Indefinite Integration of the Gamma Integral
and Related Statistical Applications

Proof of Equation (4.16)

\begin{align*}
E\left( x \right)&=\mu -\frac{{{\sigma }^{2}}}{D}\exp \left( \frac{-{{\left[ x-\mu \right]}^{2}}}{2{{\sigma }^{2}}} \right){\Bigr|}_{a}^{b}\ \\
V\left( x \right)&={{\sigma }^{2}}-\frac{{{\sigma }^{2}}}{D}\exp \left( \frac{-{{\left[ x-\mu \right]}^{2}}}{2{{\sigma }^{2}}} \right)\left( x-\mu \right)|_{a}^{b}-{{\left\{ \frac{-{{\sigma }^{2}}}{D}\exp \left( \frac{-{{\left[ x-\mu \right]}^{2}}}{2{{\sigma }^{2}}} \right){\Bigr|}_{a}^{b}\right\}}^{2}}.
\end{align*}Proof:
The mean and variance of the truncated normal variables are functions of the first and second moments
\begin{align*}
& E\left( x \right)={{m}_{1}} \\
& V\left( x \right)={{m}_{2}}-{{m}_{1}}^{2}.
\end{align*}
Bringing the result of (4.14) and (4.15), we can conclude the proof.
\begin{align*}
E\left( x \right)&=\mu -\frac{{{\sigma }^{2}}}{D}\exp \left( \frac{-{{\left[ x-\mu \right]}^{2}}}{2{{\sigma }^{2}}} \right){\Bigr|}_{a}^{b}\ \\
V\left( x \right)&={{\sigma }^{2}}-\frac{{{\sigma }^{2}}}{D}\exp \left( \frac{-{{\left[ x-\mu \right]}^{2}}}{2{{\sigma }^{2}}} \right)\left( x-\mu \right)|_{a}^{b}-{{\left\{ \frac{-{{\sigma }^{2}}}{D}\exp \left( \frac{-{{\left[ x-\mu \right]}^{2}}}{2{{\sigma }^{2}}} \right){\Bigr|}_{a}^{b}\right\}}^{2}}.
\end{align*}
$\square$

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