Problems on expected value

Solution 1:

I will answer the first question only.

Write $X = X^+ - X^-$ where $X^+ = \max \left( X, 0 \right)$ and $X^- = \max \left( -X, 0 \right)$. Assume that for integer $p \geqslant 1$, $E \left[ \left| X \right|^p \right] < \infty$. Let $f$ be the density (just to make the derivations here simpler to write). By integration by part (and truncating at the point $m > 0$) \begin{eqnarray*} E \left[ \left( X^+ \right)^p 1 \left( X^+ < m \right) \right] & = & \int_0^m x^p f \left( x \right) \mathrm{d} x\\ & = & \left[ x^p \left( F \left( x \right) - 1 \right) \right]^m_0 + p \int_0^m x^{p - 1} \left( 1 - F \left( x \right) \right) \mathrm{d} x\\ & = & - m^p P \left[ X^+ > m \right] + p \int_0^m x^{p - 1} \left( 1 - F \left( x \right) \right) \mathrm{d} x \end{eqnarray*} By Markov inequality $P \left[ X^+ > m \right] \leqslant \frac{E \left[ \left( X^+ \right)^p \right]}{m^p}$ which converges to 0 as $m \rightarrow \infty$, implying $$ E \left[ \left( X^+ \right)^p \right] = p \int_0^{\infty} x^{p - 1} \left( 1 - F \left( x \right) \right) \mathrm{d} x $$ Replace $p$ by 1 and 2 respectively to get $E \left[ X^+ \right] = \int_0^{\infty} \left( 1 - F \left( x \right) \right) \mathrm{d} x$ and $E \left[ \left( X^+ \right)^2 \right] = 2 \int_0^{\infty} x \left( 1 - F \left( x \right) \right) \mathrm{d} x$. Do the same for $X^-$.

Once what is above is grasped, it is possible to derive the rest in the same way.