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Following my previous post on optimization and mixtures (here), Nicolas told me that my idea was probably not the most clever one (there).So, we get back to our simple mixture model,In order to describe how EM algorithm works, assume first that both  and  are perfectly known, and the mixture parameter is the only one we care about.The simple model, with only one parameter that is unknownHere, the likelihood isso that we write the log likelihood aswhich might not be simple to maximize. Recall that the mixture model can interpreted through a latent variate  (that cannot be observed), taking value when  is drawn from , and 0 if it is drawn from . More generally (especially in the case we want to extend our model to 3, 4, ... mixtures),  and .With that notation, the likelihood becomesand the log likelihoodthe term on the right is useless since we only care about p, here. From here, consider the following iterative procedure,Assume that the mixture probability  is known, denoted . Then I can predict the value of  (i.e.  and ) for all observations,So I can inject those values into my log likelihood, i.e. inhaving maximum (no need to run numerical tools here)that will be denoted . And I can iterate from here.Formally, the first step is where we calculate an expected (E) value, where is the best predictor of  given my observations (as well as my belief in ). Then comes a maximization (M) step, where using , I can estimate probability .A more general framework, all parameters are now unkownSo far, it was simple, since we assumed that  and   were perfectly known. Which is not reallistic. An there is not much to change to get a complete algorithm, to estimate . Recall that we had  which was the expected value of Z_{1,i}, i.e. it is a probability that observation i has been drawn from .If , instead of being in the segment  was in , then we could have considered mean and standard deviations of observations such that =0, and similarly on the subset of observations such that =1.But we can’t. So what can be done is to consider  as the weight we should give to observation i when estimating parameters of , and similarly, 1-would be weights given to observation i when estimating parameters of .So we set, as beforeand thenand for the variance, well, it is a weighted mean again,and this is it.Let us run the code on the same data as beforeHere, the code is rather simple: let us start generating a sample&#62; X1 = rnorm(n,0,1)&#62; X20 = rnorm(n,0,1)&#62; Z  = sample(c(1,2,2),size=n,replace=TRUE)&#62; X2=4+X20&#62; X = c(X1[Z==1],X2[Z==2])then, given a vector of initial values (that I called  and then  before),&#62;  s = c(0.5, mean(X)-1, var(X), mean(X)+1, var(X))I define my function as,&#62;  em = function(X0,s) {+  Ep = s[1]*dnorm(X0, s[2], sqrt(s[4]))/(s[1]*dnorm(X0, s[2], sqrt(s[4])) ++  (1-s[1])*dnorm(X0, s[3], sqrt(s[5])))+  s[1] = mean(Ep)+  s[2] = sum(Ep*X0) / sum(Ep)+  s[3] = sum((1-Ep)*X0) / sum(1-Ep)+  s[4] = sum(Ep*(X0-s[2])^2) / sum(Ep)+  s[5] = sum((1-Ep)*(X0-s[3])^2) / sum(1-Ep)+  return(s)+  }Then I get , or . So this is it ! We just need to iterate (here I stop after 200 iterations) since we can see that, actually, our algorithm converges quite fast,&#62; for(i in 2:200){+ s=em(X,s)+ }Let us run the same procedure as before, i.e. I generate samples of size 200, where difference between means can be small (0) or large (4),Ok, Nicolas, you were right, we’re doing much better ! Maybe we should also go for a Gibbs sampling procedure ?... next time, maybe....