Problem: Let $X_1, X_2, \ldots $ be independent $C(0,1)$ and set $S_n = \sum_{k=1}^n X_k$. Show that $\frac{1}{n}\sum_{k=1}^n \frac{S_k}{k}\sim C(0,1)$.
Using the characteristic function it is easy to get that $\frac{S_k}{k}$ is $C(0,1)$. But $Y_k=\frac{S_k}{k}$ are not independent for different $k$ so that cannot be applied directly in this case.
Let $T_n=\frac1n\sum\limits_{k=1}^n\frac1kS_k$, then $T_n=\sum\limits_{k=1}^nt_{k,n}X_k$ for some coefficients $(t_{k,n})_{1\leqslant k\leqslant n}$ summing to $1$, whose values are irrelevant. Now, $u$ times a centered Cauchy random variable with parameter $v$ is a centered Cauchy random variable with parameter $uv$, and the sum of independent centered Cauchy random variables with parameters $u$ and $v$ is centered Cauchy with parameter $u+v$.
Hence, if the $(X_k)_k$ are independent and each $X_k$ is centered Cauchy with parameter $x_k$, then $T_n$ is centered Cauchy with parameter $t_n=\sum\limits_{k=1}^nt_{k,n}x_{k}$. In the present case, $x_k=1$ for every $k\geqslant1$ hence $t_n=1$ for every $n\geqslant1$.
