Finite-Horizon Ruin Calculation Workshop

Why a Fixed Horizon Changes the Ruin Question

A finite-horizon ruin probability asks whether surplus falls below zero by a specified time. It is not enough to know that a claim could exceed the current surplus: premium income may accumulate before the claim arrives, and the number and timing of claims within the horizon both matter. Conditioning on the claim count separates these effects and can turn a path question into a sequence of manageable calculations.

Throughout this tutorial, use the classical surplus model \(U(t)=w+ct-\sum_{j=1}^{N(t)}X_j\), where \(w\geq 0\) is initial surplus, \(c\geq 0\) is the premium income rate, and \(N(t)\) is a Poisson process with rate \(\lambda\). Claim amounts \(X_1,X_2,\ldots\) are independent, identically distributed, nonnegative random variables, independent of \(N(t)\). The horizon \(T\) is fixed and nonnegative. Ruin is strict: surplus must be less than zero.

Definition: The finite-horizon strict ruin probability is \(\psi(w,T)=P\bigl(\inf_{0\leq t\leq T}U(t)<0\bigr)\). The ultimate ruin probability is the corresponding probability over all future time, \(\psi(w,\infty)=P\bigl(\inf_{t\geq 0}U(t)<0\bigr)\). The first event is contained in the second, so \(\psi(w,T)\leq\psi(w,\infty)\); in general, they are not equal.

Write \(\tau_k\) for the time of the \(k\)th claim, and \(N(T)\) for the number of claims by time \(T\). Between claims, surplus increases at rate \(c\), so for \(c\geq0\) it cannot first fall below zero between claim times. Immediately after the \(k\)th claim, surplus is

U(\tau_k)=w+c\tau_k-\sum_{j=1}^{k}X_j.

Thus, for strict ruin, the relevant event is that this quantity is negative for at least one claim time \(\tau_k\leq T\). Equality to zero is not ruin under this convention.

Conditioning on the Number and Timing of Claims

For \(\lambda T>0\), the count \(N(T)\) has a Poisson distribution with mean \(\lambda T\). Conditional on \(N(T)=n\), the \(n\) claim times have the distribution of the ordered values of \(n\) independent Uniform\((0,T)\) random variables. Their joint density on \(0<t_1<\cdots<t_n<T\) is \(n!/T^n\). Claim sizes retain their original distribution and are independent of those times.

Let \(q_n(w,T)\) denote the probability of strict ruin by \(T\), conditional on \(N(T)=n\). Then the finite-horizon probability is the Poisson-weighted sum of these conditional probabilities:

\psi(w,T)=\sum_{n=0}^{\infty}P(N(T)=n)q_n(w,T) =e^{-\lambda T}\sum_{n=0}^{\infty}\frac{(\lambda T)^n}{n!}q_n(w,T).

For \(w\geq0\), \(q_0(w,T)=0\): without claims, surplus is \(w+ct\), which does not become negative. If \(\lambda T=0\), handle the case separately. There are no claims by time \(T\), and, for \(w\geq0\) and \(c\geq0\), \(\psi(w,T)=0\). In particular, conditional probabilities given \(N(T)=n\) for \(n\geq1\) are not needed in this case.

Why the decomposition holds: The events \(\{N(T)=n\}\), for \(n=0,1,\ldots\), partition the outcomes. Applying the law of total probability gives \(\psi(w,T)=\sum_n P(\text{ruin by }T\mid N(T)=n)P(N(T)=n)\). Conditional on a positive count, the ordered uniform arrival-time distribution supplies the timing probabilities needed to evaluate each term.

When the claim sizes are discrete, one can often calculate \(q_n\) by checking the surplus after each claim. When sizes are continuous, the same event definition applies, but probabilities may require integration over both ordered times and claim amounts. The examples below use discrete amounts so that the conditional calculations are transparent.

Worked Example: One Claim Can Be Safe or Ruinous Depending on Its Arrival Time

A hypothetical portfolio has initial surplus \(w=4\), earns premium at rate \(c=1\) dollar per unit time, and has claims arriving according to a Poisson process with \(\lambda=0.5\) per unit time. Every claim is exactly \(5\) dollars. Calculate the strict ruin probability over \(T=2\) time units.

Method. Condition on \(N(2)=n\). The Poisson mean is \(\lambda T=0.5(2)=1\). If there is exactly one claim at time \(t\), surplus immediately after it is \(4+t-5=t-1\). Ruin occurs precisely when \(t<1\). Conditional on one claim, its arrival time is Uniform\((0,2)\), so

q_1(4,2)=P(t<1\mid N(2)=1)=\frac{1}{2}.

