I am trying to get a good grasp on the EM algorithm, to be able to implement and use it. I spent a full day reading the theory and a paper where EM is used to track an aircraft using the position information coming from a radar. Honestly, I don't think I fully understand the underlying idea. Can someone point me to a numerical example showing a few iterations (3-4) of the EM for a simpler problem (like estimating the parameters of a Gaussian distribution or a sequence of a sinusoidal series or fitting a line).

Even if someone can point me to a piece of code (with synthetic data), I can try to step through the code.

  • 1
    k-means is very em, but with constant variance, and is relatively simple. – EngrStudent Feb 8 '16 at 11:09
  • 2
    @arjsgh21 can you please post mentioned paper about the aircraft? Sounds very interesting. Thank you – Wakan Tanka Aug 23 '16 at 11:53
  • There is a tutorial online which claims to provide a very clear mathematical understanding of the Em algorithm "EM Demystified: An Expectation-Maximization Tutorial" However, the example is so bad it borderlines the incomprehensable. – Working past 12 am is bad Dec 8 '17 at 22:24
up vote 93 down vote accepted

This is a recipe to learn EM with a practical and (in my opinion) very intuitive 'Coin-Toss' example:

  1. Read this short EM tutorial paper by Do and Batzoglou. This is the schema where the coin toss example is explained:

    enter image description here

  2. You may have question marks in your head, especially regarding where the probabilities in the Expectation step come from. Please have a look at the explanations on this maths stack exchange page.

  3. Look at/run this code that I wrote in Python that simulates the solution to the coin-toss problem in the EM tutorial paper of item 1:

    import numpy as np
    import math
    import matplotlib.pyplot as plt
    
    ## E-M Coin Toss Example as given in the EM tutorial paper by Do and Batzoglou* ##
    
    def get_binomial_log_likelihood(obs,probs):
        """ Return the (log)likelihood of obs, given the probs"""
        # Binomial Distribution Log PDF
        # ln (pdf)      = Binomial Coeff * product of probabilities
        # ln[f(x|n, p)] =   comb(N,k)    * num_heads*ln(pH) + (N-num_heads) * ln(1-pH)
    
        N = sum(obs);#number of trials  
        k = obs[0] # number of heads
        binomial_coeff = math.factorial(N) / (math.factorial(N-k) * math.factorial(k))
        prod_probs = obs[0]*math.log(probs[0]) + obs[1]*math.log(1-probs[0])
        log_lik = binomial_coeff + prod_probs
    
        return log_lik
    
    # 1st:  Coin B, {HTTTHHTHTH}, 5H,5T
    # 2nd:  Coin A, {HHHHTHHHHH}, 9H,1T
    # 3rd:  Coin A, {HTHHHHHTHH}, 8H,2T
    # 4th:  Coin B, {HTHTTTHHTT}, 4H,6T
    # 5th:  Coin A, {THHHTHHHTH}, 7H,3T
    # so, from MLE: pA(heads) = 0.80 and pB(heads)=0.45
    
    # represent the experiments
    head_counts = np.array([5,9,8,4,7])
    tail_counts = 10-head_counts
    experiments = zip(head_counts,tail_counts)
    
    # initialise the pA(heads) and pB(heads)
    pA_heads = np.zeros(100); pA_heads[0] = 0.60
    pB_heads = np.zeros(100); pB_heads[0] = 0.50
    
    # E-M begins!
    delta = 0.001  
    j = 0 # iteration counter
    improvement = float('inf')
    while (improvement>delta):
        expectation_A = np.zeros((len(experiments),2), dtype=float) 
        expectation_B = np.zeros((len(experiments),2), dtype=float)
        for i in range(0,len(experiments)):
            e = experiments[i] # i'th experiment
              # loglikelihood of e given coin A:
            ll_A = get_binomial_log_likelihood(e,np.array([pA_heads[j],1-pA_heads[j]])) 
              # loglikelihood of e given coin B
            ll_B = get_binomial_log_likelihood(e,np.array([pB_heads[j],1-pB_heads[j]])) 
    
              # corresponding weight of A proportional to likelihood of A 
            weightA = math.exp(ll_A) / ( math.exp(ll_A) + math.exp(ll_B) ) 
    
