The Cramér-Lundberg Surplus Process
The classical risk model describes the insurer’s surplus at time $t$ as:
\[U(t) = u + ct - S(t), \quad t \ge 0\]where:
- $u \ge 0$ is the initial surplus
- $c > 0$ is the premium rate (received continuously)
- $S(t) = \sum_{i=1}^{N(t)} X_i$ is the aggregate claims process
- $N(t) \sim \text{Poisson}(\lambda t)$ is the claim count (Poisson process with rate $\lambda$)
- $X_i$ are i.i.d. claim severities with distribution $F_X$, mean $\mu_X = E[X]$, independent of $N$
The net profit condition (loading condition) ensures the surplus drifts upward on average:
\[c > \lambda \mu_X \quad\Longleftrightarrow\quad \theta = \frac{c - \lambda\mu_X}{\lambda\mu_X} > 0\]where $\theta$ is the relative security loading. Without this condition, ruin is certain ($\psi(u) = 1$ for all $u$).
The time of ruin is $\tau = \inf{t \ge 0 : U(t) < 0}$ (with $\tau = \infty$ if ruin never occurs). The ruin probability starting from surplus $u$ is:
\[\psi(u) = P(\tau < \infty \mid U(0) = u)\]The Cramér-Lundberg Equation and Adjustment Coefficient
The moment generating function (MGF) of the claim severity is $M_X(r) = E[e^{rX}]$. The Cramér-Lundberg equation (also called the characteristic equation) is:
\[\lambda + cr = \lambda M_X(r)\]or equivalently, seeking the Laplace exponent:
\[\lambda M_X(r) - \lambda - cr = 0\]This equation has exactly two real roots: $r = 0$ (trivial) and $r = R > 0$, where $R$ is the adjustment coefficient (Lundberg exponent). Graphically, the convex function $\lambda M_X(r)$ starts at $\lambda$ with slope $\lambda\mu_X < c$ (by the loading condition), so it must cross the line $\lambda + cr$ at a positive point $R$.
For exponential claims $X \sim \text{Exp}(\beta)$, $M_X(r) = \beta/(\beta - r)$ for $r < \beta$, and the Cramér-Lundberg equation gives:
\[\frac{\lambda}{\beta - R} = \frac{\lambda + cR}{\lambda} \implies R = \frac{\beta\theta}{1 + \theta} = \beta - \frac{\lambda}{c}\]Ruin Probability Bounds and Exact Formulas
Lundberg’s inequality gives an exponential upper bound:
\[\psi(u) \le e^{-Ru}\]This bound is remarkably simple: the larger the adjustment coefficient $R$, the faster ruin probability decays with initial surplus. The adjustment coefficient can be maximized (over reinsurance structures) to obtain the optimal retention problem.
For exponential claim sizes $X \sim \text{Exp}(\beta)$, the exact formula is:
\[\psi(u) = \frac{\lambda}{c\beta}\,e^{-Ru} = \frac{1}{1+\theta}\,e^{-Ru}\]where $R = \beta - \lambda/c$. The prefactor $1/(1+\theta)$ is the probability of ruin starting from $u = 0$.
For mixtures of exponentials $F_X(x) = 1 - \sum_{i=1}^n \alpha_i e^{-\beta_i x}$, the ruin probability is a sum of exponentials:
\[\psi(u) = \sum_{i=1}^n A_i\,e^{-r_i u}\]where $r_1, \ldots, r_n$ are the $n$ positive roots of the Cramér-Lundberg equation (which now has $n+1$ roots in total including 0).
For general claim sizes, the Pollaczek-Khinchine formula expresses ruin probability as a geometric compound:
\[\psi(u) = (1 - \rho)\sum_{n=1}^\infty \rho^n [1 - F_e^{*n}(u)]\]where $\rho = \lambda\mu_X/c$ is the traffic intensity and $F_e(x) = \mu_X^{-1}\int_0^x [1-F_X(t)]\,dt$ is the equilibrium distribution of $F_X$. The sum $\sum_{n=1}^\infty \rho^n = \rho/(1-\rho)$ when $u \to \infty$.
