Comparing Finite-Horizon Ruin with Ultimate Bounds

Finite-Horizon and Ultimate Ruin

A finite-horizon calculation answers whether ruin occurs by a specified time $T$. An ultimate ruin probability asks whether ruin occurs at any time, however far in the future. The distinction matters: a probability for a limited operating period cannot exceed the probability of ruin over the entire future, but the ultimate probability does not say exactly how likely ruin is within that limited period.

Use the classical risk process with initial surplus $u$, premium income at constant rate $c$, and aggregate claims $S(t)$:

$$ U(t)=u+ct-S(t). $$

Assume claims arrive according to a Poisson process with rate $\lambda>0$, and their amounts are independent and identically distributed nonnegative random variables $X_1,X_2,\ldots$, independent of the arrival process. There is no interest accumulation or mortality in this surplus model. Define the time of ruin and the two ruin probabilities as follows.

Definition: The time of ruin is $\tau=\inf\{t\geq 0:U(t)<0\}$, with $\tau=\infty$ if the surplus never becomes negative. For $T\geq 0$, the finite-horizon ruin probability is $\psi_T(u)=P(\tau\leq T)$. The ultimate ruin probability is $\psi(u)=P(\tau<\infty)$.

The definition uses strict ruin, meaning that surplus must fall below zero. In the continuous-claim examples below, the chance of a claim landing exactly at a specified threshold is zero, so using a non-strict inequality would not change those calculations.

Event Inclusion Gives the First Comparison

If ruin occurs by time $T$, it has occurred at some finite time. Therefore the event of ruin by $T$ is contained in the event of ultimate ruin. For $0\leq T_1\leq T_2$, ruin by $T_1$ is also contained in ruin by $T_2$. Taking probabilities gives

\psi_{T_1}(u)\leq \psi_{T_2}(u)\leq \psi(u).

As $T$ increases without bound, the events $\{\tau\leq T\}$ increase to $\{\tau<\infty\}$. Consequently, $\psi_T(u)$ increases to $\psi(u)$. This is a limiting relationship, not a claim that the finite-horizon probability equals the ultimate probability for any particular finite $T$.

The Lundberg Exponential Bound

Let $M_X(r)=E[e^{rX}]$ be the moment-generating function of claim size. Suppose the premium rate has positive safety loading, $c>\lambda E[X]$, and there is a positive adjustment coefficient $R$ for which $M_X(R)$ is finite and

\lambda\bigl(M_X(R)-1\bigr)=cR.

Under these classical risk-model assumptions, the Lundberg inequality bounds ultimate ruin. Since finite-horizon ruin is a subset of ultimate ruin, the same bound also applies at every finite horizon.

Theorem: If $R>0$ satisfies the adjustment-coefficient equation under the stated classical model, then $$ \psi_T(u)\leq\psi(u)\leq e^{-Ru},\qquad u\geq 0,\ T\geq 0. $$ The exponential expression is an upper bound, not generally the ruin probability itself.

The key idea behind the exponential bound is that the adjustment-coefficient equation makes $e^{-RU(t)}$ a martingale. Indeed, over an interval of length $s$, the increment in aggregate claims is independent of the past and has a Poisson number of terms. Thus

E\!\left[e^{-R(U(t+s)-U(t))}\right] =e^{-Rcs}\exp\!\left(\lambda s(M_X(R)-1)\right)=1.

Stop this nonnegative martingale at ruin or at a fixed time $t$, whichever comes first. On the event $\{\tau\leq t\}$, strict ruin implies $U(\tau)<0$, so $e^{-RU(\tau)}>1$. The stopped martingale has expectation $e^{-Ru}$; its contribution from ruin by time $t$ is therefore at least $P(\tau\leq t)$. Hence $P(\tau\leq t)\leq e^{-Ru}$. Letting $t$ increase gives $\psi(u)\leq e^{-Ru}$, and event inclusion gives the finite-horizon result. This argument requires the adjustment-coefficient equation and its finiteness conditions; without an applicable positive $R$, this particular bound is not available.

Exponential Claims: A Sharper Ultimate Comparison

For exponential claim amounts with rate parameter $\beta>0$, the mean claim amount is $1/\beta$, and $M_X(r)=\beta/(\beta-r)$ for $r<\beta$. With premium rate $c=(1+\theta)\lambda/\beta$ and positive safety loading $\theta>0$, the positive adjustment coefficient is

R=\beta-\frac{\lambda}{c}=\frac{\beta\theta}{1+\theta}.

In this exponential-claim model, the exact ultimate ruin probability is $\psi(u)=\frac{1}{1+\theta}e^{-Ru}$. It is no larger than the Lundberg bound $e^{-Ru}$, because $1/(1+\theta)<1$. Both expressions bound finite-horizon ruin, but only the first is the exact ultimate probability under these specific assumptions.

