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I have two set of data that are roughly centered around zero but I suspect that they have different tails. I know a few tests to compare the distribution to a normal distribution, but I would like to compare directly the two distributions.

Is there a simple test to compare the fatness of tail of 2 distributions?

Thanks
fRed

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  • $\begingroup$ Is the "fat-tails" tag really meaningful (for future questions)? $\endgroup$
    – chl
    Nov 18 '10 at 22:22
  • $\begingroup$ @chl You tell me, I am certainly not as experienced as you in statistics. But IMO it is a classic bias to underestimate the importance of tails. Have you read the work of Mandelbrot? Fat tails are very important in applied statistics for finance and the credit crisis of 2008 came for some part from some pricing models that were assuming normality and underestimating the fat tails of some correlation distribution. We can discuss that in another thread :) $\endgroup$ Nov 19 '10 at 8:23
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    $\begingroup$ This question is potentially interesting but some clarification would be welcome. Are you concerned about one tail or both? How do you measure "fatness"? (Are you willing to shift and rescale the two distributions to make the comparison, for instance?) How do you measure deviations in "fatness"? If you contemplate a hypothesis test, then what will the alternative hypothesis be, precisely? $\endgroup$
    – whuber
    Nov 19 '10 at 13:57
  • $\begingroup$ @RockScience, I have two distributions and want to compare only the tails, did you manage how to do it? I know you can calculate kurtosis but how did you test that both tails are different? $\endgroup$ Nov 7 '16 at 12:30
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This question seems to belong to the same family as this earlier one about testing whether two samples have the same skew, so you may like to read my answer to that. I believe that L-moments would be useful here too for the same reasons (specifically L-skewnesskurtosis in this case).

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The Chi Square test (Goodness-of-Fit test) will be very good at comparing the tails of two distributions since it is structured to compare two distributions by buckets of values (graphically represented by a histogram). And, the tails will consist in the far most buckets.

Even though this test focuses on the whole distribution, not just the tail you can readily observe how much of the Chi Square value or divergence is derived by the difference in the tails's fatness.

Watch that the derived histogram may actually give you visually much more information regarding the respective fatness of the tails than any test related statistical significance. It is one thing to state that tails fatness are statistically different. It is another to visually observe it. They say a picture is worth a thousand words. Sometimes it is also worth a thousand numbers (it makes sense given that graphs encapsulate all the numbers).

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    $\begingroup$ It seems to me that the Chi Square test will be particularly poor at identifying differences in tails. If the tails are covered by many bins, then--because they are tails!--there may be few data in any of the bins, invalidating the chi-squared approximation. If the tails are covered by few bins, then you lose almost all power to discriminate their shapes, and what you do manage to discriminate might not be terribly relevant or useful. (One problem we're up against here is that "fatness of tail" has not been defined, so the question is really too vague to answer well.) $\endgroup$
    – whuber
    Nov 18 '10 at 19:22
  • $\begingroup$ @whuber, I can't say whether I concur with your comment because I don't fully understand one of your points. What do you mean exactly by "invalidating the chi-squared approximation"? $\endgroup$
    – Sympa
    Nov 18 '10 at 21:04
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    $\begingroup$ The chi-squared test is based on a Normal-theory approximation to the true distribution of the chi-squared statistic. Typically this approximation gets poor when bin populations drop below 5. $\endgroup$
    – whuber
    Nov 18 '10 at 22:47
  • $\begingroup$ @whuber, thanks for the explanation. In view of it, I feel like your initial comment's first phrase may not be as nuanced as you may have cared to ("the Chi Square test will be particularly poor at identifying difference in tails"). Maybe the more appropriate statement would have been "it depends..." This test has several merits, including forcing you to define the relevant bins. And, just as importantly facilitate the construction of a histogram. Granted if you have fewer than 5 observation in a bin, you will lose accuracy as you well explained. $\endgroup$
    – Sympa
    Nov 19 '10 at 0:16
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    $\begingroup$ @Gaetan Just to clarify a possible misapprehension: I am not advocating not graphing or exploring the data, nor have I ever said anywhere that there is no value in graphing or visualizing data--just the opposite, in fact. However, histograms often don't do a good job of displaying tails in data. Probability plots, survival curves, and related approaches do much better. A histogram is naturally associated with a chi-squared test while a probability plot is naturally associated with KS-like tests and their ilk. $\endgroup$
    – whuber
    Nov 19 '10 at 17:44
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Constructing a threshold, saying lambda, we can test equality of two means or variances of the two distributions restricted on the tail region (\lambda, infinity) based on two data sets of observations falling in this tail region. Of course, the two sample t-test or F-test may be OK but not be poweful since random variable restricted on this tail region is not normal even the original ones are.

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  • $\begingroup$ Extreme value theory studies such truncated distributions: asymptotically, the distribution of the tails usually belongs to the generalized Pareto family. One could also try to fit the data to this family of distributions and compare the parameters. $\endgroup$ Jan 31 '12 at 10:05
  • $\begingroup$ @Vincent A tail may have practically any distribution. Extreme value theory says little about the tails: it focuses on the distribution of the maxima (or minima) of iid samples, which is quite a different thing. $\endgroup$
    – whuber
    Sep 29 '15 at 23:55
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How about fitting the generalized lambda distribution and bootstrapping confidence intervals on the 3rd and 4th parameters?

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    $\begingroup$ Why would this family of distributions be particularly good for this problem and not some other family like the Pearson distributions? $\endgroup$
    – whuber
    Nov 18 '10 at 19:24

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