Why Compare Aggregate Loss Distributions?
Two hypothetical portfolios can have different aggregate loss distributions even when their losses have similar averages. For an insurer or reinsurer considering excess-loss protection, the average alone does not describe the expected payment above a chosen attachment. A useful comparison asks whether one distribution produces at least as much expected excess loss as another at every attachment.
This is the purpose of stop-loss ordering. It compares aggregate loss distributions through the expected payment above each threshold. The comparison can support decisions about which portfolio has the greater excess-loss cost, but a conclusion at one attachment does not establish an ordering across all attachments. This topic is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
For insurance attachments, \(d\) is usually nonnegative. The definition uses every real \(d\) so that the order also reflects the comparison of means. Since \(X\) and \(Y\) are nonnegative, if \(d<0\), then \((X-d)_+=X-d\), and therefore \(\Pi_X(d)=E[X]-d\). Thus an ordering over all real attachments requires \(E[X]\ge E[Y]\), as well as the required comparisons at nonnegative attachments.
What the Stop-Loss Premium Measures
For a loss \(X\) and an attachment \(d\ge0\), the payment under an unlimited excess-loss layer is \((X-d)_+\). It is zero when the loss does not exceed the attachment and equals the amount above the attachment otherwise. Its expectation, \(\Pi_X(d)\), is the layer's expected payment before expenses or any risk loading. Consequently, if \(X\ge_{\mathrm{sl}}Y\), then the expected excess-loss payment for \(X\) is at least that for \(Y\) at every attachment.
A useful identity relates this expected payment to the survival probabilities of the aggregate loss. For any integrable nonnegative loss and \(d\ge0\),
To see why, fix a realized value \(X=x\). The length of the interval from \(d\) up to \(x\), when \(x>d\), is \(x-d\); when \(x\le d\), the interval contributes zero. Therefore \((X-d)_+=\int_d^\infty \mathbf{1}_{\{X>t\}}\,dt\). Taking expectations gives the identity. It shows that a stop-loss premium measures the accumulated tail probability above the attachment, not merely the probability of exceeding that attachment.
For a discrete distribution with possible values \(x_j\) and probabilities \(p_j\), the direct calculation is \(\Pi_X(d)=\sum_j p_j\max(x_j-d,0)\). Outcomes at or below \(d\) contribute zero; outcomes above \(d\) contribute their probability multiplied by their excess over \(d\).
Worked Comparisons
Worked Example: A Shifted Aggregate Distribution
Consider two hypothetical portfolios with aggregate losses in thousands of dollars. Portfolio \(Y\) has loss \(0\) with probability \(0.5\), \(4\) with probability \(0.3\), and \(10\) with probability \(0.2\). Portfolio \(X\) has loss \(2\), \(6\), or \(12\) with those respective probabilities. Determine their stop-loss ordering and compare their expected payments at an attachment of \(5\).
Method. Pair the outcomes with the same probabilities. Under this coupling, each outcome for \(X\) is exactly \(2\) greater than the corresponding outcome for \(Y\). Since \(z\mapsto(z-d)_+\) is nondecreasing for every fixed \(d\), each paired payment for \(X\) is at least the paired payment for \(Y\). Taking expectations proves \(\Pi_X(d)\ge\Pi_Y(d)\) for every real \(d\), so \(X\ge_{\mathrm{sl}}Y\).
The means provide a check. For \(Y\), \(E[Y]=0(0.5)+4(0.3)+10(0.2)=0+1.2+2=3.2\). For \(X\), \(E[X]=2(0.5)+6(0.3)+12(0.2)=1+1.8+2.4=5.2\). The mean difference is \(5.2-3.2=2\), consistent with the shift.
At \(d=5\), the outcomes \(0\) and \(4\) for \(Y\) contribute nothing, while \(10\) contributes \(10-5=5\). Thus \(\Pi_Y(5)=0.2(5)=1.0\) thousand dollars. For \(X\), the outcomes \(2\) and \(6\) contribute \(0\) and \(6-5=1\), and \(12\) contributes \(12-5=7\). Hence \(\Pi_X(5)=0.5(0)+0.3(1)+0.2(7)=0+0.3+1.4=1.7\) thousand dollars.
