\(% Differentiation % https://tex.stackexchange.com/a/60546/ \newcommand{\Diff}{\mathop{}\!\mathrm{d}} \newcommand{\DiffFrac}[2]{\frac{\Diff #1}{\Diff #2}} \newcommand{\DiffOp}[1]{\frac{\Diff}{\Diff #1}} \newcommand{\Ndiff}[1]{\mathop{}\!\mathrm{d}^{#1}} \newcommand{\NdiffFrac}[3]{\frac{\Ndiff{#1} #2}{\Diff {#3}^{#1}}} \newcommand{\NdiffOp}[2]{\frac{\Ndiff{#1}}{\Diff {#2}^{#1}}} % Evaluation \newcommand{\LEvalAt}[2]{\left.#1\right\vert_{#2}} \newcommand{\SqEvalAt}[2]{\left[#1\right]_{#2}} % Epsilon & Phi \renewcommand{\epsilon}{\varepsilon} \renewcommand{\phi}{\varphi} % Sets \newcommand{\NN}{\mathbb{N}} \newcommand{\ZZ}{\mathbb{Z}} \newcommand{\QQ}{\mathbb{Q}} \newcommand{\RR}{\mathbb{R}} \newcommand{\CC}{\mathbb{C}} \newcommand{\PP}{\mathbb{P}} \renewcommand{\emptyset}{\varnothing} % Probabililty \DeclareMathOperator{\Cov}{Cov} \DeclareMathOperator{\Corr}{Corr} \DeclareMathOperator{\Var}{Var} \DeclareMathOperator{\Expt}{E} \DeclareMathOperator{\Prob}{P} % Distribution \DeclareMathOperator{\Binomial}{B} \DeclareMathOperator{\Poisson}{Po} \DeclareMathOperator{\Normal}{N} \DeclareMathOperator{\Exponential}{Exp} \DeclareMathOperator{\Geometric}{Geo} \DeclareMathOperator{\Uniform}{U} % Complex Numbers \DeclareMathOperator{\im}{Im} \DeclareMathOperator{\re}{Re} % Missing Trigonometric & Hyperbolic functions \DeclareMathOperator{\arccot}{arccot} \DeclareMathOperator{\arcsec}{arcsec} \DeclareMathOperator{\arccsc}{arccsc} \DeclareMathOperator{\sech}{sech} \DeclareMathOperator{\csch}{csch} \DeclareMathOperator{\arsinh}{arsinh} \DeclareMathOperator{\arcosh}{arcosh} \DeclareMathOperator{\artanh}{artanh} \DeclareMathOperator{\arcoth}{arcoth} \DeclareMathOperator{\arsech}{arsech} \DeclareMathOperator{\arcsch}{arcsch} % UK Notation \DeclareMathOperator{\cosec}{cosec} \DeclareMathOperator{\arccosec}{arccosec} \DeclareMathOperator{\cosech}{cosech} \DeclareMathOperator{\arcosech}{arcosech} % Paired Delimiters \DeclarePairedDelimiter{\ceil}{\lceil}{\rceil} \DeclarePairedDelimiter{\floor}{\lfloor}{\rfloor} \DeclarePairedDelimiter{\abs}{\lvert}{\rvert} \DeclarePairedDelimiter{\ang}{\langle}{\rangle} % Vectors \newcommand{\vect}[1]{\mathbf{#1}} \newcommand{\bvect}[1]{\overrightarrow{#1}} % https://tex.stackexchange.com/a/28213 % \DeclareMathSymbol{\ii}{\mathalpha}{letters}{"10} % \DeclareMathSymbol{\jj}{\mathalpha}{letters}{"11} % \newcommand{\ihat}{\vect{\hat{\ii}}} % \newcommand{\jhat}{\vect{\hat{\jj}}} \newcommand{\ihat}{\textbf{\^{ı}}} \newcommand{\jhat}{\textbf{\^{ȷ}}} \newcommand{\khat}{\vect{\hat{k}}} % Other Functions \DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\tr}{tr} % Other Math Symbols \DeclareMathOperator{\modulo}{mod} \newcommand{\divides}{\mid} \newcommand{\notdivides}{\nmid} \newcommand{\LHS}{\text{LHS}} \newcommand{\RHS}{\text{RHS}} \newcommand{\degree}{^{\circ}}\)

