7
$\begingroup$

Suppose I have a random variable $X\in\mathbb R$ distributed according to a smooth nonzero probability density function (PDF) $f(x)$, with cumulative distribution function (CDF) $F(x):=\int_{\infty}^x f(\bar x)\,d\bar x$. The following quantity is showing up in a calculation I'm working on: $$q(x):=\frac{F(x)\cdot[1-F(x)]}{f(x).}$$ Does this quantity have a name? And, what are the conditions under which $q(x)$ is integrable?

For what it's worth, it appears to be the product of the Mills ratio and the CDF.

$\endgroup$
2
  • $\begingroup$ $E[q(X)]$ is related to the expectations of $X$ and the largest and smallest of an iid pair of such variables, and therefore exists if and only if $E[|X|]\lt \infty].$ (This holds whether or not $X$ has a continuous distribution.) $\endgroup$
    – whuber
    Commented May 5, 2021 at 18:47
  • $\begingroup$ For what it's worth, this eventually showed up in one of our research papers: arxiv.org/pdf/2012.06958.pdf $\endgroup$ Commented Mar 17, 2022 at 13:28

1 Answer 1

2
$\begingroup$

Logistic curve

One relationship might be with logistic growth which is based on the following differential equation:

$$f'(x) = f(x)(1-f(x))$$

But then for $F(x)$ and inhomogeneous (using some variable rate $g(x)$)

$$F'(x) = g(x) F(x)(1-F(x))$$

So if we express the CDF as a logistic curve

$$F(u) = \frac{1}{1+e^{-u}}$$

where the parameter $u$ is an integral of $q(x)^{-1}$ (where $m$ is the median for which $F(m) =0.5$)

$$F(x) = \frac{1}{1+e^{-\int_{m}^x q(t)^{-1} dt}}$$

Then

$$f(x) = F'(x) = F(x)(1-F(x)) q(x)^{-1}$$

or like your expression

$$q(x) = \frac{F(x)(1-F(x))}{f(x)}$$

A related relationship is that the log odds (odds based on the CDF) are

$$\log\left(\frac{F(x)}{1-F(x)}\right) = \int_{m}^x q(t)^{-1} dt$$

And $q(x)$ is the inverse of the rate at which the log odds increase.

Order distribution

The terms like $F(x)\cdot(1-F(x))$ also occur in the distribution of order statistics.

But I am a bit puzzled how you can get this $f(x)$ in the denominator. There are not so many expression where you use $1/f(x)$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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