Free Surplus as Brownian Motion: Ruin Probability and a Reinsurance Decision
A general insurer models its free surplus, in ₹ crore, as X(t) = 800 + 60t + 200W(t), where t is in years and W is standard Brownian motion under the real-world measure, after allowing for premium growth, claims and investment return. Ruin occurs when the surplus first reaches zero. The board's risk appetite permits at most a 5% probability of ruin over the life of the company. The finance director argues that, because the expected surplus in ten years is ₹1,400 crore with a standard deviation of only ₹632 crore, ruin is a three-standard-deviation event and no action is needed. Separately, the reinsurance team proposes a quota-share treaty that would cut the drift to ₹40 crore per year and the volatility to ₹150 crore per root-year, at a one-off cost of ₹60 crore of surplus. The model is calibrated on 15 years of annual data.
1. Compute the probability of ultimate ruin with and without the treaty, and evaluate the treaty against the board's risk appetite.
Read the worked answer
For X(t) = x + μt + σW(t) with μ > 0, the probability that the surplus ever hits zero is P(τ < ∞) = exp(−2μx/σ²).
Without the treaty: 2μx/σ² = (2 × 60 × 800)/200² = 96,000/40,000 = 2.40, so P = e^(−2.40) = 0.091, i.e. about 9.1%.
With the treaty: the surplus falls to 740 by the cost, the drift to 40 and the volatility to 150, so 2μx/σ² = (2 × 40 × 740)/22,500 = 59,200/22,500 = 2.63, giving P = e^(−2.63) = 7.2%.
Assessment. The treaty is poor value against the stated objective. It costs ₹60 crore of surplus and only reduces the ruin probability from 9.1% to 7.2%, still well outside the 5% appetite. The reason is instructive: the exponent is proportional to μ/σ², so cutting σ from 200 to 150 increases the exponent by a factor of 1.78 on that account, but cutting μ from 60 to 40 nearly halves it, and the 60 crore of upfront capital reduces x. The net gain is modest.
What would work. Reducing σ without cutting μ (a stop-loss or aggregate excess-of-loss cover), raising capital by at least 200 crore, or reducing the risk appetite shortfall by a combination. Note also that the ruin formula is extremely sensitive to the exponent, so parameter error matters more than the choice of treaty.
2. Critique the finance director's three-sigma argument. What exactly does the fixed-horizon calculation measure?
Read the worked answer
The director computes X(10) ~ Normal(800 + 600, 200² × 10), so sd = 200√10 = 632.5, and P(X(10) < 0) = Φ(−1400/632.5) = Φ(−2.21) ≈ 1.3%. The arithmetic is right and the interpretation is wrong.
Point-in-time is not path. P(X(10) < 0) asks only where the surplus sits at a fixed date. Ruin is a first-passage event: it requires only that the path touch zero at some time. A path can plunge below zero in year 3, recover by year 10 and never be counted. The correct comparison is 9.1% against 1.3% - a factor of seven - and the gap widens the longer the horizon, because ultimate ruin allows infinitely many opportunities to breach.
Two further errors. First, calling it a 'three-sigma event' misstates the probability: 2.21 standard deviations is roughly 1.3%, not 0.1%. Second, and more seriously, the model's Gaussian, stationary, independent-increment structure is a modelling convenience, not a description of an insurer's surplus. Claims arrive in clusters, catastrophes produce jumps, and the parameters themselves are uncertain - the 15 years of data give a standard error on the volatility of roughly σ/√(2n) ≈ 18%, so σ could be 240 rather than 200. At σ = 240 the ruin probability becomes exp(−96,000/57,600) = e^(−1.67) = 18.8%. The director's 'no action' conclusion is an artefact of taking a fitted model literally.
3. Using Itô's lemma, show that M(t) = exp(−(2μ/σ²)X(t)) is a martingale, and use it to justify the ruin formula and to explain what the model cannot capture.
Read the worked answer
Martingale property. Let θ = −2μ/σ² and M(t) = exp(θX(t)). With dX = μ dt + σ dW, Itô gives dM = M[θμ + ½θ²σ²] dt + Mθσ dW. The drift term is θμ + ½θ²σ² = (−2μ/σ²)μ + ½(4μ²/σ⁴)σ² = −2μ²/σ² + 2μ²/σ² = 0. Hence dM = θσM dW, which has no drift and, M being bounded on [0, τ ∧ n] with X bounded below before ruin, is a martingale.
Justifying the formula. Optional stopping at τ ∧ n gives E[M(τ ∧ n)] = M(0) = exp(θx). On the ruin path M(τ) = exp(0) = 1. On the non-ruin path X → ∞ almost surely and M → 0. Letting n → ∞: P(τ < ∞) × 1 + (1 − P(τ < ∞)) × 0 = exp(−2μx/σ²), which is exactly the formula used. The martingale does the work that the 'expected surplus' argument cannot, because it is a bounded functional of the path rather than an expectation at a fixed date.
What the model cannot do. It assumes independent Gaussian increments, constant μ and σ, no jumps, no autocorrelation and no parameter uncertainty; 15 annual observations cannot identify a jump component. Real surplus has heavy tails and a volatility that rises in stress, precisely when the drift falls. It also treats the surplus as a single traded-risk-free quantity, ignoring that assets and liabilities respond differently to interest rates and equity markets. Its use should be as a capital indicator with stressed parameters, not as a literal probability.