Finite-Horizon Ruin by Claim-Count Decomposition

Introduction

A finite-horizon ruin probability asks whether an insurer’s surplus becomes negative by a specified time $T$. The number of claims by $T$ is random, but conditioning on that count separates the problem into two manageable parts: the probability of observing that many claims, and the probability that their times and sizes cause ruin. This decomposition is useful even when the conditional probability is difficult to evaluate exactly.

The approach below uses the classical compound Poisson surplus model. Its finite-horizon formula does not require the net profit condition, although the model’s long-term behavior depends on the relationship between premium income and expected claims.

Definition: Let $u\geq 0$ be initial surplus, $c\geq 0$ the premium income rate, and $N(t)$ a Poisson process with rate $\lambda>0$. Let $X_1,X_2,\ldots$ be independent, identically distributed, nonnegative claim amounts, independent of $N(t)$. The surplus at time $t$ is $$ U(t)=u+ct-\sum_{j=1}^{N(t)}X_j. $$ Strict ruin by time $T$ occurs if $U(t)<0$ for some $t\leq T$. Write $\psi(u,T)=P(\text{strict ruin by }T)$.

Claims are numbered in chronological order, and $T_{(1)}<\cdots<T_{(n)}$ denote their ordered arrival times when $N(T)=n$. With $c\geq 0$, surplus increases between claims, so it is sufficient to check surplus just after each claim. The claim at time $T_{(k)}$ causes ruin if the total claims through that claim exceed the available initial surplus and premium income:

\sum_{j=1}^{k}X_j>u+cT_{(k)}.

The order of the claim times matters. Conditional on $N(T)=n$, the $n$ unordered arrival times have the same distribution as $n$ independent Uniform$(0,T)$ random variables. Their chronological order is therefore the order-statistic sample from those uniforms. The claim sizes remain independent of these times and retain their original severity distribution.

The Claim-Count Decomposition

Theorem: For the compound Poisson surplus model, let $q_n(T)$ be the conditional probability of ruin by time $T$, given $N(T)=n$. Then $$ \psi(u,T)=\sum_{n=0}^{\infty}P(N(T)=n)q_n(T) =e^{-\lambda T}\sum_{n=0}^{\infty}\frac{(\lambda T)^n}{n!}q_n(T). $$ If $u\geq 0$, then $q_0(T)=0$. For $n\geq 1$, $q_n(T)$ is the probability that at least one $k\in\{1,\ldots,n\}$ satisfies $\sum_{j=1}^{k}X_j>u+cT_{(k)}$, using the conditional distribution of the ordered claim times and the claim sizes.

When claim sizes have density $f_X$, the conditional probability can be written as an integral. The time integral is over the ordered region $0<t_1<\cdots<t_n<T$; the claim-size integral is over nonnegative sizes. The indicator is one precisely when at least one claim-time surplus test gives strict ruin.

q_n(T)=\frac{n!}{T^n} \int_{0<t_1<\cdots<t_n<T} \int_{(0,\infty)^n} \mathbf{1}\!\left\{\exists k\leq n: \sum_{j=1}^{k}x_j>u+ct_k\right\} \prod_{j=1}^{n} f_X(x_j)\, dx_1\cdots dx_n\,dt_1\cdots dt_n.

For discrete or mixed claim sizes, the inner integration is replaced by summation or integration with respect to the claim-size distribution. The formula is a representation, not a guarantee that the integrals have a closed form.

Why the Decomposition Holds

Partition by claim count.
The events $N(T)=n$, for $n=0,1,2,\ldots$, are mutually exclusive and cover all possibilities. The law of total probability gives $\psi(u,T)=\sum_n P(N(T)=n)P(\text{ruin by }T\mid N(T)=n)$.

Use the Poisson probabilities.
Because $N(T)$ is Poisson with mean $\lambda T$, $P(N(T)=n)=e^{-\lambda T}(\lambda T)^n/n!$. Substitution gives the displayed series.

Test ruin within each count.
Given $n$ arrivals, surplus can decline only when a claim occurs. Between claims it rises at rate $c\geq0$. Thus a negative surplus occurs by $T$ exactly when at least one post-claim surplus is negative, which is the stated cumulative-claim inequality.

