Strict and Non-Strict Ruin with Discrete Claims

Why the Ruin Boundary Matters

In a surplus model, ruin is determined by whether claims exhaust the available funds. Two conventions are common: ruin may mean that surplus becomes negative, or that surplus becomes zero or negative. The distinction is invisible when surplus has no probability of landing exactly on zero. With discrete claim amounts, however, equality can have positive probability, so the convention can change a calculated ruin probability.

This tutorial uses a discrete-time model to make the boundary explicit. The insurer begins with capital $u$, receives premium $c$ per period before that period’s claims are settled, and pays claims $X_k$ at the end of period $k$. Interest is ignored, and claims are nonnegative. The period-$k$ surplus is

U_k=u+kc-\sum_{j=1}^{k}X_j,\qquad k=1,\ldots,m.

Here $m$ is the finite horizon. The timing convention matters: the premium for period $k$ is included in $U_k$ before the period’s claims are paid. The examples below use this convention throughout.

Definition: Strict ruin by period $m$ occurs if $U_k<0$ for at least one $k=1,\ldots,m$. Non-strict ruin by period $m$ occurs if $U_k\leq 0$ for at least one such $k$. Their probabilities are denoted here by $\psi_{<}(u,m)$ and $\psi_{\leq}(u,m)$, respectively. These definitions exclude any ruin at time zero; if $u=0$, a model that counts time zero would need to address that initial boundary separately.

The Boundary Probability

For a finite horizon, define the minimum period-end surplus by $M_m=\min_{1\leq k\leq m}U_k$. Strict ruin is the event $M_m<0$, while non-strict ruin is $M_m\leq0$. The latter event contains the former, and the only additional outcomes are those for which the minimum is exactly zero.

\psi_{\leq}(u,m)-\psi_{<}(u,m) =P(M_m\leq0)-P(M_m<0) =P(M_m=0).

The last equality follows because the event $M_m\leq0$ is the disjoint union of $M_m<0$ and $M_m=0$. Thus the difference is not an adjustment or approximation: it is precisely the probability that the process reaches zero without ever becoming negative during the horizon.

At period $k$, equality occurs when cumulative claims exactly match the capital and premiums available by that period:

\sum_{j=1}^{k}X_j=u+kc.

If claims have a discrete distribution, this equation may hold with positive probability. With claim amounts having a continuous distribution, equality often has probability zero under ordinary model conditions, but one should establish that from the model rather than assume it. Timing, deterministic amounts, or an atom in the surplus distribution can all make the boundary relevant.

Worked Example: One Period and an Equality Atom

A hypothetical insurer starts with $u=\$4$, collects $c=\$2$ before a single period-end claim, and has claim amount $X_1$ with the following distribution: $P(X_1=\$2)=0.4$, $P(X_1=\$4)=0.4$, and $P(X_1=\$6)=0.2$. Find strict and non-strict ruin probabilities.

Method. First calculate the amount available for claims at the settlement time:

u+c=\$4+\$2=\$6.

The period-end surplus is $U_1=\$6-X_1$. Strict ruin requires $U_1<0$, or equivalently $X_1>\$6$. None of the three possible claim amounts exceeds $\$6$, so

\psi_{<}(4,1)=P(X_1>6)=0.

Non-strict ruin requires $U_1\leq0$, equivalently $X_1\geq\$6$. The only claim amount satisfying that condition is $\$6$, which has probability $0.2$. Therefore $\psi_{\leq}(4,1)=0.2$. The boundary probability is $P(U_1=0)=P(X_1=\$6)=0.2$, and the check $0.2-0=0.2$ agrees with the boundary-probability identity.

Common mistake avoided: treating equality as strict ruin. A claim of $\$6$ leaves surplus exactly zero; it causes non-strict ruin, not strict ruin.

Tracking Ruin Across Multiple Periods

In a multi-period calculation, ruin is an event over the path, not just a test of surplus at the final period. Once a path has crossed the relevant boundary, later premiums or claims do not undo the fact that ruin has already occurred. A useful procedure is to check each period in order and retain only paths that have not already met the ruin condition.

