Averaging Cauchy data

Idea. The average of independent standard Cauchy variables is again standard Cauchy. Averaging does not concentrate anything, so the probability is the same as for a single observation.

Step 1: the mean is Cauchy. The characteristic function of $X_i$ is $e^{-|t|}$. For $\bar{X}=\frac{1}{n}\sum X_i$, $$\mathbb{E}\left[e^{it\bar{X}}\right]={\left(e^{-|t|/n}\right)}^n=e^{-|t|},$$ so $\bar{X}$ is standard Cauchy.

Step 2: one Cauchy. $$\begin{array}{rl}\mathbb{P}(-1<\bar{X}<1)& ={\int }_{-1}^1\frac{dx}{\pi (1+x^2)}\\ & =\frac{2\arctan 1}{\pi }=\frac{1}{2}.\end{array}$$

The trap. Applying the central limit theorem and answering “almost 1”. The Cauchy distribution has no mean and no variance, so neither the law of large numbers nor the central limit theorem applies: a single huge observation can move the average anywhere, and with 1000 observations such outliers are guaranteed.