The decomposition follows from conditioning and the surplus path; it does not assume exponential claim sizes. For any fixed $n$, the conditional event can depend jointly on arrival times and claim amounts. In particular, knowing $N(T)=n$ does not mean the $n$ claim times are equally spaced.

Worked Calculations

Worked Example: A One-Claim Threshold and Certain Ruin at the Second Claim

Suppose $u=\$3$, $c=\$1$ per year, $\lambda=1$ claim per year, and $T=2$ years. Every claim is exactly $\$4$, so the severity distribution is degenerate at 4. Find the probability of strict ruin by time 2.

Method. Condition on the number of claims, then evaluate the conditional ruin event. The Poisson mean over the horizon is $\lambda T=2$.

With no claims there is no ruin. Given exactly one claim, its arrival time is Uniform$(0,2)$. Ruin occurs if $4>3+T_{(1)}$, or $T_{(1)}<1$. Therefore $$ q_1(2)=\frac{1}{2},\qquad P(N(2)=1)=2e^{-2}. $$ The contribution from exactly one claim is $2e^{-2}\times\frac12=e^{-2}$.

Given at least two claims, consider the second claim. By its arrival time, cumulative claims equal $\$8$, while surplus before those claims could have been no more than $3+2=\$5$. Thus ruin has occurred by the second claim, so $q_n(2)=1$ for every $n\geq2$. The Poisson probability of at least two claims is $1-P(N=0)-P(N=1)=1-3e^{-2}$.

Adding the disjoint contributions gives $$ \psi(3,2)=e^{-2}+1-3e^{-2}=1-2e^{-2}=0.7293. $$ The probability is rounded to four decimal places. The result lies between zero and one, as required.

Common mistake avoided: The event of exactly one claim is not the same as the event of ruin. Counts of two or more also cause ruin here, and must be included.

Worked Example: Two Ordered Arrival Times Determine the Conditional Probability

Suppose $u=\$4$, $c=\$1$ per year, $\lambda=1$ claim per year, and $T=3$ years. Every claim is exactly $\$3$. Find the probability of strict ruin by time 3.

The Poisson mean is $\lambda T=3$. One claim cannot cause ruin, because $3$ is not greater than $4+T_{(1)}$. If there are exactly two claims, ruin occurs at the second claim precisely when $$ 6>4+T_{(2)},\quad\text{or}\quad T_{(2)}<2. $$ Conditional on two claims, both unordered arrival times are independent Uniform$(0,3)$. Their maximum $T_{(2)}$ is less than 2 exactly when both times are less than 2. Consequently $$ q_2(3)=\left(\frac{2}{3}\right)^2=\frac49. $$

If there are at least three claims, the third claim causes ruin: total claims are $\$9$, while $4+T_{(3)}\leq 7$. Hence the ruin probability is $$ P(N(3)\geq3)+P(N(3)=2)\frac49 =1-e^{-3}\left(1+3+\frac{9}{2}\right) +\frac{9}{2}e^{-3}\frac49 =1-\frac{13}{2}e^{-3}. $$ Since $e^{-3}=0.049787\ldots$, this is $0.6764$, rounded to four decimal places. The calculation includes the $N=2$ contribution only in proportion to its conditional ruin probability, and includes all counts $N\geq3$ because each guarantees ruin.

Common mistake avoided: $T_{(2)}$ is the later of two arrival times. Its probability of being below 2 is $(2/3)^2$, not $2/3$, which is the probability for one unordered arrival time.

Worked Example: Integrating a One-Claim Event with Exponential Severity

Suppose $u=\$0$, $c=\$1$ per year, $\lambda=1$ claim per year, and $T=1$ year. Claim sizes are independent Exponential with rate 1 per dollar, so $S_X(x)=e^{-x}$ for $x\geq0$ and $E[X]=\$1$. Calculate the probability that ruin occurs with exactly one claim by time 1.

