4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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 Jan 14 at 21:29
  • 1
    $\begingroup$ Expression 2.3 on page 27. $\endgroup$ – Ismael Ghalimi Jan 14 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 Jan 14 at 21:34
  • 1
    $\begingroup$ $\kappa$ 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 Jan 14 at 21:40
  • 1
    $\begingroup$ 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 Jan 14 at 21:44
2
$\begingroup$

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

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

$\endgroup$
  • 1
    $\begingroup$ You need a minus sign in front of the $\log x$ term. $\endgroup$ – whuber Jan 14 at 22:22
  • $\begingroup$ 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 Jan 14 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 Jan 14 at 22:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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