Let $X$ and $Y$ be two random variables (or vectors) with continuous and "smooth enough" joint distribution. Assume that the two conditional distributions $X|Y=y$ and $Y|X=x$ are log-concave for all $x$ and $y$.

We know that log-concavity implies unimodality and even strong unimodality, by a nice theorem due to I.A. Ibragimov. We also know that the joint distribution of $X$ and $Y$ might then not be log-concave. But: can we say that the joint distribution is then unimodal, or at least that the marginal distributions are unimodal?

This question arises in Bayesian inference, $X$ and $Y$ being then parameters and their distributions being conditional to the observed data say $\mathbf{Z}$. The conditional log-concavity can arise from partial conjugacy, as is the case for a normal distribution with normal and gamma prior for the mean and precision. The rephrased question then is: if the full conditionals are log-concave, is the posterior unimodal or, at least, are the marginal posteriors unimodal?


1 Answer 1


Unimodal, log-concave marginals do not imply a unimodal joint distribution.

As an example, consider uniform marginals on the interval $[0,1].$ These are (barely) log-concave. Here is a series of plots of the densities of joint distributions having these marginals.


Clearly they are multimodal.

The method of construction is to begin with the uniform distribution on the interval $[1/2,1]$ with density $f=2$ on that interval (and zero elsewhere). This is not differentiable at $1/2.$ To make it so, a sufficiently differentiable density $p$ supported on $(1/2-\epsilon/2, 1/2+\epsilon/2)$ for small $\epsilon$ (here equal to $1/4$) was integrated to form an approximate version of $f$ given by

$$\tilde f (x) = \int_0^x (p(t) + p(1-t))\mathrm{d}t.\tag{*}$$

Figure 2: plot of f-tilde

From this the joint distribution in panel 1 was constructed via

$$f_1(x,y) = (\tilde f(x)\tilde f(y) + \tilde f(1-x)\tilde f(1-y))/2 + C\tag{**}$$

for a non-negative constant $C$ and then normalized to integrate to unity. ($C$ was introduced in case you want a density that is everywhere nonzero. $C=1$ was used for the figure; after normalization this resulted in a minimum density of $(1+C)/2=1/2.$)

The marginals of $f_1$ are uniform because the symmetric definition of $\tilde f$ in $(*)$ guarantees $\tilde f(x) + \tilde f(1-x)=2,$ which in conjunction with $(**)$ yields a constant value of $f_1$ on the interval $[0,1]$ (and $0$ outside it).

The remaining panels were created by "tessellating" the first one. The tessellation of a function $g:[0,1]\to\mathbb{R}$ (in this application) scales its support down from $[0,1]\times[0,1]$ to $[0,1/2]\times[0,1/2]$ and then tacks on three reflections around the lines $x=1/2$ and $y=1/2:$

$$\operatorname{Tessellation}(g)(x,y) = \left\{ \eqalign{&g(2x,2y), & x\le 1/2,\ y \le 1/2 \\ &g(2-2x,y), & x \gt 1/2,\ y \le 1/2 \\ &g(2x,2-2y), & x \le 1/2,\ y \gt 1/2 \\ &g(2-2x,2-2y), & x \gt 1/2,\ y \gt 1/2. }\right. $$

At the "seams" along the lines $x=1/2$ and $y=1/2$ this tessellated density is constant (provided $\epsilon \lt 1$) and therefore remains as differentiable as $g$ itself.

Panel 2 plots $f_2,$ the tessellation of $f_1;$ panel 3 plots the tessellation of $f_2,$ and so on. The number of "modal patches" (where the density is greatest) at stage $n\ge 2$ is $(n-1)^2 + n^2.$ This grows without bound, demonstrating that the joint distribution may have arbitrarily many modes.

  • $\begingroup$ Thanks @whuber for this very interesting and nicely described counterexample. My question was somewhat misleading due to its second part involving margins, confusingly relating the unimodality of the margins to that of the joint distribution. Yet on an intuitive basis, having unimodal margins sounds weaker than being jointly unimodal. As for the first part, it does not involve the margins, only the conditional distributions. $\endgroup$
    – Yves
    Commented Dec 12, 2019 at 6:09

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