Past exam of the Probability, statistics and options II course of the University of Oxford / 2007 / M / 5

A Poisson process of rate $\lambda$ is a counting process starting at zero, with independent stationary increments, such that an increment over an interval of length $t$ has the Poisson distribution of parameter $\lambda t$. It has right-continuous paths with unit jumps almost surely.

Independence of the two processes makes increments of $A+B$ independent on disjoint intervals. The sum of independent Poisson random variables with parameters $\alpha t$ and $\beta t$ is Poisson with parameter $(\alpha +\beta )t$, as follows by convolution:Thus $A+B$ is a Poisson process of rate $\alpha +\beta$. Its first red and green waiting times are independent exponential random variables of rates $\alpha ,\beta$, so

The state-zero stationary distribution equation is ${\pi }_0=q{\pi }_0+q{\pi }_1$, so ${\pi }_1=(p/q){\pi }_0$. The state-one equation is ${\pi }_1=p{\pi }_0+q{\pi }_2$, giving ${\pi }_2=(p/q){\pi }_1$. More generally, the stationary probability flow across the cut $\{0,\dots ,i\}$ is balanced, so $p{\pi }_i=q{\pi }_{i+1}$. This also follows inductively from ${\pi }_i=p{\pi }_{i-1}+q{\pi }_{i+1}$.

Hence ${\pi }_i=(p/q{)}^i{\pi }_0$. Since $p<q$, the geometric series is summable, and normalization givesThese probabilities satisfy detailed balance and therefore all the stationarity equations. The Markov chain is an irreducible Markov chain, so this normalized stationary distribution is unique.