All distributions on a bounded interval $[0,1]$ satisfy:

$$\sigma^2 \le \mu (1-\mu)$$

where $\mu$ is the mean and $\sigma^2$ the variance.

Now suppose that the distribution is unimodal, in the sense that it has at most one local maximum. What's the minimum value that the following ratio can have:

$$\frac{\mu (1-\mu)}{\sigma^2}?$$

  • $\begingroup$ ...your first equation implies that the ratio cannot be smaller than 1. Are you asking what distribution makes it equal to 1? $\endgroup$ – user603 Mar 20 '15 at 22:28
  • $\begingroup$ Take a look at a Bernoulli$(p)$ with $\mu=p$. It's quite typical for solutions of these kinds of extremal problems to be discrete and on only a few points. You seem to have made several "bookwork"-like posts. Is any of this work for a subject? $\endgroup$ – Glen_b Mar 20 '15 at 23:21
  • $\begingroup$ @Glen_b The question asks for a unimodal distribution, though, which a smudged-to-be-continuous version of a Bernoulli is not. $\endgroup$ – Danica Mar 21 '15 at 0:33
  • $\begingroup$ The uniform distribution on $[0, 1]$ gives a value of 3. Beta distributions give $\alpha + \beta + 1$ and are unimodal only if $\alpha > 1$, $\beta > 1$, so also 3 (when it's also uniform). I tried several other named distribution families (from here) and never got a value better than 3. I also started writing it as an optimization problem by doing linear interpolation between points, but it looked like a hard optimization problem, and I stopped before actually coding and trying it. $\endgroup$ – Danica Mar 21 '15 at 1:06
  • 2
    $\begingroup$ Asked simultaneously on math.SE where it has already received two answers (one of which has been deleted by the author of the answer because of the perceived rudeness of the OP). $\endgroup$ – Dilip Sarwate Mar 21 '15 at 7:57

A minimum does not exist. However, an infimum does. It follows from the fact that

The supremum of the variance of unimodal distributions defined on $[0,1]$ having mean $\mu$ is $\mu(2 - 3\mu)/3$ ($0 \le \mu \le 1/2$) or $(1-\mu)(3\mu-1)/3$ ($1/2\le \mu \le 1$).

The supremum is actually attained by a distribution that--although it does not have a density function--can still (in a generalized sense) be thought of as "unimodal"; it will have an atom at $0$ (when $\mu \lt 1/2$) or an atom at $1$ (when $\mu \gt 1/2$) but otherwise be uniform.

I will sketch the argument. The question asks us to optimize a linear functional

$$\mathcal{L_{x^2}}: D[0,1] \to \mathbb{R}$$

subject to various equality and inequality constraints, where $D[0,1]$ is the set of (signed) measures on the interval $[0,1]$. For differentiable $F:[0,1]\to\mathbb{R}$ and $g:[0,1]\to\mathbb{R}$ any continuous function, define

$$\mathcal{L}_g[F] = \int_0^1 g(x) dF(x),$$

and extend $\mathcal{L}$ to all of $D[0,1]$ by continuity.

The equality constraints are

$$\mathcal{L}_1[F] = 1$$


$$\mathcal{L}_x[F] = \mu.$$

The inequality constraints are that

$$f(x) \ge 0$$

and there exists $\lambda \in [0,1]$ (a "mode") such that for all $0\le x \le y \le \lambda$ and all $\lambda \le y \le x \le 1$,

$$f(x) \le f(y).$$

These constraints determine a convex domain $\mathcal{X}\subset D[0,1]$ over which $\mathcal{L}_{x^2}$ is to be optimized.

As with any linear program in a finite dimensional space, the extrema of $\mathcal{L}_g$ will be attained at vertices of $\mathcal{X}$. These evidently are the measures, absolutely continuous with respect to Lebesgue measure, that are piecewise constant, because the vertices are where almost all the inequalities become equalities: and most of those inequalities are associated with the unimodality of $F$ (non-increasing tail behavior).

In order to satisfy the two equality constraints we need make only a single break in the graph of $f$, say at a number $0\lt \lambda \lt 1$. Letting the constant value on the interval $[0,\lambda)$ be $a$ and the constant value on $(\lambda, 1]$ be $b$, an easy calculation based on the equality constraints yields

$$a = \frac{1+\lambda-2\mu}{\lambda},\ b = \frac{2\mu-\lambda}{1-\lambda}.$$

Figure 1: Plot of a typical $f_{(\lambda,\mu)}$.

This figure says it all: it graphs the locally constant distribution function of mean $\mu$ with at most a single break at $\lambda$. (The plot of $f_{(\lambda,\mu)}$ for $\mu \gt 1/2$ looks like the reversal of this one.)

The value of $\mathcal{L}_{x^2}$ at such measures (which I will denote $f_{(\lambda, \mu)}$, the density of a distribution $F_{(\lambda, \mu)}$) is just as readily computed to be

$$\mathcal{L}_{x^2}[f_{(\lambda, \mu)}] = \frac{1}{3}\left(2\mu + (2\mu - 1)\lambda\right).$$

This expression is linear in $\lambda$, implying it is maximized at $0$ (when $\mu \lt 1/2$), $1$ (when $\mu \gt 1/2$), or at any value (when $\mu = 1/2$). However, except when $\mu=1/2$, the limiting values of the measures $f_{(\lambda, \mu)}$ are no longer continuous: the corresponding distribution $F = \lim_{\lambda\to 0} F_{(\lambda, \mu)}$ or $F = \lim_{\lambda\to 1} F_{(\lambda, \mu)}$ has a jump discontinuity at $0$ or $1$ (but not both).

Figure 2: Plot of optimal $F$ for $\mu=2/5$.

This figure graphs the optimal $F$ for a mean of $\mu \approx 2/5$.

Regardless, the optimum value is

$$\sigma^2_\mu = \sup_\lambda \mathcal{L}_{x^2}[f_{(\lambda, \mu)}] = \frac{1}{3}\mu(2 - 3\mu).$$

Consequently, the infimum of $\mu(1-\mu)/\sigma^2$ for $0\le \mu \lt 1/2$ is

$$\mu(1-\mu)/\sigma^2_\mu = \frac{3-3\mu}{2-3\mu},$$

with a comparable expression when $1/2\lt \mu \le 1$ (obtained by replacing $\mu$ by $1-\mu$).

Figure 3: Plot of the infimum versus $\mu$.

This figure plots the supremum $\mu(1-\mu)/\sigma^2_\mu$ versus $\mu$.

  • 1
    $\begingroup$ I think this is a beautiful answer. Is it based on a textbook, or paper? Is there a reference with more results like this? $\endgroup$ – becko Apr 16 '18 at 21:23
  • 2
    $\begingroup$ @becko Thank you. I wish I could help, but this is an original solution. I'm not sure where one would start looking for other such results, because I am not a specialist in distributional inequalities. $\endgroup$ – whuber Apr 16 '18 at 22:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.