Past exam of the Probability, statistics and options II course of the University of Oxford / 2009 / M / M5 / b

The necessary parameter range is $0<\lambda \le 1$: the transition probability from zero to one is $\lambda$, and this restriction also bounds all later upward transition probabilities by one.

Every upward transition has positive probability. From any positive state the transition directly to zero has positive probability, since $1-\lambda /(i+1)>0$. Thus any state can reach zero and then reach any other state by successive upward steps, proving irreducibility.

If $\lambda <1$, the holding probability at zero is positive, so its period and hence the common period of the irreducible chain is one. At $\lambda =1$ that holding probability vanishes, but zero has positive-probability return paths $0\to 1\to 0$ of length two and $0\to 1\to 2\to 0$ of length three. Their lengths have greatest common divisor one. Thus $\boxed{\text{the chain is irreducible and aperiodic for all }0<\lambda \le 1}$.

For $j\ge 1$ the only entrance is from $j-1$, so stationarity requires ${\pi }_j={\pi }_{j-1}\lambda /j$. Induction gives ${\pi }_j={\pi }_0{\lambda }^j/j!$, and normalization using the exponential series gives ${\pi }_0=e^{-\lambda }$. HenceThe remaining zero-state equation also holds: ${\sum }_i{\pi }_i\lambda /(i+1)=e^{-\lambda }(e^{\lambda }-1)=1-{\pi }_0$, so ${\sum }_i{\pi }_i(1-\lambda /(i+1))={\pi }_0$. Thus this normalized Poisson law really is stationary, not just a solution of the positive-state equations. Irreducibility makes the stationary probability law unique.