Why is the expectation of cauchy distribution not defined? (What is the intuition behind it?)

Question

Let $X$ be random variable with pdf $f_X(x) = \dfrac{1}{\pi(1+x^2)}$. I understand that mathematically, the improper integral, $\displaystyle\int\limits_{-\infty}^{\infty}\dfrac{x}{\pi(1+x^2)}dx$ does not exist ($\underset{T_1\to-\infty}{\lim}\underset{T_2\to\infty}{\lim}\displaystyle\int\limits_{T_1}^{T_2}\dfrac{x}{\pi(1+x^2)}dx = \frac{\ln(1+T_2^2) - \ln(1+T_1^2)}{2\pi} = \infty - \infty) \implies$ undefined. However I am unable to understand the intuition, why $E[X] \ne 0$ since the pdf is an even function which takes positive and negative values with equal probability. Hence large enough samples should produce mean close to 0. Please help?

Answer 1

There are several ways to look at it:

• Let $f(x):=\frac{\pi^{-1}}{1+x^2}$ so $\int_{-\infty}^c xf(x) dx=-\infty,\,\int_d^\infty xf(x) dx$ for any $c,\,d\in\mathbb{R}$. So if we choose $c<d$, you could argue the mean is $-\infty+\int_c^d xf(x) dx+\infty$. In theory, you can get any value you like if you think the infinities cancel.
• "But I'm integrating an odd function! That has to give me $0$!" Yes, if the two pieces you're cancelling are both finite. But $\infty-\infty$ is an indefinite form, so you can't use that theorem.
• The characteristic function is $\varphi(t):=\exp -\left|t\right|$. If we average $n$ samples, the result has characteristic function $\varphi^n(t/n)=\varphi(t)$. It's immune to the CLT. Nor should you expect otherwise, without a finite and well-defined mean and variance. No $\mu$, no $(X-\mu^2)$, no variance. The characteristic function provides another way to look at it: we can't very well write $\mu=\varphi'(0)$ because the modulus has undefined derivative at $0$ (the one-sided limits $\lim_{t\to 0^\pm}\frac{|t|}{t}$ differ).
• It can be shown, however, that the median of $n$ samples is asymptotically Normal for large $n$. (The proof is a bit more involved than a standard CLT argument for means; you can get an overview here.) By contrast, if you compute the mean of a gradually growing sample, it'll bounce around like crazy, because (as shown above) it's Cauchy-distributed.