Existence and Uniqueness of the Adjustment Coefficient

Why Existence and Uniqueness Matter

The adjustment coefficient is a positive solution to an equation involving the premium rate and the claim-size moment-generating function. It is important to establish that this solution exists before using it in a calculation. The equation always has a zero at \(r=0\), but that trivial root is not the positive adjustment coefficient. Under the net profit condition, convexity ensures there can be at most one positive root; whether there is one at all depends on the behavior of the moment-generating function away from zero.

The results here concern the classical compound Poisson surplus model. There is no interest accumulation or mortality assumption in this model: premiums accrue at a constant rate, and claims arrive according to a Poisson process.

Definition: Let \(N(t)\) be a Poisson process with rate \(\lambda>0\), and let the independent, identically distributed claim amounts \(X_1,X_2,\ldots\) be nonnegative with common claim amount \(X\). The surplus is \(U(t)=u+ct-\sum_{j=1}^{N(t)}X_j\), where \(u\) is initial surplus and \(c\) is the premium rate. Write \(M_X(r)=E[e^{rX}]\) wherever this expectation is finite. A positive adjustment coefficient is a number \(R>0\) satisfying \(cR=\lambda(M_X(R)-1)\).

The relevant sign and curvature are easiest to see by defining

g(r)=\lambda\bigl(M_X(r)-1\bigr)-cr.

The adjustment coefficient is a positive zero of \(g\). Since \(M_X(0)=1\), \(g(0)=0\) automatically. Thus, simply finding a zero is not enough: the question is whether another zero lies in the positive part of the moment-generating function’s domain.

What the Net Profit Condition Tells Us

Assume \(E[X]<\infty\) and that the moment-generating function is finite for some interval of positive \(r\) adjoining zero. The expected claim cost per unit time is \(\lambda E[X]\), while the premium income per unit time is \(c\). The net profit condition is \(c>\lambda E[X]\). Differentiating at zero gives

g'(0)=\lambda M_X'(0)-c=\lambda E[X]-c<0.

Consequently, \(g(r)<0\) for all sufficiently small positive \(r\). This establishes an initial direction: just to the right of zero, the graph lies below the horizontal axis. It does not, by itself, establish that the graph will return to zero at a larger positive value.

Theorem: Suppose \(X\geq 0\), \(P(X>0)>0\), \(E[X]<\infty\), and \(c>\lambda E[X]\). On an interval where \(M_X(r)\) is finite, \(g\) is strictly convex. Therefore \(g\) has at most one positive zero. If there is a positive \(r_1\) in the domain of \(M_X\) for which \(g(r_1)>0\), then there is exactly one positive zero between 0 and \(r_1\).

Why Convexity Gives Uniqueness

For a fixed claim amount \(x\geq 0\), the function \(e^{rx}\) is convex in \(r\), and is strictly convex when \(x>0\). Averaging over \(X\) shows that \(M_X(r)\) is strictly convex when \(P(X>0)>0\). Subtracting the linear function \(1+(c/\lambda)r\) and multiplying by \(\lambda>0\) preserves strict convexity, so \(g\) is strictly convex wherever the moment-generating function is finite.

Here is a direct uniqueness argument. Suppose there were two positive zeros \(r_1<r_2\). Because \(g(0)=g(r_2)=0\), strict convexity requires the value at every point strictly between 0 and \(r_2\) to be below the straight line joining those endpoint values. In particular,

g(r_1)<\frac{r_2-r_1}{r_2}g(0)+\frac{r_1}{r_2}g(r_2)=0.

This contradicts \(g(r_1)=0\). Thus there cannot be two positive zeros. For existence, the net profit condition gives \(g(r)<0\) for sufficiently small \(r>0\). If \(g(r_1)>0\) at a point \(r_1>0\) where the moment-generating function is finite, continuity and the intermediate value theorem give a zero between that small positive value and \(r_1\). The uniqueness argument then makes it the only positive zero.

Check the model and net profit condition.
Confirm \(\lambda>0\), nonnegative claims with \(P(X>0)>0\), finite \(E[X]\), and \(c>\lambda E[X]\). This makes the slope of \(g\) at zero negative.

