Why Compare Losses at Every Attachment?
Two aggregate loss distributions can have the same expected loss but differ substantially in their variability and tail behavior. A stop-loss contract with attachment \(d\) pays \((X-d)^+\), where \(X\) is the loss and \(z^+=\max(z,0)\). Comparing the expected payments at every attachment gives a precise way to compare how much loss remains above different thresholds.
The resulting comparison is not limited to stop-loss contracts. For equal-mean losses, it is equivalent to convex order: every convex function of the loss has a no-greater expectation for the loss that is lower in convex order. This includes many measures of risk that penalize dispersion or large losses more heavily.
Convex order requires equal means. Indeed, the functions \(\phi(x)=x\) and \(\phi(x)=-x\) are both convex, so the two inequalities together imply \(E[X]=E[Y]\). The characterization below states the converse in terms of stop-loss premiums.
The theorem uses every real \(d\). If losses are nonnegative, it is enough to check \(d\ge0\): for \(d<0\), \((X-d)^+=X-d\), so \(\Pi_X(d)=E[X]-d=E[Y]-d=\Pi_Y(d)\). For general real-valued random variables, the negative attachments cannot be omitted without an additional assumption.
Why Stop-Loss Comparisons Are Convex Comparisons
The function \(x\mapsto(x-d)^+\) is convex for every \(d\). Thus, if \(X\le_{cx}Y\), applying the definition of convex order to this particular function gives \(\Pi_X(d)\le\Pi_Y(d)\). This proves one direction directly.
For the converse, the key fact is that a convex function can be built from an affine function and nonnegative amounts of “hinge” functions. A right-facing hinge is \((x-t)^+\), the payment above a threshold. A left-facing hinge is \((t-x)^+\), which measures how far \(x\) falls below a threshold. Stop-loss comparisons directly compare the first kind; equal means let us compare the second kind as well.
Here is the full passage from hinges to a general convex function. A finite convex function \(\phi\) has a locally finite, nonnegative curvature measure \(\mu\), its distributional second derivative. Let \(m=\phi'_-(0)\), the left derivative at zero. The integral representation of a convex function is
The measure’s mass at zero, if any, represents a jump in slope at zero and is included in the right-facing hinges. On either side of zero, the hinges add the nonnegative changes in slope described by \(\mu\). This representation therefore has the same value at zero and the same slopes and curvature as \(\phi\).
To justify the expectation comparison when the support is unbounded, first restrict each integral to \([-n,n]\), splitting at zero as in the display. These are integrals of nonnegative hinges. The hinge comparisons just proved give the desired inequality for each truncated integral. As \(n\) increases, the truncated integrals increase pointwise to the full integrals. Tonelli’s theorem allows the expectation and the nonnegative-measure integral to be interchanged; the monotone convergence theorem then passes the inequalities to the full integrals. The affine parts have equal expectations because the means are equal. Consequently \(E[\phi(X)]\le E[\phi(Y)]\), including when an expectation is interpreted as \(+\infty\); in particular the inequality holds whenever both expectations are finite. This establishes the converse without assuming bounded support.
Worked Comparisons
Worked Example: A Constant Loss Versus a Variable Loss
A hypothetical portfolio has aggregate loss \(X=2\) with certainty. A second portfolio has aggregate loss \(Y=0\) or \(Y=4\), each with probability \(1/2\). Determine whether \(X\le_{cx}Y\).
Method. First check the means, then compare the stop-loss premiums by attachment intervals.
Means. \(E[X]=2\). Also, \(E[Y]=(1/2)(0)+(1/2)(4)=2\). The required equal-mean condition holds.
For \(d<0\), both losses are nonnegative, so \(\Pi_X(d)=2-d\) and \(\Pi_Y(d)=2-d\). At \(d=0\), the same expressions give 2. For \(0<d<2\), \(\Pi_X(d)=2-d\), while only the outcome \(Y=4\) contributes, giving \(\Pi_Y(d)=(1/2)(4-d)=2-d/2\). Their difference is \(\Pi_Y(d)-\Pi_X(d)=d/2\ge0\).
For \(2\le d<4\), \(\Pi_X(d)=0\) and \(\Pi_Y(d)=2-d/2\ge0\). For \(d\ge4\), both premiums are zero. Thus the stop-loss inequality holds for every attachment, and the theorem gives \(X\le_{cx}Y\).
As a direct check for any convex \(\phi\), convexity gives \(\phi(2)=\phi((0+4)/2)\le(\phi(0)+\phi(4))/2=E[\phi(Y)]\), while \(E[\phi(X)]=\phi(2)\). Common mistake avoided: equal means alone would not establish the order; the premium comparison at all attachments is the needed check.
Worked Example: Equal Means Do Not Determine the Order
Consider two hypothetical aggregate losses. \(X\) is 0 or 10, each with probability \(1/2\); \(Y\) is 4 or 6, each with probability \(1/2\). A reviewer claims \(X\le_{cx}Y\) because both have mean 5. Test the claim and identify the actual ordering.
Means. \(E[X]=(1/2)(0)+(1/2)(10)=5\), and \(E[Y]=(1/2)(4)+(1/2)(6)=5\).