Conditional on exactly one claim, its arrival time $T_{(1)}$ is Uniform$(0,1)$. At time $t$, ruin occurs if $X_1>t$, with conditional probability $e^{-t}$. Thus the conditional ruin probability is $$ q_1(1)=\int_0^1 e^{-t}\,dt=1-e^{-1}=0.6321. $$ The probability of exactly one claim is $P(N(1)=1)=e^{-1}$. Therefore the probability of both exactly one claim and ruin is $$ P(N(1)=1)q_1(1)=e^{-1}(1-e^{-1}) =e^{-1}-e^{-2}=0.2325. $$ This is the contribution from the $n=1$ term, rounded to four decimal places; it is not the total finite-horizon ruin probability, since counts of two or more have not been included. The calculation uses strict ruin, $X_1>t$; equality has probability zero under this continuous severity model.

Common mistake avoided: Using $P(X_1>1)$ as the conditional probability would treat the claim as if it always arrived at the horizon. Its actual arrival time is random, so the required probability is averaged over $0\leq t\leq1$.

Interpreting and Using the Series

The Poisson weights and conditional ruin probabilities play different roles. The weight $P(N(T)=n)$ describes how likely a count is; $q_n(T)$ describes how dangerous that count is under the given surplus and severity model. A large conditional probability for some $n$ need not have a large effect on the total if that count is very unlikely.

In practical calculations, one can evaluate $q_n(T)$ analytically for simple severities and thresholds, integrate over ordered arrival times when feasible, or estimate the conditional event by simulation. If a numerical calculation truncates the infinite sum at $n=m$, the omitted contribution is nonnegative and no larger than $P(N(T)>m)$, because $0\leq q_n(T)\leq1$. This gives a direct check on the error due solely to dropping the Poisson tail.

Common Exam Traps

Ignoring the time of a claim. Premium income accumulates before the claim. The condition at claim $k$ is $\sum_{j=1}^kX_j>u+cT_{(k)}$, not simply $\sum_{j=1}^kX_j>u$.
Treating the ordered times as independent uniforms. The unordered times are independent uniforms conditional on the count; the ordered times are dependent. For example, $P(T_{(2)}<a)=(a/T)^2$ when $0\leq a\leq T$, since both arrival times must be below $a$.
Checking only the final aggregate amount. Ruin is a path event. A surplus can become negative at an early claim even if later premium income would have raised it, so check each post-claim surplus.
Calling a partial count contribution the total. A term $P(N(T)=n)q_n(T)$ accounts only for that count. The total requires summing all count contributions, or explicitly establishing that other counts have zero or certain ruin.
Changing the ruin convention silently. This tutorial uses strict ruin, $U(t)<0$. With discrete claim amounts, equality at zero can have positive probability, so changing to a non-strict convention can change the answer.

Key takeaway: Condition on $N(T)=n$, use the ordered-uniform distribution of arrival times and the claim-size distribution to determine $q_n(T)$, then weight by the Poisson probability and sum over $n$. The formula is exact; the main work is identifying the conditional path event correctly.

Check Your Understanding

For each problem, use strict ruin and state whether the requested probability is a total ruin probability or only one claim-count contribution.

A hypothetical insurer has $u=\$2$, premium rate $c=\$1$ per year, claim rate $\lambda=2$ per year, and horizon $T=1$ year. Every claim is $\$5$. Find the probability of ruin by time 1. (Answer: $1-e^{-2}=0.8647$.)
Let $u=\$1$, $c=\$1$ per year, $\lambda=1$ per year, and $T=2$ years. Every claim is $\$2$. Find the total probability of ruin by time 2. (Answer: $1-2e^{-2}=0.7293$.)
Let $u=\$5$, $c=\$1$ per year, $\lambda=1$ per year, and $T=3$ years. Every claim is $\$3$. Find the total probability of ruin by time 3. (Answer: $1-8e^{-3}=0.6017$.)
Let $u=\$1$, $c=\$2$ per year, $\lambda=2$ per year, and $T=1$ year. Claim sizes are Exponential with rate 1 per dollar. Find the probability of ruin with exactly one claim by time 1. (Answer: $e^{-3}-e^{-5}=0.0430$.)

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