The theorem in my notes says that:
If $|X - a|^\alpha$ has expectation for any $a \in \mathbb{R}$, and any $\alpha > 0$, then $|X - b|^\beta$ has expectation for any $b \in \mathbb{R}$ and any $\beta$ such that $0 \leq \beta \leq \alpha$.
So if we know that the second moment $E \, X^2$ is not finite while $\operatorname{Var} X$ is, are $E\,X$ and $E(X - E(X \mid Y))^2$ finite? Is there any example or counter-example for this case?
Update: From the theorem, it seems that $|X−a|^\alpha$ can have expectation for only a set of $a$ (say $a\in A$) at a specific $\alpha$. So is it possible that $|X−E\,X|^2$ has expectation while $X^2$ doesn't?
Every random variable with finite variance has finite second moment. Variance is only defined for random variables with finite first moment, and as you probably know, $$E [X^2] = \operatorname{Var}(X) + E[X]^2.$$