Expectation of a squared random variable and of its absolute value Is it true that if $\mathbb{E}[X^2]<\infty$ then $\mathbb{E}[|X|]<\infty$? If so, why?
 A: The most direct way to see this, without referring to a single theorem, is to consider the set $E=\{|X| \leq 1\}$. Then if $1_E$ is the indicator function of $E$,
$$\mathbb{E}[|X|] = \mathbb{E}[|X|\cdot 1_E] + \mathbb{E}[|X| \cdot 1_{E^c}] \leq \mathbb{E}[1 \cdot1_E] + \mathbb{E}[X^2 \cdot 1_{E^c}] \leq1+\mathbb{E}[X^2] < \infty.$$
The more clever arguments using Jenson's or Holder's inequalities provide a much sharper bound, however.
A: $\newcommand{\+}{^{\dagger}}%
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\begin{align}
0 &\leq \angles{\pars{\vphantom{\Large A}x - \angles{x}}^{2}}=
\angles{x^{2} - 2x\angles{x} + \angles{x}^{2}}
=
\angles{x^{2}} - 2\angles{x}\angles{x} + \angles{x}^{2}
=
\angles{x^{2}} - \angles{x}^{2}
\end{align}

$$\imp\qquad\color{#0000ff}{\large%
\verts{\vphantom{\LARGE A}\angles{x}}\ \leq\ \angles{x^{2}}^{1/2}}
$$

A: Yes, it is true. Apply Jensen inequality to the convex function $x \mapsto x^2$ or Cauchy-Schwarz inequality $\Bbb E[|XY|] \leq \sqrt{\Bbb E[|X|^2] \Bbb E[|Y|^2]}$ with $Y = 1$. 