Compute available funds.
At period $k$, compare cumulative claims with $u+kc$, using the stated premium timing.

Apply the selected boundary.
Use cumulative claims $>u+kc$ for strict ruin and $\geq u+kc$ for non-strict ruin.

Combine path probabilities.
Count a path at its first ruin period only, so that paths meeting the condition in more than one period are not counted twice.

Check the difference.
Confirm that non-strict ruin probability is at least strict ruin probability; their difference is the probability that the minimum surplus is exactly zero.

Worked Example: Two Periods with Discrete Claims

A hypothetical insurer has initial capital $u=\$3$, receives $c=\$1$ before claims in each of two periods, and has independent period claims with $P(X_k=0)=0.25$, $P(X_k=2)=0.50$, and $P(X_k=5)=0.25$. Find the strict and non-strict ruin probabilities by period 2.

At period 1, available funds are $u+c=\$4$. A claim of $\$5$ causes strict ruin, with probability $0.25$. No claim amount equals $\$4$, so there is no period-1 equality outcome.

At period 2, cumulative available funds are $u+2c=\$5$. First consider paths that survived period 1, which means $X_1$ is $0$ or $2$. If $X_1=0$, none of the possible second claims makes cumulative claims exceed $\$5$. If $X_1=2$, a second claim of $\$5$ makes cumulative claims $\$7$, exceeding $\$5$. This path has probability $0.50(0.25)=0.125$. It follows that

\psi_{<}(3,2)=0.25+0.125=0.375.

For non-strict ruin, the period-1 claim of $\$5$ still counts. Among paths surviving period 1, period-2 cumulative claims must be at least $\$5$. If $X_1=0$, only $X_2=5$ qualifies, with probability $0.25(0.25)=0.0625$. If $X_1=2$, only $X_2=5$ qualifies, with probability $0.50(0.25)=0.125$. Thus

\psi_{\leq}(3,2)=0.25+0.0625+0.125=0.4375.

The difference is $0.4375-0.375=0.0625$. It is exactly the probability of $X_1=0,\ X_2=5$, the only surviving path that reaches zero surplus at period 2. The calculation also checks the path logic: outcomes with $X_1=5$ are counted once, at period 1.

Common mistake avoided: adding period-specific ruin probabilities without restricting later-period calculations to paths that survived earlier periods. That can count the same path more than once.

A Compound Claim Count Can Create the Boundary Atom

Discrete claim sizes can create a boundary atom even when the number of claims is random. In a one-period aggregate-settlement model, let $N$ be the number of claims, and suppose each claim has a fixed amount $b$. The total claim amount is $bN$. If the available funds equal $bn$ for an integer $n$, then $N=n$ produces exactly zero surplus. The probability of that claim count contributes to the difference between the two ruin conventions.

Worked Example: A Poisson Number of Fixed Claims

In a hypothetical annual aggregate-settlement model, an insurer starts with $u=\$4$, receives an annual premium $c=\$2$ before settlement, and pays all claims at the end of the year. The number of claims satisfies $N\sim\operatorname{Poisson}(0.5)$, and each claim is exactly $\$2$. There is no interest or mortality assumption in this one-period model. Find both ruin probabilities.

Available funds are $u+c=\$6$; aggregate claims are $2N$ dollars. Strict ruin requires $2N>6$, or $N\geq4$. Non-strict ruin requires $2N\geq6$, or $N\geq3$. For a Poisson random variable with mean $0.5$, $P(N=k)=e^{-0.5}(0.5)^k/k!$. Therefore,

\begin{aligned} P(N\geq4) &=1-P(N=0)-P(N=1)-P(N=2)-P(N=3)\\ &=1-e^{-0.5}\left(1+0.5+0.125+\frac{0.125}{6}\right)\\ &=1-e^{-0.5}\left(\frac{79}{48}\right)\\ &\approx 0.0018. \end{aligned}