Worked Example: Comparing a Finite-Horizon Lower Bound with Ultimate Bounds

A hypothetical insurer has Poisson claim arrivals at rate $\lambda=2$ per year. Claim amounts are exponential with rate $\beta=0.001$ per dollar, so the mean is $1/\beta=\$1{,}000$. The safety loading is $\theta=0.5$, the initial surplus is $u=\$3{,}000$, and the horizon is $T=0.5$ year. There is no interest or mortality assumption in this surplus model. Find an upper bound for finite-horizon ruin, the exact ultimate probability, and a lower bound for ruin by $T$ from the event that the first claim causes ruin.

Method and adjustment coefficient. The premium rate is $c=(1+\theta)\lambda/\beta=1.5(2)/0.001=\$3{,}000$ per year. The exponential-claim adjustment coefficient is $R=\beta\theta/(1+\theta)=0.001(0.5)/1.5=0.000333333$ per dollar.

Ultimate comparisons. Here $Ru=0.000333333(3{,}000)=1$. Thus the Lundberg upper bound is $e^{-1}=0.3679$, and the exact exponential-claim ultimate probability is $(1/1.5)e^{-1}=0.2453$. Event inclusion gives $\psi_{0.5}(3{,}000)\leq0.2453$, as well as the looser bound $0.3679$.

First-claim lower bound. Before the first claim at time $t$, surplus is $u+ct$. A first claim larger than this amount causes immediate ruin. The first arrival has density $\lambda e^{-\lambda t}$, and the exponential severity tail is $S_X(x)=e^{-\beta x}$ for $x\geq0$. Integrating the probability of this event gives

\begin{aligned} P(\text{first claim causes ruin by }T) &=\int_0^T\lambda e^{-\lambda t}e^{-\beta(u+ct)}\,dt\\ &=\lambda e^{-\beta u}\frac{1-e^{-(\lambda+\beta c)T}}{\lambda+\beta c}\\ &=2e^{-3}\frac{1-e^{-2.5}}{5}\\ &=0.0183. \end{aligned}

Therefore $0.0183\leq\psi_{0.5}(3{,}000)\leq0.2453$. The lower bound counts only ruin caused by the first claim; ruin after two or more claims can add probability. Probabilities are rounded to four decimal places. The mistake to avoid is reporting $e^{-1}=0.3679$ as the finite-horizon probability: it is only an upper bound, and in this example the exact ultimate probability is already smaller.

Worked Example: Choosing Capital for a Finite-Horizon Target

A hypothetical insurer has Poisson arrivals at rate $\lambda=4$ per year and exponential claim amounts with rate $\beta=0.002$ per dollar, hence mean $\$500$. Its safety loading is $\theta=0.25$. Find the smallest whole-dollar initial surplus for which the finite-horizon ruin probability, for any chosen horizon, is guaranteed to be at most $0.05$ using the exact ultimate ruin probability. There is no interest or mortality in this model.

Method. The premium rate is $c=(1.25)(4)/0.002=\$2{,}500$ per year, and $R=0.002(0.25)/1.25=0.0004$ per dollar. The exact ultimate probability is $0.8e^{-0.0004u}$, so the event-inclusion comparison guarantees the target whenever

0.8e^{-0.0004u}\leq0.05 \quad\Longleftrightarrow\quad u\geq\frac{\ln(0.8/0.05)}{0.0004} =\frac{\ln(16)}{0.0004} =6{,}931.4718\ldots.

The smallest whole-dollar amount meeting this condition is $\$6{,}932$. To check the rounding, $0.0004(6{,}932)=2.7728>\ln(16)$, so the bound is below $0.05$ at $\$6{,}932$; at $\$6{,}931$, the exponent $2.7724<\ln(16)$, so it is above $0.05$. The result applies to every finite horizon because each such probability is no larger than the ultimate probability.

If one used only the Lundberg bound, the sufficient condition would instead be $e^{-0.0004u}\leq0.05$, requiring $u\geq\ln(20)/0.0004=7{,}489.3306\ldots$, or $\$7{,}490$ in whole dollars. That is a valid but less sharp capital requirement. The correction to the common mistake of using the looser bound as an exact probability is to use the exact ultimate formula when its exponential-claim assumptions hold, while still describing the resulting finite-horizon guarantee as a bound.

Worked Example: Comparing Two Horizons Without Treating a Lower Bound as Exact

Consider a hypothetical model with Poisson arrivals at $\lambda=1$ per year, exponential claim amounts with rate $\beta=1$ per dollar (mean $\$1$), safety loading $\theta=1$, and initial surplus $u=\$2$. Compare bounds for horizons $T=1$ and $T=2$. There is no interest or mortality assumption.

