 = 4，x5 = 7'

从上面的数据考虑，第一枚硬币的概率为p，第二枚硬币的概率为1-p。连续迭代(r)和(r+1)之后EM算法的终止条件为

$(p^(r+1) -p^(r))^2 + ( theta1^(r+1) - theta1^(r)  )^2 + 
                                    (theta2^(r+1)-theta2^(r))^2<=10^(-10)$

例如，我编写了这段代码，但是只需要1次迭代即可收敛。


EMflipcoin = function(p,n,theta1,theta2,x) {

   likelihood = function(p,x,theta,n){
   p * dbinom(x, n, theta) / (p * dbinom(x,n,theta) + 
                                   (1 - p) * (dbinom(x,n,1-theta)))}

   M = function(p,x,theta,n){pi.est = likelihood(p,x,theta,n)
   mean(pi.est)}

 tol=10^(-10)  
 maxiter=1000
 diff=1
 iter=0
 while(diff>tol & iter<maxiter){
   ## E-step: 
   d1 = likelihood(p=p,x=x,theta=theta1,n=n)
   d2 = likelihood(p=p,x=x,theta=theta2,n=n)
   p1 = likelihood(p=p,x=x,theta=p,n=n)
   ## M-step:
   m1    = M(p,x,theta1,n)
   m2    = M(p,x,theta2,n)
   p.new = M(p,x,p,n)
   ## calculate diff to check convergence 
   diff= (mean(p1)-p.new)^2+(mean(d1)-m1)^2+ (mean(d2)-m2)^2
   d1=m1; d2=m2; 
   iter=iter+1;
   cat("Iter", iter, 
       ", p=", p.new, 
       ": theta1=", m1, 
       ", theta2=",m2,
       ", diff=", diff, "\n")}
}
x=c(8,5,9,4,7)
EMflipcoin(p=0.5,n=10,theta1=0.6,theta2=0.4,x=x)

期望最大化

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