Finite-Horizon and Ultimate Ruin

Why the Time Horizon Matters

A portfolio can avoid ruin for several years and still be ruined later. Conversely, a probability of eventual ruin does not say how much risk occurs within the next year or any other specified period. Those are different questions, so the time horizon must be explicit whenever a ruin probability is reported.

This tutorial distinguishes ruin by a specified time from ruin at any future time. The key relationship is simple but important: ruin by time $t$ is an event contained within ruin at some finite future time. As the horizon grows, the finite-horizon probability cannot decrease, and its limit is the ultimate ruin probability.

Definition: In the classical surplus model, let $U(s)=u+cs-S(s)$, where $u\geq 0$ is initial surplus, $c\geq 0$ is the premium rate, and $S(s)$ is cumulative claims by time $s$. Define the ruin time by $T=\inf\{s\geq 0: U(s)<0\}$, taking $T=\infty$ if the set is empty. The finite-horizon ruin probability is $\psi(u,t)=P(T\leq t)$. The ultimate ruin probability is $\psi(u)=P(T<\infty)$.

The model is often specified further by assuming that claims arrive according to a Poisson process of rate $\lambda$, with independent, identically distributed claim amounts independent of the arrival process. The definitions of $\psi(u,t)$ and $\psi(u)$ do not themselves require those particular assumptions. In any application, state the claim model and the convention for ruin. Here ruin means that surplus becomes strictly negative; if a model allows a positive probability of surplus being exactly zero, a different ruin convention can change the answer.

How the Two Probabilities Are Related

For any fixed $t$, the event $\{T\leq t\}$ is a subset of $\{T<\infty\}$. Therefore,

0\leq \psi(u,t)\leq \psi(u).

Also, if $0\leq t_1\leq t_2$, then $\{T\leq t_1\}$ is a subset of $\{T\leq t_2\}$. Hence $\psi(u,t_1)\leq\psi(u,t_2)$: extending the observation period cannot reduce the probability that ruin has occurred by its end.

The limit statement follows from continuity of probability for increasing events. Consider horizons $1,2,3,\ldots$ in the chosen time unit. The events $\{T\leq n\}$ increase as $n$ increases, and their union is exactly $\{T<\infty\}$. Thus,

\lim_{n\to\infty}\psi(u,n) =P\left(\bigcup_{n=1}^{\infty}\{T\leq n\}\right) =P(T<\infty) =\psi(u).

The same limit holds as a real-valued horizon $t$ increases without bound, since the finite-horizon probability is nondecreasing and the integer horizons grow without bound. This result identifies a limit; it does not provide a rate of convergence. In particular, a finite-horizon probability need not equal the ultimate probability at any finite time.

Theorem: For a fixed initial surplus, $\psi(u,t)$ is nondecreasing in $t$, is no greater than $\psi(u)$, and converges to $\psi(u)$ as $t\to\infty$. A finite horizon measures short-term risk; the ultimate probability includes ruin at every finite future time.

An Exact Case and a Useful Bound

Some models make the distinction especially clear. If claims have a fixed positive size, there is no premium income, and every claim exceeds initial surplus, then ruin occurs at the first claim. For a Poisson claim process, the finite-horizon probability is the probability of at least one claim by that horizon. In more general models, ruin may occur after several claims or after a claim that does not itself cause ruin. A probability based only on the first claim causing ruin is then a lower bound, not necessarily the full finite-horizon probability.

A useful calculation is available when arrivals are Poisson and claim amounts are exponential. Suppose claim amounts, measured in thousands of dollars, have exponential rate $\beta$, so $P(X>x)=e^{-\beta x}$ for $x\geq0$. Let $W$ be the waiting time to the first claim. Since $W$ is exponential with rate $\lambda$, the probability that the first claim occurs by $t$ and causes ruin is

\begin{aligned} P(W\leq t,\ X_1>u+cW) &=\int_0^t \lambda e^{-\lambda s}P(X_1>u+cs)\,ds\\ &=\int_0^t \lambda e^{-\lambda s}e^{-\beta(u+cs)}\,ds\\ &=\frac{\lambda e^{-\beta u}}{\lambda+\beta c} \left(1-e^{-(\lambda+\beta c)t}\right). \end{aligned}

Here $\lambda$ is measured per year, $u$ and $X_1$ in thousands of dollars, $c$ in thousands of dollars per year, and $\beta$ per thousand dollars. The exponent $(\lambda+\beta c)t$ is dimensionless. This event is included in ruin by time $t$, so the expression is a lower bound for $\psi(u,t)$. It does not count ruin caused by later claims when the first claim leaves the surplus nonnegative.