With no claims, ruin cannot occur. With at least two claims, the second claim alone brings cumulative claims to at least \(10\). At its arrival time, which is less than \(2\) with probability one, surplus is at most \(4+2-10=-4<0\). Therefore \(q_n(4,2)=1\) for every \(n\geq2\).

The required Poisson probabilities are \(P(N(2)=0)=e^{-1}\), \(P(N(2)=1)=e^{-1}\), and \(P(N(2)\geq2)=1-2e^{-1}\). Hence

\psi(4,2)=0\cdot e^{-1}+\frac{1}{2}e^{-1}+1\cdot(1-2e^{-1}) =1-\frac{3}{2}e^{-1} =0.4482.

The final probability is rounded to four decimal places. As a check, the three count cases contribute \(0\), approximately \(0.1839\), and approximately \(0.2642\), respectively; their sum is \(0.4482\). The common mistake is to treat a single \(5\)-dollar claim as automatically ruinous because it exceeds the initial surplus of \(4\). Premium earned before arrival changes the surplus, and only arrivals before time \(1\) cause strict ruin in the one-claim case.

Worked Example: Two Claim Sizes and a Time-Dependent Threshold

A hypothetical insurer starts with \(w=5\), receives premium at rate \(c=1\), and observes claims from a Poisson process with \(\lambda=1.5\) per unit time. Independently, each claim is \(2\) dollars or \(5\) dollars, each with probability \(1/2\). Find the strict ruin probability for \(T=1\).

Method. The Poisson mean is \(\lambda T=1.5\). If there is one claim, even a \(5\)-dollar claim leaves surplus \(5+t-5=t\geq0\) at its arrival time \(t\), so \(q_1=0\). With two claims, the second claim causes ruin whenever at least one of the two claims is \(5\): their total is then at least \(7\), whereas surplus just before the second claim is at most \(5+1=6\). If both claims are \(2\), their total is \(4\), so there is no ruin. The probability that at least one of two independent claims is \(5\) is \(1-(1/2)^2=3/4\). Thus \(q_2=3/4\).

For \(n\geq3\), the total of the first three claims is at least \(6\). Since the third arrival time satisfies \(\tau_3<1\) with probability one, surplus immediately after that claim is at most \(5+\tau_3-6<0\). Therefore \(q_n=1\) for all \(n\geq3\).

The count probabilities needed are \(P(N=0)=e^{-1.5}\), \(P(N=1)=1.5e^{-1.5}\), and \(P(N=2)=(1.5)^2e^{-1.5}/2=1.125e^{-1.5}\). It follows that

\begin{aligned} \psi(5,1) &=\frac{3}{4}P(N=2)+P(N\geq3)\\ &=\frac{3}{4}(1.125e^{-1.5})+ \left[1-e^{-1.5}(1+1.5+1.125)\right]\\ &=1-2.78125e^{-1.5}\\ &=0.3794. \end{aligned}

The final result is rounded to four decimal places. The count cases are exhaustive: zero and one claim cannot cause ruin, two claims cause ruin with conditional probability \(3/4\), and three or more claims cause ruin. A common trap is to assume that three claims always cause ruin without checking the third arrival time; here the strict inequality follows because \(\tau_3<1\) almost surely, not merely because the claim total is at least \(6\).

Finite-Time Probability Is Not Ultimate Probability

The finite-horizon calculation stops at \(T\). Claims after \(T\) have no effect on \(\psi(w,T)\), even though they remain relevant to ultimate ruin. A useful way to see the distinction is to choose a model in which the claim-count threshold for ruin is easy to identify.

Worked Example: A Finite-Horizon Probability Below Ultimate Ruin

Consider a hypothetical portfolio with initial surplus \(w=5\), no premium income (\(c=0\)), and deterministic claim amount \(3\). Claims arrive according to a Poisson process with \(\lambda=0.4\) per unit time. Calculate the strict ruin probability by \(T=3\), then compare it with the ultimate ruin probability.

