Is a mixture of two uniform distributions more complex than a single distribution? I'm a psychologist studying perception of visual ensembles (e.g., lots of lines with different orientations drawn on a screen) that have different underlying probability distributions. One of the reviewer's for our paper has asked us to justify a statement that seems intuitively correct to me but I wasn't able to find a proper reference. The statement in question says that a mixture of two uniform distributions is more complex than a normal or a uniform one. The mixture distribution here consists of two non-intersecting uniform distributions with equal ranges but different means. Intuitively it seems to me that it should be more complex as its probability density function has more parameters than the functions of a single uniform or a normal distribution hence it could be said that it has lower "description length". 
Am I correct in saying that this mixture distribution is more complex? And if so, could you please provide any reference supporting this?
 A: in one of your comments you mentioned the paper "Measuring the complexity of continuous distributions". In this paper Complexity is defined as a function of Shannon's entropy. Shannon's entropy provide a measure of the average uncertainty of a system given a probability distribution. Thus, this Complexity determines the "balance" between emergence of new patterns, and the self-organization of the system.
Intuitively, Emergence can be understood as the uniformization of a probability distribution (for instance, the white noise which has a uniform distribution has the highest Emergence). Self-organization can be understood as the concentration of probability around a specific state(s) of the probability distribution (a delta Dirac has the highest self-organization, thus, the lowest emergence since there is no change). Then, a distribution with high complexity has one or few states which concentrate a large proportion of the probability, and many others with very low probability. 
Thus, using these measures, it is not conclusive if a mixture of distributions will be more complex than only 1 distribution. For instance, compare a power-law distribution with the second mixture provided by @BruceET. Given a suitable x_min value, the power-law will surely display more complexity than the mixture of normals, since the latter is more equiprobable than the former.
Best regards,
Guillermo
A: I am puzzled about your distributions. In the Question you say 'uniform', but in one of the Comments you say 'normal'. I suspect you may mean 'normal' throughout.
Supposing that you mean 'normal' distributions, here is an Answer to your
question:  If the means $\mu_1$ and $\mu_2$ of two normal distributions
are separated by several standard deviations, then their mixture is bi-modal (has two 'humps').
For a specific example, suppose the two normal distributions are
$Norm(\mu_1 = 80, \sigma = 10)$ and $Norm(\mu_2=120, \sigma = 10).$
Also suppose that the first distribution is randomly selected 60% of
the time and the second 40% of the time.
Below is a simulation in R statistical software that simulates this
experiment with $n = 500$ subjects making random choices between the
distributions as above to get values $Z_i$ for $ i = 1, 2, \dots, 500.$
n = 500;  x = rnorm(n, 80, 10);  y = rnorm(n, 120, 10)
choice = rbinom(n, 1, .6)
z = choice*x + (1-choice)*y
hist(z, prob=T, ylim=c(0,.03), col="skyblue2", main="Simulated Mixture of 2 Normals")
  curve(.6*dnorm(x,80,10) + .4*dnorm(x,120,10), lwd=2, col="blue", add=T)

The density curve shown is found as follows: Let $\varphi_1(z)$ and 
$\varphi_2(z)$ be the density functions of the two normal distributions. Then the density function of the mixture distribution is
$$\varphi_{\text{mix}}(z) = .6\varphi_1(z) + .4\varphi_2(z).$$
Each simulation run will produce a slightly different histogram, but the
density curve is the same as long as the parameters remain the same.

If the means of the two normal distributions being mixed are closer
together, the density function of the mixture distribution may not
show two distinct modes, but this does not mean that the mixture
distribution is an ordinary normal distribution. The figure below
shows the simulated mixture distribution for $\mu_1 = 90,\, \mu_2 = 110,$
and $\sigma = 10.$ In particular, the mixture distribution is not symmetrical.
As you suggest in your comments it is more 'complex' in that it takes
four parameters to describe: the two means, the common standard deviation,
and the probability with which the first distribution is chosen.

