Does $\mathbb{E}(X-a)^2$ finite imply $\mathbb{E}(X^2)$ finite? Can we say that, if for some arbitrary constant  $a$, $$\mathbb{E}(X-a)^2< \infty$$ $$\implies \mathbb{E}(X^2)< \infty$$
I think that we can say, yes, because $\mathbb{E}(X-a)^2= \mathbb{E}(X^2-2aX+a^2)=a^2-2a\mathbb{E}X+\mathbb{E}X^2$
Hence, if we can show that $\mathbb{E}X$ is finite, then we would be done. I think that if $\mathbb{E}X$ was not finite, then $\mathbb{E}(X-a)^2$  would not be finite, however I am not sure how to show that this would be true. 

The triangle inequality for $L^2$ spaces implies that
$$
\lVert X\rVert_2\leq \lVert X-a\rVert_2+\lVert a\rVert_2<\infty
$$
where $\lVert X\rVert_2=(EX^2)^{1/2}$.