              # corresponding weight of B proportional to likelihood of B
            weightB = math.exp(ll_B) / ( math.exp(ll_A) + math.exp(ll_B) ) 
    
            expectation_A[i] = np.dot(weightA, e) 
            expectation_B[i] = np.dot(weightB, e)
    
        pA_heads[j+1] = sum(expectation_A)[0] / sum(sum(expectation_A)); 
        pB_heads[j+1] = sum(expectation_B)[0] / sum(sum(expectation_B)); 
    
        improvement = ( max( abs(np.array([pA_heads[j+1],pB_heads[j+1]]) - 
                        np.array([pA_heads[j],pB_heads[j]]) )) )
        j = j+1
    
    plt.figure();
    plt.plot(range(0,j),pA_heads[0:j], 'r--')
    plt.plot(range(0,j),pB_heads[0:j])
    plt.show()
    
  • 2
    @Zhubarb: can you please explain the loop termination condition (i.e. to determine when the algorithm converges)? What does the "improvement" variable calculate? – stackoverflowuser2010 Oct 24 '14 at 21:31
  • @stackoverflowuser2010, improvement looks at two deltas: 1) the change between pA_heads[j+1] and pA_heads[j] and 2) the change between pB_heads[j+1] and pB_heads[j]. And it takes the max of the two changes. For instance if Delta_A=0.001 and Delta_B=0.02, improvement from step j to j+1 will be 0.02. – Berkan Oct 27 '14 at 8:31
  • 1
    @Zhubarb: Is that a standard approach for computing convergence in EM, or is that something you came up with? If it's a standard approach, can you please cite a reference? – stackoverflowuser2010 Oct 27 '14 at 22:52
  • Here is a reference on the convergence of EM. I wrote the code sometime ago so cannot remember too well. I believe what you see in the code is my convergence criterion for this particular case. The idea is to stop iterations when the max of improvements for A and B is less than delta. – Berkan Oct 28 '14 at 10:05
  • An alternative, obviously, is to enforce the improvement threshold delta on the loglikelihood of the maximised model at each iteration. – Berkan Feb 18 '15 at 9:32

It sounds like your question has two parts: the underlying idea and a concrete example. I'll start with the underlying idea, then link to an example at the bottom.


EM is useful in Catch-22 situations where it seems like you need to know $A$ before you can calculate $B$ and you need to know $B$ before you can calculate $A$.

The most common case people deal with is probably mixture distributions. For our example, let's look at a simple Gaussian mixture model:

You have two different univariate Gaussian distributions with different means and unit variance.

You have a bunch of data points, but you're not sure which points came from which distribution, and you're also not sure about the means of the two distributions.

And now you're stuck:

  • If you knew the true means, you could figure out which data points came from which Gaussian. For example, if a data point had a very high value, it probably came from the distribution with the higher mean. But you don't know what the means are, so this won't work.

  • If you knew which distribution each point came from, then you could estimate the two distributions' means using the sample means of the relevant points. But you don't actually know which points to assign to which distribution, so this won't work either.

So neither approach seems like it works: you'd need to know the answer before you can find the answer, and you're stuck.

What EM lets you do is alternate between these two tractable steps instead of tackling the whole process at once.

You'll need to start with a guess about the two means (although your guess doesn't necessarily have to be very accurate, you do need to start somewhere).

If your guess about the means was accurate, then you'd have enough information to carry out the step in my first bullet point above, and you could (probabilistically) assign each data point to one of the two Gaussians. Even though we know our guess is wrong, let's try this anyway. And then, given each point's assigned distributions, you could get new estimates for the means using the second bullet point. It turns out that, each time you do loop through these two steps, you're improving a lower bound on the model's likelihood.

That's already pretty cool: even though the two suggestions in the bullet points above didn't seem like they'd work individually, you can still use them together to improve the model. The real magic of EM is that, after enough iterations, the lower bound will be so high that there won't be any space between it and the local maximum. As a result, and you've locally optimized the likelihood.

So you haven't just improved the model, you've found the best possible model one can find with incremental updates.


This page from Wikipedia shows a slightly more complicated example (two-dimensional Gaussians and unknown covariance), but the basic idea is the same. It also includes well-commented R code for implementing the example.

In the code, the "Expectation" step (E-step) corresponds to my first bullet point: figuring out which Gaussian gets responsibility for each data point, given the current parameters for each Gaussian. The "Maximization" step (M-step) updates the means and covariances, given these assignments, as in my second bullet point.

