Why Severity Tails Matter
In the classical ruin model, the adjustment coefficient is determined jointly by the premium rate, claim frequency, and claim-severity distribution. The net profit condition ensures that expected premium income exceeds expected claim outgo. It does not, by itself, guarantee that the adjustment-coefficient equation has a positive solution. Whether that equation can be used depends importantly on how quickly the severity tail decreases.
A distribution can have a finite expected claim amount but still have a tail too heavy for its moment-generating function to exist at any positive argument. In that case there is no positive adjustment coefficient, even when the net profit condition holds. Conversely, when the moment-generating function is finite near zero, the premium margin and the behavior of that function determine whether a positive root exists and where it lies.
All rates and amounts must use compatible units. For example, if \(X\) is measured in thousands of dollars, \(c\) must be in thousands of dollars per year, and \(R\) is measured per thousand dollars. The Poisson rate \(\lambda\) and premium rate \(c\) must refer to the same time unit.
From Tail Behavior to the MGF Domain
For a nonnegative claim amount, \(M_X(r)\) is finite at \(r=0\), where it equals 1. For positive \(r\), the factor \(e^{rX}\) gives increasing weight to large claims. Thus, the existence of \(M_X(r)\) for some \(r>0\) requires the severity tail to decrease sufficiently quickly.
This identity holds with either side allowed to be infinite. To see why, write \(e^{rX}=1+\int_0^X re^{rt}\,dt\). Taking expectations and interchanging the nonnegative integral and expectation gives \(M_X(r)=1+\int_0^\infty re^{rt}P(X>t)\,dt\), which is the stated formula because \(S_X(t)=P(X>t)\). The identity makes the tail connection explicit: a tail that decreases more slowly than exponential can make the integral diverge for every positive \(r\).
The interval of positive \(r\) for which \(M_X(r)\) is finite is called its positive domain of finiteness. Some distributions have an MGF finite for every real \(r\); others have it finite only below a positive boundary; and some have no positive values in this domain. Having a finite MGF in a neighborhood to the right of zero is a useful starting condition for an adjustment coefficient, but existence of a root still depends on how the MGF behaves across its domain.
Why the Root Depends on the Whole MGF
Define \(g(r)=\lambda(M_X(r)-1)-cr\) wherever \(M_X(r)\) is finite. The adjustment-coefficient equation asks for a positive \(r\) with \(g(r)=0\). At zero, \(g(0)=0\). If the MGF is finite in a neighborhood of zero and \(E[X]\) is finite, its right derivative at zero is \(g'(0)=\lambda E[X]-c<0\) under the net profit condition. Therefore \(g(r)\) is initially negative just to the right of zero. A positive root requires \(g\) eventually to return to zero within the MGF’s domain.
For a nondegenerate nonnegative severity with a finite MGF on an open interval, \(M_X\) is strictly convex there; consequently \(g\) is strictly convex. Its initial negative slope means there can be at most one positive root on that interval. To conclude that a root exists, however, one must also establish that \(g\) becomes positive somewhere in the domain. For example, divergence of \(M_X(r)\) at a finite right endpoint ensures that it eventually exceeds the linear term \(cr\). If the MGF remains finite at the edge of its domain, a positive crossing is not automatic.
Confirm the claim rate, premium rate, severity units, and time units are consistent.
Compute \(\lambda E[X]\) and verify that \(c>\lambda E[X]\).
Determine whether \(M_X(r)\) is finite for any positive \(r\), and identify the range in which the equation can be considered.
Use \(g(0)=0\), \(g'(0)<0\), and the behavior of \(g\) farther into its domain to determine whether a positive root exists.
Worked Calculations and Tail Comparisons
Worked Example: Exponential Severity and the Positive Root
A hypothetical insurer has Poisson claim frequency \(\lambda=2\) claims per year. Claim amounts are exponentially distributed with mean 1 thousand dollars, so the exponential rate is 1 per thousand dollars. Premium income is \(c=2.2\) thousand dollars per year. Determine whether the net profit condition holds and calculate the adjustment coefficient, if it exists.
Method and condition. The mean claim amount is \(E[X]=1\) thousand dollars. Expected claim outgo per year is therefore \(\lambda E[X]=2(1)=2\) thousand dollars. Since \(c=2.2>2\), the net profit condition holds.
For an exponential distribution with mean 1 thousand dollars, the MGF is \(M_X(r)=1/(1-r)\) for \(0\le r<1\), with \(r\) measured per thousand dollars. Substitute into the adjustment-coefficient equation:
For a positive root, divide both sides by \(r\), using \(r>0\), and simplify:
The result is inside the MGF domain \(r<1\). The adjustment coefficient is \(R=0.0909\) per thousand dollars, rounded to four decimal places. Checking in the original equation, \(M_X(R)-1=(1/11)/(10/11)=0.1\), so the left side is \(2(0.1)=0.2\), while the right side is \(2.2(1/11)=0.2\). The two sides agree. This calculation avoids treating the MGF’s domain boundary as the root: the domain limit is 1, whereas the equation gives \(R=1/11\).
