Finite-Horizon Ruin by First-Claim Conditioning

Why Condition on the First Claim?

Finite-horizon ruin depends on the entire sequence of claims that may arrive before a fixed time $T$. Conditioning on the first claim organizes that sequence into a manageable recursion: first determine whether a claim arrives before $T$, then consider its time and amount, and finally evaluate the remaining ruin risk from the reserve it leaves behind. The resulting integral equation is a starting point for analysis and numerical calculation, not generally a closed-form solution.

We use the classical surplus model. An insurer starts with reserve $u\geq 0$, receives premium continuously at rate $c>0$, and pays claims from a compound Poisson process with arrival rate $\lambda>0$. Claim amounts are independent and identically distributed copies of a nonnegative random variable $X$, independent of the Poisson arrival process. The surplus is $U(t)=u+ct-\sum_{j=1}^{N(t)}X_j$, where $N(t)$ is Poisson with rate $\lambda$.

Definition: Under the strict ruin convention, the ruin time is $\tau=\inf\{t\geq 0:U(t)<0\}$. The finite-horizon ruin probability from reserve $u$ is $\psi(u,T)=P(\tau\leq T\mid U(0)=u)$. Thus, a claim that leaves surplus exactly zero is not immediate ruin; subsequent claims may still cause ruin. We take $\psi(u,0)=0$ for $u\geq0$.

Let $F_X$ be the distribution function of a claim amount and let $S_X(y)=P(X>y)$. The claim distribution may be continuous or may have atoms. Keeping that possibility in view matters at the boundary: with strict ruin, a claim equal to the pre-claim reserve does not cause immediate ruin.

Conditioning on the First Claim

The first claim arrival time has density $\lambda e^{-\lambda t}$ for $t>0$. The probability that no claim arrives by time $T$ is $e^{-\lambda T}$; on that event the reserve only increases, so ruin does not occur. If the first claim arrives at time $t$, the reserve just before it is $u+ct$. A claim of amount $x>u+ct$ causes immediate ruin. If $x\leq u+ct$, the reserve immediately after the claim is $u+ct-x$, and the remaining time is $T-t$.

After the first arrival, the future Poisson increments are independent of the past and have the same rate $\lambda$. The future claim amounts are also independent copies of $X$. Consequently, conditional on the first claim leaving reserve $u+ct-x$, the probability of ruin by the horizon is $\psi(u+ct-x,T-t)$.

Theorem: For $u\geq0$, $T\geq0$, $c>0$, and a compound Poisson claim process as specified above, the finite-horizon ruin probability satisfies $$ \psi(u,T)=\int_0^T \lambda e^{-\lambda t} \left[ S_X(u+ct)+\int_{[0,u+ct]}\psi(u+ct-x,T-t)\,dF_X(x) \right]dt. $$ The first term inside the brackets is immediate ruin at the first claim. The Stieltjes integral is ruin after a first claim that leaves nonnegative reserve. The no-claim event contributes zero to the probability of ruin.

For a claim distribution with density $f_X$, the continuation term can instead be written as $\int_0^{u+ct}\psi(u+ct-x,T-t)f_X(x)\,dx$. If $F_X$ has atoms, the Stieltjes-integral form avoids incorrectly dropping their probability mass. The integration includes $x=u+ct$, because that claim leaves zero surplus and belongs to the continuation case under the strict ruin convention.

Why the Integral Equation Holds

The equation follows by partitioning outcomes according to the first claim time and amount. The density of a first arrival at $t$ is $\lambda e^{-\lambda t}$: the factor $e^{-\lambda t}$ means no claim arrived earlier, and $\lambda\,dt$ is the arrival probability in a short interval around $t$. Conditional on that arrival, the two claim-amount cases are disjoint and exhaustive: either $X>u+ct$, or $X\leq u+ct$.

Allow for no arrival.
The probability of no claim by $T$ is $e^{-\lambda T}$, and ruin does not occur on this event when $u\geq0$ and $c>0$.

Weight each first-arrival time.
For $0<t\leq T$, a first arrival at $t$ has density $\lambda e^{-\lambda t}$.

Split by the first claim amount.
Amounts greater than $u+ct$ cause immediate ruin. Amounts no greater than $u+ct$ leave reserve $u+ct-x$.

