"Expectation exists" vs "Expectation is finite" 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.
 A: They are not equivalent, but this is somewhat a matter of opinion. There are three classes of random variables:


*

*Variables with finite expected value.

*Variables with infinite expected value. (This could be $+\infty$ or $-\infty$).

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