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Typically when one takes random sample averages of a distribution (with sample size greater than 30) one obtains a normal distribution centering around the mean value. However, I heard that the Cauchy distribution has no mean value. What distribution does one obtain then when obtaining sample means of the Cauchy distribution?

Basically for a Cauchy distribution $\mu_x$ is undefined so what is $\mu_{\bar{x}}$ and what is the distribution of $\bar{x}$?

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    $\begingroup$ From the Wikipedia page, it looks like the sample mean of i.i.d. Cauchy variables would have the same distribution as the samples themselves. $\endgroup$ – GeoMatt22 Oct 3 '16 at 20:33
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If $X_1, \ldots, X_n$ are i.i.d. Cauchy$(0, 1)$ then we can show that $\bar{X}$ is also Cauchy$(0, 1)$ using a characteristic function argument:

\begin{align} \varphi_{\bar{X}}(t) &= \text{E} \left (e^{it \bar{X}} \right ) \\ &= \text{E} \left ( \prod_{j=1}^{n} e^{it X_j / n} \right ) \\ &= \prod_{j=1}^{n} \text{E} \left ( e^{it X_j / n} \right ) \\ &= \text{E} \left (e^{it X_1 / n} \right )^n \\ &= e^{- |t|} \end{align}

which is the characteristic function of the standard Cauchy distribution. The proof for the more general Cauchy$(\mu, \sigma)$ case is basically identical.

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    $\begingroup$ To help out those who might have trouble connecting some of the details, the step from the second to third line uses independence, the next one uses "identically distributed", the next one can be done in several ways, but the easiest is to see that the expectation inside the power is the same integral as that for a Cauchy cf but in $t/n$, so (if you already know the cf for a Cauchy) you get $[e^{-|t/n|}]^n$ and then bringing the $n$th power down the $n$ terms cancel. $\endgroup$ – Glen_b Oct 3 '16 at 21:36
  • $\begingroup$ I liked that the other answer also explained that this means it is a stable distribution. $\endgroup$ – Apollys supports Monica Nov 23 '19 at 0:19
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Typically when one takes random sample averages of a distribution (with sample size greater than 30) one obtains a normal distribution centering around the mean value.

Not exactly. You're thinking of the central limit theorem, which states that given a sequence $X_n$ of IID random variables with finite variance (which itself implies a finite mean $μ$), the expression $\sqrt{n}[(X_1 + X_2 + \cdots + X_n)/n - μ]$ converges in distribution to a normal distribution as $n$ goes to infinity. There is no guarantee that the sample mean of any finite subset of the variables will be normally distributed.

However, I heard that the Cauchy distribution has no mean value. What distribution does one obtain then when obtaining sample means of the Cauchy distribution?

Like GeoMatt22 said, the sample means will be themselves Cauchy distributed. In other words, the Cauchy distribution is a stable distribution.

Notice that the central limit theorem doesn't apply to Cauchy distributed random variables because they don't have finite mean and variance.

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  • $\begingroup$ My comment was intended to be a bit stronger than "sample mean is also Cauchy", because the sample mean will have the same parameters. That is, like for a normal distribution, the location parameter will be the same, but unlike the normal case, the scale parameter will also be the same (whereas for the normal case, the scale decreases as $1/\sqrt{N}$). At least, this is my interpretation of the first 2 transformation properties listed at my link. $\endgroup$ – GeoMatt22 Oct 3 '16 at 20:57
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    $\begingroup$ You said: "the sample mean of the first n elements convergences in distribution to a normal distribution as n goes to infinity" ... not exactly. Under weaker conditions than you need for the CLT, the mean itself converges to the constant $\mu$ (by the weak law of large numbers). You have to standardize the mean to get convergence to a normal distribution. $\endgroup$ – Glen_b Oct 3 '16 at 21:40
  • $\begingroup$ @DilipSarwate Corrected. Don't forget that you can edit other people's answers. $\endgroup$ – Kodiologist Oct 3 '16 at 21:48

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