As you can see in the animation, these updates quickly allow the algorithm to go from a set of terrible estimates to a set of very good ones: there really do seem to be two clouds of points centered on the two Gaussian distributions that EM finds.

Here's an example of Expectation Maximisation (EM) used to estimate the mean and standard deviation. The code is in Python, but it should be easy to follow even if you're not familiar with the language.

The motivation for EM

The red and blue points shown below are drawn from two different normal distributions, each with a particular mean and standard deviation:

enter image description here

To compute reasonable approximations of the "true" mean and standard deviation parameters for the red distribution, we could very easily look at the red points and record the position of each one, and then use the familiar formulae (and similarly for the blue group).

Now consider the case where we know that there are two groups of points, but we cannot see which point belongs to which group. In other words, the colours are hidden:

enter image description here

It's not at all obvious how to divide the points into two groups. We are now unable to just look at the positions and compute estimates for the parameters of the red distribution or the blue distribution.

This is where EM can be used to solve the problem.

Using EM to estimate parameters

Here is the code used to generate the points shown above. You can see the actual means and standard deviations of the normal distributions that the points were drawn from. The variables red and blue hold the positions of each point in the red and blue groups respectively:

import numpy as np
from scipy import stats

np.random.seed(110) # for reproducible random results

# set parameters
red_mean = 3
red_std = 0.8

blue_mean = 7
blue_std = 2

# draw 20 samples from normal distributions with red/blue parameters
red = np.random.normal(red_mean, red_std, size=20)
blue = np.random.normal(blue_mean, blue_std, size=20)

both_colours = np.sort(np.concatenate((red, blue)))

If we could see the colour of each point, we would try and recover means and standard deviations using library functions:

>>> np.mean(red)
2.802
>>> np.std(red)
0.871
>>> np.mean(blue)
6.932
>>> np.std(blue)
2.195

But since the colours are hidden from us, we'll start the EM process...

First, we just guess at the values for the parameters of each group (step 1). These guesses don't have to be good:

# estimates for the mean
red_mean_guess = 1.1
blue_mean_guess = 9

# estimates for the standard deviation
red_std_guess = 2
blue_std_guess = 1.7

enter image description here

Pretty bad guesses - the means look like they are a long way from any "middle" of a group of points.

To continue with EM and improve these guesses, we compute the likelihood of each data point (regardless of its secret colour) appearing under these guesses for the mean and standard deviation (step 2).

The variable both_colours holds each data point. The function stats.norm computes the probability of the point under a normal distribution with the given parameters:

likelihood_of_red = stats.norm(red_mean_guess, red_std_guess).pdf(both_colours)
likelihood_of_blue = stats.norm(blue_mean_guess, blue_std_guess).pdf(both_colours)

This tells us, for example, that with our current guesses the data point at 1.761 is much more likely to be red (0.189) than blue (0.00003).

We can turn these two likelihood values into weights (step 3) so that they sum to 1 as follows:

likelihood_total = likelihood_of_red + likelihood_of_blue

red_weight = likelihood_of_red / likelihood_total
blue_weight = likelihood_of_blue / likelihood_total

With our current estimates and our newly-computed weights, we can now compute new, probably better, estimates for the parameters (step 4). We need a function for the mean and a function for the standard deviation:

def estimate_mean(data, weight):
    return np.sum(data * weight) / np.sum(weight)

def estimate_std(data, weight, mean):
    variance = np.sum(weight * (data - mean)**2) / np.sum(weight)
    return np.sqrt(variance)

These look very similar to the usual functions to the mean and standard deviation of data. The difference is the use of a weight parameter which assigns a weight to each data point.

This weighting is the key to EM. The greater the weight of a colour on a data point, the more the data point influences the next estimates for that colour's parameters. Ultimately, this has the effect of pulling each parameter in the right direction.

The new guesses are computed with these functions:

# new estimates for standard deviation
blue_std_guess = estimate_std(both_colours, blue_weight, blue_mean_guess)
red_std_guess = estimate_std(both_colours, red_weight, red_mean_guess)

# new estimates for mean
red_mean_guess = estimate_mean(both_colours, red_weight)
blue_mean_guess = estimate_mean(both_colours, blue_weight)

The EM process is then repeated with these new guesses from step 2 onward. We can repeat the steps for a given number of iterations (say 20), or until we see the parameters converge.

