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I found empirically a density function that becomes linear when one takes logarithm two times. So, the density function is something of the form $$\alpha e^{\beta e^{-\gamma x}}.$$ I cannot find out whether it is something well-known or not. Does this distribution appear in any applied context?

Hypsoline
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2 Answers2

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Looks like the Gumbel (https://en.wikipedia.org/wiki/Gumbel_distribution) distribution, or something related to it.

aleshing
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    It looks like a Gumbel cdf; that won't do for a density unless you restrict the range of values it can take. – Glen_b Dec 02 '17 at 04:18
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I agree it looks like a Gumbel distribution, and extreme value distribution, but here is a thing. What is the support of X such as $f_X(x) = \alpha e^{\beta e^{-\gamma x}}$ is well-defined?

Then, if you found, then since it is a continuous random variable, you can compute by taking integral from the support. For example, for integration convenience, we take x is defined from $-\infty$ to $\infty$, then I try

$$\int_{-\infty}^\infty \alpha e^{\beta e^{-\gamma x}}dx$$ Then it looks like the integration method of doing Gamma distribution, by u-substitution, let $u = \beta e^{-\gamma x}$, $du = -\beta \gamma e^{-\gamma x}dx = -\gamma u dx$, plug in the original formula, we get $$\int_{-\infty}^\infty \alpha e^{\beta e^{-\gamma x}}dx = \int_0^\infty -\frac{\alpha}{\gamma u} e^{-u}du=-\frac{\alpha}{\gamma}\int_0^\infty u^{0-1} e^{-u}du=-\frac{\alpha}{\gamma} \cdot \Gamma(0)=-\frac{\alpha}{\gamma}$$

Here I assumed $\Gamma(0) = 1$, and you should double check whether I'm right. But generally it is the idea.

son520804
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    This analysis has multiple problems: you get a negative answer for a positive integral and you incorrectly suppose $\Gamma(0)$ is finite, among other things. – whuber Dec 01 '17 at 19:10