Worked Example: A Gamma Severity with a Finite MGF Boundary
A hypothetical portfolio has Poisson claim frequency \(\lambda=1\) per year. Claim severity \(X\), measured in thousands of dollars, has a gamma distribution with shape 2 and scale 0.5. Premium income is \(c=14/9\) thousand dollars per year. Determine whether a positive adjustment coefficient exists and verify its value.
Method and condition. For this gamma distribution, \(E[X]=2(0.5)=1\) thousand dollars. Expected claim outgo is \(\lambda E[X]=1\) thousand dollars per year, and \(c=14/9>1\), so the net profit condition holds. The MGF is \(M_X(r)=(1-0.5r)^{-2}\) for \(0\le r<2\). Its domain ends at \(r=2\).
Test \(r=0.5\), which is within the domain. At this value, \(1-0.5r=1-0.25=0.75=3/4\), so \(M_X(0.5)=(3/4)^{-2}=16/9\). The left side of the adjustment-coefficient equation is \(\lambda(M_X(0.5)-1)=1(16/9-1)=7/9\). The right side is \(c(0.5)=(14/9)(1/2)=7/9\). Thus the equation holds, and \(R=0.5\) per thousand dollars.
The root is within the MGF domain and satisfies the equation by direct substitution. Strict convexity of \(g\), together with its negative initial slope, means this is the unique positive root in the domain. The important check is not just that the net profit condition holds; the proposed root must also be a point where the MGF is finite.
Worked Example: Finite Mean but No Positive MGF
A hypothetical insurer has Poisson claim frequency \(\lambda=1\) per year. Claim amounts, in thousands of dollars, have a Lomax (Pareto type II) distribution with survival function \(S_X(x)=(1+x/0.5)^{-2}\) for \(x\ge0\). Premium income is \(c=1.25\) thousand dollars per year. Does a positive adjustment coefficient exist?
Method and condition. For this distribution, the shape parameter is 2 and the scale parameter is 0.5. Its mean is \(0.5/(2-1)=0.5\) thousand dollars. Thus expected claim outgo is \(\lambda E[X]=0.5\) thousand dollars per year, and \(1.25>0.5\); the net profit condition holds.
The density is \(f_X(x)=2(0.5)^{-1}(1+x/0.5)^{-3}\) for \(x\ge0\). For any \(r>0\), the MGF integral includes \(e^{rx}f_X(x)\). As \(x\) grows, the density decreases at a polynomial rate proportional to \(x^{-3}\), while \(e^{rx}\) grows exponentially. Therefore \(\int_0^\infty e^{rx}f_X(x)\,dx=\infty\) for every \(r>0\). Equivalently, the tail-integral identity shows that \(r\int_0^\infty e^{rx}(1+x/0.5)^{-2}\,dx\) diverges for each positive \(r\).
Since \(M_X(r)\) is infinite for every \(r>0\), the adjustment-coefficient equation has no positive finite solution. The mean is finite and the net profit condition holds, but neither fact supplies a positive MGF value. This example avoids the mistaken inference that a profitable model must always have an adjustment coefficient.
Common Exam Trap: Confusing a Scale with a Root
An exponential severity has an MGF domain boundary at the reciprocal of its rate parameter. That boundary is not generally the adjustment coefficient. The coefficient is found by solving the premium-and-frequency equation inside the MGF domain. Likewise, a finite mean alone is insufficient: in the Lomax example, the mean exists but the MGF is infinite at every positive argument.
A useful algebra check is to substitute a proposed positive root into the original equation, not merely into a rearranged expression. For instance, if rearrangement involved dividing by \(r\), first remember that \(r=0\) is always a root of the unreduced equation but is not the positive adjustment coefficient. Then verify that the positive candidate lies in the MGF domain and makes both sides equal.
Check Your Understanding
For each problem, check the net profit condition and MGF domain before solving or deciding whether a positive adjustment coefficient exists. Amounts are in thousands of dollars unless stated otherwise.
- A hypothetical model has \(\lambda=2\) claims per year, exponential claim severity with mean \(0.5\), and \(c=1.25\) per year. Find the positive adjustment coefficient. (Answer: \(R=0.4\) per thousand dollars.)
- A hypothetical model has \(\lambda=1\) claim per year, exponential severity with mean \(0.25\), and \(c=0.5\) per year. Find the positive adjustment coefficient and verify that it is inside the MGF domain. (Answer: \(R=2\) per thousand dollars.)
- A hypothetical model has \(\lambda=1\) per year, Lomax severity with shape 3 and scale 0.5, and \(c=1\) per year. The severity mean is finite. Does a positive adjustment coefficient exist? (Answer: No; the MGF is infinite for every positive argument.)
- A hypothetical model has \(\lambda=1\) per year, gamma severity with shape 2 and scale 0.5, and \(c=14/9\) per year. Find the positive adjustment coefficient. (Answer: \(R=0.5\) per thousand dollars.)