Are there any cases where $\mathbb E(|X|)<\infty$ and $\mathbb E(X)<\infty$ aren't equivalent?

Question

I often see $\mathbb E(|X|)<\infty$ among the givens in a statement. That made me wonder: why not just demand $\mathbb E(X)<\infty$? In the light of the theorem

Let $f$ be measurable. Then the following conditions are equivalent:

a) $f\in\mathcal L^1(\mu)$

b) $f^+, f^- \in\mathcal L^1(\mu)$

c) $|f|\in\mathcal L^1(\mu)$

I was expecting $\mathbb E(|X|)<\infty$ and $\mathbb E(X)<\infty$ to be equivalent as well. Are there any cases where they aren't?

Answer 1

The statement $E(|X|)\lt\infty$ is definitely preferable because $E(|X|)$ always exists as an extended nonnegative real number, that is, an element of $[0,\infty]$, since it is the expectation of a random variable with values in $[0,\infty]$. On the other hand, $E(X)$ the expectation of a random variable with real values, exists only when $X$ is integrable.

To sum up, when $X$ is not integrable, $E(|X|)=+\infty$ is correct but the statement $E(X)=+\infty$ only makes sense when $X\geqslant0$ almost surely.

Answer 2

In some constructions of the integral the value $\mathbb E(X)$ is defined if $f=f^+-f^-$ and at least one of the integrals of $f^\pm$ is finite ($\pm\infty$ are allowed as values). With such convention you can have $\mathbb E(X)<\infty$, but $\mathbb E(|X|)=\infty$. There are also situations when $\mathbb E(X)$ can be some improper integral defined even if $f\not\in L^1(\mu)$, e.g. a conditionally convergent series or an improper Riemann integral. For clarity $\mathbb E(|X|)<\infty$ is used to emphasize that $f\in L^1(\mu)$.