For exponential claims, the ultimate ruin probability has a convenient exact form when the premium rate satisfies the positive net profit condition $c>\lambda/\beta$:

\psi(u)=\frac{\lambda}{c\beta} \exp\left[-\left(\beta-\frac{\lambda}{c}\right)u\right].

This formula is specific to the compound Poisson model with exponential claim amounts and the stated condition. It is not a general formula for finite-horizon ruin, and it should not be used if the model assumptions or positive net profit condition fail.

Worked Applications

Worked Example: Exact Finite-Horizon Ruin With No Premiums

A hypothetical fund starts with $u=\$2{,}000$, receives no premium income, and pays a fixed $\$5{,}000$ at each claim. Claims arrive according to a Poisson process with rate $\lambda=0.3$ per year. Find the probability of ruin within four years and the ultimate ruin probability.

Method. With $c=0$, the surplus remains at $\$2{,}000$ between claims. A single $\$5{,}000$ claim makes surplus negative, so ruin occurs if and only if at least one claim occurs.

The number of claims in four years is Poisson with mean $\lambda t=0.3(4)=1.2$. Therefore,

\begin{aligned} \psi(2{,}000,4) &=P(N(4)\geq1)\\ &=1-P(N(4)=0)\\ &=1-e^{-1.2}\\ &=0.6988. \end{aligned}

For the ultimate probability, the chance of no claims by time $t$ is $e^{-0.3t}$, which tends to zero as $t$ tends to infinity. Thus a claim eventually occurs with probability one, and each claim causes ruin. Hence $\psi(2{,}000)=1$. The check $0.6988\leq1$ agrees with the general finite-versus-ultimate inequality.

Common mistake avoided: The four-year probability is not the probability of ultimate ruin. It counts only claims during the first four years. This example has no premium income and is used for its exact calculation; it is not a positive-loading portfolio.

Worked Example: Bounding Ruin for an Exponential-Claim Portfolio

A hypothetical insurer has initial surplus $u=0.5$ thousand dollars. Claims arrive at rate $\lambda=0.5$ per year, and claim amounts are exponential with rate $\beta=1$ per thousand dollars, so the mean claim is $\$1{,}000$. The premium rate is $c=0.75$ thousand dollars per year. Find a lower bound for ruin within two years and the exact ultimate ruin probability.

Method. First calculate the probability that the first claim arrives by year two and by itself causes ruin. Then use the exponential-claim ultimate formula. The positive net profit condition holds because $c=0.75>\lambda/\beta=0.5$, with both rates expressed in thousands of dollars per year.

For the first-claim lower bound, $\lambda+\beta c=0.5+1(0.75)=1.25$ per year and $\beta u=1(0.5)=0.5$. Therefore,

\begin{aligned} P(W\leq2,\ X_1>u+cW) &=\frac{0.5e^{-0.5}}{1.25} \left(1-e^{-1.25(2)}\right)\\ &=\frac{0.5(0.6065)}{1.25}(1-e^{-2.5})\\ &=\frac{0.3033}{1.25}(1-0.0821)\\ &=0.2227. \end{aligned}

Thus $\psi(0.5,2)\geq0.2227$. For ultimate ruin, the adjustment exponent in this exponential model is $\beta-\lambda/c=1-0.5/0.75=1/3$ per thousand dollars, and the prefactor is $\lambda/(c\beta)=0.5/(0.75)=2/3$. Hence

\psi(0.5)=\frac{2}{3}e^{-(1/3)(0.5)} =\frac{2}{3}e^{-1/6} =0.5643.

Combining the results gives $0.2227\leq\psi(0.5,2)\leq0.5643$. These bounds do not determine the exact two-year ruin probability: the lower bound counts first-claim ruin, while the upper bound includes ruin at all future times.

Common mistake avoided: Substituting $t=2$ into the ultimate formula would not produce a finite-horizon probability. The ultimate formula has no time argument and is used here only as an upper bound.

Worked Example: How a Longer Horizon Changes the Bound

Use the same model as in the preceding example, but extend the horizon from two years to four years. Calculate the first-claim lower bound at four years, compare it with the two-year lower bound, and state what can be concluded about the exact probabilities.

