Is the Mixture of Gaussians model (an example of latent class analysis) gauranteed to converge on a viable solution even on Unimodal data using the Expectation Maximization algorithm to estimate the classes?

EM guarantees it will find a local optimum, but what does that necessarily mean with regard to the parameters?

E.g. If I have (singlevariate or multivariate) gaussian data, and I try to fit a mixture model of two Gaussians on it, what is theoretically supposed to happen?

  • 1
    $\begingroup$ There do exist algorithms for fitting Gaussian mixtures with convergence guarantees (given some assumptions on separation of the true mixture components, or similar), following a method of moments (sometimes called spectral learning) approach rather than EM. See e.g. Vempala and Wang (2004). $\endgroup$
    – Danica
    May 20, 2015 at 23:43

1 Answer 1


Assuming no restrictions on the variables $\mu_k$ and $\sigma_k$ (the means and standard deviations of the separate components of the mixture model), the EM algorithm may not converge on a local maximum, as the likelihood function is unbounded in this case.

To illustrate, suppose we observe data points $x_1, ..., x_n$ and we fit a two component Gaussian mixture model. Then the log likelihood can be written as

$ \displaystyle \sum_{i = 1}^n \log\left( p_1 \times f(x_i | \mu_1, \sigma_1) + p_2 \times f(x_i|\mu_2, \sigma_2)\right)$

where $f(x|\mu, \sigma)$ is the pdf of the normal distribution with mean = $\mu$ and standard deviation = $\sigma$. Consider now the case if we set $\mu_1 = \bar x$ (mean of observed data) and $\sigma_1 = \hat \sigma$ (standard deviation of observed data), with $p_1 = p_2 = 0.5$ (side note: these values are somewhat arbitrary, used only getting a lower bound). Then the contribution to the likelihood function for each point will be at least

$\log(0.5 \times f(x_i|\bar x, \hat \sigma) )$

i.e. the contribution for each observation can be bounded from below for all values of $\mu_2$ and $\sigma_2$. Then consider if we set $\mu_2 = x_1$. As $\sigma_2 \rightarrow 0$, $f(x_1 | x_1, \sigma_2) \rightarrow \infty$. Of course, this implies

$\log(0.5 \times f(x_1| x_1, \sigma_2) + 0.5 \times f(x_1 | \bar x, \hat \sigma)) \rightarrow \infty$

as well. Since the contributions of all the other observations are bounded from below by $\log(0.5 \times f(x_i|\bar x, \hat \sigma) )$, this implies that the likelihood function is unbounded.

Since the likelihood function is unbounded, if the EM algorithm approaches this infinite "peak", it will not converge. However, this likelihood typically does have non-degenerate local maximum, assuming $n$ is sufficiently large relative to $k$ (number of components). So the EM algorithm may find these non-degenerate local maximum, which are undoubtably better estimators than the degenerate solution. In fact, if $n$ is large relative to $k$, it is very likely to find a non-degenerate local max rather than to approach the degenerate solution.

  • $\begingroup$ Just to clarify a little: the reason the likelihood approaches $\infty$ when $\epsilon$ approaches 0 is that the likelihood contribution of $x_1$ approaches $\infty$ as component 2 "closes in" on it. The likelihood contributions of the other observations are prevented from approaching $-\infty$ by component 1. $\endgroup$
    – Cliff AB
    May 20, 2015 at 21:21
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    $\begingroup$ I was not able to follow this argument, because the nature of the likelihood function must depend on the data but you make no reference to them at all. I suspect you might also need to make clearer distinctions between the true parameters and potential estimates of them. The wildly varied and inconsistent notation (bars, hats, subscripts, no subscripts, greek letters, and latin letters) provides no apparent help in deciphering what you really mean. Do you think you could clarify this answer? $\endgroup$
    – whuber
    May 20, 2015 at 22:05
  • $\begingroup$ I am bit confused at what $x_1$, $\hat s$, and $\bar x$ are referring to as well. Are you suggesting that the second gaussian is collapsing on one point, approaching a delta function? $\endgroup$
    – user27886
    May 20, 2015 at 23:02
  • $\begingroup$ Sorry, just cleaned it up a little. Does that make more sense? $\endgroup$
    – Cliff AB
    May 20, 2015 at 23:10
  • $\begingroup$ And yes, the issue is that one of the components can collapse around a single point. $\endgroup$
    – Cliff AB
    May 20, 2015 at 23:19

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