Bonus-Malus System

Study guide: Exam MAS-I

A Bonus-Malus System (BMS) is a merit-rating scheme, most common in motor insurance, in which each policyholder occupies one of a finite set of premium levels and moves to a cheaper level (bonus) after a claim-free year or a dearer one (malus) after claims. When a policyholder's annual claim counts are independent with a fixed distribution, the level is a Markov Chain.

$P_{ij} = \sum_{k \,:\, s_i(k) = j} a_k$

$\text{Long-run average premium} = \sum_j \pi_j\, b_j$

$s_i(k)$ is the level reached from level $i$ after a year with $k$ claims, $a_k$ the probability of $k$ claims (for a Poisson rate $\lambda$ , $a_k = e^{-\lambda}\lambda^k/k!$ ), $b_j$ the premium at level $j$ and $\pi_j$ the stationary probability of level $j$ . The rules and the claim distribution together fix the Transition Probability Matrix.
A BMS is a coarse form of Experience Rating: it reacts to claim counts, not amounts, and is justified by heterogeneity in $\lambda$ across drivers — a Mixing Distribution. For one driver with a fixed $\lambda$ and independent yearly claim counts, the level is a homogeneous Markov chain.
Hunger for bonus: a reported claim costs future discounts, so policyholders rationally pay small losses themselves. The insurer's data then under-count small claims, which the pricing actuary has to allow for.
Separation can be weak. Because levels respond only to counts and reset quickly, long-run premiums differ far less than claim frequencies do — see the first example.

Long-Run Premium in a Three-Level System Levels 0, 1 and 2 carry premiums $\$500$ , $\$400$ and $\$300$ . A claim-free year moves a driver up one level (to a maximum of 2); any claim returns it to level 0. Claims are Poisson with $\lambda = 0.2$ per year. Find the stationary distribution and the long-run average premium.
Answer Let $p = e^{-0.2} = 0.8187$ (claim-free) and $q = 1 - p = 0.1813$ :

$\mathbf{P} = \begin{array}{c|ccc} & 0 & 1 & 2 \\ \hline 0 & q & p & 0 \\ 1 & q & 0 & p \\ 2 & q & 0 & p \end{array}$

Every row sends probability $q$ to level 0, so $\pi_0 = q$ ; level 1 is reached only from level 0; level 2 takes the rest:
 $\begin{align*} \pi_0 &= q \\ &= 0.1813 \\ \pi_1 &= p\,q \\ &= 0.1484 \\ \pi_2 &= p^2 \\ &= 0.6703 \end{align*}$ 
 $\begin{align*} \text{Average premium} &= 500(0.1813) + 400(0.1484) + 300(0.6703) \\ &= 90.63 + 59.36 + 201.10 \\ &= \$351.09 \end{align*}$ 
Repeating with $\lambda = 0.5$ gives $\boldsymbol{\pi} = (0.3935, 0.2387, 0.3679)$ and $\$402.56$ . A driver with $2.5$ times the claim frequency pays only about $15\%$ more in the long run.

Should a Small Loss Be Reported? In the same system ( $\lambda = 0.2$ ), a driver at level 2 has an at-fault loss of $\$250$ on the last day of an otherwise claim-free year. Reporting it puts the driver at level 0 next year; absorbing it keeps level 2. Ignoring interest, compare the expected extra premium from reporting with the loss.
Answer Year 1: reporting costs $500 - 300 = \$200$ extra for certain.

Year 2: if year 1 is claim-free (probability $p$ ), the reporting path is at level 1 ( $\$400$ ) and the other at level 2 ( $\$300$ ); if year 1 has a claim, both paths are at level 0. From year 3 on, the paths coincide.
 $\begin{align*} E[\text{extra premium}] &= 200 + 100\,p \\ &= 200 + 100(0.8187) \\ &= \$281.87 \end{align*}$ 
Reporting the $\$250$ loss would cost about $\$282$ in expected premium, so the driver keeps it. Any loss below roughly $\$282$ goes unreported — hunger for bonus in one line.