After five iterations, we see our initial bad guesses start to get better:

enter image description here

After 20 iterations, the EM process has more or less converged:

enter image description here

For comparison, here are the results of the EM process compared with the values computed where colour information is not hidden:

          | EM guess | Actual 
----------+----------+--------
Red mean  |    2.910 |   2.802
Red std   |    0.854 |   0.871
Blue mean |    6.838 |   6.932
Blue std  |    2.227 |   2.195

Note: this answer was adapted from my answer on Stack Overflow here.

Following Zhubarb's answer, I implemented the Do and Batzoglou "coin tossing" E-M example in GNU R. Note that I use the mle function of the stats4 package - this helped me to understand more clearly how E-M and MLE are related.

require("stats4");

## sample data from Do and Batzoglou
ds<-data.frame(heads=c(5,9,8,4,7),n=c(10,10,10,10,10),
    coin=c("B","A","A","B","A"),weight_A=1:5*0)

## "baby likelihood" for a single observation
llf <- function(heads, n, theta) {
  comb <- function(n, x) { #nCr function
    return(factorial(n) / (factorial(x) * factorial(n-x)))
  }
  if (theta<0 || theta >1) { # probabilities should be in [0,1]
    return(-Inf);
  }
  z<-comb(n,heads)* theta^heads * (1-theta)^(n-heads);
  return (log(z))
}

## the "E-M" likelihood function
em <- function(theta_A,theta_B) {
  # expectation step: given current parameters, what is the likelihood
  # an observation is the result of tossing coin A (vs coin B)?
  ds$weight_A <<- by(ds, 1:nrow(ds), function(row) {
    llf_A <- llf(row$heads,row$n, theta_A);
    llf_B <- llf(row$heads,row$n, theta_B);

    return(exp(llf_A)/(exp(llf_A)+exp(llf_B)));
  })

  # maximisation step: given params and weights, calculate likelihood of the sample
  return(- sum(by(ds, 1:nrow(ds), function(row) {
    llf_A <- llf(row$heads,row$n, theta_A);
    llf_B <- llf(row$heads,row$n, theta_B);

    return(row$weight_A*llf_A + (1-row$weight_A)*llf_B);
  })))
}

est<-mle(em,start = list(theta_A=0.6,theta_B=0.5), nobs=NROW(ds))
  • 1
    Shouldn't the last line of the code take theta_A = 0.6? The R code above. – sbik Feb 28 '15 at 21:37
  • 1
    thanks for pointing it out, I fixed it - choosing such an unreasonable starting value did not affect the result though. – user3096626 Mar 2 '15 at 13:27
  • 1
    @user3096626 Can you please explain why in maximization step you multiply likelihood of an A coin (row$weight_A) by a log probability (llf_A)? Is there a special rule or reason we do it? I mean one would just multiply the likelihoods or loglikehoods but not mixing hem together. I also opened a new topic – Tonja Sep 21 '16 at 17:16

All of the above look like great resources, but I must link to this great example. It presents a very simple explanation for finding the parameters for two lines of a set of points. The tutorial is by Yair Weiss while at MIT.

http://www.cs.huji.ac.il/~yweiss/emTutorial.pdf
http://www.cs.huji.ac.il/~yweiss/tutorials.html

The answer given by Zhubarb is great, but unfortunately it is in Python. Below is a Java implementation of the EM algorithm executed on the same problem (posed in the article by Do and Batzoglou, 2008). I've added some printf's to the standard output to see how the parameters converge.

thetaA = 0.71301, thetaB = 0.58134
thetaA = 0.74529, thetaB = 0.56926
thetaA = 0.76810, thetaB = 0.54954
thetaA = 0.78316, thetaB = 0.53462
thetaA = 0.79106, thetaB = 0.52628
thetaA = 0.79453, thetaB = 0.52239
thetaA = 0.79593, thetaB = 0.52073
thetaA = 0.79647, thetaB = 0.52005
thetaA = 0.79667, thetaB = 0.51977
thetaA = 0.79674, thetaB = 0.51966
thetaA = 0.79677, thetaB = 0.51961
thetaA = 0.79678, thetaB = 0.51960
thetaA = 0.79679, thetaB = 0.51959
Final result:
thetaA = 0.79678, thetaB = 0.51960

