3
$\begingroup$

Log-log seems more conventional to plot a probability distribution to look for evidence of a heavy tail. Why is this the case? For data with a heavier tail than an exponential distribution, wouldn't log-linear be best, i.e., $log(f(x))$ vs. $x$? On that plot, an exponential function would be linear.

$\endgroup$
3
  • $\begingroup$ Do you want to<lot data or a theoretical distribution? $\endgroup$ Commented Mar 7, 2022 at 23:43
  • $\begingroup$ I'm not sure what you're asking, but I mean I want to compare sample data to a theoretical distribution and am wondering why the axes should be log-log to look for evidence of a heavy tail, rather than log-linear. $\endgroup$
    – user142054
    Commented Mar 7, 2022 at 23:47
  • 1
    $\begingroup$ If the comparison is at all apt, then both distributions would have heavy tails. Moreover, a log-linear plot when used to compare two distributions will be quite difficult to interpret. You seem to be thinking of graphing a density. For some purposes a log-linear plot will be fine: see the examples I posted at stats.stackexchange.com/a/168918/919. But when the tail is truly heavy, you will need a log-log plot (or an even more severe transformation of the $x$ axis). Consider, e.g, the distribution with survival function $(x^{(1/\log(x)^2)}-1)/(e-1)$ for $x\gt e^1.$ $\endgroup$
    – whuber
    Commented Mar 8, 2022 at 0:06

1 Answer 1

1
$\begingroup$

The classical concept of "heavy tails" that we are usually interested in is when the tails are sufficiently heavy to give infinite variance. This occurs under power-law tail behaviour when one or both tails decay at a rate that is no faster than cubic decay. (Conversely, the variance will be finite if both tails decay faster than cubic decay.) You can find a detailed explanation of the log-log properties of a distribution with power-law tails in this related question.

If a distribution has a power-law tail of the form $f(x) \rightarrow c x^{-\omega-1}$ then it can be shown that the log-tail-probability and log-deviation from a fixed point $x_0$ are related in the tail by:

$$\ln \mathbb{P}(X>x) \rightarrow \text{const} - \omega \ln(x-x_0) \quad \quad \text{as } x \rightarrow \infty.$$

Now, if we let $x_{(1)} \geqslant \cdots \geqslant x_{(n)}$ be the ordered sample values in our data, we can estimate the log-tail-probability by $\ln \hat{\mathbb{P}}(X>x_{(i)}) = \ln(2i-1)- \ln(2n)$. For values in the tails of the distribution we should therefore expect the values to follow the relationships:

$$\begin{matrix} \text{Right tail } & & & & \quad \quad \quad \ln(2i-1) \approx \text{const} - \omega \ln|x_{(i)}-x_0|, \\ \text{Left tail} \quad & & & & \text{ } \ln(2(n-i)-1) \approx \text{const} - \omega \ln|x_{(i)}-x_0|. \\ \end{matrix}$$

As you can see, this gives us a log-log comparison that allows us to see whether the data exhibits power-law decay in the tails, and lets us estimate the rate of decay. If $\omega \leqslant 2$ then there is decay no faster than cubic decay, which means the distribution has infinite variance --- i.e., it is heavy tailed.


Generating the tail-plot: You can generate the tail-plot for a set of data using the tailplot function in the utilities package in R. This function takes in a set of data and produces the tail-plot for one or both tails. (The function can also produce the Hill-plot and/or the DeSousa-Michailidis-plot, both of which are also useful for estimating rates of tail decay. Below we generate some data from an exponential distribution and show the tail-plot for both tails. We can see that the left-tail decays extremely rapidly (which is unsurprising, since it is bounded) and the right-tail also decays faster than cubic decay. This plot provides strong empirical evidence that the distribution is not heavy tailed ---i.e., it has finite variance.

#Generate data from an exponential distribution
set.seed(1)
DATA <- rexp(1000, rate = 1)

#Generate the tail-plot for the data (both tails)
library(utilities)
tailplot(DATA)

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.