Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 2 / 26I / e / Solution

If $\vartheta >1$, the general level solution is a constant multiple of $(1,1{)}^T$ plus a multiple of ${\vartheta }^i(b,1{)}^T$. Boundedness of probabilities removes the growing component, and $h_{W_0}=1$ then fixes the constant to one. If $\vartheta =1$, the nonzero Jordan part produces linear growth unless the initial vector lies in the one-dimensional eigenspace; the same conclusion follows. ThusStarting at $W_0$, the chain leaves to $C_0$ and returns with probability one. An irreducible chain with a recurrent state has all states recurrent, so the chain is recurrent in both cases.