Java code follows below:

import java.util.*;

/*****************************************************************************
This class encapsulates the parameters of the problem. For this problem posed
in the article by (Do and Batzoglou, 2008), the parameters are thetaA and
thetaB, the probability of a coin coming up heads for the two coins A and B.
*****************************************************************************/
class Parameters
{
    double _thetaA = 0.0; // Probability of heads for coin A.
    double _thetaB = 0.0; // Probability of heads for coin B.

    double _delta = 0.00001;

    public Parameters(double thetaA, double thetaB)
    {
        _thetaA = thetaA;
        _thetaB = thetaB;
    }

    /*************************************************************************
    Returns true if this parameter is close enough to another parameter
    (typically the estimated parameter coming from the maximization step).
    *************************************************************************/
    public boolean converged(Parameters other)
    {
        if (Math.abs(_thetaA - other._thetaA) < _delta &&
            Math.abs(_thetaB - other._thetaB) < _delta)
        {
            return true;
        }

        return false;
    }

    public double getThetaA()
    {
        return _thetaA;
    }

    public double getThetaB()
    {
        return _thetaB;
    }

    public String toString()
    {
        return String.format("thetaA = %.5f, thetaB = %.5f", _thetaA, _thetaB);
    }

}


/*****************************************************************************
This class encapsulates an observation, that is the number of heads
and tails in a trial. The observation can be either (1) one of the
observed observations, or (2) an estimated observation resulting from
the expectation step.
*****************************************************************************/
class Observation
{
    double _numHeads = 0;
    double _numTails = 0;

    public Observation(String s)
    {
        for (int i = 0; i < s.length(); i++)
        {
            char c = s.charAt(i);

            if (c == 'H')
            {
                _numHeads++;
            }
            else if (c == 'T')
            {
                _numTails++;
            }
            else
            {
                throw new RuntimeException("Unknown character: " + c);
            }
        }
    }

    public Observation(double numHeads, double numTails)
    {
        _numHeads = numHeads;
        _numTails = numTails;
    }

    public double getNumHeads()
    {
        return _numHeads;
    }

    public double getNumTails()
    {
        return _numTails;
    }

    public String toString()
    {
        return String.format("heads: %.1f, tails: %.1f", _numHeads, _numTails);
    }

}

/*****************************************************************************
This class runs expectation-maximization for the problem posed by the article
from (Do and Batzoglou, 2008).
*****************************************************************************/
public class EM
{
    // Current estimated parameters.
    private Parameters _parameters;

    // Observations from the trials. These observations are set once.
    private final List<Observation> _observations;

    // Estimated observations per coin. These observations are the output
    // of the expectation step.
    private List<Observation> _expectedObservationsForCoinA;
    private List<Observation> _expectedObservationsForCoinB;

    private static java.io.PrintStream o = System.out;

    /*************************************************************************
    Principal constructor.
    @param observations The observations from the trial.
    @param parameters The initial guessed parameters.
    *************************************************************************/
    public EM(List<Observation> observations, Parameters parameters)
    {
        _observations = observations;
        _parameters = parameters;
    }

    /*************************************************************************
    Run EM until parameters converge.
    *************************************************************************/
    public Parameters run()
    {

        while (true)
        {
            expectation();

            Parameters estimatedParameters = maximization();

            o.printf("%s\n", estimatedParameters);

            if (_parameters.converged(estimatedParameters)) {
                break;
            }

            _parameters = estimatedParameters;
        }

        return _parameters;

    }

    /*************************************************************************
    Given the observations and current estimated parameters, compute new
    estimated completions (distribution over the classes) and observations.
    *************************************************************************/
    private void expectation()
    {

        _expectedObservationsForCoinA = new ArrayList<Observation>();
        _expectedObservationsForCoinB = new ArrayList<Observation>();

        for (Observation observation : _observations)
        {
            int numHeads = (int)observation.getNumHeads();
            int numTails = (int)observation.getNumTails();

            double probabilityOfObservationForCoinA=
                binomialProbability(10, numHeads, _parameters.getThetaA());

            double probabilityOfObservationForCoinB=
                binomialProbability(10, numHeads, _parameters.getThetaB());

            double normalizer = probabilityOfObservationForCoinA +
                                probabilityOfObservationForCoinB;

            // Compute the completions for coin A and B (i.e. the probability
            // distribution of the two classes, summed to 1.0).

            double completionCoinA = probabilityOfObservationForCoinA /
                                     normalizer;
            double completionCoinB = probabilityOfObservationForCoinB /
                                     normalizer;

            // Compute new expected observations for the two coins.

