Stationary law of a two-state chain

The matrix $P$ is a transition matrix: all four entries are nonnegative because $0<a,b\le 1$, and each row sums to one, $(1-a)+a=1$ and $b+(1-b)=1$.

The chain is irreducible: $P(0,1)=a>0$ and $P(1,0)=b>0$, so $0$ and $1$ communicate in one step each way.

The vector $\pi :=(b\mathrm{/}(a+b),a\mathrm{/}(a+b))$ is a probability vector: $a+b>0$ and both coordinates are positive, with $\frac{b}{a+b}+\frac{a}{a+b}=1$.

The vector $\pi$ is invariant. At state $0$: $(\pi P)(0)=\pi (0)(1-a)+\pi (1)b=\pi (0)-a\pi (0)+b\pi (1)$ and $a\pi (0)=ab\mathrm{/}(a+b)=b\pi (1)$, so this equals $\pi (0)$; at state $1$: $(\pi P)(1)=\pi (0)a+\pi (1)(1-b)=\pi (1)+(a\pi (0)-b\pi (1))=\pi (1)$. Hence $\pi P=\pi$, which is invariance by [F2].

Uniqueness: by [F3] the finite chain has an invariant probability, so by the equivalence [F4] the irreducible chain is positive recurrent, and [F5] then gives that it has exactly one invariant probability.

Combining steps 2.1 and 2.2, the unique stationary law of the chain is $\pi =(b\mathrm{/}(a+b),a\mathrm{/}(a+b))$.

Boundary and scope cases: at $a=b=1$ the matrix is $\left(\begin{array}{cc}0& 1\\ 1& 0\end{array}\right)$, the chain alternates deterministically, $\pi =(1\mathrm{/}2,1\mathrm{/}2)$, and the formula is unaffected by the period; at $a=1$, $b<1$ the matrix has $P(0,1)=1$ and the formula still gives a positive probability vector; if $a$ or $b$ were $0$ the chain would fail to be irreducible and the argument for uniqueness through [F5] would not apply, so the strict positivity of $a$ and $b$ is used exactly in step 1.2; the verification checks both rows of the stationarity equations rather than only the first; and the objects are determined by the two given parameters, so steps 1.1–2.1 are choice-free while the uniqueness argument of step 2.2 spends the axiom [A1] exactly through the AC-carrying suppliers [F4] and [F5], whose statements assume Choice.