Check the moment-generating function’s domain.
Only evaluate \(g(r)\) at positive values for which \(M_X(r)\) is finite. A finite claim mean alone does not guarantee a finite moment-generating function at any positive argument.

Establish a positive sign.
Find a domain value \(r_1>0\) with \(g(r_1)>0\). Together with the negative sign just to the right of zero, this proves existence.

Conclude uniqueness.
Strict convexity proves that the positive zero is unique. Do not count the zero at \(r=0\) as an adjustment coefficient.

Worked Applications

Worked Example: Proving a Positive Root for Finite Claims

A hypothetical insurer has Poisson claim frequency \(\lambda=2\) per year. Each claim is 1 or 3 monetary units, each with probability \(1/2\), independently of other claims. The premium rate is \(c=5\) monetary units per year. There is no interest accumulation or mortality. Establish whether a unique positive adjustment coefficient exists; do not solve for its numerical value.

Method. Compute the mean and the sign of \(g\) at a convenient positive argument, then apply the convexity result. The common mistake avoided is to infer existence from the net profit condition alone.

The mean claim amount is

E[X]=\tfrac12(1)+\tfrac12(3)=2.

Thus \(\lambda E[X]=2(2)=4\), and \(c=5>4\), so the net profit condition holds. The claim-size moment-generating function is \(M_X(r)=\tfrac12(e^r+e^{3r})\). Hence

g(r)=2\left(\tfrac12(e^r+e^{3r})-1\right)-5r =e^r+e^{3r}-2-5r.

At zero, \(g(0)=1+1-2=0\), and \(g'(0)=\lambda E[X]-c=4-5=-1\). To check a positive value, take \(r_1=1\):

g(1)=e+e^3-7 \approx 2.7183+20.0855-7 =15.8038>0.

The moment-generating function is finite at \(r=1\), and claims are positive, so \(g\) is strictly convex. It follows that there is exactly one positive zero between 0 and 1. This establishes existence and uniqueness without finding the zero itself.

Worked Example: Net Profit Without a Positive Adjustment Coefficient

A hypothetical insurer has Poisson claim frequency \(\lambda=1\) per year and claim amount \(X\) with density \(f_X(x)=3x^{-4}\) for \(x\geq1\), and zero otherwise. Its premium rate is \(c=2\) monetary units per year. There is no interest accumulation or mortality. Determine whether a positive adjustment coefficient exists.

Method. Check both the net profit condition and the positive-argument moment-generating function. The common mistake avoided is treating net profit as a sufficient condition for existence.

The density integrates to one because \(\int_1^\infty 3x^{-4}\,dx=1\). Its mean is

E[X]=\int_1^\infty x(3x^{-4})\,dx =3\int_1^\infty x^{-3}\,dx =3\left(\tfrac12\right)=1.5.

Therefore \(\lambda E[X]=1.5\), and \(c=2>1.5\): the net profit condition holds. But for every \(r>0\),

M_X(r)=\int_1^\infty 3e^{rx}x^{-4}\,dx=\infty.

The integral diverges because the exponential factor eventually grows faster than the polynomial denominator. Thus the adjustment equation does not have a finite value at any positive \(r\), so there is no positive adjustment coefficient as defined here. The negative slope at zero still holds, but there is no finite positive argument at which to perform the required sign test.

Worked Example: Break-Even Premium and the Zero Root

A hypothetical insurer has Poisson claim frequency \(\lambda=3\) per year. Each claim is 1 or 3 monetary units, each with probability \(1/2\), and the premium rate is \(c=6\) monetary units per year. There is no interest accumulation or mortality. Determine whether there is a positive adjustment coefficient.

Method. Compare premium income with expected claim cost, then use strict convexity. The common mistake avoided is counting the ever-present zero root as a positive adjustment coefficient.

The mean is \(E[X]=\tfrac12(1)+\tfrac12(3)=2\), so \(\lambda E[X]=3(2)=6=c\). The net profit condition is not strict. Here

g(r)=3\left(\tfrac12(e^r+e^{3r})-1\right)-6r, \qquad g'(0)=\lambda E[X]-c=0.

