# Why are all known distributions unimodal?

I do not know any multimodal distributions.

Why are all known distributions unimodal? Is there any "famous" distribution that have more than one mode?

Of course, mixtures of distributions are often multimodal, but I would like to know whether there exist any "non-mixture" distributions that have more than one mode.

• You're talking about "standard" distributions rather than "known' distributions. – Stéphane Laurent Mar 27 '14 at 11:30
• How about beta with $\alpha=\beta=0.5$? – amoeba Mar 27 '14 at 11:32
• If you don't mind bounded bimodal distributions, Wikipedia also mentions the U-quadratic and the arcsine distribution. I think these are just special cases of the beta distribution though...Wikipedia also mentions some examples of natural occurrences of multimodal distributions. – Nick Stauner Mar 27 '14 at 12:13
• @StéphaneLaurent: I like "brand-name distributions", for conveying that having been named doesn't in itself imply any special status for a distribution. "Known" distributions makes it sound like the rest may be out there somewhere waiting to be discovered, like the Loch-Ness monster or dark matter. – Scortchi Mar 27 '14 at 13:00
• Excellent @Scortchi, great vocabulary! Many non-mathematicians scientists I have encountered are under the impression that a distribution without name does not exist. Perhaps there's a related deeper philosophical fact behind that, the confusion of a name and of the thing denoted by this name (as said Russell, "The word 'dog' bears no resemblance to a dog,") – Stéphane Laurent Mar 27 '14 at 13:16

The first part of the question is answered in comments to the question: plenty of "brand-name" distributions are multimodal, such as any Beta$(a,b)$ distribution with $a\lt 1$ and $b\lt 1$. Let's turn, then, to the second part of the question.

All discrete distributions are clearly mixtures (of atoms, which are unimodal).

I will show that most continuous distributions are also mixtures of unimodal distributions. The intuition behind this is simple: we can "sand off" bumps from a bumpy graph of a PDF, one by one, until the graph is horizontal. The bumps become the mixture components, each of which is obviously unimodal.

Consequently, except perhaps for some unusual distributions whose PDFs are highly discontinuous, the answer to the question is "none": all multimodal distributions that are absolutely continuous, discrete, or a combination of those two are mixtures of unimodal distributions.

Consider continuous distributions $F$ whose PDFs $f$ are continuous (these are the "absolutely continuous" distributions). (Continuity is not much of a limitation; it can be further relaxed by more careful analysis, assuming merely that the points of discontinuity are discrete.)

To cope with "plateaus" of constant values that might occur, define a "mode" to be an interval $m = [x_l, x_u]$ (which might be a single point where $x_l=x_u$) such that

1. $f$ has a constant value on $m,$ say $y$.

2. $f$ is not constant on any interval that strictly contains $m$.

3. There exists a positive number $\epsilon$ such that the maximum value of $f$ attained on $[x_l-\epsilon, x_u+\epsilon]$ equals $y$.

Let $m = [x_l, x_u]$ be any mode of $f$. Because $f$ is continuous, there are intervals $[x_l^\prime, x_u^\prime]$ containing $m$ for which $f$ is nondecreasing in $[x_l^\prime, x_l]$ (which is a proper interval, not just a point) and nonincreasing in $[x_u, x_u^\prime]$ (which is also a proper interval). Let $x_l^\prime$ be the infinimum of all such values and $x_u^\prime$ the supremum of all such values.

This construction has defined one "hump" on the graph of $f$ extending from $x_l^\prime$ to $x_u^\prime$. Let $y$ be the larger of $f(x_l^\prime)$ and $f(x_u^\prime)$. By construction, the set of points $x$ in $[x_l^\prime, x_u^\prime]$ for which $f(x)\ge y$ is a proper interval $m^\prime$ strictly containing $m$ (because it contains either the whole of $[x_l^\prime, x_l]$ or $[x_u, x_u^\prime]$).

In this illustration of a multimodal PDF, a mode $m=[0,0]$ is identified by a red dot on the horizontal axis. The horizontal extent of the red portion of the fill is the interval $m^\prime$: it is the base of the hump determined by the mode $m$. The base of that hump is at height $y\approx 0.16$. The original PDF is the sum of the red fill and the blue fill. Notice that the blue fill only has one mode near $2$; the original mode at $[0,0]$ has been removed.

Writing $|m^\prime|$ for the length of $m^\prime$, define

$$p_m = {\Pr}_F(m^\prime) - y|m^\prime|$$

and

$$f_m(x) = \frac{f(x) - y}{p_m}$$

when $x \in m^\prime$ and $f_m(x)=0$ otherwise. (This makes $f_m$ a continuous function, incidentally.) The numerator is the amount by which $f$ rises above $y$ and the denominator $p_m$ is the area between the graph of $f$ and $y$. Thus $f_m$ is non-negative and has total area $1$: it is the PDF of a probability distribution. By construction it has a unique mode $m$.