The result at \(d=5\) agrees with the ordering, but the proof by paired outcomes is what establishes the comparison at every attachment. A common mistake is to test only one attachment and treat that result as proof of a global ordering.
Worked Example: A Higher Mean Does Not Guarantee Ordering
Portfolio \(A\) has aggregate loss \(6\) with probability \(1\). Portfolio \(B\) has aggregate loss \(0\) with probability \(0.5\) and \(10\) with probability \(0.5\). Determine whether either portfolio is greater in stop-loss order than the other.
First calculate the means: \(E[A]=6\), while \(E[B]=0(0.5)+10(0.5)=5\). For nonnegative \(d\), the premiums are \(\Pi_A(d)=6-d\) for \(0\le d<6\), and \(0\) for \(d\ge6\). For \(B\), \(\Pi_B(d)=0.5(10-d)=5-0.5d\) for \(0\le d<10\), and \(0\) for \(d\ge10\).
At attachment \(d=0\), \(\Pi_A(0)=6\) and \(\Pi_B(0)=5\), so \(A\)'s premium is larger. At \(d=4\), \(\Pi_A(4)=6-4=2\), whereas \(\Pi_B(4)=0.5(10-4)=3\), so \(B\)'s premium is larger. Thus \(A\not\ge_{\mathrm{sl}}B\), because the inequality fails at \(d=4\); and \(B\not\ge_{\mathrm{sl}}A\), because it fails at \(d=0\).
The curves are equal at \(d=2\), since \(6-2=5-0.5(2)=4\). For attachments between \(2\) and \(6\), \(B\)'s premium exceeds \(A\)'s; beyond \(6\) and below \(10\), \(A\)'s premium is zero while \(B\)'s remains positive. This example shows why the larger mean does not establish greater excess-loss cost at every attachment. The common mistake is to use the mean comparison as a substitute for checking the premiums.
Worked Example: Verifying an Order from a Finite Probability Table
Two hypothetical portfolios have aggregate losses, in thousands of dollars, only at \(0,3,6,\) and \(9\). Portfolio \(M\) has probabilities \(0.2,0.2,0.3,0.3\), respectively. Portfolio \(N\) has probabilities \(0.4,0.2,0.2,0.2\), respectively. Determine whether \(M\ge_{\mathrm{sl}}N\).
The probabilities for \(M\) sum to \(0.2+0.2+0.3+0.3=1\), and those for \(N\) sum to \(0.4+0.2+0.2+0.2=1\). Calculate the premiums at the support points, where the formula's contributing outcomes change:
| Attachment \(d\) | \(\Pi_M(d)\) | \(\Pi_N(d)\) |
|---|---|---|
| 0 | \(0.2(0)+0.2(3)+0.3(6)+0.3(9)=5.1\) | \(0.4(0)+0.2(3)+0.2(6)+0.2(9)=3.6\) |
| 3 | \(0.3(6-3)+0.3(9-3)=0.9+1.8=2.7\) | \(0.2(6-3)+0.2(9-3)=0.6+1.2=1.8\) |
| 6 | \(0.3(9-6)=0.9\) | \(0.2(9-6)=0.6\) |
| 9 | \(0\) | \(0\) |
At each listed attachment, \(\Pi_M(d)\ge\Pi_N(d)\). Between consecutive support points, each premium is linear in \(d\): the set of outcomes above the attachment does not change within that interval. The difference \(\Pi_M(d)-\Pi_N(d)\) is therefore linear there. A linear function that is nonnegative at both endpoints of an interval is nonnegative throughout that interval. This verifies the inequality for every \(d\) from \(0\) through \(9\); for \(d\ge9\), both premiums are zero.
