In my self-study, I consider a Gaussian mixture distribution:

$$p(x)= p(k=1) N(x|\mu_1,\sigma^2_1) + p(k=0) N(x|\mu_0,\sigma^2_0)$$

where $p(k=1)+p(k=0)=\pi_1+\pi_0=1$. I am now asked to do three things:

  1. Write down the likelihood of the observations as a product over $n$ observations

  2. Write down the likelihood as a product over the likelihoods for ${K_0}$ and $K_1$, where $K$ is the set of indices for $k=1$ and $k=0$, respectively.

  3. Compute the log-likelihood and maximize for $\mu_0$ and $\sigma_0$.

I am not really sure what I am asked to do. I believe the likelihood is given by:

$$p(x|\pi_0, \pi_1, \mu_0, \mu_1, \sigma_0^2, \sigma_1^2) = \prod_{i=1}^n \bigg[ \pi_1 N(x_i|\mu_1,\sigma^2_1) + \pi_0 N(x_i|\mu_0,\sigma^2_0) \bigg]$$

So the log-likelihood is the sum of the logarithm of the sum in the parentheses. This seems correct. However no closed form solution exists of the derivative nor maximizer.

First, I am not sure which of no. 1 or 2 my likelihood expression solves. I think no. 1 but then, second, I am not sure what I am asked to do in no. 2. I suppose the solution to 2 is easier to maximize in 3. Third, it seems there are two different expressions for the likelihood then, but shouldn't there be just one?

Note: I was thinking for no. 2 along the lines of

$$p(x|\pi_0, \pi_1, \mu_0, \mu_1, \sigma_0^2, \sigma_1^2) = \prod_{i=1}^n \bigg[(\pi_1N(x_i|\mu_1,\sigma^2_1))^{k_i} (\pi_0N(x_i|\mu_0,\sigma^2_0))^{1-k_i} \bigg]$$

but got stuck.


2 Answers 2


This is the proper start but I wonder at the wording of the exercise. I would have asked the following:

  1. Write the likelihood of the sample $(x_1,\ldots,x_n)$ when the $X_i$'s are iid from$$p(x)= \mathbb{P}(K=1) N(x|\mu_1,\sigma^2_1) + \mathbb{P}(K=0) N(x|\mu_0,\sigma^2_0)\qquad\qquad(1)$$and conclude at the lack of closed-form expression for the maximum likelihood estimator.
  2. Introducing the latent variables $K_i$ associated with the component of each $X_i$, namely$$\mathbb{P}(K_i=1)=\pi_1=1-\mathbb{P}(K=0)$$and$$X_i|K_i=k\sim N(x|\mu_k,\sigma^2_k)$$show that the marginal distribution of $X_i$ is indeed (1).
  3. Give the density of the pair $(X_i,K_i)$ and deduce the density of the completed sample $((x_1,k_x),\ldots,(x_n,k_n))$, acting as if the $k_i$'s were also observed. We will call this density the completed likelihood.
  4. Derive the maximum likelihood estimator of the parameter $(\pi_0,\mu_0,\mu_1,\sigma_0,\sigma_1)$ based on the completed sample $((x_1,k_x),\ldots,(x_n,k_n))$.
  • $\begingroup$ Your (justified) confusion emerged because the exercise does not state explicitly that $k_i$ are observed. This is implicit in the formulation in no. 2 about the sets. In particular the exercise says here "we know that $n \in {K_0}$ are the indices for the disease free patients and $n \in K_1$ are the indices for the patients with the disease (i.e. $K_0$ and $K_1$ are non-intersecting sets of indices from 1 to N)". I believe this is supposed to tell me that $k$ is observed, but I am still not sure. $\endgroup$
    – tomka
    Sep 19, 2016 at 20:07
  • 2
    $\begingroup$ My confusion is about the wording, not the problem: This is rather standard stuff about mixtures, leading towards the EM algorithm. For instance, this is how we proceed in our book (Chapter 5). In a mixture model, the $K_i$'s are not observed, otherwise it would not be a mixture but an aggregate of two normal samples. $\endgroup$
    – Xi'an
    Sep 19, 2016 at 20:09
  • $\begingroup$ Please see my suggested solution below. Comments welcome. $\endgroup$
    – tomka
    Sep 19, 2016 at 20:24

I continued working on this exercise and came up with a solution. I'd be glad about comments.

Let $\theta=[\pi_0,\pi_1,\mu_0,\mu_1,\sigma_0^2,\sigma_1^2]$

  1. The likelihood over N observations is given by: $$ P(x|\theta) = \prod_{i=1}^n \bigg[\pi_0 N(x_i|\mu_0,\sigma_0^2)+\pi_1 N(x_i|\mu_1,\sigma_1^2) \bigg]$$

  2. The likelihood written as product over sets $K_0$ and $K_1$ is given by

$$ P(x|\theta) = \prod_{i=1}^n \bigg[ (\pi_0 N(x_i|\mu_0,\sigma_0^2))^{1-k_i}(\pi_1 N(x_i|\mu_1,\sigma_1^2))^{k_i} \bigg]$$

  1. The log-likelihood is given by

$$ \ln P(x|\theta) = \sum_{i=1}^n \bigg[ (1-k_i) (\ln \pi_0 + \ln N(x_i|\mu_0,\sigma_0^2))+k_i(\ln \pi_1 + \ln N(x_i|\mu_1,\sigma_1^2)) \bigg] $$

Consequently we can find the MLE for $\mu_0$ and $\sigma^2_0$ in a nearly standard way finding:

$$\hat{\mu}_0 = \frac{1}{\sum_{i_1}^n (1-k_i)} \sum_{i_1}^n (1-k_i) x_i$$

$$\hat{\sigma}^2_0 = \frac{1}{\sum_{i_1}^n (1-k_i)} \sum_{i_1}^n (1-k_i)(x_i - \hat{\mu}_0)^2$$

  • 3
    $\begingroup$ The answer is correct and constitutes the penultimate step to the derivation of the EM algorithm. $\endgroup$
    – Xi'an
    Sep 19, 2016 at 20:32
  • $\begingroup$ Out of curiosity, why is the likelihood here taken to be simply the multiplication of the individual Gaussian functions? I can understand the probability density function is given by a Gaussian, and if these variables are normally distributed and independent, then the joint density function, i.e. $p(x_1,x_2)$, is equal to $p(x_1)p(x_2)$. But why is the likelihood equal to the joint density function here? Shouldn't it be multiplied by a prior (which I suppose may simply be uniform here) to be equivalent? $\endgroup$
    – Mathews24
    Apr 25, 2019 at 2:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.