Also by construction, the function

$$f_m^\prime(x) = \frac{f(x) - p_mf_m(x)}{1 - p_m}$$

is a PDF provided $p_m\lt 1$. (Obviously if $p_m=1$ there is nothing left of $f,$ which must have been unimodal to begin with.) Moreover, it has no modes in the interval $m^\prime$ (where it is constant, which is why the previous careful definition of a mode as an interval was necessary). Furthermore,

$$f(x) = p_m f_m(x) + (1-p_m)f_m^\prime(x)$$

is a mixture of the unimodal PDF $f_m$ and the PDF $f_m^\prime$.

Iterate this procedure with $f_m^\prime$ (which as a linear combination of continuous functions is still a continuous function, enabling us to proceed as before), producing a sequence of modes $m=m_1, m_2, \ldots$; corresponding sequences of weights $p_1=p_m, p_2=p_{m_2}, \ldots$; and PDFs $f_1=f_m, f_2=f_{m_2}, \ldots.$ The limiting result exists because (a) the interval where $f_i$ is flattened includes a proper interval that had not been flattened in the preceding $i-1$ operations and (b) the real numbers cannot be decomposed into more than a countable number of such intervals. The limit cannot have any modes and therefore is constant, which must be zero (for otherwise its integral would diverge). Consequently, $f$ has been expressed (perhaps not uniquely, because the order in which modes were selected will matter) as a mixture

$$f(x) = \sum_i p_i f_i(x)$$

of unimodal distributions, QED.

By unimodal, I think the OP plainly means that there is just one interior mode (i.e. excluding corner solutions). The question is thus really asking ... $$\text{why is it that brand name distributions do NOT have more than one interior mode?}$$

i.e. why do most brand name distributions look something like this:

... plus or minus some skewness or some discontinuities? When the question is posed thus, the Beta distribution would not be a valid counter example.

It appears the OP's conjecture has some validity: most common brand name distributions do not allow for more than one interior mode. There may be theoretical reasons for this. For example, any distribution that is a member of the Pearson family (which includes the Beta) will necessarily be (interior) unimodal, as a consequence of the parent differential eqn that defines the entire family. And the Pearson family nests most of the best-known brand names.

Nevertheless, here are some brand name counter examples ...

Counter example

One brand-name counter-example is the $\text{Sinc}^2$ distribution with pdf:

$$f(x)=\frac{\sin ^2(x)}{\pi x^2}$$

defined on the real line. Here is a plot of the $\text{Sinc}^2$ pdf:

We could also perhaps add the family of cardiod and distributions related to this class ... with pdf plots such as:

The family of reflected brand name distributions would also perhaps be possible brand name contenders (though, these might be thought of as a 'cheat solution' ... but they are still brand names) such as the Reflected Weibull shown here:

• My, that plot of $\text{Sinc}^2$ sure looks like it has some negative values! (Could that be a plotting artifact?) ... And the cardioid distributions look like they have only one interior mode each. – whuber Jun 6 '16 at 20:40
• Hi @whuber ... must agree re the plotting artefact (I will take that up on Mathematica SE !). Re cardiod family: idea is that one can extend the domain of such families as one please, and like a sine wave, it keeps on giving :) – wolfies Jun 7 '16 at 4:46
• (+1) It is a strange artifact: your last plot (of reflected distributions) does not seem to exhibit it. You might trace the generation of plot points in the $\text{Sinc}^2$ plot to see where they lie; I suspect the slight negative values might be an overshooting of a spline of a small number of points. – whuber Jun 7 '16 at 14:06
• I think it is just because the plotted line is thicker than the axis line, so appears to 'overshoot' the axis when close to zero. If the line is plotted thinner, the artefact disappears. – wolfies Jun 7 '16 at 14:15
• But there is no such artifact in your bottom figure, which also has lines thicker than the axis. – whuber Jun 7 '16 at 16:24

That you mightn't think of any doesn't mean there aren't any.

I can name "known" distributions that aren't unimodal.

For example, a Beta distribution with $\alpha$ and $\beta$ both $<1$.

http://en.wikipedia.org/wiki/Beta_distribution

also see

(This isn't a special case of the beta distribution, in spite of the comment that says it is. The two families have some overlap, however.)

Mixture distributions are certainly known, and many of those are multimodal.

• The U-quadratic is a truncated Beta distribution. – becko Apr 11 '18 at 13:26

Alpha-skew-normal distribution (Elal-Olivero 2010) has a PDF:

$$\frac{\left(1-\alpha\frac{x-\mu}{\sigma}\right)^2+1}{2+\alpha^2} \varphi\left(\frac{x-\mu}{\sigma}\right),$$

where $$\varphi$$ is the PDF of a standard Gaussian.

For $$|\alpha|>1.34$$ the distribution is bimodal. Examplary plot for $$\mu=1,\sigma=0.5,a=2$$: