Why Discrete Claim Sizes Lead to a Finite Sum
In a classical surplus model, the adjustment coefficient summarizes how the claim process and premium income combine in the exponential bound for ultimate ruin. When claim sizes have a discrete distribution, the central calculation is explicit: the moment-generating function is a probability-weighted sum of exponentials. The challenge is to construct that sum correctly and solve the resulting equation for its positive root.
The coefficient is not found by substituting the mean claim size into the exponential-claim formula. Except in special cases, the entire claim-size distribution matters. In particular, larger possible claims contribute exponentially more to the moment-generating function when the argument is positive.
For a claim amount with finite discrete support \(x_1,\ldots,x_m\), where \(P(X=x_j)=p_j\), the probabilities satisfy \(p_j\geq 0\) and \(\sum_{j=1}^m p_j=1\). Its moment-generating function is \(M_X(r)=\sum_{j=1}^m p_j e^{rx_j}\). Thus the adjustment coefficient equation becomes \(\lambda\left(\sum_{j=1}^m p_j e^{Rx_j}-1\right)=cR\). The quantities \(x_j\) must use the same claim-size unit used for \(c\): for example, dollars of claim amount require a premium rate in dollars per unit time.
Why This Equation Holds
The finite sum follows directly from the definition of expectation. For discrete \(X\), the random variable \(e^{rX}\) takes the value \(e^{rx_j}\) with probability \(p_j\), so its expectation is the sum of those values weighted by their probabilities.
The adjustment equation also expresses a balance between the exponential effect of aggregate claims and premium income. To see this, condition on the number of claims in a time interval. The compound Poisson moment-generating function is \(E[e^{rS(t)}]=\exp\{\lambda t(M_X(r)-1)\}\). Since \(U(t)=u+ct-S(t)\),
The exponential factor in time is neutral when \(\lambda(M_X(r)-1)-cr=0\), which is the adjustment coefficient equation. The adjustment coefficient is the positive solution, not the always-present solution \(r=0\). Under the usual classical-model conditions and the net profit condition, existence and uniqueness of the positive solution are addressed in the earlier material on existence and uniqueness of the adjustment coefficient. For finite claim support, \(M_X(r)\) is finite for every real \(r\).
Worked Calculations
Worked Example: Two Possible Claim Amounts and a Numerical Root
A hypothetical insurer has a compound Poisson claim process with rate \(\lambda=1\) claim per year. Each claim is either \(1\) thousand dollars or \(2\) thousand dollars, each with probability \(1/2\). The premium rate is \(c=2.5\) thousand dollars per year. Find the adjustment coefficient in inverse thousand-dollar units.
Method and model check. The mean claim amount is \(E[X]=(1/2)(1)+(1/2)(2)=1.5\) thousand dollars. The expected claim cost per year is \(\lambda E[X]=1.5\) thousand dollars per year, which is less than \(c=2.5\). The net profit condition holds. The claim-size moment-generating function is \(M_X(r)=(e^r+e^{2r})/2\), with \(r\) measured in inverse thousand-dollar units.
Substitute into the adjustment equation:
Define \(g(r)=(e^r+e^{2r})/2-1-2.5r\). The root at zero is not the adjustment coefficient, so solve for a positive root. At \(r=0.55000\), \(e^r=1.733253\) and \(e^{2r}=3.004166\). Therefore \(g(0.55000)=(1.733253+3.004166)/2-1-2.5(0.55000)=-0.006290\).
Newton's method uses \(r_{\text{new}}=r-g(r)/g'(r)\), where \(g'(r)=(e^r+2e^{2r})/2-2.5\). At \(r=0.55000\), \(g'(r)=(1.733253+2(3.004166))/2-2.5=1.370793\). The next iterate is \(0.55000-(-0.006290)/1.370793=0.55459\). A further Newton step gives approximately \(0.55454\). Substitution at that value gives a residual close to zero; the positive root, to four decimal places, is \(R=0.5545\) per thousand dollars.
Check. At the reported root, both sides of the equation are approximately \(1.386\): the left side is \((e^{0.5545}+e^{1.1090})/2-1\), and the right side is \(2.5(0.5545)\). The close agreement confirms the numerical solution. A common mistake is to report the root at zero; checking the equation alone does not distinguish it from the required positive root.
Worked Example: Verifying a Calibrated Positive Root
A hypothetical insurer has claim rate \(\lambda=1\) per year and claim amounts of \(1\) or \(2\) thousand dollars, each with probability \(1/2\). Its premium rate is \(c=2/\ln 2\) thousand dollars per year. Verify the positive adjustment coefficient and calculate the premium loading relative to expected claims.
Method. First calculate the mean claim amount: \(E[X]=(1/2)(1)+(1/2)(2)=1.5\) thousand dollars. Hence expected claims cost \(1.5\) thousand dollars per year. The premium rate is \(2/\ln 2\approx2.885390\) thousand dollars per year, so it exceeds expected claims cost and the net profit condition holds.
The moment-generating function is \(M_X(r)=(e^r+e^{2r})/2\). At \(r=\ln 2\), \(e^r=2\) and \(e^{2r}=4\), giving \(M_X(\ln 2)=(2+4)/2=3\). The left side of the adjustment equation is therefore \(\lambda(M_X(\ln 2)-1)=1(3-1)=2\). Its right side is \(c\ln 2=(2/\ln 2)\ln 2=2\). The equation is satisfied, so the positive adjustment coefficient is \(R=\ln 2\), approximately \(0.6931\) per thousand dollars.
The relative safety loading is the premium rate divided by expected claims cost, minus one: \[ \frac{c}{\lambda E[X]}-1 =\frac{2/\ln 2}{1.5}-1 \approx 0.9236. \] Thus the premium rate is about \(92.36\%\) above expected claims cost. This is a verification of a deliberately calibrated root, rather than a numerical search.
Check. The units agree: both \(\lambda(M_X(R)-1)\) and \(cR\) are per year, since \(R\) is inverse thousand dollars and \(c\) is thousand dollars per year. A common error is to use \(c=2\) instead of \(2/\ln 2\); that changes the equation and no longer gives \(R=\ln 2\).
Worked Example: Three Claim Amounts and Probability Weights
A hypothetical portfolio has claim rate \(\lambda=1\) per year and claim amounts \(1\), \(2\), or \(4\) thousand dollars with respective probabilities \(1/2\), \(1/3\), and \(1/6\). The premium rate is \(c=4/\ln 2\) thousand dollars per year. Verify the adjustment coefficient and check the input distribution.
Method. The probabilities sum to \(1/2+1/3+1/6=3/6+2/6+1/6=1\). The mean claim amount is \[ E[X]=\frac12(1)+\frac13(2)+\frac16(4) =\frac12+\frac23+\frac23 =\frac{11}{6} \] thousand dollars. The expected claims cost is \(11/6\) thousand dollars per year, while \(c=4/\ln 2\approx5.770780\) thousand dollars per year. The net profit condition holds.
The finite-sum moment-generating function is \(M_X(r)=\frac12e^r+\frac13e^{2r}+\frac16e^{4r}\). At \(r=\ln 2\), its three weighted terms are \((1/2)(2)=1\), \((1/3)(4)=4/3\), and \((1/6)(16)=8/3\). Therefore \(M_X(\ln 2)=1+4/3+8/3=5\). The left side of the adjustment equation equals \(1(5-1)=4\), while the right side equals \((4/\ln 2)\ln 2=4\). Thus the positive adjustment coefficient is \(R=\ln 2\), or \(0.6931\) per thousand dollars.
Check. The large claim size contributes \(8/3\) to \(M_X(\ln 2)\), even though its probability is only \(1/6\). Omitting its probability weight, or replacing the distribution by its mean before evaluating the exponential, would produce a different and incorrect equation.
Practical Checks and Common Traps
A reliable calculation keeps the distribution, premium rate, and time unit together from the first line. For finite support, a useful setup is to make a table of each possible claim amount, its probability, and its contribution \(p_j e^{Rx_j}\). This makes missing probabilities and exponent errors easier to detect.
Confirm that the claim probabilities sum to one, calculate \(E[X]=\sum_jp_jx_j\), and verify \(c>\lambda E[X]\).
Use \(M_X(r)=\sum_jp_je^{rx_j}\). Keep claim amounts in the unit that matches the premium rate.
Set \(\lambda(M_X(r)-1)=cr\). The solution \(r=0\) is always present; the adjustment coefficient is the positive solution.
Substitute the positive root into both sides, check that the values agree to the stated precision, and report the coefficient in inverse claim-size units.
Two errors are particularly costly. First, using \(M_X(r)\) in place of \(M_X(r)-1\) changes the equation: the subtraction of one is part of the compound Poisson exponent. Second, mixing units changes the numerical root. If claim amounts are changed from thousands of dollars to dollars, the numerical value of \(R\) must change by the reciprocal scale factor. The adjustment equation itself remains consistent only when the premium rate and claim amounts use matching money units.
The coefficient is useful because it captures features of the full claim-size distribution that a mean alone cannot represent. In a discrete distribution, each possible claim amount contributes according to both its probability and the exponential factor at the candidate root. This is why the largest possible claims can have an important effect on the equation even when they are relatively unlikely. Numerical roots should therefore be checked against the original weighted sum, not just a rounded or simplified intermediate calculation.
Check Your Understanding
For each problem, write the finite-sum moment-generating function before using the adjustment equation. Unless stated otherwise, amounts are in thousands of dollars.
- A hypothetical insurer has \(\lambda=0.6\) per year, claim amounts \(1\) and \(3\) with probabilities \(3/4\) and \(1/4\), and premium rate \(c=1.5/\ln 2\) per year. Verify the positive root and state the relative safety loading. (Answer: \(R=\ln 2\approx0.6931\) per thousand dollars; loading \(=1.5/(\ln 2\cdot0.6\cdot1.5)-1\approx0.4427\).)
- Claims arrive at rate \(1\) per year and have amounts \(2\) and \(5\) with probabilities \(0.7\) and \(0.3\). What premium rate makes \(R=\ln 2\) satisfy the adjustment equation? (Answer: \(c=20/\ln 2\approx28.8539\) thousand dollars per year.)
- Claim amounts are \(1\) and \(2\) with probabilities \(1/3\) and \(2/3\), \(\lambda=1.5\) per year, and \(c=3.5/\ln 2\) thousand dollars per year. Verify \(R=\ln 2\) and find the relative safety loading. (Answer: \(R=\ln 2\); loading \(=(1.4/\ln 2)-1\approx1.0190\), or \(101.90\%\).)
- Claim amounts \(1\), \(2\), and \(3\) occur with probabilities \(0.2\), \(0.5\), and \(0.3\), respectively. If \(\lambda=0.5\) per year, what premium rate makes \(R=\ln 2\) satisfy the equation? (Answer: \(c=1.9/\ln 2\approx2.7414\) thousand dollars per year.)
- A model has claim amounts \(1\) and \(2\) with equal probability, \(\lambda=1\) per year, and \(c=2.5\) thousand dollars per year. If the claim amounts are instead recorded in dollars, what is the adjustment coefficient to five decimal places? (Answer: approximately \(0.00055\) per dollar; specifically, \(0.55454/1000\approx0.00055454\).)