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I just can’t seem to find the name of this distribution:

$$\frac{e^{-x}}{(1+e^{-x}) ^2}.$$

From my understanding, it is generally applied to pandemics/epidemics.

None of the statistics books that I have looked at contained any information on it. I would really appreciate if someone can guide me about it.

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    $\begingroup$ This is a density, not a distribution. See en.wikipedia.org/wiki/Logistic_distribution. Writing it as $p(1-p)$ with $p=1/(1+e^{-x})$ is very suggestive ;-). $\endgroup$
    – whuber
    Commented Nov 28, 2023 at 20:53
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    $\begingroup$ @whuher I would have taken the derivative of the logistic function as the cumulative distribution function to come up with this density. How would p(1-p) have any importance here beyond a "numerical accident". I don't see the connection $\endgroup$
    – Ggjj11
    Commented Nov 28, 2023 at 21:58
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    $\begingroup$ @Ggjj11 The connections are many and deep. You might, for instance, recognize $p(1-p)$ as the variance of a Bernoulli$(p)$ variable as well as the Beta$(2,2)$ density. Moreover, the differential relationship $(1/p)^\prime=p(1-p)$ leads to many nice algebraic formulas for quantities associated with the logistic distribution. These all can be interpreted in terms of quantities meaningful in epidemics as well as more generally. $\endgroup$
    – whuber
    Commented Nov 28, 2023 at 22:39
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    $\begingroup$ Note that $\dfrac{e^{-x}}{(1+e^{-x}) ^2} = \dfrac{e^{x}}{(1+e^{x}) ^2}$ demonstrating the symmetry about $0$ of the standard logistic distribution $\endgroup$
    – Henry
    Commented Nov 29, 2023 at 15:28

1 Answer 1

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This is the PDF of a logistic distribution with its location parameter set to zero and scale parameter set to one.

In general, a logistic distribution has a PDF as follows, where $\mu$ is the mean and $s$ is the scale (related to, but not the same as, standard deviation and variance).

$$ f_X(x) = \dfrac{ \exp\left( -\left(x - \mu\right)/s \right) }{ s\left( 1 + \exp\left( -\left(x - \mu\right)/s \right) \right)^2 } $$

If you set $\mu = 0$ and $s = 1$, you get the expression given in the OP.

The CDF is as follows.

$$ F_X(x) = \dfrac{ 1 }{ 1 + \exp\left( -\left(x - \mu\right)/s\right) } $$

(Note that there is a technical difference between a PDF and a distribution, even if people often use the terms as synonyms in statistical slang. If you look at the edit history, I did it, too.)

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  • $\begingroup$ Tldr: start with the logistic function as cdf, then take the derivative to get the PDF in the form of the question $\endgroup$
    – Ggjj11
    Commented Nov 28, 2023 at 21:59
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    $\begingroup$ Thank you so much for the help. $\endgroup$ Commented Nov 29, 2023 at 7:01

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