【the EM algorithm】自己動手，豐衣足食。

1. 首先，用MATLAB生成符合雙峯正態分佈的隨機數。

r=0.5;
mu1=50;
sigma1=20;
mu2=200;
sigma2=20;
x=zeros(10000,1);
for i=1:10000
r1=rand;
x(i,1)=(mu2+sigma2*randn)*heaviside(r1-r)+(mu1+sigma1*randn)*heaviside(r-r1);
end
hist(x)

2. 然後就是廢話少說，上代碼。

import java.io.BufferedReader;
import java.io.FileReader;
import java.io.IOException;

public class MatlabData {
	double[] d = new double[10005];
	double[] r = new double[10005];
	int N=10000;
	private double normal(double mu, double sigma, double x)
	{
		double exp = -(x-mu)*(x-mu)/(2*sigma*sigma);
		return 1.0/(sigma*Math.sqrt(2*Math.PI))*Math.pow(Math.E, exp);
	}
	//http://blog.163.com/huai_jing@126/blog/static/17186198320119231094873/
	private void EM(double pai, double mu1, double mu2, double s1, double s2)
	{
		int i,cnt=0;
		for(i=1;i<=N;i++)
			r[i] = 0;
		double eps = 1e-6;
		double opai=Integer.MIN_VALUE, omu1=Integer.MIN_VALUE, omu2=Integer.MIN_VALUE, os1=Integer.MIN_VALUE, os2=Integer.MIN_VALUE;
		do{
			prt("====================== " + (++cnt) + "\n");
			for(i=1;i<=N;i++)
			{
				r[i] = pai*normal(mu2, s2, d[i])/((1-pai)*normal(mu1, s1, d[i])+pai*normal(mu2, s2, d[i]));
			}
			double up=0, down=0;
			for(i=1;i<=N;i++)
			{
				up+=((1-r[i])*d[i]);
				down+=(1-r[i]);
			}
			omu1 = mu1;
			mu1 = up/down;
			up = 0;
			for(i=1;i<=N;i++)
			{
				up+=((1-r[i])*(d[i]-omu1)*(d[i]-omu1));
			}
			os1 = s1;
			s1 = Math.sqrt(up/down);
			
			up=0; down=0;
			for(i=1;i<=N;i++)
			{
				up+=((r[i])*d[i]);
				down+=(r[i]);
			}
			omu2 = mu2;
			mu2 = up/down;
			up = 0;
			for(i=1;i<=N;i++)
			{
				up+=((r[i])*(d[i]-omu2)*(d[i]-omu2));
			}
			os2 = s2;
			s2 = Math.sqrt(up/down);
			
			opai = pai;
			pai = down / N;
			prt("mu1 : "+mu1+"\n");
			prt("mu2 : "+mu2+"\n");
			prt("s1  : "+s1+"\n");
			prt("s2  : "+s2+"\n");
			prt("pai : "+pai+"\n");
		}while(Math.abs(mu1-omu1)>=eps
				||Math.abs(mu2-omu2)>=eps
				||Math.abs(s1-os1)>=eps
				||Math.abs(s2-os2)>=eps
				||Math.abs(pai-opai)>=eps);
	}
	public static void main(String[] args) throws IOException
	{
		MatlabData m = new MatlabData();
		String iptPath = "data/data.txt";
		BufferedReader br = new BufferedReader(new FileReader(iptPath));
		String temp;
		int i = 0;
		while((temp = br.readLine())!=null)
		{
			m.d[++i] = Double.parseDouble(temp);
		}
		br.close();
		prt("end of read : "+i+"!\n");
		m.EM(0.1, 125.2, 125.4, 6023.4, 6023.4);
		prt("end of the EM !\n");
	}
	private static void prt(String string)
	{
		System.out.print(string);
	}
}

結果如下：

end of read : 10000!
====================== 1
...
====================== 9
mu1 : 50.04014214593736
mu2 : 199.9913405589886
s1  : 20.07997207253577
s2  : 19.995272104999646
pai : 0.5016524895149193
end of the EM !

可以跟第一步的參數對比一下，還是蠻準的嘛！

參考文獻：

matlab中如何生成符合雙峯正態分佈的隨機數

EM算法（expectation-maximization algorithm）