Method. With one claim, surplus becomes \(5-3=2\), so there is no ruin. With two claims, it becomes \(5-3-3=-1\), so ruin occurs. Since all claims are positive and \(c=0\), the same is true for every count \(n\geq2\). Thus ruin by time \(3\) occurs exactly when \(N(3)\geq2\). The Poisson mean is \(0.4(3)=1.2\), giving

\begin{aligned} \psi(5,3) &=P(N(3)\geq2)\\ &=1-P(N(3)=0)-P(N(3)=1)\\ &=1-e^{-1.2}(1+1.2)\\ &=0.3374. \end{aligned}

The final probability is rounded to four decimal places. Over an unlimited time horizon, a Poisson process with positive rate eventually produces at least two claims with probability one. Therefore \(\psi(5,\infty)=1\) for this particular model. The comparison checks the inequality \(\psi(5,3)\leq\psi(5,\infty)\), and shows why the finite-time result cannot be substituted for an ultimate ruin probability.

This example deliberately has no premium income; it is used to isolate the effect of the horizon, not to represent a premium-loaded portfolio. The finite-horizon decomposition itself does not require a positive safety loading.

A Practical Calculation Sequence

For a finite-horizon problem, first translate the words “ruin by time \(T\)” into a claim-time event. Then condition on the number of claims, evaluate the ruin condition for each relevant count, and combine the conditional results using Poisson probabilities. The following sequence keeps timing, claim amounts, and the horizon separate.

State the surplus model and ruin convention.
Record \(w\), \(c\), \(\lambda\), \(T\), the claim-size distribution, and whether ruin means surplus less than zero or less than or equal to zero.

Inspect surplus at claim times.
For each \(k\), test whether \(w+c\tau_k-\sum_{j=1}^{k}X_j<0\) for some \(\tau_k\leq T\). Between claims, surplus cannot decrease when \(c\geq0\).

Find the conditional probabilities.
For each count \(n\) that can cause ruin, calculate \(q_n(w,T)\), using ordered uniform arrival times conditional on \(N(T)=n\) and the specified claim-size law.

Weight by the Poisson count probabilities.
Sum \(e^{-\lambda T}(\lambda T)^nq_n/n!\) over all counts. Check that omitted counts have conditional ruin probability zero or are included in a grouped tail.

Interpret the horizon correctly.
Report the answer as ruin by \(T\). Do not call it an ultimate ruin probability unless the event has explicitly been evaluated over all future time.

Common Exam Traps

Ignoring premium earned before a claim. A claim larger than initial surplus need not cause ruin if it arrives after enough premium has accumulated.
Using an unconditional arrival-time distribution. Conditional on exactly \(n\) arrivals by \(T\), the ordered times have the ordered-uniform distribution; the \(n\)th claim is not an independent uniform time.
Stopping at one or two claims without checking larger counts. Establish why all remaining counts have zero ruin probability or group them into a tail with a justified conditional probability.
Changing the strictness convention mid-calculation. Under strict ruin, a surplus of exactly zero does not qualify. Boundary events may have positive probability in discrete models.
Confusing finite and ultimate ruin. Ruin by a fixed horizon is a subset of ruin at any future time, so the finite-horizon probability cannot exceed the ultimate probability.

Key takeaway: A finite-horizon ruin calculation combines the Poisson distribution of the claim count with the conditional distribution of claim times and amounts. The event must be checked along the surplus path, and the time horizon must remain explicit throughout.

Check Your Understanding

For each problem, identify the claim counts that can cause ruin before combining their probabilities. Use strict ruin and round final probabilities to four decimal places.

A hypothetical portfolio has \(w=3\), \(c=2\), deterministic claims of \(4\), \(\lambda=0.5\), and \(T=2\). Calculate the finite-horizon ruin probability. (Answer: \(1-\frac{7}{4}e^{-1}=0.3562\))
A portfolio has \(w=7\), \(c=0\), \(\lambda=2\), and \(T=0.5\). Claims are independently \(4\) or \(8\), each with probability \(1/2\). Find the probability of ruin by the horizon. (Answer: \(1-\frac{3}{2}e^{-1}=0.4482\))
A portfolio has \(w=4\), \(c=1\), deterministic claims of \(3\), \(\lambda=0.5\), and \(T=2\). Calculate the finite-horizon ruin probability, accounting for the claim arrival times. (Answer: \(1-2e^{-1}=0.2642\))
A hypothetical portfolio has \(w=2\), \(c=0\), deterministic claims of \(3\), and \(\lambda=0.4\). Find its ruin probability by \(T=1\), and its ultimate ruin probability. (Answer: finite horizon \(1-e^{-0.4}=0.3297\); ultimate \(1\))

What You'll Learn