# The Expectation Maximisation (EM) Algorithm Algoritmo esperanza-maximización (EM)

In this post I will briefly introduce the EM algorithm with two simple examples. The EM algorithm uses an iterative approach to find the Maximum Likelihood estimate for a model with latent variables.

Note I will not provide a thorough coverage of the mathematics but rather focus on two examples of Gaussian Mixture Models.

## General idea of EM

The objective of the EM algorithm is to obtain Maximum Likelihood estimates of the parameters and the values of latent variables, for a model with latent variables. This is achieved by iteratively updating the values of the latent variables (e.g. in this example, our best estimate of which Gaussian the data point came from) given the current estimate of the model parameters, and then updating the parameters of the model (e.g. the parameters of the Gaussians) given the values of the latent variables.

Note this bootstrap approach makes sense because it is simple to infer missing data given the parameters of the model, and to find the Maximum Likelihood estimates of the parameters for the model, given the data. However note that this can lead to problems with local optima and initialisation.

The likelihood is defined as follows (i.e. probability of the data given the model), but note we usually cannot marginalise over $\mathbf{Z}$ directly:

$L(\boldsymbol \theta; \mathbf{X} ) = P(\mathbf{X} | \boldsymbol \theta ) = \sum_z P(\mathbf{X}, \mathbf{Z} | \boldsymbol \theta )$

So instead we take this iterative updating approach. We first update our estimate of the expected value of the log-likelihood with respect to the latent variables, $\mathbf{Z}$, given our current estimate of the model parameters. Then, using this soft assignment, we update our estimates of the parameters.

Assigning $\mathbf{Z}$ (i.e. the Gaussian assignments in this example - technically we are calculating the conditional distribution $P(\mathbf{Z}|\mathbf{X},\boldsymbol \theta)$ ):

$Q(\boldsymbol \theta | \boldsymbol \theta^t ) = \mathbb{E}_{Z|X,\theta^t} \left [ \log ( L(\boldsymbol \theta; \mathbf{X}, \mathbf{Z} )) \right ]$

Updating the parameters $\boldsymbol \theta$:

$\boldsymbol \theta ^{t+1} = \arg\max_\theta Q(\boldsymbol \theta | \boldsymbol \theta^t )$

## Application to Gaussian Mixture Models

Now let's apply this to the problem of fitting a Gaussian Mixture Model to data. In general the problem is as follows, we wish to fit $k$ Gaussians to describe the mixture of data we observe. The data has $D$ dimensions.

We have a data matrix $\mathbf{X}$ of size $N \times D$, a matrix of assignment probabilities $\mathbf{Z}$ of size $N \times k$, a vector of marginal probabilities of assignment to the Gaussians (i.e. the proportion of points that come from this Gaussian) $\boldsymbol \tau$ which is of size $k \times 1$. We also have the parameters of the Gaussians: their means $\boldsymbol \mu_j$ of each of size $D \times 1$ and their covariances $\boldsymbol \Sigma_j$ each of size $D \times D$.

For the sake of brevity, we will first consider fitting two Gaussians in a one-dimensional space. This means that $D=1$, $k=2$ and that the means, $\mu_1$ and $\mu_2$, and the variances, $\sigma^2_1$ and $\sigma^2_2$ are all scalars. An example of an EM solution to a 1D mixture of 2 Gaussians.
A full example is included later.

### The Likelihood

In this case we can write the likelihood as follows:

\begin{align}L(\boldsymbol \theta; \mathbf{X}, \mathbf{Z}) &= P(\mathbf{X}, \mathbf{Z} | \boldsymbol \theta) = \prod_{i=1}^n P(Z_i = z_i ) \ \mathcal{N} (x_i | (\mu_j, \sigma^2_j)) \cr &= \prod_{i=1}^n \sum_{j=1}^2 \mathbb{I} (z_i = j) \tau_j \ \mathcal{N} (x_i, \mu_j, \sigma^2_j) \end{align}

Or in exponential form (so we can take the log):

\begin{align}L(\boldsymbol \theta ; \mathbf{X}, \mathbf{Z}) = & \exp \left ( \sum_{i=1}^n \sum_{j=1}^2 \mathbb{I}(Z_i=j)\right . \cr & \left . \left [ \log (\tau_j) - \log(\sigma_j )- \frac{1}{2\sigma^2_j} (x_i - \mu_j)^2 - \frac{d}{2} \log(2\pi) \right ] \right )\end{align}

Where $\mathbb{I}(Z_i=j)$ is the indicator function (i.e. returns 1 iff. $Z_i = j$, else returns 0).