Continue from the post-claim state.
Independence of future increments makes the remaining ruin probability $\psi(u+ct-x,T-t)$. Integrate over claim amounts and then over first-arrival times.

The equation is an integral equation because the unknown function $\psi$ appears inside the integral at reserves and horizons different from the starting values. In general, evaluating the first-claim integral is not the same as calculating the full ruin probability: the continuation term captures all possible later claims. The horizon argument $T-t$ is essential; after a claim at time $t$, only that much time remains.

Worked Example: A First Claim Always Causes Ruin

A hypothetical insurer has initial reserve $u=1$, premium rate $c=1$, claim arrival rate $\lambda=1$, and horizon $T=1$. Each claim is exactly $2$. Find the finite-horizon ruin probability by first-claim conditioning.

Method and assumptions. Claim arrivals form a Poisson process of rate $1$, and every claim amount is $2$. At time $t\leq1$, the reserve immediately before the first claim is $1+t\leq2$. For $t<1$, a claim of size $2$ therefore causes strict ruin. The equality at $t=1$ has probability zero because the arrival time has a continuous distribution.

The continuation term is zero: there is no first claim of size $2$ that leaves positive surplus before the horizon. Thus

\begin{aligned} \psi(1,1) &=\int_0^1 e^{-t}\,dt\\ &=\left[-e^{-t}\right]_0^1\\ &=1-e^{-1}\\ &\approx 0.6321. \end{aligned}

The no-claim probability is $e^{-1}$, so the answer also equals one minus the probability of no arrival. The check confirms that ruin occurs precisely when at least one claim arrives by time $1$. The common mistake avoided here is to add a separate no-claim term to the ruin probability; it represents survival, not ruin.

Worked Example: A First Claim Can Leave Time for Later Ruin

A hypothetical insurer has $u=0.5$, $c=1$, $\lambda=1$, and $T=1$. Every claim is exactly $1$. Find the probability of ruin by time $1$, including ruin caused by a later claim.

Method and assumptions. If the first claim arrives at $t<0.5$, the pre-claim reserve $0.5+t$ is less than $1$, so the first claim causes immediate ruin. If $0.5\leq t\leq1$, the first claim leaves reserve $t-0.5$. A second claim arriving at time $s\leq1$ sees pre-claim reserve $(t-0.5)+(s-t)=s-0.5\leq0.5$, so a second claim of size $1$ causes ruin. Given the first arrival at $t$, the probability of a second arrival by time $1$ is $1-e^{-(1-t)}$.

The first-claim equation therefore reduces to

\begin{aligned} \psi(0.5,1) &=\int_0^{0.5}e^{-t}\,dt +\int_{0.5}^1e^{-t}\left(1-e^{-(1-t)}\right)dt\\ &=(1-e^{-0.5})+ \left[(e^{-0.5}-e^{-1})-\frac{1}{2}e^{-1}\right]\\ &=1-\frac{3}{2}e^{-1}\\ &\approx 0.4482. \end{aligned}

The second integral is not optional: it counts paths where the first claim leaves nonnegative surplus but a subsequent claim causes ruin. The result is between zero and one, and its decomposition is consistent with the event logic: immediate first-claim ruin plus a first claim in the later interval followed by another claim before the horizon.

Common mistake and correction. Counting only $\int_0^{0.5}e^{-t}\,dt=1-e^{-0.5}$ gives approximately $0.3935$, but that is only the probability of immediate ruin at the first claim. Add the continuation integral, which is $e^{-0.5}-\frac{3}{2}e^{-1}$, to obtain the full probability $1-\frac{3}{2}e^{-1}\approx0.4482$.

Worked Example: Exponential Claim Amounts

A hypothetical insurer has $u=1$, $c=1$, $\lambda=2$, and horizon $T=1$. Claim amounts are exponential with mean $1$, so $f_X(x)=e^{-x}$ for $x\geq0$. Write the first-claim equation and calculate the immediate-ruin contribution.

Method and assumptions. The survival function is $S_X(y)=e^{-y}$ for $y\geq0$. At first-arrival time $t$, the reserve before the claim is $1+t$. The first-claim equation is

\psi(1,1)=\int_0^1 2e^{-2t} \left[ e^{-(1+t)} +\int_0^{1+t}\psi(1+t-x,1-t)e^{-x}\,dx \right]dt.