Cramér-Lundberg Asymptotics
As $u \to \infty$, the ruin probability satisfies:
\[\psi(u) \sim C\,e^{-Ru}\]where the constant $C$ is:
\[C = \frac{\lambda\mu_X - c + c}{\ldots} = \frac{1 - \rho}{M_X'(R) - c/\lambda} \cdot \frac{1}{\lambda}\]More precisely, if $M_X(R) < \infty$ (light-tailed claims), then:
\[\lim_{u \to \infty} e^{Ru}\,\psi(u) = \frac{c - \lambda\mu_X}{c\,R\,\text{Var}_e(X)/2 + \ldots} = \frac{\theta}{(1+\theta)\,(\lambda M_X''(R)c^{-1} - 1)}\]For heavy-tailed claims (no MGF exists, e.g., Pareto), there is no adjustment coefficient and the bound $\psi(u) \le e^{-Ru}$ breaks down. Instead:
\[\psi(u) \sim \frac{\lambda}{c-\lambda\mu_X}\int_u^\infty [1 - F_X(x)]\,dx = \frac{1}{\theta\mu_X}\bar{F}_e(u)\]The Gerber-Shiu Expected Discounted Penalty Function
The Gerber-Shiu function (1998) unifies many ruin-related quantities in a single expectation:
\[\phi(u) = E\!\left[e^{-\delta\tau}\,w(U(\tau^-), |U(\tau)|)\,\mathbf{1}(\tau < \infty) \;\Big|\; U(0) = u\right]\]where:
- $\delta \ge 0$ is a discounting rate (or Laplace parameter)
-
$w(x, y)$ is a penalty function depending on the surplus just before ruin $U(\tau^-)$ and the deficit at ruin $ U(\tau) $ -
$U(\tau^-)$ is the “last record” surplus; $ U(\tau) $ is the “severity of ruin”
Special cases:
| Choice of $w$ and $\delta$ | $\phi(u)$ equals | ||
|---|---|---|---|
| $w \equiv 1$, $\delta = 0$ | $\psi(u)$ — ruin probability | ||
| $w \equiv 1$, $\delta > 0$ | $E[e^{-\delta\tau}\mathbf{1}(\tau<\infty)]$ — Laplace transform of ruin time | ||
| $w(x,y) = y$, $\delta = 0$ | $E[ | U(\tau) | \mathbf{1}(\tau<\infty)]$ — expected deficit |
| $w(x,y) = \mathbf{1}(y > d)$, $\delta = 0$ | $P(\tau < \infty,\, | U(\tau) | > d)$ |
The Gerber-Shiu function satisfies an integro-differential equation:
\[c\phi'(u) = (\lambda + \delta)\phi(u) - \lambda\int_0^u \phi(u-x)f_X(x)\,dx - \lambda\int_u^\infty w(u, x-u)f_X(x)\,dx\]This reduces to an IDE that can be solved by Laplace transforms for exponential and phase-type claim distributions, or by numerical methods for general $F_X$.
Finite-Time Ruin Probability
The finite-horizon ruin probability $\psi(u, t) = P(\tau \le t \mid U(0) = u)$ is harder to compute. For the classical model, it satisfies the partial integro-differential equation:
\[\frac{\partial}{\partial t}\psi(u,t) = c\frac{\partial}{\partial u}\psi(u,t) - \lambda\psi(u,t) + \lambda\int_0^u \psi(u-x,t)f_X(x)\,dx + \lambda[1-F_X(u)]\]Analytical solutions exist only for exponential claims:
\[\psi(u, t) = 1 - (1+\theta)e^{-Ru}\Phi\!\left(\sqrt{\frac{\lambda t}{1+\theta}}\!\cdot\!\frac{u + (c-\lambda\mu_X)t}{\sqrt{\lambda\mu_X^{(2)} t / 2}}\right) + \ldots\](exact formula involves a series; see Seal, 1969). In practice, finite-time ruin probabilities are computed by Monte Carlo simulation or by discretizing the PIDE on a grid.