By Central Limit Theorem, the probability density function of the the sum of a large independent random variables tends to a Normal. Therefore can we say that the sum of a large number of independent Cauchy random variables is also Normal?

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    $\begingroup$ What are the hypohteses of the version of the Central Limit Theorem that you've learned? $\endgroup$ May 20, 2016 at 18:40

1 Answer 1



You're missing one of the central assumptions of the central limit theorem:

... random variables with finite variances ...

The Cauchy distribution does not have a finite variance.

The Cauchy distribution is an example of a distribution which has no mean, variance or higher moments defined.

In fact

If $X_1, \ldots, X_n$ are independent and identically distributed random variables, each with a standard Cauchy distribution, then the sample mean $\frac{X_1 + \cdots + X_n}{n}$ has the same standard Cauchy distribution.

So the situation in your question is quite clear cut, you just keep getting back the same Cauchy distribution.

This is the concept of a stable distribution right?

Yes. A (strictly) stable distribution (or random variable) is one for which any linear combination $a X_1 + b X_2$ of two i.i.d copies is distributed proportionally to the original distribution. The Cauchy distribution is indeed strictly stationary.

(*) Quotations from wikipedia.

  • $\begingroup$ wow. I should go brush up my CLT concept. thanks alot for the answer. $\endgroup$ May 20, 2016 at 18:49
  • $\begingroup$ The cauchy is a really good example in this space. There is just enough mass in the tails that the averaging does not pull it in towards the mean, but not enough that outliers cause mass to accumulate in the tails. Its right on the boundary where the CLT fails. $\endgroup$ May 20, 2016 at 18:54
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    $\begingroup$ "It's right on the boundary where the CLT fails." Not quite--a $t$ distribution with 2 degrees of freedom would have $E(|X|)$ finite, but $E(X^2)$ infinite, whereas the Cauchy has neither. For the Cauchy, the law of large numbers doesn't even apply! $\endgroup$
    – Andrew M
    May 20, 2016 at 19:51
  • $\begingroup$ Ohhh, interesting! I suppose I really skimmed over some nuance there. $\endgroup$ May 20, 2016 at 20:25
  • $\begingroup$ If I remember right there's actually a corresponding limit theorem for the t2 and the Cauchy. If I recall correctly an appropriate choice of standardization as a function of $n$ of t2's $\bar{X}-\mu$ converges to normality -- very slowly -- while for the Cauchy we have that sample means are the same Cauchy we started with. $\endgroup$
    – Glen_b
    Sep 10, 2017 at 5:59

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