In order for the CLT to hold we need the distribution we wish to approximate to have mean $\mu$ and finite variance $\sigma^2$. Would it be true to say that for the case of the Cauchy distribution, the mean and the variance of which, are undefined, the Central Limit Theorem fails to provide a good approximation even asymptotically?

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    $\begingroup$ Yes, it fails. The sample mean of iid Cauchy's is again Cauchy with the same spread. Thus if you multiply the sample mean by root $n$ as in the CLT, you get a distribution with infinite spread instead of a nice Gauss curve. $\endgroup$
    – Michael M
    Commented Oct 31, 2013 at 17:44

1 Answer 1


The distribution of the mean of $n$ i.i.d. samples from a Cauchy distribution has the same distribution (including the same median and inter-quartile range) as the original Cauchy distribution, no matter what the value of $n$ is.

So you do not get either the Gaussian limit or the reduction in dispersion associated with the Central Limit Theorem.

  • $\begingroup$ Well it is the standardized variable that follows the CLT not the variable in itself. $\endgroup$
    – JohnK
    Commented Oct 31, 2013 at 18:27
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    $\begingroup$ @Ioannis: that is true and you cannot standardise a Cauchy distribution to have a mean of $0$ and standard deviation of $1$. My second paragraph was aimed more at the common interpretation of the Central Limit Theorem as suggesting a Gaussian approximation with dispersion $1/\sqrt{n}$ times the original $\endgroup$
    – Henry
    Commented Oct 31, 2013 at 18:36
  • $\begingroup$ This question (stats.stackexchange.com/questions/91512/…) has graphs showing what happens to the cumulative sample averages as $n$ increases: most of the time they are moving towards the center of symmetry, but they occasionally are kicked away by extreme outliers. $\endgroup$ Commented Jun 22, 2020 at 21:42

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