Let $\mathcal{A} = \{a,b\}$ be an alphabet, and define the random substitution $$ \begin{align*} \theta(a) &= \begin{cases} aa & \text{with probability }p, \\ bb & \text{with probability }1-p, \end{cases}\\ \theta(b) &= aa \qquad \text{with probability }1. \end{align*}$$
The associated incidence matrix is then $$M = \begin{pmatrix} 2p & 2(1-p) \\ 2 & 0 \end{pmatrix}$$
which is primitive and expanding. It has eigenvalues $2$ and $2(1-p)$ so the Perron-Frobenius eigenvalue is $2$ and the Perron-Frobenius eigenvector, normalised to be a probability vector, is $$\pi = \begin{pmatrix} \frac{1}{2} \\ \frac{1}{2} \end{pmatrix}.$$
This means that as $n \to \infty$, the proportion of $a$s in $\theta^n(a)$ tends to $\frac{1}{2}$, and same for $b$ (see https://arxiv.org/abs/1810.11122 for a proof of this).
But this is clearly nonsense: the proportion of $a$s should be $p$. Set $p=0.99$ and do some trial runs on a computer and it should be clear that almost all of the letters after $n$ iterations are $a$, with the proportion tending towards 99%.
Where have I gone wrong?
My goal with this exercise was to try to find a random substitution whose incidence matrix depends on the probabilities but whose Perron-Frobenius eigenvector doesn't. I have succeeded with the substitution above, but the frequency as $n \to \infty$ just doesn't make sense.
It's just a guess (I glanced at the paper but did not read it carefully: the pain/gain ratio seemed unfavorable) but it seems likely that the matrix in question should be transposed, so "right" eigenvector appears where one might expect left eigenvector, and so on.