No claims discount system

A policyholder's future insurance premiums depend on a finite discount level, updated according to whether a claim is submitted in each year. With independent yearly accidents and repair costs, and a claim decision depending only on the current level and cost, the level is a Discrete-time Markov chain. Accident probability and submitted-claim probability differ when the policyholder pays small repair costs without claiming.

In a no claims discount system with discounts $0,\alpha ,\beta$, a reported claim resets next year's discount to zero. With no later accidents, the next two premiums after reporting are $C+(1-\alpha )C$. At the zero level, not reporting instead gives $(1-\alpha )C+(1-\beta )C$; at either other level it gives $2(1-\beta )C$. Their differences give the displayed repair-cost thresholds. A fully reimbursed loss above its threshold should be reported under this undiscounted two-year rule. Current premiums cancel. Different thresholds generally produce state-dependent submitted-claim probabilities, even with a common accident mechanism.

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.

Suppose annual claim counts are independent with geometric distribution mass $p{q}^j$, $j\ge 0$. In a no claims discount system with discounts $0,\alpha ,\beta$, a claim returns either of the first two levels to zero; at the top, one claim sends the level to $\alpha$ and two or more send it to zero. The transition matrix is $\left(\begin{array}{ccc}q& p& 0\\ q& 0& p\\ q^2& pq& p\end{array}\right)$. Solving $\pi P=\pi$ gives the displayed stationary distribution. With full premium $c$, the stationary expected value of the premium is $c[1-(\alpha pq+\beta p^2)/(1-p^{2}q)]$. Independent annual sampling is essential; a persistent random intensity does not justify this marginal transition matrix.

For a three-level birth-death chain whose down-or-stay claim probabilities are $p_1,p_2,p_3$, with each in $(0,1)$, detailed balance is ${\pi }_1(1-p_1)={\pi }_2{p}_2$ and ${\pi }_2(1-p_2)={\pi }_3{p}_3$. Normalizing gives the displayed stationary distribution. The finite chain is irreducible and has endpoint holding probabilities, so it is aperiodic and its level proportions converge to this distribution from any initial level.

Let $b$ be the undiscounted annual premium and $d_i$ the discount fraction at level $i$. Compare the next two levels $i_r^0$ after no current claim with $i_r^1$ after a current claim, assuming no later accidents. Their total additional premium is the displayed threshold. A repair cost above it makes a claim advantageous under this two-year rule. For discounts $0,\alpha ,\beta$ and one-level moves with reflecting endpoints, the thresholds at levels one, two, and three are respectively $b\beta$, $b(2\beta -\alpha )$, and $b(\beta -\alpha )$. The premium scale must be specified or fixed as the monetary unit.

For a no claims discount system with discounts $0,\alpha ,\beta$, assume a policyholder starts at $\beta$, the first claim lowers next year's discount to $\alpha$, and a second claim in the same year lowers it to zero. With no further losses and undiscounted future premiums, claiming for the first loss costs $c(\beta -\alpha )$ in extra premium. Once that claim is submitted, the second claim adds $c\alpha$ next year and $c(\beta -\alpha )$ the year after, a total $c\beta$. Thus the second decision uses the incremental threshold, not the cost of both claims relative to no claim. For independent unit-rate exponential distribution losses the probabilities are $e^{-c(\beta -\alpha )}$ and $e^{-c\beta }$, respectively; the latter remains the conditional probability given submission of the first claim.