Similarly,

\begin{aligned} P(N\geq3) &=1-P(N=0)-P(N=1)-P(N=2)\\ &=1-e^{-0.5}(1+0.5+0.125)\\ &\approx 0.0144. \end{aligned}

Thus, to four decimal places, strict ruin is $0.0018$ and non-strict ruin is $0.0144$. Their difference is $P(N=3)=e^{-0.5}(0.5)^3/3!\approx0.0126$, the probability of exactly $\$6$ in claims and hence exactly zero surplus. The result is checked by the event inclusions: $N\geq4$ is contained in $N\geq3$, and the gap is precisely $N=3$.

Common mistake avoided: using $N\geq4$ for both conventions. Three claims exhaust the $\$6$ available; they do not make surplus negative, but they do satisfy non-strict ruin.

Interpretation and Exam Traps

The choice of ruin convention must be read from the definition in the problem, not inferred from the word “ruin” alone. The practical distinction is the inequality at the boundary:

Convention	Surplus condition	Cumulative-claims condition at period $k$
Strict ruin	$U_k<0$	$\sum_{j=1}^{k}X_j>u+kc$
Non-strict ruin	$U_k\leq0$	$\sum_{j=1}^{k}X_j\geq u+kc$

A second trap is applying a continuous-time intuition to a discrete settlement model. In the examples here, claims are settled at specified period ends, so the amount available at each settlement is well defined and equality can have positive probability. A continuous-time model with premium income accumulating between claims has a different path structure; in particular, a claim’s arrival time affects the surplus just before it occurs. Do not transfer a period-end threshold to that setting without deriving the appropriate condition.

For calculations, a reliable check is to identify the equality event separately. If its probability is zero, both conventions give the same result. If it is positive, adding that probability to strict ruin must produce non-strict ruin. This provides both a conceptual explanation and a numerical check.

Key takeaway: Strict ruin tests whether surplus becomes negative; non-strict ruin tests whether it reaches zero or below. In a discrete model, their probability difference is exactly the probability that the minimum surplus over the specified horizon equals zero.

Check Your Understanding

For each problem, apply the stated timing convention and distinguish equality from an actual deficit.

A hypothetical insurer has $u=\$7$, receives $c=\$1$ before one period-end claim, and $P(X_1=\$4)=0.2$, $P(X_1=\$8)=0.5$, $P(X_1=\$9)=0.3$. Find strict and non-strict ruin probabilities. (Answer: strict $=0.3000$; non-strict $=0.8000$.)
In a two-period model, $u=\$3$, $c=\$1$ per period before claims, and independent claims satisfy $P(X_k=0)=0.25$, $P(X_k=2)=0.50$, $P(X_k=5)=0.25$. Find the probability of reaching exactly zero surplus at period 2 without earlier ruin. (Answer: $0.0625$.)
A one-period aggregate-settlement model has available funds of $\$6$, a Poisson claim count with mean $0.5$, and fixed claim size $\$2$. Find strict and non-strict ruin probabilities to four decimal places. (Answer: strict $=0.0018$; non-strict $=0.0144$.)
In a one-period model, available funds are $\$5$, and the aggregate claim amount is $\$3N$, where $N\sim\operatorname{Poisson}(0.4)$. Find strict and non-strict ruin probabilities to four decimal places. (Answer: strict $=P(N\geq2)=0.0616$; non-strict $=P(N\geq2)=0.0616$, since no integer $N$ gives aggregate claims exactly $\$5$.)
For any finite horizon with positive initial capital, explain the event whose probability equals non-strict ruin probability minus strict ruin probability. (Answer: the probability that the minimum period-end surplus over the horizon is exactly zero.)

Convention	Surplus condition	Cumulative-claims condition at period \(k\)
Strict ruin	\(U_k<0\)	\(\sum_{j=1}^{k}X_j>u+kc\)
Non-strict ruin	\(U_k\leq0\)	\(\sum_{j=1}^{k}X_j\geq u+kc\)

What You'll Learn