Today I realized a quite known fact. The log transformation of a random variable, drawn from a fat tail distribution, maps into an exponential tail distribution. My question is very simple:

Is the logarithm sufficient to tame every distribution?

I don't know distributions that are more extreme of the Pareto distribution, then I think so, but I don't know how to prove it. This doubt came from the observation, that peoples in finance tame their random variables with logarithms, but seems they have very bad times during financials earthquakes.


The answer is NO. You can construct such distributions that are untamed by log following the example of Log-Cauchy distribution.

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    $\begingroup$ More: For whatever fixed transformation you propose, there is a distribution it cannot "tame". $\endgroup$ – kjetil b halvorsen Feb 10 '16 at 20:28
  • $\begingroup$ @kjetilbhalvorsen, agreed, as long as you have an inverse, you can do the same trick as with exp/log in Log-Cauchy example. $\endgroup$ – Aksakal Feb 11 '16 at 14:01

No. Consider the following situations:

  • If your distribution is discrete and only takes a few different values, taking logs generally doesn't really "tame" anything. e.g. 99.9% chance of 2.8 and 0.1% chance of ten billion, take natural logs and you have 99.9% chance of 1.03 and 0.1% chance of 23(ish). Take logs again and you have 99.9% chance of 0.03 and 0.1% chance of ~3.14. Take logs again and you have 99.9% chance of -3.53 and 0.1% chance of ~1.14. In each case your distribution remains a scaled Bernoulli, so it has exactly the same skewness ($\gamma_1 \approx 31.5$), and exactly the same proportion of the distribution beyond 3,10, and 30 sd's above the mean.

  • If your distribution is symmetric or only mildly right skew, taking logs will often make it distinctly left skew (and if your distribution is left-skew, taking logs will generally make it more left skew).

  • take any random variable, $X$, with a distribution you regard as "just" inside the boundary of "tame" (set up so that $e^X$ is definitely not tame), however you want to measure it. Exponentiate twice ($Y=e^{(e^X)}$). Taking logs only once leaves it as "not tame" by that measure of tame.

  • As Nick Cox points out in comments, you can't take logs of values that aren't positive -- consider a symmetric distribution on the real line that's "not tame" in the tail (it doesn't have to be centered at 0, but let's do that anyway). You can't even take logs of the values that aren't positive, so trying to take logs won't work.

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    $\begingroup$ Perhaps too elementary to mention, but logarithms are not helpful with zero or negative values. $\endgroup$ – Nick Cox Feb 11 '16 at 1:16
  • $\begingroup$ @Nick I think it is worth mentioning, thanks. Actually, it raises a point I had in mind when I first started to type the post but then left out. (Edit: I've now added a point that relates to this) $\endgroup$ – Glen_b Feb 11 '16 at 1:17
  • $\begingroup$ Too elementary too, but you can take the absolute value of a number. $\endgroup$ – emanuele Feb 11 '16 at 14:28
  • $\begingroup$ I using the word "tame" in the meaning that the final distribution has mean and variance finites. $\endgroup$ – emanuele Feb 11 '16 at 14:31
  • $\begingroup$ i) A Pareto can have finite mean and variance; it's only ones with small values of the index that don't -- so I wouldn't have guessed you intended that meaning from your question. ii) Taking the absolute value of something that can be both positive and negative is not order-preserving (and non-invertible), so that would be an odd transformation to choose; one could also say "why not replace every value with 7?" but it's not very useful to choose transformations that do things like that. $\endgroup$ – Glen_b Feb 11 '16 at 20:44

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