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If $X$ follows a Cauchy distribution then $Y = \bar{X} = \frac{1}{n} \sum_{i=1}^n X_i$ also follows exactly the same distribution as $X$; see this thread.

  • Does this property have a name?

  • Are there any other distributions for which this is true?

EDIT

Another way of asking this question:

let $X$ be a random variable with probability density $f(x)$.

let $Y=\frac 1 n\sum_{i=1} ^n X_i$, where $X_i$ denotes the ith observation of $X$.

$Y$ itself can be considered as a random variable, without conditioning on any specific values of $X$.

If $X$ follows a Cauchy distribution, then the probability density function of $Y$ is $f(x)$

Are there any other kinds of (non trivial*) probability density functions for $f(x)$ that result in $Y$ having a probability density function of $f(x)$?

*The only trivial example I can think of is a Dirac delta. i.e. not a random variable.

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  • $\begingroup$ Your title makes little sense, because the "expected value of a sample" is a number. Do you mean the arithmetic mean of the sample instead? The question is also vague: by "distribution" do you mean a specific distribution or do you mean--as is suggested by the term "Cauchy"--a family of distributions? That's not some minor subtlety: the answer completely changes depending on what you mean. Please edit your post to clarify it. $\endgroup$
    – whuber
    Sep 10, 2018 at 13:39
  • $\begingroup$ @whuber, I added a second part to the question which hopefully tightens down the range of possible interpretations. $\endgroup$ Sep 10, 2018 at 14:06
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    $\begingroup$ Thank you; that clears most of it up. However, there are different answers depending on whether you fix $n$ or if you want this result to hold for all $n.$ If it's the latter, the condition on the cf or cgf is severe and leads to a ready solution. If it's the former, then potentially there are additional solutions. $\endgroup$
    – whuber
    Sep 10, 2018 at 14:11
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    $\begingroup$ I was thinking for all $n$ but if anyone wants to provide an analysis on a fixed $n$ also, that would be welcome. $\endgroup$ Sep 10, 2018 at 14:13

2 Answers 2

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This is not really an answer, but at least it does not seem to be easy to create such an example from a stable distribution. We would need to produce a r.v. whose characteristic function is the same as that of its average.

In general, for an iid draw, the c.f. of the average is

$$ \phi_{\bar{X}_n}(t)=[\phi_X(t/n)]^n $$ with $\phi_X$ the c.f. of a single r.v. For stable distributions with location parameter zero, we have $$ \phi_X(t)=\exp\{-|ct|^\alpha(1-i\beta \text{sgn}(t)\Phi)\}, $$ where $$ \Phi=\begin{cases}\tan\left(\frac{\pi\alpha}{2}\right)&\alpha\neq1\\-\frac{2}{\pi}\log|t|&\alpha=1\end{cases} $$ The Cauchy distribution corresponds to $\alpha=1$, $\beta=0$, so that $\phi_{\bar{X}_n}(t)=\phi_X(t)$ indeed for any scale parameter $c>0$.

In general, $$ \phi_{\bar{X}_n}(t)=\exp\left\{-n\left|c\frac{t}{n}\right|^\alpha\left(1-i\beta \text{sgn}\left(\frac{t}{n}\right)\Phi\right)\right\}, $$ To get $\phi_{\bar{X}_n}(t)=\phi_X(t)$, $\alpha=1$ seems called for, so \begin{eqnarray*} \phi_{\bar{X}_n}(t)&=&\exp\left\{-n\left|c\frac{t}{n}\right|\left(1-i\beta \text{sgn}\left(\frac{t}{n}\right)\left(-\frac{2}{\pi}\log\left|\frac{t}{n}\right|\right)\right)\right\}\\ &=&\exp\left\{-\left|ct\right|\left(1-i\beta \text{sgn}\left(t\right)\left(-\frac{2}{\pi}\log\left|\frac{t}{n}\right|\right)\right)\right\}, \end{eqnarray*} but $$ \log\left|\frac{t}{n}\right|\neq\log\left|t\right| $$

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  • $\begingroup$ So is it fair to say that based on your analysis, Cauchy is the only solution for a = 1? $\endgroup$ Sep 10, 2018 at 12:37
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    $\begingroup$ That is my impression from these results, but I am pretty sure that there are people more knowledgeable around here w.r.t. stable distributions. $\endgroup$ Sep 10, 2018 at 12:46
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    $\begingroup$ You don't need to invoke the theory of stable distributions. Letting $\psi=\log \phi$ be the cgf, your equation is $$\psi(t/n) = \psi(t)/n$$ for $n=1,2,3,\ldots.$ Since $\psi$ is an even continuous function and zero at the origin, this immediately implies that the germ of $\psi$ at the origin is $-|ct|.$ $\endgroup$
    – whuber
    Sep 10, 2018 at 14:53
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    $\begingroup$ Should this be the accepted answer? Besides $\alpha = 1$ the only way I can see to solve this is with $\alpha = 0$, which (I think) is the Dirac delta. $\endgroup$ Sep 11, 2018 at 5:22
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Normal distribution and shifted Poisson are examples. The shifted Poisson is $s=x-\lambda$, where $\lambda$ is Poisson intensity. There's a whole family of distribution such that the linear combination (not just the sample mean) of variables follows the same distribution, it's called stable distribution.

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  • $\begingroup$ The Poisson cannot possibly be a solution, because the support of the mean of a sample of $n$ is the set of non-negative rational numbers of the form $i/n,$ $i\in\mathbb Z,$ many of which are not in the support of the Poisson distribution. $\endgroup$
    – whuber
    Jun 22, 2020 at 21:56
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    $\begingroup$ Note that the question is not about the result being stable (same distribution up to location and scale parameters), it is about being exactly the same distribution with same location and scale. The analysis in the accepted answer strongly suggests the only possible solutions are Cauchy and the Dirac delta. $\endgroup$ Jun 23, 2020 at 6:39

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