So the log-likelihood, which we want to maximise is:

$\log ( L(\boldsymbol \theta ; \mathbf{X}, \mathbf{Z}) ) = \sum_{i=1}^n \sum_{j=1}^2 \mathbb{I}(Z_i=j) \left [ \log (\tau_j) - \log(\sigma_j) - \frac{1}{2\sigma^2_j} (x_i - \mu_j)^2 - \frac{d}{2} \log(2\pi) \right ]$

### The Expectation step

Recall that the Gaussian assignment probabilities are in $\mathbf{T}$, this is what we wish to return in the E-step. From Bayes Rule:

$\mathbf{T}_{j,i}^t = P(Z_i = j | X_i=x_i, \boldsymbol \theta^t) = \frac{\tau_j^t \ \mathcal{N}(x_i; \mu_j^t, \sigma_j^t)}{\tau_1^t \ \mathcal{N}(x_i; \mu_1^t, \sigma_1^t) + \tau_2^t \ \mathcal{N}(x_i; \mu_2^t, \sigma_2^t) }$

This can also be expressed in terms of $Q$:

\begin{align}Q(\boldsymbol \theta | \theta^t ) &= \mathbb{E} \left [ \log L(\boldsymbol \theta; \mathbf{X}, \mathbf{Z} ) \right ] = \mathbb{E} \left [ \log \prod_{i=1}^n L(\boldsymbol \theta; \mathbf{X}_i, \mathbf{Z}_i ) \right ] \cr &= \mathbb{E} \left [ \sum_{i=1}^n \log L(\boldsymbol \theta; \mathbf{X}_i, \mathbf{Z}_i ) \right ] = \sum_{i=1}^n \mathbb{E} \left [ \log L(\boldsymbol \theta; \mathbf{X}_i, \mathbf{Z}_i ) \right ]\end{align} $Q(\boldsymbol \theta | \theta^t )= \sum_{i=1}^n \sum_{j=1}^2 \mathbf{T}_{j,i}^t \left [ \log \tau_j - \log ( \sigma_j ) - \frac{1}{2\sigma_j^2} (X_i - \mu_j) - \frac{d}{2} \log(2\pi ) \right ]$

But we only need $\mathbf{T}$ to compute the updates.

### The Maximisation step

In this step we update our estimates for all of the parameters.<//p>

First let's update our estimates for $\boldsymbol \tau$, the marginal probabilities for assignment to the Gaussians. Recall that $\tau_1 + \tau_2 = 1$. The update rule is as follows:

\begin{align}\tau^{t+1} &= \arg\max_{\boldsymbol \tau} Q(\boldsymbol \theta | \boldsymbol \theta^t ) \cr &= \arg\max_\tau \left ( \left [ \sum_{i=1}^n \mathbf{T}_{i,1}^t \right ] \log \tau_1 + \left [ \sum_{i=1}^n T_{2,i}^t \right ] \log \tau_2 \right ) \end{align}

This has the same for as the Maximum Likelihood Estimate for the binomial distribution, so the update rule is:

$\tau_j^{t+1} = \frac{\sum_{i=1}^n \mathbf{T}_{j,i}^t }{\sum_{i=1}^n \left ( \mathbf{T}_{1,i}^t + \mathbf{T}_{2,i}^t \right )} = \frac{1}{n} \sum_{i=1}^n \mathbf{T}_{j,i}^t$

We now wish to estimate $\mu_1$ and $\sigma_1$:

\begin{align}(\mu_1^{t+1} , \sigma_1^{t+1} ) &= \arg\max_{\mu_1, \sigma_1} Q ( \boldsymbol \theta | \boldsymbol \theta^t) \cr &= \arg\max_{\mu_1 , \sigma_1} \sum_{i=1}^n \mathbf{T}_{1,i}^t \left ( - \log(\sigma_1 ) - \frac{1}{2 \sigma_1 ^2 } (x_i-\mu_1)^2 \right )\end{align}

This has the same form as a weighted Maximum Likelihood Estimator for a normal distribution, so the update rules are as follows:

$\mu_1^{t+1} = \frac{\sum_{i=1}^n \mathbf{T}_{1,i}^t X_i}{\sum_{i=1}^n T_{1,i}^t }$ $\sigma_1^{t+1} = \frac{\sum_{i=1}^n \mathbf{T}_{1,i}^t \left (X_i - \mu_1^{t+1} \right ) ^2}{\sum_{i=1}^n \mathbf{T}_{1,i}^t}$

And equally for the second Gaussian (note the updates are independent):

$\mu_2^{t+1} = \frac{\sum_{i=1}^n \mathbf{T}_{2,i}^t X_i}{\sum_{i=1}^n T_{2,i}^t }$ $\sigma_2^{t+1} = \frac{\sum_{i=1}^n \mathbf{T}_{2,i}^t \left (X_i - \mu_2^{t+1} \right ) ^2}{\sum_{i=1}^n \mathbf{T}_{2,i}^t}$

We carry out these updates alternately until the log-likelihood converges.

## Interactive example for 2 clusters in 1D 