What density does $Z=\frac{a_n-\sum_{i=1}^n\xi_i}{\sqrt n}$, $\xi_i\sim\operatorname{Cauchy}(0,1)$, converge to?

Question

Let $\{\xi_i\}_{i=1}^n$ be a sequence of independent Cauchy random variables with density $$ p(x)=\frac{1}{\pi}\frac{1}{1+x^2}. $$ Is there $a_n,\:n\in\mathbb R$ such that the density of $$ Z=\cfrac{a_n-\sum_{i=1}^n\xi_i}{\sqrt n} $$ converges to some density function? And if yes then -- to which one?

The statement of question makes me assume that the answer is 'yes'. My idea was to use the local limit theorem for densities but $E\xi_n$ (an expected value) does not exist. I know no more ways. Can you please help me?

Answer 1

There is a generalized central limit theorem (GCLT) that deals with the limiting distributions of sums of random variables in the case where the random variables may have unusual properties like undefined moments and thus to not converge to the normal distribution via CLT.

One example of application of the GCLT is for reciprocals of random variables, say $X^{-1}$, where $X$ has a density $f_X$ that is continuous in a neighborhood of the origin and $f_X(0)>0$. In this case, we say $X^{-1}$ lies within the domain of attraction (DOA) of the Cauchy Law and have the following "CLT like" relationship as $n\to\infty$: $$ \frac{\frac{1}{n}\sum_{i=1}^nX_i^{-1}-\operatorname{PVE}(X^{-1})}{\pi f_X(0)}\overset{d}{\to}\operatorname{Cauchy}(0,1), $$ where $f_X(0)$ is the density of $X$ at the origin and $$ \operatorname{PVE}(X^{-1})=\operatorname{p.v.}\int \frac{f_X(x)}{x}\,\mathrm dx $$ is the Cauchy principal value expectation of $X^{-1}$.

For the case $\xi_i\sim\operatorname{Cauchy}(0,1)$ recall that $\xi_i$ can be written as the ratio of independent standard normal variables and thus falls within the DOA of the Cauchy law. Furthermore, $\operatorname{PVE}(\xi)=0$, and $f_{\xi^{-1}}(0)=1/\pi$ so $$ \frac{1}{n}\sum_{i=1}^n\xi_i\overset{d}{\to}\operatorname{Cauchy}(0,1), $$ as $n\to\infty$ (in fact for this very specific example the relation holds for any finite $n$ because the Cauchy distribution is an example of a Stable distirbution and falls within is own DOA).

Edit:

The OP asked about more elementary means to prove the result in the question. Note that the Cauchy distribution admits a characteristic function $$ \varphi_\xi(t)=\mathsf Ee^{it\xi}=e^{-|t|}. $$ So it follows that if $Z=\frac{1}{n}\sum_{i=1}^n\xi_i$ then $$ \varphi_Z(t)=\mathsf Ee^{\frac{it}{n}\sum_{i=1}^n\xi_i}=(e^{-\frac{1}{n}|t|})^n=e^{-|t|}. $$ What this shows is $$ \frac{\sum_{i=1}^n\xi_i}{n}\overset{d}{=}\operatorname{Cauchy}(0,1), $$ for all $n\in\Bbb N$.