Q:

If $\mathrm{E} |X|^2$ exists, then $\mathrm{E} X$ also exists

Is the following claim correct:

If $\mathrm{E} |X|^2$ exists, then $\mathrm{E} X$ also
  exists, because   $$ \mathrm{E} X  \leq \mathrm{E} |X| \leq  \sqrt{\mathrm{E} |X|^2} $$ by Jensen's inequality.

Is the following claim correct:

If $\mathrm{E} |X|^2$ exists and is finite, then $\mathrm{E} X$ also
  exists and is finite, because   $$ \mathrm{E} X  \leq \mathrm{E} |X| \leq  \sqrt{\mathrm{E} |X|^2} $$ by Jensen's inequality.

I guess the arguments are not right, because Jensen's inequality assumes $X$ to be integratable, i.e. $\mathrm E |X| < \infty$.
Thanks.

The first statement itself is wrong, and the Second is correct.

A:

Inequality $\left|X\right|\leq1+X^{2}$ gives you $\mathbb{E}\left|X\right|\leq1+\mathbb{E}X^{2}$

A:

$\newcommand{\E}{\operatorname{E}}$
Suppose $\E(X^2)<\infty$.  Then
$$
\E |X| \le \E\left.\begin{cases} 1 & \text{if }|X|\le 1 \\ X^2 & \text{if }|X|>1 \end{cases}\right\} \le 1 + \E(X^2)<\infty.
$$
Hence $\E(X)$ exists (and is finite).