Method. Apply the first-claim expression with the same parameters, changing only $t$. The factor $\lambda e^{-\beta u}/(\lambda+\beta c)$ remains $0.3033/1.25$, or approximately $0.2426$.

\begin{aligned} P(W\leq4,\ X_1>u+cW) &=\frac{0.5e^{-0.5}}{1.25} \left(1-e^{-1.25(4)}\right)\\ &=\frac{0.3033}{1.25}(1-e^{-5})\\ &=0.2426(1-0.0067)\\ &=0.2410. \end{aligned}

The first-claim lower bound rises from $0.2227$ at two years to $0.2410$ at four years. The exact four-year probability is at least $0.2410$ and no greater than the ultimate probability $0.5643$. In addition, the exact four-year probability is at least the exact two-year probability, whether or not either exact value is known. The two calculated lower bounds alone do not establish the exact increase in the full ruin probabilities.

Common mistake avoided: It would be incorrect to call $0.2410$ the exact four-year probability. Ruin after a second or later claim is not included in this first-claim calculation.

Interpreting Results and Avoiding Traps

The distinction matters in practice because a short-term solvency question and a long-term solvency question have different time scales. A finite-horizon probability is the relevant quantity when the period is specified, such as the next year. Ultimate ruin is a long-run measure. Neither number can be substituted for the other without a stated reason.

Keep the horizon attached to the probability. Write $\psi(u,t)$ for a specified horizon and $\psi(u)$ for ultimate ruin. A number without its time period can be ambiguous.
Use event inclusion carefully. Any valid calculation of a particular type of ruin by time $t$, such as first-claim ruin, gives a lower bound for $\psi(u,t)$. The ultimate probability gives an upper bound.
Do not confuse monotonicity with equality. A larger horizon cannot lower the exact finite-horizon probability, but a lower-bound calculation for a larger horizon need not equal that probability.
Check the model before using an exact formula. The exponential-claim ultimate formula requires the compound Poisson and exponential assumptions and $c>\lambda/\beta$. Its validity does not extend automatically to other claim distributions or to finite horizons.
Check units in exponents. Terms such as $\beta u$ and $(\lambda+\beta c)t$ must be dimensionless. If claim amounts are measured in thousands, use $\beta$ per thousand dollars and $c$ in thousands of dollars per year.

Key takeaway: Finite-horizon ruin is the probability that ruin has occurred by a specified time; ultimate ruin is the probability that ruin occurs at any finite future time. The former increases with the horizon and converges to the latter, but generally is not equal to it at a finite horizon.

Check Your Understanding

For each problem, state clearly whether the requested quantity is exact or a bound.

A hypothetical fund has initial surplus $\$1{,}000$, no premium income, fixed claims of $\$3{,}000$, and Poisson claim arrivals at rate $0.2$ per year. Find the exact probability of ruin within five years and the ultimate ruin probability. (Answer: $1-e^{-1}=0.6321$ within five years; ultimate probability $1$.)
In a fixed surplus model, explain why the probability of ruin within three years cannot exceed the probability of ruin within five years, and why the five-year probability cannot exceed the ultimate probability. (Answer: $\{T\leq3\}\subseteq\{T\leq5\}\subseteq\{T<\infty\}$, so $\psi(u,3)\leq\psi(u,5)\leq\psi(u)$.)
Claims are Poisson with rate $\lambda=1$ per year and exponential with $\beta=1$ per thousand dollars. Let $u=0$, $c=1.5$ thousand dollars per year, and $t=1$ year. Find the first-claim ruin lower bound. (Answer: $(1-e^{-2.5})/2.5=0.3672$, a lower bound for $\psi(0,1)$.)
Claims are Poisson with rate $\lambda=0.4$ per year and exponential with $\beta=2$ per thousand dollars. The premium rate is $c=0.4$ thousand dollars per year and initial surplus is $u=1.5$ thousand dollars. Find the ultimate ruin probability and give the resulting upper bound for ruin within two years. (Answer: $\psi(1.5)=0.5e^{-1.5}=0.1116$, so $0\leq\psi(1.5,2)\leq0.1116$.)
A calculation gives $\psi(u,2)=0.18$ and $\psi(u,6)=0.31$ for the same model. What can be said about $\psi(u)$, and what can be said about the probability of ruin after year six? (Answer: $\psi(u)\geq0.31$; the probability of ruin after year six is $\psi(u)-0.31$, which is nonnegative but cannot be determined from the two finite-horizon values alone.)

What You'll Learn