Gaussian Mixture Model (GMM) has a maximum likelihood when two clusters one includes the other What do people do when you perform GMM to cluster and the result comes to out to be a bimodal distribution. However, one cluster is broad enough that it includes the other cluster and this results in a situation where the very end of the broader cluster may fit observed data better than the smaller cluster creating two cutoffs. How would you account for this? Ideally you would want one clean cutoff that discerns one cluster from the other. Thank you! 
 A: The situation of overlapping clusters with similar locations but very different spreads comes about when your data has a much higher and narrower peak than a Gaussian distribution.  A Gaussian mixture model will try to fit the peak with a Gaussian that has relatively little spread and fit the tails with a Gaussian that has a lot more spread.  In this case, it isn't really "clustering" in the traditional sense, just trying to fit a distribution that is significantly different from a Gaussian (although it may be that the mixture model does identify some clusters as well as trying to fit leptokurtotic data in one or more of the clusters.)
In the case of the graph linked to in a comment to the OP, it appears to me that you have two clusters: one represented, more or less, by the leftmost (green) curve that probably has a fair amount of negative skew, and one represented, more or less, by a mixture of the three rightmost curves, which is probably somewhat "blocky" with a fair amount of positive skew.  
The case for a mixture of Gaussians to represent a single cluster can be illustrated with a simple example.  I generate 5,000 observations from a $t$ distribution with two degrees of freedom, throwing out observations such that $|x_i| > 20$ to make the plotting a little easier.  This distribution is sufficiently spread out that it doesn't have a variance, consequently any peak is too much peak relative to a Gaussian.  I then fit a very simple mixture of two Gaussians with the same mean (equal to the mean of the observations) but different spreads:
x <- rt(5000,2)
x <- x[x>-20 & x<20]

lnl <- function(theta) {
  p1 <- theta[1]
  sd1 <- theta[2]
  sd2 <- theta[3]

  -sum(log(p1*dnorm(x,mean(x),sd1) +
          (1-p1)*dnorm(x,mean(x),sd2)))
}

tmp <- optim(fn=lnl, par=c(0.5,1,2), method="L-BFGS-B",
             lower=c(0.0001,0.1,0.1),
             upper=c(0.9999,10,10))

I now plot a histogram of the data, the MLE of a single Gaussian (the black line), and the two Gaussians in the mixture model, weighted by the mixture probability:
hist(x, freq=FALSE, xlim=c(-7,7),
     breaks=c(-20,seq(-7,7,by=1), 20))
x.val <- seq(-7,7,by=0.01)
lines(dnorm(x.val, mean(x), sd(x))~x.val)
lines(tmp$par[1]*dnorm(x.val,mean(x),tmp$par[2])~x.val, 
      lwd=2, col=2)
lines((1-tmp$par[1])*dnorm(x.val,mean(x),tmp$par[3])~x.val, 
      lwd=2, col=4)

which results in:

You can see that the single Gaussian (black line) simply cannot fit the peak, because it has to be spread out enough to fit the tails (which do in fact extend out to $\pm 20$, although this isn't shown on the plot so that we can see the detail in the middle of the distribution.)
If we had plotted the two Gaussians without multiplying by the probability of cluster membership, the nature of the mixture would be clearer:
 
and the effect described above is pretty clear.
And a final plot compares the single Gaussian fit with the mixture model fit:

which is evidently far better, even though the underlying distribution is not a mixture of two Gaussians.
A: What I would do depends on what the initial goal was.
@jbowman ably demonstrated that you may have a distribution that is off from a normal distribution. In that demonstration case, one legitimate way to model that situation is a mixture of two Gaussian distributions. This is just in terms of statistical modeling.
If you're actually trying to infer that two distinct populations exist and you're trying to provide evidence that they do, then from what you describe, you've probably failed to demonstrate that. Probably. Depends on the data and what you measured, though.
What you described sounds like a case of very low entropy. From what I read, entropy alone shouldn't be used to determine how many clusters or latent classes you have; you would rely on BIC and/or (modified) likelihood ratio tests. However, if the classes were poorly separated on all the measures you had, then you would probably not be able to reject the 1-class solution in favor of the 2-class one using either criteria.
A: If your overlapping clusters reflect a heavy-tailed distribution (as jbowman showed), a possible solution would be to use heavy-tailed distributions for your mixture components instead of Gaussians. For example, student T mixture models might be appropriate. Here are some relevant papers:


*

*Peel and McLachlan (2000). Robust mixture modelling using the t distribution.

*Svensen and Bishop (2005). Robust Bayesian Mixture Modelling.

*Archambeau and Verleysen (2007). Robust Bayesian clustering.

