6

For many common continuous unimodal distribution function $F$ with density $F'$, I find that the derivative (wrt $x$) of

$$(1)\quad \frac{F'(x)}{F''(x)}$$

i.e.

$$\frac {d}{dx}\left(\frac{F'(x)}{F''(x)}\right)$$

is positive (this is the case for the Weibull, normal, log-normal, exponential, gamma and probably others as well).

In one theorem I am working on, I need that the derivative of (1) is positive and $F$ unimodal and continuous. I am wondering how restrictive it is to assume that the derivative of (1) is positive for unimodal continuous distributions $F$.

Edit:

Maybe the set of all distributions for which the derivative of (1) is positive has a name, or has been studied. I would like to know those things.

Community
  • 1
user603
  • 22,585
  • 3
  • 83
  • 149
  • 2
    In what sense do you want to analyze or quantify "restrictive"? Of course this criterion excludes some distributions, because not all unimodal ones are log convex. – whuber Jun 17 '15 at 18:28
  • @whuber: it is not clear to me how this relates to log-convexity. Could you make your comment more explicit? – user603 Jun 17 '15 at 18:36
  • 3
    Let $f(x)=F^\prime(x)$. Since

    $$\frac{d}{dx} \frac{F^\prime(x)}{F^{\prime\prime}(x)} = \frac{d}{dx}\left(1/\frac{f^\prime(x)}{f(x)}\right)= \frac{d}{dx}\left(1/\frac{d}{dx}\log(f(x))\right)=-\frac{\frac{d^2}{dx^2}\log(x)}{{(\cdots)}^2},$$ the positivity of the left hand side assures the negativity of the second derivative of $\log(f)$. A standard Calculus theorem asserts $\log(f)$ is convex (on an open set) provided its second derivative is everywhere positive. That's what "log-convex" means.

     – whuber Jun 17 '15 at 18:48

2 Answers2

12

The OP really seeks the distributions with log-concave density functions -the quotient he differentiates is the reciprocal of the one we examine in order to determine the log-concavity or log-convexity of a function.

Specifically: For $f$ to be log-concave, it means that $\ln f$ is concave, for which we require that

$$\frac {d^2}{dx^2} \ln f \leq 0 \implies \frac {d}{dx}\left (\frac {f'}{f}\right) \leq 0 \implies f''\cdot f-(f')^2 \leq 0$$

From his part, the OP wants

$$\frac {d}{dx} \left (\frac {F'}{F''}\right) \geq 0 \implies \frac {d}{dx} \left (\frac {f}{f'}\right) \geq 0 \implies (f')^2 - f\cdot f'' \geq 0 $$

and upon re-arranging, the OP wants

$$ f\cdot f'' - (f')^2 \leq 0$$

which is the condition for log-concavity, not log-convexity. It is the expression of the condition through the use of the reciprocal quotient that may cause some confusion.

A good free resource on some of the "named" distributions that have log-concave densities is

Mark Bagnoli and Ted Bergstrom. "Log-concave Probability and its Applications" 2004

It focuses on log-concavity but it also contains results on log-convexity. We see that the OP's assertion that "Weibull, normal, log-normal, exponential, gamma" have log-concave densities is not fully correct: for example, the normal and the exponential distribution do have log-concave densities, while Weibull and gamma have also log-concave densities for some of their incarnations, while for others they have log-convex densities. The log-normal has a density that is neither log-concave, nor log-convex.

Another resource that examines log-concavity and log-convexity more abstractly and rigorously is

An, M. Y. (1998). Logconcavity versus logconvexity: a complete characterization. Journal of Economic theory, 80(2), 350-369.

Alecos Papadopoulos
  • 58,953
6

Let $f(x)=F^\prime(x)$. Since

$$\frac{d}{dx} \frac{F^\prime(x)}{F^{\prime\prime}(x)} = \frac{d}{dx}\left(1/\frac{f^\prime(x)}{f(x)}\right)= \frac{d}{dx}\left(1/\frac{d}{dx}\log(f(x))\right)=-\frac{\frac{d^2}{dx^2}\log(x)}{{(\cdots)}^2},$$ the positivity of the left hand side assures the negativity of the second derivative of $\log(f)$. A standard Calculus theorem asserts $\log(f)$ is concave (on an open set) provided its second derivative is everywhere negative. That's what "log-concave" means.

An important technical condition is that the domain of the function be convex. Convex sets of real numbers are intervals, potentially infinite ones. The "support" of a continuous density function $f$ is the closure of the set of numbers where it is nonzero. By extension, we may also call that set the support of the distribution function $F$.

Your functions therefore are all the distribution functions of interval support whose density functions are second-differentiable and log-concave.

Why the technicalities about the support? To rule out situations where this local characterization is not a global one. For instance, let $X$ be a random variable with density function equal to $2x$ on the interval $(0,1)$ and zero elsewhere. Consider the distribution function $F$ of an equal mixture of $X$ and $X+2$. Its density is defined, nonzero, and continuously differentiable in $(0,1)\cup(2,3)$ where its log is either $\log(x)$ or $\log(x-2)$. Both are concave functions--but $F$ should not qualify as a log-concave distribution because $f$ itself is obviously not concave and log-concavity is intended to be even stronger than concaveity.

Couldn't we not bother with this and just require $F$ to be supported on the entire real line? Sure--but that would rule out interesting and important examples like Gamma distributions.

Alecos Papadopoulos
  • 58,953
whuber
  • 322,774
  • @Alecos Thank you, you're right: I got them mixed up. (The terminology gets reversed in certain situations: for instance, the lower envelope of the graph of a concave function--which is what one would naturally look at in a density function-- is a convex set.) I will correct the terminology. – whuber Jun 17 '15 at 20:40
  • Yeah, there is this unfortunate "convex set" and "convex function" terminology issue. – Alecos Papadopoulos Jun 17 '15 at 20:43