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

All entries must lie between zero and one. The tightest restriction comes from $p_{01}=\lambda$, so together with the given positivity the parameter range is

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.

For $j\ge 1$, the only incoming transition is from $j-1$, so a stationary distribution must satisfyNormalization gives ${\pi }_0{e}^{\lambda }=1$, henceThe zero-state balance also holds: ${\sum }_i{\pi }_i(1-\lambda /(i+1))=1-{\sum }_{j\ge 1}{\pi }_j={\pi }_0$. Thus the Poisson distribution is indeed stationary, including when $\lambda =1$. Irreducibility gives uniqueness.