 Solution

Normalization uses ${\int }_{-\infty }^{\infty }(1+x^2{)}^{-1}dx=\pi$, henceEach variable therefore has the standard Cauchy distribution. Let $S_n=X_1+\cdots +X_n$. We prove by induction that its probability density function isIt holds for $n=1$. If it holds for $n=m$, independence and convolution of probability densities giveThe supplied integral identity makes this $(m+1)/\{\pi [(m+1{)}^2+x^2]\}$, completing the induction. Thus addition increases the Cauchy scale to $n$.

For ${\overline{X}}_n=S_n/n$, the density transformation is $f_{{\overline{X}}_n}(x)=n{f}_{S_n}(nx)$, givingThe Cauchy sample mean has an unchanged distribution. In particular, for every $\delta >0$,independently of $n$, so the average does not converge in probability to zero.

There is no contradiction with the usual weak law of large numbers, whose finite-first-moment hypothesis is absent: $\mathbb{E}|X_1|=(2/\pi ){\int }_0^{\infty }x/(1+x^2)dx=\infty$. The positive and negative parts of $X_1$ both have infinite expectations, so $\mathbb{E}{X}_1$ is undefined. Its symmetric principal value is zero, but a principal value is not an expected value. The law therefore supplies no conclusion that these sample averages must approach zero.