\(% 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}}\)
Since \(Z_1\) and \(Z_2\) are independent, we have \[ \Cov (Z_1, Z_2) = 0, \] and hence \[ \Corr (Z_1, Z_2) = \frac {\Cov (Z_1, Z_2)}{\sqrt {\Var (Z_1) \Var (Z_2)}} = \frac {0}{\sqrt {1 \cdot 1}} = 0. \]
For \(Y_2\), we have \begin {align*} \Expt (Y_2) & = \Expt \left (\rho _{12} Z_1 + \sqrt {1 - \rho _{12}^2} Z_2\right ) \\ & = \rho _{12} \Expt (Z_1) + \sqrt {1 - \rho _{12}^2} \Expt (Z_2) \\ & = \rho _{12} \cdot 0 + \sqrt {1 - \rho _{12}^2} \cdot 0 \\ & = 0, \end {align*}
\begin {align*} \Var (Y_2) & = \Var (\rho _{12} Z_1 + \sqrt {1 - \rho _{12}^2} Z_2) \\ & = \rho _{12}^2 \Var (Z_1) + (1 - \rho _{12}^2) \Var (Z_2) \\ & = \rho _{12}^2 \cdot 1 + (1 - \rho _{12}^2) \cdot 1 \\ & = 1, \end {align*}
and hence \begin {align*} \Corr (Y_1, Y_2) & = \frac {\Cov (Y_1, Y_2)}{\sqrt {\Var (Y_1) \Var (Y_2)}} \\ & = \frac {\Expt (Y_1 Y_2) - \Expt (Y_1) \Expt (Y_2)}{\sqrt {1 \cdot 1}} \\ & = \Expt \left (\rho _{12} Z_1^2 + \rho _{12} \sqrt {1 - \rho _{12}^2} Z_1 Z_2\right ) - 0 \cdot 0 \\ & = \rho _{12} \Expt (Z_1^2) + \rho _{12} \sqrt {1 - \rho _{12}^2} \Expt (Z_1 Z_2) \\ & = \rho _{12} \left (\Var (Z_1) + \Expt (Z_1)^2\right ) + \rho _{12} \sqrt {1 - \rho _{12}^2} \Expt (Z_1) \Expt (Z_2) \\ & = \rho _{12} \left (1 + 0^2\right ) + \rho _{12} \sqrt {1 - \rho _{12}^2} \cdot 0 \cdot 0 \\ & = \rho _{12}. \end {align*}
For \(Y_3\), we have \begin {align*} \Var (Y_3) & = \Var (a Z_1 + b Z_2 + c Z_3) \\ & = a^2 \Var (Z_1) + b^2 \Var (Z_2) + c^2 \Var (Z_3) \\ & = a^2 + b^2 + c^2 \\ & = 1, \end {align*}
and hence \(a^2 + b^2 + c^2 = 1\).
For the correlation, we have \begin {align*} \Corr (Y_1, Y_3) & = \frac {\Cov (Y_1 Y_3)}{\sqrt {\Var (Y_1) \Var (Y_3)}} \\ & = \frac {\Expt (Y_1 Y_3) - \Expt (Y_1) \Expt (Y_3)}{\sqrt {1 \cdot 1}} \\ & = \frac {\Expt (aZ_1^2 + bZ_1Z_2 + cZ_1Z_3) - 0 \cdot 0}{1} \\ & = a\Expt (Z_1^2) + b\Expt (Z_1 Z_2) + c\Expt (Z_1 Z_3) \\ & = a (\Var (Z_1) + \Expt (Z_1)^2) + b \Expt (Z_1) \Expt (Z_2) + c \Expt (Z_1) \Expt (Z_3) \\ & = a (1 + 0^2) + b \cdot 0 \cdot 0 + c \cdot 0 \cdot 0 \\ & = a \\ & = \rho _{13}, \end {align*}
and \begin {align*} \Corr (Y_2, Y_3) & = \frac {\Cov (Y_2 Y_3)}{\sqrt {\Var (Y_2) \Var (Y_3)}} \\ & = \frac {\Expt (Y_2 Y_3) - \Expt (Y_2) \Expt (Y_3)}{\sqrt {1 \cdot 1}} \\ & = \frac {\Expt \left ((aZ_1 + bZ_2 + cZ_3) \cdot \left (\rho _{12} Z_1 + \sqrt {1 - \rho _{12}^2 Z_2}\right )\right ) - 0 \cdot 0}{1} \\ & = \Expt \left (a\rho _{12}Z_1^2 + b\sqrt {1 - \rho _{12}^2} Z_2^2\right ) \\ & = a \rho _{12} (\Var (Z_1) + \Expt (Z_1)^2) + b\sqrt {1 - \rho _{12}^2} (\Var (Z_2) + \Expt (Z_2)^2) \\ & = a \rho _{12} + b\sqrt {1 - \rho _{12}^2} \\ & = \rho _{23}, \end {align*}
since all the cross-term expectation is \(0\), i.e. for \(i \neq j\), \(\Expt (Z_i Z_j) = \Expt (Z_i) \Expt (Z_j) = 0\). Hence, \[ b = \frac {\rho _{23} - \rho _{12} \rho _{13}}{\sqrt {1 - \rho _{12}^2}}, \] and therefore, \[ c = \sqrt {1 - a^2 - b^2} = \sqrt {1 - \rho _{13}^2 - \frac {\left (\rho _{23} - \rho _{12} \rho _{13}\right )^2}{1 - \rho _{12}^2}}. \]
We could have \(X_i = \mu _{i} + \sigma _{i} Y_i\) for \(i = 1, 2, 3\), since \[ \Expt (X_i) = \mu _{i} + \sigma _{i} \Expt (Y_i) = \mu _{i} + \sigma _{i} \cdot 0 = \mu _{i}, \] and \[ \Var (X_i) = \sigma ^2_{i} \Var (Y_i) = \sigma ^2_{i} \cdot 1 = \sigma ^2_{i}. \]
As for correlation, we notice that for any random variables \(U, V\), we have \begin {align*} \Corr (aU + b, cU + d) & = \frac {\Cov (aU + b, cV + d)}{\sqrt {\Var (aU + b) \Var (cV + d)}} \\ & = \frac {\Expt ((aU+b)(cV+d)) - \Expt (aU+b)\Expt (cV+d)}{\sqrt {a^2 \Var (U) c^2 \Var (V)}} \\ & = \frac {ac\Expt (UV) + bc\Expt (V) + ad\Expt (U) + bd - (a\Expt (U) + b)(c\Expt (V) + d)}{ac\sqrt {\Var (U)\Var (V)}} \\ & = \frac {ac\Expt (UV) + bc\Expt (V) + ad\Expt (U) + bd - ac\Expt (U)\Expt (V) - bc\Expt (V) - ad\Expt (U) - bd}{ac\sqrt {\Var (U)\Var (V)}} \\ & = \frac {ac (\Expt (UV) - \Expt (U)\Expt (V))}{ac\sqrt {\Var (U) \Var (V)}} \\ & = \frac {\Cov (U, V)}{\sqrt {\Var (U) \Var (V)}} \\ & = \Corr (U, V), \end {align*}
which shows linear coding does not affect the correlation. This implies \[ \Corr (X_i, X_j) = \Corr (Y_i, Y_j) = \rho _{ij} \] for \(i \neq j\). Therefore, \(X_i = \mu _{i} + \sigma _{i} Y_i\) for \(i = 1, 2, 3\) satisfies the desired.