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A log-log-normal distribution is a continuous probability distribution of a random variable whose logarithm logarithm $\ln(\ln(x))$ is normally distributed.

What is the Probability Density Function for a log-log-normal distribution?

I could find an equation on page 27 (expression 2.3) of this PhD thesis but I am not sure about the $\kappa$ parameter related to attenuation. Is it always there, and what is it called? Also, the variable in the example is between 0 and 1, but I wonder what the function would be for variables greater than 1.

I also found this dissertation, but it is not available online. I wonder if there are other online materials that could be useful to study this distribution.

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  • $\begingroup$ An exhaustive search turns up nothing in your first reference that mentions a "log-log-normal" distribution, so please tell us explicitly what you are referring to. $\endgroup$
    – whuber ♦
    Commented Jan 14, 2019 at 21:29
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    $\begingroup$ Expression 2.3 on page 27. $\endgroup$
    – Ismael Ghalimi
    Commented Jan 14, 2019 at 21:31
  • $\begingroup$ That explains why I couldn't find it--it is called an "LLN" distribution! But since equation (2.3) gives the PDF, why are you asking what it is? The function isn't defined for $g \ge 1,$ as you can check (from the fact that $\kappa$ is negative). Evidently $\kappa$ is a unit conversion factor to decibels. $\endgroup$
    – whuber ♦
    Commented Jan 14, 2019 at 21:34
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    $\begingroup$ $\kappa$ is not a "parameter" in the statistical sense: it merely is a unit conversion used for engineering. It's superfluous as far as statistical or mathematical analysis might be concerned. (After all, it can be absorbed into the definition of $m_c$.) There is no "alternative definition" because of the limited domain of the logarithm. $\endgroup$
    – whuber ♦
    Commented Jan 14, 2019 at 21:40
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    $\begingroup$ I don't think so: all the minus signs I see are needed. You could define a distribution on $g\gt 1$ because $\log\log(g)$ would then be defined, but this wouldn't be an essentially different one because $\log\log(g) = \log(-\log(1/g)),$ allowing you always to assume $g\lt 1.$ $\endgroup$
    – whuber ♦
    Commented Jan 14, 2019 at 21:44

1 Answer 1

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With $\mu$ and $\sigma$ being the mean and standard deviation of the underlying normal process:

$f(x) = \displaystyle \frac{1}{\sqrt{2\pi\sigma}x\ln(x)}\exp\Bigg({\frac{-\big(\ln(\ln(x)) - \mu\big)^2}{2\sigma^2}}\Bigg) \quad x \geq 1$

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  • $\begingroup$ Now, I wonder if the “standard” definition should not just get rid of the minus sign in front of the log x term by defining x as greater or equal to 1, just for simplicity sake. $\endgroup$
    – Ismael Ghalimi
    Commented Jan 14, 2019 at 22:30
  • $\begingroup$ And I am still wondering about the minus sign on the numerator. If x is greater or equal to 1, don’t you want f(x) to be positive always? I have updated the answer accordingly. Please let me know if I got it wrong. It’s kinda late where I am right now... $\endgroup$
    – Ismael Ghalimi
    Commented Jan 14, 2019 at 22:31
  • $\begingroup$ Small typo: the "$\sigma$" in the first fraction should not be under the square root sign. $\endgroup$
    – whuber ♦
    Commented Aug 1, 2021 at 17:24

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