Exercise: When the CLT Fails — Cauchy Samples

Prerequisites: Central Limit Theorem, Expectation and Variance

Problem

The standard Cauchy distribution has density

f(x) = \pi^{-1}(1 + x^2)^{-1}

for

x \in \mathbb{R}

. It has no mean (the integral

\int x f(x)\,dx

is conditionally convergent to any desired value) and no variance (the integral of

x^2 f(x)

diverges).

Simulate $N = 100{,}000$ independent samples of $\bar X_n = (X_1 + \cdots + X_n)/n$ with $X_i$ i.i.d. standard Cauchy, for $n \in \{1, 10, 100, 1000\}$ . Plot histograms of $\bar X_n$ on the same horizontal scale. What does the histogram of $\bar X_n$ look like as $n$ grows — is it concentrating near zero (as the LLN would suggest) or spreading out, or staying the same?
A remarkable fact: the average of $n$ i.i.d. standard Cauchy random variables is itself standard Cauchy, for every $n$ . Verify this numerically by overlaying the standard Cauchy density on each histogram.
Using the result in part 2, explain in one sentence why both the Law of Large Numbers and the Central Limit Theorem fail for Cauchy samples.
A risk manager proposes using sample means of intraday returns as unbiased estimators of expected return. If intraday returns were truly Cauchy-distributed (no mean), what would happen to the sample mean as they collected more data? Why does this matter even though real returns aren't literally Cauchy?

Use rng.standard_cauchy(size=(N, n)) in NumPy. The stability of the Cauchy under averaging follows from its characteristic function

\varphi(t) = e^{-|t|}

— a fact you can cite without proof.

Jump to the solution when you're ready.