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Mix the conditional Poisson distribution over the rate-$\nu$ exponential distribution intensity. The Gamma integral giveswhere $p=\nu /(1+\nu )$ and $q=1/(1+\nu )$. Thus the annual claim count has the zero-based geometric distribution, with probabilities $\mathbb{P}(N=0)=p$, $\mathbb{P}(N=1)=pq$ and $\mathbb{P}(N\ge 2)=q^2$.

For the intended Markov chain calculation, take these annual counts to be independent. Order the states by discounts $0,\alpha ,\beta$. From either of the first two states, a positive count returns the policy to zero, while a zero count raises it by one level. From the top, two or more claims send it to zero, exactly one sends it to $\alpha$, and no claim keeps it at $\beta$. Hence the transition matrix isWrite its stationary distribution as $({\pi }_0,{\pi }_{\alpha },{\pi }_{\beta })$. The final two balance equations give $q{\pi }_{\beta }=p{\pi }_{\alpha }$ and ${\pi }_{\alpha }=p{\pi }_0+pq{\pi }_{\beta }$. It follows that ${\pi }_{\alpha }=p{\pi }_0/[q(1+p)]$ and ${\pi }_{\beta }=p^2{\pi }_0/[q^2(1+p)]$. Normalizing yields the three-level discount equilibrium with geometric annual counts:The Markov chain is irreducible for $0<p,q<1$ and has a self-loop, so it is aperiodic and this equilibrium is the limiting distribution. Weighting the three premiums $c,c(1-\alpha ),c(1-\beta )$ givesThere is a necessary qualification to the temporal model: the one-year mixture alone does not specify independence between years. A persistent claim intensity can destroy the discount Markov property. If one fixed $\Lambda$ is drawn for the policyholder and annual counts are conditionally independent given it, then histories $0,\alpha ,\beta ,\beta$ and $0,0,\alpha ,\beta$ both end at the top after three years but give different next-year no-claim probabilities. The first history is three zero counts, so its posterior intensity has exponential rate $\nu +3$ and next no-claim probability $(\nu +3)/(\nu +4)$. The second is a positive count followed by two zero counts; its posterior density is proportional to $(1-e^{-\lambda })e^{-(\nu +2)\lambda }$, giving next no-claim probability $(\nu +2)/(\nu +4)$. Thus the displayed matrix and premium require independent annual marginal sampling, for example a fresh intensity each year. They are not valid for a persistent unobserved policy intensity without conditioning on that intensity.

Under the persistent-intensity interpretation, one can instead condition on $\lambda$. Put $a_0=e^{-\lambda }$ and $b_1=\lambda e^{-\lambda }$. The conditional transition matrix has rows $(1-a_0,a_0,0)$, $(1-a_0,0,a_0)$ and $(1-a_0-b_1,b_1,a_0)$. Its stationary distribution isThe corresponding population limiting expected value of the premium is $c{\int }_0^{\infty }[1-\alpha {\pi }_{\alpha }(\lambda )-\beta {\pi }_{\beta }(\lambda )]\nu e^{-\nu \lambda }d\lambda$. This gives the correct alternative when the intensity persists, and makes the extra assumption behind the requested marginal matrix explicit.