The claim amount is positive with probability one, so \(g\) is strictly convex. Since \(g(0)=0\) and its initial derivative is zero, strict convexity implies \(g(r)>0\) for every \(r>0\) in its domain. For example,

g(1)=3\left(\tfrac12(e+e^3)-1\right)-6 \approx 3(11.4019-1)-6 =25.2057>0.

There is no positive zero: the only zero at the start is \(r=0\), and the function is strictly positive thereafter. Thus a positive adjustment coefficient does not exist at break-even.

When a Positive Sign Is Guaranteed

A convenient sufficient condition for existence is that the moment-generating function be finite for every \(r\geq0\) and that \(P(X>0)>0\). Choose \(a>0\) with \(P(X\geq a)=p>0\). Then

M_X(r)\geq p e^{ar},

so \(\lambda(M_X(r)-1)-cr\) eventually becomes positive as \(r\) increases: exponential growth eventually exceeds linear growth. Under the net profit condition, the earlier negative sign near zero and this eventual positive sign imply exactly one positive root.

If the moment-generating function is finite only on a restricted positive interval, this argument cannot automatically be extended to arbitrarily large \(r\). In that case, test the sign of \(g\) at a point inside its finite domain. A negative initial slope and strict convexity still imply at most one positive root, but neither property alone supplies the positive sign needed for existence.

Common Exam Traps

Assuming net profit proves existence. It proves \(g'(0)<0\), not that \(g\) later becomes positive. Check the moment-generating function’s domain and find a positive sign, or give a reason it must eventually occur.
Using a positive argument where the moment-generating function is infinite. The adjustment equation requires a finite \(M_X(r)\). A finite mean does not ensure a positive moment-generating function exists.
Counting \(r=0\) as the answer. Since \(M_X(0)=1\), zero always solves the equation. The adjustment coefficient, if it exists, must be strictly positive.
Claiming uniqueness without positive claim amounts. Strict convexity requires \(P(X>0)>0\). If claims are zero with probability one, the usual strict-convexity argument does not apply.
Ignoring units. If claim amounts are measured in thousands of dollars, the premium rate must also be in thousands of dollars per unit time; the adjustment coefficient is then in inverse thousands of dollars.

Key takeaway: Under the net profit condition, the adjustment equation starts below zero immediately to the right of \(r=0\). Strict convexity guarantees at most one positive root. To prove there is exactly one, also establish a positive value of \(g(r)\) at a finite point in the moment-generating function’s domain, or establish that \(g\) eventually becomes positive.

Check Your Understanding

For each hypothetical classical surplus model, determine what the stated information implies about a positive adjustment coefficient. Assume no interest accumulation or mortality in every question.

Claims arrive at rate \(\lambda=1\) per year. Claim amounts are 1 and 2 monetary units with equal probabilities, and \(c=1.8\) monetary units per year. At \(r=0.5\), \(g(r)=\tfrac12(e^{0.5}+e^1)-1-0.9\approx0.2835>0\). State the conclusion. (Answer: Exactly one positive adjustment coefficient exists, and it lies between 0 and 0.5.)
Claims arrive at rate \(\lambda=2\) per year. Each claim is 0 with probability 0.2 or 4 with probability 0.8, and \(c=7\) monetary units per year. State the conclusion using the net profit condition and the finite-support claim distribution. (Answer: Exactly one positive adjustment coefficient exists.)
Claims arrive at rate \(\lambda=3\) per year, with survival function \(P(X>x)=(1+x)^{-4}\) for \(x\geq0\), and \(c=1.5\) monetary units per year. The claim mean is \(1/3\). Determine whether a finite positive adjustment coefficient exists. (Answer: No; although the net profit condition holds, the moment-generating function is infinite for every positive argument.)
Claims arrive at rate \(\lambda=1\) per year and every claim is exactly 2 monetary units; \(c=2\) monetary units per year. Determine whether a positive adjustment coefficient exists. (Answer: No; this is break-even, and \(g(r)=e^{2r}-1-2r>0\) for every \(r>0\), leaving only the zero root.)

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