Integrability of random variable- theoretic aspect

Question

in some topics (for example in martingale theory, strong law of large numbers), we have to check integrability of a given random variable $X$, i.e. $E|X|<\infty$. Okay, so assume that I have some random variable $X$ and I know only that $EX=3$. I cant compute $E|X|$, but do I need it to confirm integrability? Random variable is integrable if $EX^+$ and $EX^-$ are finite and I think that until now we know only that $EX^+-EX^-=3$ and I think that it implies that integrability properties mentioned above, because we cant get the $3$ number as a difference of two infinities. Or maybe we can?

Answer 1

When we say $X$ must be integrable we mean that $E[X]$ exists and is finite.

So consider the four cases:

1.) $E[X^+] < \infty$ and $E[X^-]<\infty \Rightarrow E[X]<\infty$ AND $E|X|<\infty$

2.) $E[X^+] = \infty$ and $E[X^-]=\infty\Rightarrow E[X]$ d.n.e AND $E|X| = \infty$

3.) $E[X^+] < \infty$ and $E[X^-]=\infty\Rightarrow E[X]=-\infty$ AND $E|X| = \infty$

4.) $E[X^+] = \infty$ and $E[X^-]<\infty\Rightarrow E[X]=\infty$ AND $E|X| = \infty$

Noting that $E[X] = E[X^+]-E[X^-]$ and $E|X| = E[X^+]+E[X^-]$.

So when you consider $E|X|$ you narrow checking X being integrable down to whether $E|X|$ is infinite or not. This alongside the non-negativity of $E|X|$ is why this definition is used as it's convenient.

Since you know E[X] = 3 you know it is integrable and that it integrates to 3.