I am reading the paper and try to understand the example solution in p. 14.

In particular, if the Markov chain has stationary distribution $\pi$ and $a$-step transition distribution $P^a$, then $$β(a) = \int π(dx)||P^a(x) − π||_{TV}. \tag{18} $$ Consider first the two-state Markov chain $S_t$ pictured in Figure 2. By direct calculation using (18), the mixing coefficients for this process are $β(a) = \frac {4}{9} (\frac{1}{2})^a$. enter image description here

Also, I cited the

Definition 1 ($β$-mixing). For each $a \in N$ and any $t \in Z$, the coefficient of absolute regularity, or $β$-mixing coefficient, $β(a)$, is $$β(a) :=||P^t_{−∞} ⊗ P^∞_{t+a} - P_{t,a}||_{TV} \tag{1}$$ where $|| · ||_{TV}$ is the total variation norm, and $P_{t,a}$ is the joint distribution of $(X^t_{−∞}, X^∞_{t+a})$. A stochastic process is said to be absolutely regular, or $β$-mixing, if $β(a) → 0$ as $a → ∞$.

My question. How to obtain the $β(a) = \frac {4}{9} (\frac{1}{2})^a$?

My attempt is.

P = matrix(c(1/2, 1, 1/2, 0),,2) # transition matrix
MyStates <- c("A", "B")
M <- matrix(data = P, byrow = FALSE, nrow = 2, 
            dimnames = list(MyStates, MyStates))

mcM <- new("markovchain", transitionMatrix = P)
is.irreducible(mcM) # TRUE
steadyStates(mcM)   # stationary distribution
#              1         2
# [1,] 0.6666667 0.3333333


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