2012.3.13 Question 13

1. We have \begin {align*} \Expt (Z \mid a < Z < b) & = \frac {\int _{a}^{b} z \Phi '(z) \Diff z}{\int _{a}^{b} \Phi '(z) \Diff z} \\ & = \frac {\int _{a}^{b} z e^{-\frac {z^2}{2}}\Diff z}{\sqrt {2\pi } \left (\Phi (b) - \Phi (a)\right )} \\ & = \frac {\left [-e^{-\frac {z^2}{2}}\right ]_{a}^{b}}{\sqrt {2\pi } \left (\Phi (b) - \Phi (a)\right )} \\ & = \frac {e^{-\frac {a^2}{2}} - e^{-\frac {b^2}{2}}}{\sqrt {2\pi } \left (\Phi (b) - \Phi (a)\right )} \end {align*}

2. Since \(X = \mu + \sigma Z\) \begin {align*} \Expt (X \mid X > 0) & = \Expt (\mu + \sigma Z \mid (\mu + \sigma Z) > 0) \\ & = \mu + \sigma \Expt (Z \mid (\mu + \sigma Z) > 0) \\ & = \mu + \sigma \Expt \left (Z \mid Z > - \frac {\mu }{\sigma }\right ), \end {align*}

   as desired.

   Hence, \begin {align*} m & = \Expt (\abs *{X}) \\ & = \Expt (\abs *{X} \mid X > 0) \cdot \Prob (X > 0) + \Expt (\abs *{X} \mid X < 0) \cdot \Prob (X < 0) \\ & = \Expt (X \mid X > 0) \cdot \Prob (X > 0) - \Expt (X \mid X < 0) \cdot \Prob (X < 0) \\ & = \left [\mu + \sigma \Expt \left (Z \mid Z > - \frac {\mu }{\sigma }\right )\right ] \cdot \Prob (\mu + \sigma Z > 0) \\ & \phantom {=} - \left [\mu + \sigma \Expt \left (Z \mid Z < - \frac {\mu }{\sigma }\right )\right ] \cdot \Prob (\mu + \sigma Z < 0) \\ & = \left [\mu + \sigma \cdot \frac {\exp \left (-\frac {1}{2} \left (- \frac {\mu }{\sigma }\right )^2\right )}{\sqrt {2\pi } \left (1 - \Phi \left (- \frac {\mu }{\sigma }\right )\right )}\right ] \cdot \left [1 - \Phi \left (- \frac {\mu }{\sigma }\right )\right ] - \left [\mu + \sigma \cdot \frac {- \exp \left (-\frac {1}{2} \left (- \frac {\mu }{\sigma }\right )^2\right )}{\sqrt {2\pi } \Phi \left (- \frac {\mu }{\sigma }\right )}\right ] \cdot \Phi \left (- \frac {\mu }{\sigma }\right ) \\ & = \mu \left [1 - \Phi \left (- \frac {\mu }{\sigma }\right ) - \Phi \left (- \frac {\mu }{\sigma }\right )\right ] + \frac {\sigma \exp \left (-\frac {1}{2} \left (- \frac {\mu }{\sigma }\right )^2\right )}{\sqrt {2\pi }} \cdot (1 + 1) \\ & = \mu \left [1 - 2 \Phi \left (-\frac {\mu }{\sigma }\right )\right ] + \frac {\sqrt {2}\sigma \exp \left (-\frac {1}{2} \cdot \frac {\mu ^2}{\sigma ^2}\right )}{\sqrt {\pi }}, \end {align*}

   as desired.

   To find the variance of \(\abs *{X}\), we would like to find \(\Expt (\abs *{X}^2)\). But this is precisely \(\Expt (\abs *{X}^2) = \Expt (X^2) = \Var (X) + \Expt (X)^2 = \sigma ^2 + \mu ^2\). Hence, \begin {align*} \Var (\abs *{X}) & = \Expt (\abs *{X}^2) - \Expt (\abs *{X})^2 \\ & = \sigma ^2 + \mu ^2 - m^2. \end {align*}