What does it mean to not have an expected value?

Question

My professor showed us that the Cauchy distribution does not have an expected value, that is, the integral $\int_{-\infty}^{\infty} x f(x) \text{d} x$ is not defined ($f(x)$ is the p.d.f. of the Cauchy distribution). I find that very counterintuitive. What does it actually mean, in the context of probability, to not have a defined expected value?

Answer 1

The Law of Large Numbers fails for distributions without an expected value. So empirically, if you were to average many samples from the Cauchy distribution, you won't be able to say that the average converges to zero (as one might think).

A property of the Cauchy distribution is that if $X_1,...,X_n$ and $Z$ are i.i.d. Cauchy random variables, then $\frac{1}{n}\sum_{i=1}^nX_i \ \stackrel{d}{=} \ Z$.

With a standard normal random variable, the distribution of the LHS would be $\frac{1}{\sqrt n}$ times as 'spread out' as a single random variable, whereas in this case, averaging a bunch of i.i.d. Cauchy variables changes nothing about their distribution.

Our intuitive interpretation of the expected value is tied to the law of large numbers. In this case, it doesn't apply.

Answer 2

For a random variable $X$, write $X=X^+-X^-$, where $X^+=\max\{X,0\}$ and $X^-=\max\{-X,0\}$. Then $|X|=X^++X^-$, so $\mathbb E[|X|]<\infty$ if and only if $\mathbb E[X^+]<\infty$ and $\mathbb E[X^-]<\infty$. When this condition holds, we define $$ \mathbb E[X] = \mathbb E[X^+-X^-]= \mathbb E[X^+]-\mathbb E[X^-]. $$ If $X$ has standard Cauchy distribution with density $f(x) = \frac1{\pi(1+x^2)}$ then for each $L>0$ we have \begin{align} \mathbb E\left[X^+\mathsf 1_{(0,L]}\right](X) = \int_0^L \frac x{\pi(1+x^2)}\ \mathsf dx = \frac{\log\left(1+L^2\right)}{2\pi}\stackrel{L\to\infty}\longrightarrow +\infty, \end{align} so that $\mathbb E[X^+]=+\infty$. By symmetry, $\mathbb E[X^-] = -\infty$. It follows that the expression $\mathbb E[X^+]-\mathbb E[X^-]$ is not well-defined, and neither is $\mathbb E[X]$.