Generating function of a stochastic matrix

This is an example demonstrating my answer to this math.stackexchange post: Let {\{X_n:n\in\mathbb N_0\}} be a Markov chain on {\{0,1\}} with transition matrix

\displaystyle P = \begin{pmatrix}1-\alpha&\alpha\\\beta&1-\beta\end{pmatrix}

where {\alpha,\beta\in(0,1)}. Then {P} is irreducible and aperiodic, so there exists a unique stationary distribution {\pi} satisfying

\displaystyle \lim_{n\rightarrow\infty}(P^n)_{i\cdot} = \pi,\quad i=1,2

where {(P^n)_{i\cdot} } denotes the {i^{\mathrm{th}}} row of {P^n}. From the global balance equation {\alpha\pi_0 = \beta\pi_1} we see that {\pi_1 = \frac\alpha\beta\pi_0}, and normalization yields

\displaystyle \pi = \left(\frac\beta{\alpha+\beta},\frac\alpha{\alpha+\beta} \right).

To actually compute {P^n}, we could write {P^n=AD^nA^{-1}} where the columns of {A} are linearly independent eigenvectors of {P} and {D} is a diagonal matrix whose entries are the corresponding eigenvalues. This is possible because the Perron-Frobenius theorem implies that the eigenspace corresponding to the eigenvalue {1} is one-dimensional, and the rank of {P} is two. Instead, we will compute the generating function of {P}. Let

\displaystyle P(s) = \sum_{n=0}^\infty P^n s^n,


\displaystyle  \begin{aligned} P(s) &= (I-sP)^{-1}\\ &= \frac{1}{\det (I-sP)}\mathrm{adj} A\\ &= \frac1{(1-s)(1-(1-\alpha-\beta)s)}\begin{pmatrix}1-(1-\beta)s&\alpha s\\\beta s& 1-(1-\alpha)s\end{pmatrix}. \\ &= \begin{pmatrix}P_{00}(s) & P_{01}(s)\\ P_{10}(s) & P_{11}(s), \end{pmatrix} \end{aligned}


\displaystyle P_{ij}(s):=\sum_{n=0}^\infty \mathbb P(X_n=j\mid X_0=i)s^n.

Partial fraction decomposition yields

\displaystyle  \begin{aligned} &\quad\frac1{(1-s)(1-(1-\alpha-\beta)s}\\ &= (\alpha+\beta)^{-1}\frac1{1-s} + \left(1-(\alpha+\beta)^{-1}\right)\frac1{1-(1-\alpha-\beta)s}, \end{aligned}

and so we have

\displaystyle  \begin{aligned} P_{00}(s) &= \left(\frac{1-(1-\beta)s}{\alpha+\beta}\right)\left(\frac1{1-s} + \frac1{1-(1-\alpha-\beta)s}\right)\\ &= \left(\frac{1-(1-\beta)s}{\alpha+\beta}\right) \left( \sum_{n=0}^\infty s^n + \sum_{n=0}^\infty (1-\alpha-\beta)^ns^n\right)\\ &= \sum_{n=0}^\infty \left(\frac\beta{\alpha+\beta}+\frac{\alpha(1-\alpha-\beta)^n}{\alpha+\beta}\right) s^n. \end{aligned}

Similar computations yield

\displaystyle  \begin{aligned} P_{01}(s) &= \sum_{n=0}^\infty \left(\frac\alpha{\alpha+\beta}-\frac{\alpha(1-\alpha-\beta)^n}{\alpha+\beta}\right) s^n,\\\\ P_{10}(s) &= \sum_{n=0}^\infty \left(\frac\beta{\alpha+\beta}-\frac{\beta(1-\alpha-\beta)^n}{\alpha+\beta}\right) s^n,\\\\ P_{11}(s) &= \sum_{n=0}^\infty \left(\frac\alpha{\alpha+\beta}+\frac{\beta(1-\alpha-\beta)^n}{\alpha+\beta}\right) s^n, \end{aligned}

and it follows that

\displaystyle  P(s) = \begin{pmatrix} \pi_0(1+(1-\alpha-\beta)^n) & \pi_1 (1-(1-\alpha-\beta)^n) \\ \pi_0(1-(1-\alpha-\beta)^n) & \pi_1(1+(1-\alpha-\beta)^n)\end{pmatrix} s^n.

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