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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?

Tim
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JohnK
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    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. – Michael M Oct 31 '13 at 17:44

1 Answers1

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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.

Henry
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  • Well it is the standardized variable that follows the CLT not the variable in itself. – JohnK Oct 31 '13 at 18:27
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    @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 – Henry Oct 31 '13 at 18:36
  • This question (https://stats.stackexchange.com/questions/91512/how-can-a-distribution-have-infinite-mean-and-variance/472380#472380) 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. – Thomas Lumley Jun 22 '20 at 21:42