            Observation expectedObservationForCoinA =
                new Observation(numHeads * completionCoinA,
                                numTails * completionCoinA);

            Observation expectedObservationForCoinB =
                new Observation(numHeads * completionCoinB,
                                numTails * completionCoinB);

            _expectedObservationsForCoinA.add(expectedObservationForCoinA);
            _expectedObservationsForCoinB.add(expectedObservationForCoinB);
        }
    }

    /*************************************************************************
    Given new estimated observations, compute new estimated parameters.
    *************************************************************************/
    private Parameters maximization()
    {

        double sumCoinAHeads = 0.0;
        double sumCoinATails = 0.0;
        double sumCoinBHeads = 0.0;
        double sumCoinBTails = 0.0;

        for (Observation observation : _expectedObservationsForCoinA)
        {
            sumCoinAHeads += observation.getNumHeads();
            sumCoinATails += observation.getNumTails();
        }

        for (Observation observation : _expectedObservationsForCoinB)
        {
            sumCoinBHeads += observation.getNumHeads();
            sumCoinBTails += observation.getNumTails();
        }

        return new Parameters(sumCoinAHeads / (sumCoinAHeads + sumCoinATails),
                              sumCoinBHeads / (sumCoinBHeads + sumCoinBTails));

        //o.printf("parameters: %s\n", _parameters);

    }

    /*************************************************************************
    Since the coin-toss experiment posed in this article is a Bernoulli trial,
    use a binomial probability Pr(X=k; n,p) = (n choose k) * p^k * (1-p)^(n-k).
    *************************************************************************/
    private static double binomialProbability(int n, int k, double p)
    {
        double q = 1.0 - p;
        return nChooseK(n, k) * Math.pow(p, k) * Math.pow(q, n-k);
    }

    private static long nChooseK(int n, int k)
    {
        long numerator = 1;

        for (int i = 0; i < k; i++)
        {
            numerator = numerator * n;
            n--;
        }

        long denominator = factorial(k);

        return (long)(numerator / denominator);
    }

    private static long factorial(int n)
    {
        long result = 1;
        for (; n >0; n--)
        {
            result = result * n;
        }

        return result;
    }

    /*************************************************************************
    Entry point into the program.
    *************************************************************************/
    public static void main(String argv[])
    {
        // Create the observations and initial parameter guess
        // from the (Do and Batzoglou, 2008) article.

        List<Observation> observations = new ArrayList<Observation>();
        observations.add(new Observation("HTTTHHTHTH"));
        observations.add(new Observation("HHHHTHHHHH"));
        observations.add(new Observation("HTHHHHHTHH"));
        observations.add(new Observation("HTHTTTHHTT"));
        observations.add(new Observation("THHHTHHHTH"));

        Parameters initialParameters = new Parameters(0.6, 0.5);

        EM em = new EM(observations, initialParameters);

        Parameters finalParameters = em.run();

        o.printf("Final result:\n%s\n", finalParameters);
    }
}
% Implementation of the EM (Expectation-Maximization)algorithm example exposed on:
% Motion Segmentation using EM - a short tutorial, Yair Weiss, %http://www.cs.huji.ac.il/~yweiss/emTutorial.pdf
% Juan Andrade, jandrader@yahoo.com

clear all
clc

%% Setup parameters
m1 = 2;                 % slope line 1
m2 = 6;                 % slope line 2
b1 = 3;                 % vertical crossing line 1
b2 = -2;                % vertical crossing line 2
x = [-1:0.1:5];         % x axis values
sigma1 = 1;             % Standard Deviation of Noise added to line 1
sigma2 = 2;             % Standard Deviation of Noise added to line 2

%% Clean lines
l1 = m1*x+b1;           % line 1
l2 = m2*x+b2;           % line 2

%% Adding noise to lines
p1 = l1 + sigma1*randn(size(l1));
p2 = l2 + sigma2*randn(size(l2));

%% showing ideal and noise values
figure,plot(x,l1,'r'),hold,plot(x,l2,'b'), plot(x,p1,'r.'),plot(x,p2,'b.'),grid