Ultimate bounds. The premium rate is $c=(1+1)(1)/1=\$2$ per year and $R=1(1)/(1+1)=0.5$ per dollar. Thus the exact ultimate probability is $0.5e^{-1}=0.1839$, while the Lundberg bound is $e^{-1}=0.3679$. Both are upper bounds for either finite horizon.

First-claim lower bounds. For a first claim to cause ruin by $T$, it must arrive at time $t\leq T$ and exceed $u+ct$. Integrating as in the preceding example gives

P(\text{first claim causes ruin by }T) =\frac{e^{-2}}{3}\left(1-e^{-3T}\right).

At $T=1$, this is $e^{-2}(1-e^{-3})/3=0.0429$. At $T=2$, it is $e^{-2}(1-e^{-6})/3=0.0450$. Therefore the comparisons are $0.0429\leq\psi_1(2)\leq0.1839$ and $0.0450\leq\psi_2(2)\leq0.1839$, with probabilities rounded to four decimal places. The finite-horizon probabilities themselves cannot decrease as the horizon grows, by event inclusion. These particular lower bounds also increase here, but a lower bound based on a selected subset of ruin events need not generally reveal the size of the full probability or equal it. The mistake to avoid is treating the first-claim calculation as the exact finite-horizon probability; it omits ruin after multiple claims.

Using the Comparisons Correctly

The event-inclusion comparison and the Lundberg inequality answer different questions. Event inclusion is exact and requires only the definitions of the two ruin events. The Lundberg inequality adds model conditions and supplies an upper bound. In an exponential-claim model, an exact ultimate formula may sharpen that upper bound, but it still does not give the exact probability of ruin by a finite time.

Specify the model.
State the claim arrival process, severity distribution, premium rate, initial surplus, and the horizon. Check that the adjustment-coefficient assumptions apply before using the Lundberg inequality.

Find the applicable ultimate result.
Use $e^{-Ru}$ as an upper bound when a valid positive adjustment coefficient exists. Use an exact ultimate formula only when its additional distributional assumptions are satisfied.

Transfer the upper bound to the finite horizon.
Apply $\psi_T(u)\leq\psi(u)$. If useful, construct a lower bound by identifying a specific event that causes ruin by $T$, and verify that this event is contained in $\{\tau\leq T\}$.

These comparisons are useful for screening capital levels and checking finite-horizon calculations. They also provide immediate consistency checks: a reported finite-horizon probability above the applicable ultimate probability is impossible, and a probability above $e^{-Ru}$ violates the Lundberg bound when its assumptions hold. Conversely, being below an upper bound does not establish that a calculation is exact.

Check Your Understanding

Use event inclusion and the stated model assumptions. Probabilities in the answers are rounded to four decimal places.

A classical risk model has $\lambda=3$ per year, exponential claim amounts with rate $\beta=0.004$ per dollar, safety loading $\theta=0.5$, and $u=\$1{,}500$. Find the adjustment coefficient and compare the exponential Lundberg bound with the exact ultimate probability. (Answer: $R=0.0013333$ per dollar; $e^{-Ru}=0.1353$, exact ultimate probability $=0.0902$, and every finite-horizon probability is at most $0.0902$.)
For $\lambda=2$ per year, exponential claim rate $\beta=0.002$ per dollar, safety loading $\theta=0.5$, $u=\$1{,}000$, and $T=0.4$ year, calculate the lower bound from the event that the first claim causes ruin by $T$. (Answer: $0.0468$.)
For exponential claims with $\lambda=1$ per year, $\beta=0.001$ per dollar, and $\theta=1$, find the smallest whole-dollar surplus that guarantees ultimate ruin probability, and hence every finite-horizon ruin probability, at most $0.02$. (Answer: $R=0.0005$ per dollar and the minimum is $\$6{,}438$, using $0.5e^{-0.0005u}\leq0.02$.)
Explain why a finite-horizon ruin probability cannot be larger at $T=1$ than at $T=2$, for the same model and initial surplus. (Answer: $\{\tau\leq1\}\subseteq\{\tau\leq2\}$, so $\psi_1(u)\leq\psi_2(u)$.)
For $\lambda=1$ per year, exponential claims with $\beta=1$ per dollar, $\theta=1$, and $u=\$1$, calculate the first-claim lower bounds at $T=0.5$ and $T=1$, and compare them with the exact ultimate probability. (Answer: lower bounds $0.0952$ and $0.1165$, respectively; exact ultimate probability $0.3033$, so each finite-horizon probability is at most $0.3033$.)

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