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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.

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  • $\begingroup$ $E[q(X)]$ 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 2 hours ago
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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 $1/f(x)$.

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