The first term in brackets gives the probability contribution from a first claim larger than the reserve at time $t$. Its total contribution is

\begin{aligned} \int_0^1 2e^{-2t}e^{-(1+t)}\,dt &=2e^{-1}\int_0^1e^{-3t}\,dt\\ &=\frac{2}{3e}(1-e^{-3})\\ &\approx 0.2330. \end{aligned}

The remaining integral in the equation is the contribution from first claims that leave nonnegative surplus, followed by possible ruin before time $1$. The value $0.2330$ is only the immediate-ruin component, not the total finite-horizon ruin probability. In general, the continuation integral must also be evaluated or solved for before the full probability is known.

Assumptions, Boundaries, and Exam Traps

The first-claim equation relies on the model assumptions stated above: Poisson arrivals, independent and identically distributed claim amounts, and independence between arrival times and claim amounts. A time-varying claim arrival rate would change the first-arrival density. Dependence between claim sizes and arrival times, or dependence between successive claims, would also require a different conditioning model.

Do not omit continuation ruin. A first claim that leaves nonnegative surplus does not guarantee survival. The future ruin probability from the remaining reserve and time belongs in the equation.
Use the remaining horizon. After a first claim at $t$, use $T-t$, not $T$, for the future probability.
Keep the survival convention consistent. Under strict ruin, immediate ruin means $X>u+ct$, while equality is included in the continuation case. If a model instead defines ruin as nonpositive surplus, the boundary allocation changes, particularly when claim amounts have atoms.
Do not count the no-claim event as ruin. Its probability is $e^{-\lambda T}$, but it contributes zero to $\psi(u,T)$ for a nonnegative initial reserve and positive premium rate.

Finite-horizon first-claim conditioning is useful because it turns the pathwise ruin event into a decomposition over a single arrival time, a single claim amount, and a continuation probability. It makes the role of each model component explicit and provides a foundation for numerical methods. It does not by itself imply that the resulting equation has a simple closed-form solution.

Key takeaway: Integrate the first-arrival density over $0\leq t\leq T$, split claim amounts into immediate ruin and non-ruin cases, and carry the post-claim reserve and remaining horizon into the continuation probability. No arrival by $T$ is survival, not ruin.

Check Your Understanding

For each problem, use strict ruin and the stated compound Poisson assumptions. Keep immediate ruin distinct from ruin after a surviving first claim.

A hypothetical insurer has $u=1.2$, $c=0.8$, $\lambda=1.5$, and horizon $T=0.6$, with claim distribution $F_X$. Write the first-claim equation, including its claim-size integration limits. (Answer: $\psi(1.2,0.6)=\int_0^{0.6}1.5e^{-1.5t}\left[S_X(1.2+0.8t)+\int_{[0,1.2+0.8t]}\psi(1.2+0.8t-x,0.6-t)\,dF_X(x)\right]dt$.)
A hypothetical insurer has $u=2$, $c=1$, $\lambda=0.5$, and $T=1$; every claim is exactly $4$. Find the ruin probability. (Answer: $1-e^{-0.5}\approx0.3935$.)
A hypothetical insurer has $u=0.8$, $c=1$, $\lambda=1$, and $T=1$; every claim is exactly $1.4$. For a first claim at $t\geq0.6$, ruin requires a second claim before time $1$. Find the full ruin probability. (Answer: $1-1.4e^{-1}\approx0.4850$.)
A hypothetical insurer has $u=0.5$, $c=0.5$, $\lambda=2$, and $T=1$, with exponential claim amounts of rate $0.5$. Write the first-claim equation and find the immediate-ruin contribution, without claiming it is the full ruin probability. (Answer: $\psi(0.5,1)=\int_0^1 2e^{-2t}\left[e^{-0.25-0.25t}+\int_0^{0.5+0.5t}\psi(0.5+0.5t-x,1-t)\,0.5e^{-0.5x}\,dx\right]dt$; immediate contribution $=\frac{2e^{-0.25}}{2.25}(1-e^{-2.25})$.)
A hypothetical insurer has $u=1$, $c=1$, $\lambda=0.7$, and $T=2$; every claim is exactly $2.5$. A first claim before time $1.5$ causes immediate ruin; a first claim later than $1.5$ requires another claim before time $2$. Find the full ruin probability. (Answer: $1-1.35e^{-1.4}\approx0.6671$.)

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