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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.

Record the discount level at the start of each policy year as the state. A no claims discount system is a Discrete-time Markov chain when, conditional on this state, the next year's claim behavior and update rule do not depend on earlier history. Independent annual accident and loss mechanisms with a fixed state-based reporting rule give this property; one should not equate the probability of an accident with the probability of a submitted claim.

The density identifies $Z=(\log L-\mu )/\sigma$ as a standard normal random variable: substitute $x=e^{\mu +\sigma z}$ and $dx=\sigma xdz$. Therefore, for $a>0$,Here $L$ is the numerical repair cost. The character preceding $L$ in the PDF is a currency symbol, not a separate multiplier.

Order states as $0\%,20\%,30\%$. With the stipulated common annual probability $p$ of no submitted claims, the transition matrix isA claim resets any state to zero, while a no-claim year raises the discount one level or keeps the top level fixed. If claim probabilities depend on the state, the corresponding row instead uses $p_i=\mathbb{P}(\text{no submitted claim}\mid \text{state }i)$; the same transition mechanism applies. The supplied single $p$ is a simplifying model assumption.

For a policyholder currently at zero discount, the current premium $C$ has the same value under either decision and cancels in the comparison. If the repair is self-funded and no later accidents occur, the next two premiums are $0.8C$ and $0.7C$, giving a total cost $L+2.5C$ including the current premium. If a claim is made, the next two premiums are $C$ and $0.8C$, giving total cost $2.8C$, with the repair paid by the insurer. Thus the two-year no-claims threshold with reset to zero isAt equality the holder is indifferent; it has probability zero under the continuous density. This assumes full reimbursement and no excess or deductible, as in the stated model.

Put $a=0.3C$. The conditional expected claim is $\mathbb{E}[L\mid L>a]$. Completing the square in the normal integral gives the truncated lognormal momentDividing by the reporting probability gives the expected size of a submitted claim for the full-premium holder:All monetary amounts must use the same unit despite the PDF's mixed currency notation. The other two current levels both have next-two-year no-claim premiums $0.7C+0.7C$, versus $C+0.8C$ after a claim, so their thresholds are $0.4C$. Consequently incorporating this behavioral decision into claim probabilities generally produces state-specific $p_i$, rather than deriving a common $p$ from the severity density alone; no accident-frequency law is provided.

Let $Z_m$ be the discount level during policy year $m$, with $Z_0=1$. Under the usual assumption of independent successive annual accidents and repair costs with the same laws each year, the submitted-claim decision depends only on $Z_m$ and that year's cost. Consequently the conditional distribution of $Z_{m+1}$ depends on the past only through $Z_m$: this is the Markov property. The no claims discount system is therefore a time-homogeneous Discrete-time Markov chain. Stationarity of accident probabilities alone, without this independence assumption, would not by itself establish the Markov property.

At level one a claim keeps the policyholder at one, while no submitted claim moves them to two. At level two, a submitted claim moves them to one and no claim to three. At level three, a claim moves them to two and no claim keeps them at three. Thus $p_i$ in the specified transition matrix is the annual probability of a submitted claim at current level $i$, not merely of an accident. A small accident paid privately counts as no claim for these transitions.

Let $b>0$ denote the undiscounted annual premium. It is not specified numerically in the source; costs can instead be measured in units of that premium, giving $b=1$. The actual premium payments at the three levels are $b$, $b(1-\alpha )$, and $b(1-\beta )$. Compare the two future paths, assuming no further accidents, to obtain the two-year premium-loss claim threshold. From level one, a current claim leads to future levels $(1,2)$, whereas no claim leads to $(2,3)$, giving extra premiumFrom level two the two paths are $(1,2)$ and $(3,3)$, and from level three they are $(2,3)$ and $(3,3)$. HenceThese are all strictly positive. The PDF's discount fractions are $\alpha$ and $\beta$; the converted TeX has lost these factors in the percentages.

For a repair cost $X$ with the specified lognormal distribution, $\log X\sim N(\mu ,{\sigma }^2)$. Putwhere $\Phi$ is the standard normal distribution function. Conditional on an accident, the submitted-claim probability is $\mathbb{P}(X>h_i)=\overline{\Phi }(z_i)$. Multiplying by the accident probability givesSet $b=1$ for the premium-unit convention. Without a monetary premium scale, discount percentages alone cannot determine a numerical threshold for a monetary repair cost.

This nearest-neighbour Markov chain is a finite birth-death chain. Its stationary distribution $\pi$ satisfies the endpoint balance equationsThese detailed balance equations also imply the middle-state balance equation. The three-level no claims discount equilibrium is thereforewhereSince each $p_i$ lies strictly between zero and one, this is an irreducible Markov chain. Positive holding probabilities at the endpoints also make it an aperiodic Markov chain. Its unique stationary distribution is the limiting distribution even from the initial level one, and gives the steady-state proportions for a homogeneous population of such policyholders.

A submitted claim at level one conditions the repair cost on $X>h_1$. The accident probability cancels in this conditional expectation. For $z_1=(\log h_1-\mu )/\sigma$, the truncated lognormal moment follows by completing the square:Dividing by $\mathbb{P}(X>h_1)=\overline{\Phi }(z_1)$ yields the expected size of an actual submitted claim:The unconditional expected amount paid per year from level one would instead multiply its tail first moment by $p$; it is not the conditional claim size requested here.

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.

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.