Estimating Finite-Horizon Ruin by Claim-Path Simulation

Finite-Horizon Ruin as a Path Event

A finite-horizon ruin probability concerns whether surplus becomes negative at any time during a specified period, not just whether the insurer ends that period with negative surplus. A claim can cause an insurer to cross below zero and later premium income can restore positive surplus. A simulation must therefore retain the order and timing of claims.

Use the classical surplus model with initial surplus $u$, constant premium income at rate $c$ per unit time, and aggregate claims $S(t)$ by time $t$. Assume a homogeneous Poisson claim process with rate $\lambda$, independent and identically distributed claim severities $X_1,X_2,\ldots$, and independence between claim arrivals and severities. There is no investment income, expense, tax, or other cash flow in this model. For a horizon $T>0$, the surplus is

U(t)=u+ct-S(t), \qquad 0\le t\le T.

Write the ordered claim times as $0<T_1<\cdots<T_N\le T$. The number of claims $N$ has a Poisson distribution with mean $\lambda T$; each claim at time $T_j$ reduces surplus by severity $X_j$. Define ruin as surplus becoming strictly negative at some time in the horizon.

Definition: The finite-horizon ruin probability is $\psi(u,T)=P\bigl(\min_{0\le t\le T}U(t)<0\bigr)$. With the strict inequality convention, surplus equal to zero is not ruin. The initial value $U(0)=u$ is part of the path and must be checked.

Under the stated model, premium income accumulates continuously between claims. If $c\ge 0$, surplus does not decrease between claims, so the minimum on each interval before the next claim occurs at the interval's starting point. It is therefore sufficient to check $U(0)$ and the surplus immediately after each claim. If $u\ge0$ as well as $c\ge0$, checking post-claim surplus alone is sufficient: the initial surplus cannot already be negative. This condition matters even on a simulated path with no claims.

Generating and Evaluating a Claim Path

One direct simulation method first generates $N\sim\operatorname{Poisson}(\lambda T)$. Conditional on $N=n$, generate $n$ independent uniform times on $(0,T)$ and sort them. These sorted times have the distribution of the arrival times of a homogeneous Poisson process conditional on its count. Generate $n$ independent severities from the specified severity distribution, independently of the times. Alternatively, generate exponential interarrival times with rate $\lambda$, stopping when the next arrival would exceed $T$.

For a generated path, update surplus recursively. Set $T_0=0$ and $U(T_0)=u$. Immediately after claim $j$, calculate

U(T_j)=U(T_{j-1})+c(T_j-T_{j-1})-X_j.

Mark the path as ruined as soon as $U(0)<0$ or a post-claim surplus is negative. A path with no claims is not automatically safe: its initial surplus still needs checking. When $c\ge0$, if no ruin has occurred by the last claim, the remaining premium income cannot create ruin before $T$.

Specify the model.
Set $u,c,T,\lambda$, the severity distribution and its parameters, and any assumptions about cash flows.

Generate one path.
Generate the number and times of claims in $[0,T]$, then generate a severity for each claim.

Track surplus and classify ruin.
Check $U(0)$; update the surplus at each claim and mark whether it ever becomes negative.

Repeat and summarize.
Repeat independently for $n$ paths. Estimate the ruin probability by the fraction of paths classified as ruined.

Worked Example: Detecting Ruin Within a Claim Path

A hypothetical insurer has $u=\$500$, earns premium continuously at $c=\$60$ per year, and is assessed over $T=2$ years. For the simulation model, arrivals are Poisson with $\lambda=1.5$ per year and claim severities are independent exponential random variables with mean $\$200$. Assume no interest or other cash flows. One generated path has claims at years $0.4,1.1,1.8$, with severities $\$100,\$250,\$300$. Determine whether the path is ruined.

Method. Check the initial surplus and then update it at each claim using the elapsed time since the previous claim. This path-by-path check avoids the mistake of looking only at the horizon-end surplus.

Initially, $U(0)=\$500$, so there is no initial ruin. Before the first claim, premium income is $60(0.4)=\$24$. Thus $U(0.4)=500+24-100=\$424$.

The second claim is $0.7$ years later. The premium earned in that interval is $60(0.7)=\$42$, giving $U(1.1)=424+42-250=\$216$.

The third claim is also $0.7$ years after the preceding claim, so $U(1.8)=216+42-300=-\$42$. The path is ruined at year $1.8$, the first time a checked surplus is negative.

The remaining $0.2$ years produce $60(0.2)=\$12$ of premium, so $U(2)=-42+12=-\$30$. The horizon-end surplus is negative as well, but that is not what establishes the first ruin time. All values are stated to the cent; here the arithmetic is exact.

Estimating the Probability and Its Simulation Error

For each simulated path $r$, define $I_r=1$ if ruin occurs during $[0,T]$ and $I_r=0$ otherwise. With $n$ independent simulated paths and $K=\sum_{r=1}^{n}I_r$ ruined paths, the Monte Carlo estimate is

\widehat{\psi}(u,T)=\frac{K}{n}.

Each $I_r$ is a Bernoulli random variable with success probability $\psi(u,T)$. Consequently, $E[I_r]=\psi(u,T)$ and $\operatorname{Var}(I_r)=\psi(u,T)(1-\psi(u,T))$. Independence gives $E[\widehat{\psi}]=\psi(u,T)$ and

\operatorname{Var}(\widehat{\psi})=\frac{\psi(u,T)(1-\psi(u,T))}{n}.