%% initial guess
m11(1) = -1;            % slope line 1
m22(1) = 1;             % slope line 2
b11(1) = 2;             % vertical crossing line 1
b22(1) = 2;             % vertical crossing line 2

%% EM algorithm loop
iterations = 10;        % number of iterations (a stop based on a threshold may used too)

for i=1:iterations

    %% expectation step (equations 2 and 3)
    res1 = m11(i)*x + b11(i) - p1;
    res2 = m22(i)*x + b22(i) - p2;
    % line 1
    w1 = (exp((-res1.^2)./sigma1))./((exp((-res1.^2)./sigma1)) + (exp((-res2.^2)./sigma2)));

    % line 2
    w2 = (exp((-res2.^2)./sigma2))./((exp((-res1.^2)./sigma1)) + (exp((-res2.^2)./sigma2)));

    %% maximization step  (equation 4)
    % line 1
    A(1,1) = sum(w1.*(x.^2));
    A(1,2) = sum(w1.*x);
    A(2,1) = sum(w1.*x);
    A(2,2) = sum(w1);
    bb = [sum(w1.*x.*p1) ; sum(w1.*p1)];
    temp = A\bb;
    m11(i+1) = temp(1);
    b11(i+1) = temp(2);

    % line 2
    A(1,1) = sum(w2.*(x.^2));
    A(1,2) = sum(w2.*x);
    A(2,1) = sum(w2.*x);
    A(2,2) = sum(w2);
    bb = [sum(w2.*x.*p2) ; sum(w2.*p2)];
    temp = A\bb;
    m22(i+1) = temp(1);
    b22(i+1) = temp(2);

    %% plotting evolution of results
    l1temp = m11(i+1)*x+b11(i+1);
    l2temp = m22(i+1)*x+b22(i+1);
    figure,plot(x,l1temp,'r'),hold,plot(x,l2temp,'b'), plot(x,p1,'r.'),plot(x,p2,'b.'),grid
end
  • 3
    Are you able to add some discussion or explanation to the raw code? It would be useful to many readers to at least mention the language you're writing in. – Glen_b Sep 24 '14 at 8:40
  • 1
    @Glen_b - this is MatLab. I wonder how polite it is considered to more extensively annotate someones code in their answer. – EngrStudent Feb 8 '16 at 11:12

Well, I would suggest you to go through a book on R by Maria L Rizzo. One of the chapters contain the use of EM algorithm with a numerical example. I remember going through the code for better understanding.

Also, try to view it from a clustering point of view in the beginning. Work out by hand, a clustering problem where 10 observations are taken from two different normal densities. This should help.Take help from R :)

Just in case, I have written a Ruby implementation of the above mentioned coin toss example by Do & Batzoglou and it produces exactly the same numbers as they do w.r.t. the same input parameters $\theta_A = 0.6$ and $\theta_B = 0.5$.

# gem install distribution
require 'distribution'

# error bound
EPS = 10**-6

# number of coin tosses
N = 10

# observations
X = [5, 9, 8, 4, 7]

# randomly initialized thetas
theta_a, theta_b = 0.6, 0.5

p [theta_a, theta_b]

loop do
  expectation = X.map do |h|
    like_a = Distribution::Binomial.pdf(h, N, theta_a)
    like_b = Distribution::Binomial.pdf(h, N, theta_b)

    norm_a = like_a / (like_a + like_b)
    norm_b = like_b / (like_a + like_b)

    [norm_a, norm_b, h]
  end

  maximization = expectation.each_with_object([0.0, 0.0, 0.0, 0.0]) do |(norm_a, norm_b, h), r|
    r[0] += norm_a * h; r[1] += norm_a * (N - h)
    r[2] += norm_b * h; r[3] += norm_b * (N - h)
  end

  theta_a_hat = maximization[0] / (maximization[0] + maximization[1])
  theta_b_hat = maximization[2] / (maximization[2] + maximization[3])

  error_a = (theta_a_hat - theta_a).abs / theta_a
  error_b = (theta_b_hat - theta_b).abs / theta_b

  theta_a, theta_b = theta_a_hat, theta_b_hat

  p [theta_a, theta_b]

  break if error_a < EPS && error_b < EPS
end

protected by Community Aug 8 '16 at 11:57

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