Solution
= Solution
{c}
Within the class $\{1,2\}$, the two-step <matrix> has <eigenvalues> one and $1/9$, with stationary distribution $(1/2,1/2)$. Diagonalizing this symmetric <matrix>, or solving the recursion $p_{n+1}=4/9+p_n/9$ from $p_0=0$, gives
$$
\mathbb P(X_{2n}=2\mid X_0=1)=\frac12\left[1-\left(\frac19\right)^n\right].
$$
Consequently \b[the limit is $1/2$]. Restriction to even times is essential because the original chain has period two.