This justifies the usual estimated Monte Carlo standard error:

\widehat{\operatorname{SE}}\bigl(\widehat{\psi}\bigr) =\sqrt{\frac{\widehat{\psi}(1-\widehat{\psi})}{n}}.

When the number of ruined paths and the number of non-ruined paths are both sufficiently large, an approximate 95% confidence interval is $\widehat{\psi}\pm1.96\,\widehat{\operatorname{SE}}$. The approximation is unreliable when the estimate is near zero or one, particularly if there are very few or no observed ruined paths. In that setting, a binomial-based interval is preferable. A confidence interval describes simulation error conditional on the specified model; it does not account for uncertainty in $\lambda$, the severity distribution, or other model assumptions.

Worked Example: Reporting a Ruin Estimate and Its Precision

For a hypothetical portfolio, suppose the model has $u=\$400$, $c=\$220$ per year, horizon $T=1$ year, Poisson claim rate $\lambda=2$ per year, and independent exponential claim severities with mean $\$100$. There is no interest or other cash flow. In $n=1{,}000$ independent simulated paths, 80 are classified as ruined. Estimate the finite-horizon ruin probability and give an approximate 95% confidence interval.

Method. Use the sample proportion of ruined paths and the Bernoulli standard-error formula. The estimate is $\widehat{\psi}=80/1{,}000=0.0800$.

The estimated standard error is \[ \sqrt{\frac{0.0800(1-0.0800)}{1{,}000}} =\sqrt{\frac{0.0736}{1{,}000}} =\sqrt{0.0000736} \approx0.008579. \] The approximate 95% margin is $1.96(0.008579)\approx0.016815$. Therefore the interval is \[ 0.0800-0.016815=0.063185,\qquad 0.0800+0.016815=0.096815. \] To four decimal places, the estimate is $0.0800$, the standard error is $0.0086$, and the approximate interval is $(0.0632,0.0968)$. The interval quantifies Monte Carlo error for this model, not the possibility that the chosen claim model is wrong.

Initial Ruin, Strict Inequalities, and Other Traps

A ruin indicator must follow the definition of the path event, not a shortcut based on the final balance. Two details commonly change the classification: whether initial surplus is negative and whether a balance of exactly zero counts as ruin. With the strict-negative definition used here, a path that reaches zero but never falls below zero is not ruined. A different problem may define ruin using a nonpositive surplus; follow the stated convention.

Worked Example: Why the Initial Surplus Must Be Checked

Consider a hypothetical insurer with initial surplus $u=-\$25$, premium rate $c=\$100$ per year, horizon $T=1$ year, Poisson arrival rate $\lambda=0.5$ per year, and independent exponential severities with mean $\$100$. There is no interest or other cash flow. On one generated path, no claims occur. Determine whether ruin occurs.

Incorrect calculation. Checking only post-claim surplus finds no claims and therefore no post-claim values to test. It would incorrectly classify the path as not ruined.

Correction. The path starts at $U(0)=u=-\$25<0$, so ruin has already occurred at time zero. The surplus at year one is $-25+100(1)=\$75$, but later recovery does not undo an earlier ruin event. The path is ruined. The initial check is necessary regardless of how the later surplus evolves.

If $u\ge0$ and $c\ge0$, checking post-claim values alone is sufficient under this model: $U(0)\ge0$, and between claims surplus is nondecreasing. That shortcut does not apply when initial surplus may be negative. Also avoid replacing the path event by a test of $U(T)<0$: the example with a negative post-claim balance followed by premium recovery shows why those events are not equivalent.

Key takeaway: Simulate and classify the entire surplus path through the horizon. Check $U(0)$, then update at every claim; use the fraction of ruined paths as the probability estimate and report its Monte Carlo uncertainty separately from model uncertainty.

Check Your Understanding

For each problem, use strict-negative ruin: surplus equal to zero is not ruin. Unless stated otherwise, interest and other cash flows are absent.

A hypothetical insurer starts with $u=\$300$, receives premium at $c=\$50$ per year, and is followed for two years. Claims occur at years $0.5$ and $1.5$, with severities $\$100$ and $\$180$. Calculate the post-claim surpluses and horizon-end surplus, and classify the path. (Answer: post-claim surpluses are $\$225$ and $\$95$; ending surplus is $\$120$; no ruin.)
A hypothetical insurer has $u=\$120$, $c=\$40$ per year, and $T=1$. A single claim of $\$140$ occurs at year $0.5$, with no other claims. Does reaching zero at the claim time count as ruin? (Answer: No; surplus reaches exactly $\$0$, then ends at $\$20$, so it never becomes negative.)
In $400$ independent simulated paths, $36$ are ruined. Calculate the estimated probability, estimated standard error, and approximate 95% interval. (Answer: estimate $0.0900$; standard error $\sqrt{0.09(0.91)/400}\approx0.0143$; interval approximately $(0.0620,0.1180)$.)
A simulated path has $u=-\$10$, premium rate $c=\$50$ per year, horizon $T=0.5$, and no claims. What is the ruin classification, even though the horizon-end surplus is positive? (Answer: Ruin occurs at time zero because $U(0)=-\$10$; the ending surplus of $\$15$ does not change that classification.)
Among $1{,}000$ simulated paths, none is ruined. What is the estimated probability, and why is the ordinary estimated standard error not a useful confidence statement here? Using the one-sided 95% binomial upper bound $1-0.05^{1/n}$, calculate that bound. (Answer: estimate $0$; the Wald standard error is $0$ and its interval is uninformative; the stated upper bound is $1-0.05^{1/1000}\approx0.0030$.)

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