Past exam of the options II course of the University of Oxford / 2009 / M / M5 / b / i / Solution

Every finite upward path has positive probability, since each upward transition probability is positive. Every state $i\ge 1$ has a positive transition probability to zero because $\lambda /(i+1)<1$. From zero one can therefore reach every state, and from any state return to zero and then reach any other. Hence the chain is irreducible.

If $\lambda <1$, state zero has a positive self-loop, so its period is one. If $\lambda =1$, that self-loop disappears, but the paths $0\to 1\to 0$ and $0\to 1\to 2\to 0$ have positive probabilities and lengths two and three. Their greatest common divisor is one. Thus zero has period one in this boundary case too; irreducibility makes the whole chain aperiodic.