"Expectation exists" vs "Expectation is finite"

Question

Are the statements

• "Expectation exists"
• "Expectation is finite"

equivalent? If not, could someone please provide a counterexample.

In case it's relevant, I don't know measure theory, but am confortable with probability theory & statistics at the undergraduate level.

Answer 1

They are not equivalent, but this is somewhat a matter of opinion. There are three classes of random variables:

1. Variables with finite expected value.
2. Variables with infinite expected value. (This could be $+\infty$ or $-\infty$).
3. Variables whose expectation does not exist.

An example of variable in class (2) is the St. Petersburg random variable, which is equal to $2^k$ with probability $2^{-k}$ for $k\ge 1$. Another example is, letting $X_i$ be an iid sequence of random variables equal to $\pm1$ with equal probabilities, the random variable $\tau$ equal to the smallest positive integer for which $X_1+X_2+\dots+X_\tau=1$.

For variables in class (3), there is the Cauchy distribution with pdf proportional to $\frac1{1+x^2}$.

The question is whether or not you consider variables in category (2) to have an expectation which "exists" or not. I say the expectation exists, so that an expectation can exist but not be finite. However, I think some people disagree, just like there is disagreement as to whether $$ \lim_{x\to0}\frac1{x^2} $$ is either non-existent, or existent and equal to $+\infty$. I think most professional mathematicians have no qualms saying the limit exists, but it is often taught in entry level calculus courses that such limits do not exist.

Answer 2

A counterexample is the Cauchy distribution. It is symmetric around $0$ but both tails of the distribution are 'too' heavy: $$ \begin{align} E[X]&=\int_{-\infty}^{\infty} x\cdot\frac{1}{\pi(1+x^2)}\,\mathrm{d}x\\ &= \int_{-\infty}^{0} x\cdot\frac{1}{\pi(1+x^2)}\,\mathrm{d} x + \int_{0}^{\infty} x\cdot\frac{1}{\pi(1+x^2)}\,\mathrm{d} x \\ &=\infty -\infty \end{align} $$

Answer 3

Contra Stan Tendijck's answer, the Cauchy distribution's mean doesn't exist because $\infty-\infty$ is undefined (technically, it's an indeterminate form). However, the modulus of a Cauchy-distributed variable has a mean, but that mean is infinite. So no, a mean existing isn't equivalent to its being finite. In fact whenever $X$ has finite mean but infinite variance it's because $X^2$ has infinite mean. An example is when $X=\sqrt{|Y|}$ with a Cauchy $Y$.