No-Claims Discount: Stationary Distribution Versus a One-Year Forecast
A UK motor insurer operates a three-level no-claims discount (NCD) scheme. Level 1 carries a 0% discount, level 2 a 25% discount and level 3 a 50% discount. At each annual renewal a policyholder makes at most one claim. A policyholder who makes no claim in a year moves up one level, or stays at level 3; a policyholder who claims moves immediately to level 1. The pricing team estimates the probability of a claim in any year as 0.15, independently for every policyholder and every year, and confirms that claims are settled before renewal. New business always enters at level 1. The undiscounted premium is £600 and expenses are 20% of the discounted premium. On 1 January the book contains 10,000 policies, currently distributed as 3,000 at level 1, 3,000 at level 2 and 4,000 at level 3. The pricing team asks for the long-run distribution across levels and wishes to use it to forecast next year's premium income.
1. Write down the one-year transition matrix for a single policyholder and derive the stationary distribution across the three levels. State whether the distribution is reached from any starting state, and why.
Read the worked answer
Order the states 1, 2, 3 (level 1 = 0% discount). A claim always sends the policyholder to state 1 with probability 0.15; no claim sends them one level up (0.85), or keeps them at level 3.
$$P=\begin{pmatrix}0.15&0.85&0\0.15&0&0.85\0.15&0&0.85\end{pmatrix}$$
Solve $\pi P=\pi$ with $\sum\pi_i=1$. Since a claim in any year resets to state 1, $\pi_1=0.15(\pi_1+\pi_2+\pi_3)=0.15$. Then $\pi_2=0.85\pi_1=0.1275$, and $\pi_3=0.85(\pi_2+\pi_3)$, giving $0.15\pi_3=0.85\pi_2$, so $\pi_3=0.7225$. Check: $0.15+0.1275+0.7225=1$.
The chain is irreducible (a claim returns any state to 1; successive no-claims walk upward) and aperiodic (state 3 has a self-loop, and state 1 can be revisited in one or two steps). So the limiting distribution exists and equals $\pi$ whatever the initial state. In fact the chain forgets its past very quickly: $P^2$ row 1 already equals $\pi$, because the state after two years depends only on the claim indicators of those two years, not on the starting level.
2. Using the current distribution of 3,000 / 3,000 / 4,000 across levels 1 to 3, compute the expected premium income for next year. Compute what the answer would be using the stationary distribution instead, and explain the discrepancy.
Read the worked answer
Next year's level depends only on next year's claim indicator and, for levels 2 and 3, on the current level:
- $P(\text{level }1)=0.15$, whatever the current level (a claim resets to 1).
- $P(\text{level }2)=0.85\times0.30=0.255$ (currently at level 1 and no claim).
- $P(\text{level }3)=0.85\times(0.30+0.40)=0.595$ (currently at level 2 or 3 and no claim).
Discounted premiums are £600, £450 and £300, so the average is $0.15(600)+0.255(450)+0.595(300)=90+114.75+178.50=£383.25$. For 10,000 policies, premium income is £3,832,500.
The stationary approach gives $0.15(600)+0.1275(450)+0.7225(300)=90+57.375+216.75=£364.125$, i.e. £3,641,250 - about £191,000 (5%) lower, because $\pi$ puts 72.25% of policyholders at maximum discount whereas the book currently has only 59.5% there after one year. $\pi$ is a limit, not a one-year forecast.
The 20% expense load is not part of premium income; it is £766,500 of outgo, so the premium is only adequate on a claims basis if claims cost less than £3,066,000.
3. The pricing team wants to replace this model with the stationary distribution permanently, arguing that the scheme has been unchanged for six years. Give the strongest objections and say what you would model instead.
Read the worked answer
Three objections matter.
- The book is not a closed chain. New business enters at level 1 every year and lapses exit from whatever level they are on. The level mix of an open book is driven by entry, exit and cohort ageing, so the proportion at level 1 will be higher than the stationary 15% so long as new entrants keep arriving. The stationary distribution applies to a single policyholder followed indefinitely, not to a portfolio with churn.
- The transition is not homogeneous across policyholders. Claim frequencies differ by driver age, area, vehicle and prior claims. If the portfolio is a mixture of chains with different claim probabilities, the portfolio distribution converges to a mixture of the individual stationary distributions, which is not $\pi$ evaluated at the average $p=0.15$; because high-risk policyholders claim more often and sit at level 1 more, the mixture puts more mass at level 1 than the naive calculation suggests. A latent-class or random-effects Markov model is the appropriate structure.
- $p=0.15$ is an estimate, not a constant. If frequency drifts (pricing, weather, claims inflation), $\pi$ shifts at once: $\pi_1=p$, $\pi_2=0.85p$ and $\pi_3=1-p-0.85p$. Quoting a stationary distribution without a sensitivity range gives false precision.
I would model the level mix as an open Markov chain with explicit new-business and lapse vectors, estimated from the last six years of movements, and test homogeneity by fitting separate transition matrices to risk groups.