\(% 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}}\)
This setup gives a Markov Chain. Let the column vector \(\vect {x}_n\) represent a state \[ \vect {x}_n = \begin {pmatrix} A_n \\ B_n \\ C_n \\ D_n \end {pmatrix}, \] and hence we have the components of the column vector must sum to \(1\). The initial state is defined by \[ \vect {x}_0 = \begin {pmatrix} 1 \\ 0 \\ 0 \\ 0 \end {pmatrix}, \] and the state transition matrix is \[ \vect {M} = \begin {pmatrix} 1/2 & 1/4 & 0 & 1/4 \\ 1/4 & 1/2 & 1/4 & 0 \\ 0 & 1/4 & 1/2 & 1/4 \\ 1/4 & 0 & 1/4 & 1/2 \end {pmatrix} = \frac {1}{4} \begin {pmatrix} 2 & 1 & 0 & 1 \\ 1 & 2 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 1 & 0 & 1 & 2 \end {pmatrix}, \] which gives \[ \vect {x}_{n + 1} = \vect {M} \vect {x}_{n}. \]
Notice that \[ \vect {x}_1 = \vect {M} \vect {x}_0 = \frac {1}{4} \begin {pmatrix} 2 & 1 & 0 & 1 \\ 1 & 2 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 1 & 0 & 1 & 2 \end {pmatrix} \begin {pmatrix} 1 \\ 0 \\ 0 \\ 0 \end {pmatrix} = \frac {1}{4} \begin {pmatrix} 2 \\ 1 \\ 0 \\ 1 \end {pmatrix}, \] and hence \(A_1 = \frac {1}{2}, B_1 = \frac {1}{4}, C_1 = 0, D_1 = \frac {1}{4}\).
Also, \[ \vect {x}_2 = \vect {M} \vect {x}_1 = \frac {1}{4} \begin {pmatrix} 2 & 1 & 0 & 1 \\ 1 & 2 & 1 & 0 \\ 0 & 1 & 2 & 1 \\ 1 & 0 & 1 & 2 \end {pmatrix} \cdot \frac {1}{4} \begin {pmatrix} 2 \\ 1 \\ 0 \\ 1 \end {pmatrix} = \frac {1}{16} \begin {pmatrix} 6 \\ 4 \\ 2 \\ 4 \end {pmatrix} = \frac {1}{8} \begin {pmatrix} 3 \\ 2 \\ 1 \\ 2 \end {pmatrix}, \] and hence \(A_2 = \frac {3}{8}, B_2 = \frac {1}{4}, C_2 = \frac {1}{8}, D_2 = \frac {1}{4}\).
We claim that \(B_n = D_n\) for all \(n\) by symmetry, and notice that \[ B_{n + 1} = \frac {1}{4} \cdot (A_n + 2 B_n + C_n) = \frac {1}{4} \cdot (A_n + B_n + C_n + D_n) = \frac {1}{4}, \] and \[ D_{n + 1} = \frac {1}{4} \cdot (A_n + C_n + 2 D_n) = \frac {1}{4} \cdot (A_n + B_n + C_n + D_n) = \frac {1}{4}, \] so that \(B_n = D_n = \frac {1}{4}\) for all \(n \geq 1\). (For \(n = 0\), \(B_n = D_n = 0\)).
Hence, for \(n \geq 1\), we have \[ A_{n + 1} = \frac {1}{4} (2 A_n + B_n + D_n) = \frac {1}{4} \left (2 A_n + \frac {1}{2}\right ) = \frac {1}{2} A_n + \frac {1}{8}, \] which means \[ A_{n + 1} - \frac {1}{4} = \frac {1}{2} \left (A_n - \frac {1}{4}\right ), \] which shows that \(A_n - \frac {1}{4}\) is a geometric sequence with common ratio \(\frac {1}{2}\). The initial term of the geometric sequence is \(A_1 - \frac {1}{4} = \frac {1}{4}\), and hence \[ A_n - \frac {1}{4} = \frac {1}{2^{n + 1}}, \] which shows \(A_n = \frac {1}{4} + \frac {1}{2^{n + 1}}\) for \(n \geq 1\).
Also, \(C_n\) has the same inductive relationship as \(A_n\), the only difference being that the initial term is \(C_1 - \frac {1}{4} = -\frac {1}{4}\), and hence \[ C_n - \frac {1}{4} = -\frac {1}{2^{n + 1}}, \] which shows \(C_n = \frac {1}{4} - \frac {1}{2^{n + 1}}\) for \(n \geq 1\).
Hence, we have \[ \vect {x}_n = \begin {pmatrix} A_n \\ B_n \\ C_n \\ D_n \end {pmatrix} = \begin {cases} \left ( 1, 0, 0, 0 \right )^\intercal , & n = 0, \\ \left ( \frac {1}{4} + \frac {1}{2^{n + 1}}, \frac {1}{4}, \frac {1}{4} - \frac {1}{2^{n + 1}}, \frac {1}{4} \right )^\intercal , & \text {otherwise}. \end {cases} \]