Let $X_{1}, \cdots, X_{n}$ be i.i.d. Cauchy random variables with parameters $\alpha=0$ and $\beta=1$. (That is, their density is $f(x)=\frac{1}{\pi\,(1+x^{2})}$, $-\infty < x < \infty$.) Show that $\frac{1}{n} \sum_{j=1}^{n}X_{j}$ also has a Cauchy distribution.
We are given the hint to use characteristic functions, but I am wondering if there is an easier way to approach this problem.
I tried some stuff using the fact that the $X_{i}$ are independent, but it really didn't go anywhere, so any help you could give would be most appreciated!
$\varphi_{\frac{1}{n}\sum_{j=1}^{n}X_{j}}(u)=\varphi_{\sum_{j=1}^{n}X_{j}}(\frac{1}{n}u) = \varphi_{X_{1}+X_{2}+\cdots +X_{n}}(\frac{1}{n}u)=\varphi_{X_{1}}(\frac{1}{n}u)\varphi_{X_{2}}(\frac{1}{n}u)\cdots\varphi_{X_{n}}(\frac{1}{n}u)$
And since each of the $X_{i}$ is Cauchy distributed,
$=\exp(-|\frac{1}{n}u|)\cdot\exp(-|\frac{1}{n}u|)\cdots\exp(-|\frac{1}{n}u|) = \exp(-\frac{1}{n}|u|)\cdot\exp(-\frac{1}{n}|u|) \cdots\exp(-\frac{1}{n}|u|)=\exp(-\frac{1}{n}\sum_{j=1}^{n}|u|)=\exp(-\frac{1}{n}\cdot n \cdot |u|)=\exp(-|u|) $.
By the uniqueness of characteristic functions, we have that $\frac{1}{n}\sum_{j=1}^{n}X_{j}$ is also Cauchy-distributed.