\(% 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}}\)
We have \begin {align*} V(x) & = \Expt [(X-x)^2] \\ & = \Expt (X^2 - 2xX + x^2) \\ & = \Expt (X^2) - 2x\Expt (X) + x^2 \\ & = \sigma ^2 + \mu ^2 - 2x\mu + x^2. \end {align*}
Therefore, if \(Y = V(X)\), then \begin {align*} \Expt (Y) & = \Expt (V(X)) \\ & = \Expt (\sigma ^2 + \mu ^2 - 2X\mu + X^2) \\ & = \sigma ^2 + \mu ^2 - 2\mu \Expt (X) + \Expt (X^2) \\ & = \sigma ^2 + \mu ^2 - 2\mu ^2 + \mu ^2 + \sigma ^2 \\ & = 2\sigma ^2. \end {align*}
Let \(X \sim U[0, 1]\), we have \(\mu = \Expt (X) = \frac {1}{2}\), and \(\sigma ^2 = \Var (X) = \frac {1}{12}\). Therefore, \begin {align*} V(x) & = \frac {1}{12} + \frac {1}{4} - x + x^2 \\ & = x^2 - x + \frac {1}{3}. \end {align*}
The c.d.f. of \(X\) is \(F\), defined as \[ \Prob (X \leq x) = F(x) = \begin {cases} 0, & x \leq 0, \\ x, & 0 < x \leq 1, \\ 1, & 1 < x \end {cases} \]
Let the c.d.f. of \(Y\) be \(G\), we have \(G(y) = \Prob (Y \leq y)\).
Since \(V([0, 1]) = \left [\frac {1}{12}, \frac {1}{3}\right ]\), we must have \(G(y) = 0\) for \(y \leq \frac {1}{12}\) and \(G(y) = 1\) for \(y > \frac {1}{3}\).
For \(y \in \left (\frac {1}{12}, \frac {1}{3}\right ]\), we have \begin {align*} G(y) = \Prob (Y \leq y) & = \Prob (V(X) \leq y) \\ & = \Prob \left (\left (x - \frac {1}{2}\right )^2 + \frac {1}{12} \leq y\right ) \\ & = \Prob \left (\abs *{x - \frac {1}{2}} \leq \sqrt {y - \frac {1}{12}}\right ) \\ & = \Prob \left (\frac {1}{2} - \sqrt {y - \frac {1}{12}} \leq x \leq \frac {1}{2} + \sqrt {y - \frac {1}{12}}\right ) \\ & = F\left (\frac {1}{2} + \sqrt {y - \frac {1}{12}}\right ) - F\left (\frac {1}{2} - \sqrt {y - \frac {1}{12}}\right ) \\ & = \left (\frac {1}{2} + \sqrt {y - \frac {1}{12}}\right ) - \left (\frac {1}{2} - \sqrt {y - \frac {1}{12}}\right ) \\ & = 2 \sqrt {y - \frac {1}{12}}. \end {align*}
Therefore, the p.d.f. of \(y\), \(g\) satisfies that for \(y \in \left (\frac {1}{12}, \frac {1}{3}\right ]\), \[ g(y) = G'(y) = \frac {1}{\sqrt {y - \frac {1}{12}}} \] and \(0\) everywhere else.
Hence, we have \begin {align*} \Expt (Y) & = \int _{\RR } y f(y) \Diff y \\ & = \int _{\frac {1}{12}}^{\frac {1}{3}} \frac {y}{\sqrt {y - \frac {1}{12}}} \Diff y \\ & = \int _{y = \frac {1}{12}}^{y = \frac {1}{3}} 2y \Diff \sqrt {y - \frac {1}{12}} \\ & = \left [2y\sqrt {y - \frac {1}{12}}\right ]_{\frac {1}{12}}^{\frac {1}{3}} - 2\int _{\frac {1}{12}}^{\frac {1}{3}} \sqrt {y - \frac {1}{12}} \Diff y \\ & = \left [2y\sqrt {y - \frac {1}{12}} - \frac {4}{3} \left (y - \frac {1}{12}\right )^{\frac {3}{2}}\right ]_{\frac {1}{12}}^{\frac {1}{3}} \\ & = 2 \cdot \frac {1}{3} \cdot \frac {1}{2} - \frac {4}{3} \cdot \frac {1}{8} \\ & = \frac {1}{6}. \end {align*}
Also, \(2 \sigma ^2 = 2 \cdot \frac {1}{12} = \frac {1}{6} = \Expt (Y)\), so the formula we derived holds in this case.