Existence of $k$th moment doesn't imply existence of absolute $k$th moment but what about the expected value?

Question

I know and I am able to prove (from the theory of Lebesgue integral) that, given a random variable $X$ and for a generic $k=0,1,2,3,\dots$, it is true that $$\mathbb{E}(|X|^k)<\infty\,\Rightarrow \mathbb{E}(X^k)<\infty$$ (and also $\mathbb{E}(X^r)<\infty$ for every moment of order $r<k$).

On the contrary I know that isn't true that $$\mathbb{E}(X^k)<\infty\,\Rightarrow \mathbb{E}(|X|^k)<\infty$$ and I am able to show it with some counterexamples.

On the other hand, I know and I am able to prove that for a very important case - $k=1$, the moment of order 1 AKA the expected value - it is true that $\mathbb{E}(X)$ exists finite if and only if $\mathbb{E}(|X|)<\infty$: $$\mathbb{E}(X)<\infty\,\iff \mathbb{E}(|X|)<\infty$$

Is this "double implication" (if and only if instead of if) an exception only true for the 1st moment? If so, why is that and/or where am I reasoning wrong?

Thanks in advance for your help!

Answer 1

If $k$ is odd, use $X_\text{new}=X_\text{old}^k$. If $k$ is even, $X^k=|X|^k$.

Answer 2

A basic fact in measure theory says $\int |f| d\mu <\infty$ iff $\int f^{+} d\mu$ and $\int f^{-} d\mu$ are both finite which is true if and only if $\int f d\mu$ is well defined and its value is a real number (not $+\infty$ or $-\infty$). In this sense $EX^{k}$ is well defined and it is a real number if and only if $E|X|^{k} <\infty$ so the equivalence does hold for every positive integer $k$.