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According to my professor, we can divide $X$ into $X_+=\max(X,0)$ and $X_-=-\min(X,0)$, both nonnegative.
And we have $EX=EX_+-EX_-$ and $E|X|=EX_++EX_-$
For $E\{X_+\}$ finite and $E\{X_-\}$ finite, we have both $E\{X\}$ and $E\{|X|\}$ finite. If only of $E\{X_+\}$ and $E\{X_-\}$ is finite and the other one is infinite, then both of $E\{X\}$ and $E\{|X|\}$ are infinite.
However, for both $E\{X_+\}$ and $E\{X_-\}$ being infinite, $E\{X\}$ doesn't exist.
My quesition is: Why $E\{X\}$ doesn't exist for the last case? Can't it be a finite number? Could you prove this? If it's wrong, could you find a counterexample?
Thanks!
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Thanks for the comments! How about this question, can $EX$ be finite when $E|X|$ is infinite? How to prove or give a counterexample?
It's just one of those things and not particular to probability. Consider $\int_{\mathbb{R}}\frac{dx}{x}$. This function is not integrable, but there are various ways to get finite numbers out of it, e.g. $$ \lim_{\epsilon\to0^+}\left(\int_{\epsilon}^{1/\epsilon}\frac{dx}{x}+\int_{-1/\epsilon}^{-\epsilon}\frac{dx}{x}\right)=0. $$ Or consider that a conditionally convergent series can be rearranged to sum to any real number, e.g. $$ \sum_n\frac{1}{2n}=\infty, \quad\sum_n\frac{1}{2n+1}=\infty,\quad \sum_n\frac{(-1)^n}{n}=\log2. $$
More generally, there's a whole industry of getting finite answers from divergent phenomenon. The definition of $L^1$ is the "unambiguous" case where there's really only one way to make sense of the thing.
By $L^1$ or integrable I mean $\int|f|<\infty$. In other words $\int f_+$ and $\int f_-$ are both finite using your notation. If exactly one of them is infinite, you might assign $\pm\infty$ to the result without ambiguity, but $\infty-\infty$ doesn't make unambiguous sense; you would need some way to interpret it like the examples above.