For negative \(d\), compare the means: \(E[M]=5.1\) and \(E[N]=3.6\), so \(\Pi_M(d)-\Pi_N(d)=5.1-3.6=1.5\). This is positive for every \(d<0\). Therefore \(M\ge_{\mathrm{sl}}N\). The method is to check all support-point endpoints, justify the interval behavior, and separately account for negative attachments when using the all-real definition.
How to Make and Interpret the Comparison
For discrete aggregate distributions, the support points divide the attachment axis into intervals on which each stop-loss premium is linear. This makes an all-attachment comparison manageable: calculate both premiums at the support points of the combined distributions, compare them, and check the negative-attachment region if the definition is over all real \(d\). Between adjacent support points, nonnegative endpoint differences guarantee a nonnegative difference throughout the interval. A negative difference at even one attachment disproves the proposed order.
The practical interpretation is specific: if \(X\ge_{\mathrm{sl}}Y\), then the expected payment for unlimited excess-loss coverage on \(X\) is no smaller than the corresponding expected payment on \(Y\) at every attachment. It does not say that \(X\) exceeds \(Y\) in every paired outcome, nor does a comparison at one attachment prove the order. When the premium curves cross, neither distribution is greater in stop-loss order than the other.
- Compare premiums, not just means. Means are the premiums at attachment zero; they do not settle comparisons at positive attachments.
- Keep the direction consistent. \(X\ge_{\mathrm{sl}}Y\) means \(\Pi_X(d)\ge\Pi_Y(d)\), not the reverse.
- Check the full attachment range. A single favorable comparison cannot rule out a crossing elsewhere.
- Use consistent units. If aggregate losses are stated in thousands of dollars, premiums computed from them are also in thousands of dollars.
Check Your Understanding
For each problem, use the stop-loss premium definition and compare at the required attachments or over the full range as requested.
- Portfolio \(U\) has aggregate loss \(2\) with certainty. Portfolio \(V\) has loss \(0\) or \(6\), each with probability \(0.5\). Which distribution is greater in stop-loss order? (Answer: \(V\ge_{\mathrm{sl}}U\).)
- Portfolio \(C\) has loss \(0\) or \(8\), each with probability \(0.5\). Portfolio \(D\) has loss \(3\) with certainty. Determine whether \(C\ge_{\mathrm{sl}}D\), \(D\ge_{\mathrm{sl}}C\), or neither. (Answer: \(C\ge_{\mathrm{sl}}D\).)
- Portfolio \(R\) has loss \(5\) with certainty. Portfolio \(S\) has loss \(0\) with probability \(0.4\) and loss \(10\) with probability \(0.6\). Compare their premiums at \(d=0\) and \(d=6\), and state whether those comparisons point in the same direction. (Answer: At \(d=0\), \(\Pi_R=5\) and \(\Pi_S=6\); at \(d=6\), \(\Pi_R=0\) and \(\Pi_S=2.4\). \(S\)'s premium is greater at both attachments.)
- Portfolio \(M\) has loss \(0,2,5\) with probabilities \(0.2,0.5,0.3\). Portfolio \(N\) has loss \(0,2,5\) with probabilities \(0.4,0.2,0.4\). Compare their premiums at \(d=0\) and \(d=2\), then determine whether either is greater in stop-loss order. (Answer: At \(d=0\), \(\Pi_M=2.5\) and \(\Pi_N=2.4\); at \(d=2\), \(\Pi_M=0.9\) and \(\Pi_N=1.2\). Neither is greater in stop-loss order.)
- Portfolio \(E\) has losses \(0,4,8\) with probabilities \(0.2,0.3,0.5\). Portfolio \(F\) has the same possible losses with probabilities \(0.4,0.3,0.3\). Determine whether \(E\ge_{\mathrm{sl}}F\), using the support points to justify the comparison. (Answer: \(E\ge_{\mathrm{sl}}F\); premiums at \(d=0,4,8\) are \(5.2,2.0,0\) for \(E\) and \(3.6,1.2,0\) for \(F\), and the difference is linear between consecutive support points.)