= Persistent claim intensity can destroy the discount Markov property
Let a policyholder have one fixed <exponential distribution> intensity $\Lambda$ of rate $\nu$, with conditionally independent annual <Poisson distribution> counts. Its one-year marginal count is geometric, but averaging that count does not make the discount process a <Markov chain>. In the three-level rule, start at zero discount. At time three, the histories $0,\alpha,\beta,\beta$ and $0,0,\alpha,\beta$ both end at the same top level. The first means three no-claim years, giving next no-claim probability $(\nu+3)/(\nu+4)$. The second means a positive count followed by two zero counts, giving $(\nu+2)/(\nu+4)$: divide the integrals of $(1-e^{-\lambda})e^{-(\nu+3)\lambda}$ and $(1-e^{-\lambda})e^{-(\nu+2)\lambda}$. These are distinct, so the <Markov property> fails when only the current discount is recorded. Conditioning on the intensity, or recording its <Bayesian posterior>, restores the required predictive information.
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