At \(d=5\), \(\Pi_X(5)=(1/2)(0)+(1/2)(10-5)=2.5\). For \(Y\), only the outcome 6 contributes, so \(\Pi_Y(5)=(1/2)(6-5)=0.5\). Since \(2.5>0.5\), the claimed order \(X\le_{cx}Y\) fails.
In fact, the reverse order holds. For \(d<0\), both premiums equal \(5-d\). For \(0\le d<4\), \(\Pi_X(d)=5-d/2\) and \(\Pi_Y(d)=5-d\), so \(\Pi_Y(d)\le\Pi_X(d)\). For \(4\le d<6\), \(\Pi_X(d)=5-d/2\) and \(\Pi_Y(d)=3-d/2\), again giving \(\Pi_Y(d)\le\Pi_X(d)\). For \(6\le d<10\), \(\Pi_Y(d)=0\le(10-d)/2=\Pi_X(d)\); for \(d\ge10\), both are zero. Therefore \(Y\le_{cx}X\).
Check. Both means are 5, and the premium comparison holds in the reverse direction at every attachment. Common mistake avoided: equal means do not imply equal risk or either convex order.
Worked Example: Aggregate Stop-Loss Curves Can Cross
A hypothetical insurer compares aggregate losses \(X\) and \(Y\). The distribution of \(X\) assigns probabilities \(1/4,1/2,1/4\) to losses 0, 4, 8, respectively. The distribution of \(Y\) assigns probabilities \(1/3,2/3\) to losses 0 and 6. Determine whether either loss is smaller in convex order.
Means. \(E[X]=(1/4)(0)+(1/2)(4)+(1/4)(8)=4\). Also, \(E[Y]=(1/3)(0)+(2/3)(6)=4\).
For \(d<0\), both stop-loss premiums equal \(4-d\). For \(0\le d<4\), the positive outcomes of \(X\) contribute: \(\Pi_X(d)=(1/2)(4-d)+(1/4)(8-d)=4-3d/4\). For \(4\le d<8\), only the outcome 8 contributes, so \(\Pi_X(d)=2-d/4\); for \(d\ge8\), it is zero.
For \(0\le d<6\), only the outcome 6 contributes for \(Y\), giving \(\Pi_Y(d)=(2/3)(6-d)=4-2d/3\). For \(d\ge6\), \(\Pi_Y(d)=0\).
At \(d=2\), \(\Pi_X(2)=4-3(2)/4=2.5\), while \(\Pi_Y(2)=4-2(2)/3=8/3\). Thus \(\Pi_X(2)<\Pi_Y(2)\). But at \(d=5\), \(\Pi_X(5)=2-5/4=0.75\), while \(\Pi_Y(5)=4-10/3=2/3\), so \(\Pi_X(5)>\Pi_Y(5)\). The curves cross, and neither convex-order comparison holds.
The crossing is consistent with equal means: neither aggregate loss is uniformly less costly at every stop-loss attachment. Common mistake avoided: checking a single attachment cannot establish convex order.
Interpreting Convex Order for Aggregate Risk
If \(X\le_{cx}Y\), then for each attachment \(d\), the expected payment on a stop-loss layer beginning at \(d\) is no greater under \(X\) than under \(Y\). This comparison holds even though the two portfolios have the same expected aggregate loss. It says that \(Y\) is more dispersed in the specific sense that every convex penalty on the loss has an expectation at least as large under \(Y\).
For nonnegative aggregate losses, attachments are commonly nonnegative. The theorem’s all-attachment condition then follows from checking nonnegative attachments, because negative attachments give equal premiums when means agree. The result is a comparison of risk distributions, not a claim that one portfolio has a smaller loss in every outcome or that the premiums at any one chosen attachment determine the entire ordering.
Check Your Understanding
For each problem, use the equal-mean condition and the relevant stop-loss comparisons to decide what follows.
- A hypothetical loss \(X\) is constantly 3. Loss \(Y\) is 1 or 5, each with probability \(1/2\). Find the means and determine the convex-order relationship. (Answer: \(E[X]=E[Y]=3\), and \(X\le_{cx}Y\).)
- A loss \(X\) is 0 or 8, each with probability \(1/2\). A loss \(Y\) is 3 or 5, each with probability \(1/2\). Compare their stop-loss premiums at \(d=4\), and state which proposed order that comparison rules out. (Answer: \(\Pi_X(4)=2\), \(\Pi_Y(4)=0.5\); this rules out \(X\le_{cx}Y\).)
- A loss \(X\) has probabilities \(1/2,1/4,1/4\) at 0, 2, 6. A loss \(Y\) is 0 or 4, each with probability \(1/2\). Find both means and compare premiums at \(d=3\). (Answer: both means are 2; \(\Pi_X(3)=0.75>\Pi_Y(3)=0.5\), so \(X\not\le_{cx}Y\).)
- A loss \(X\) is 0 or 8, each with probability \(1/2\); \(Y\) is 2 or 6, each with probability \(1/2\). Show which loss is smaller in convex order by comparing the premiums over the attachment intervals \(d<0\), \(0\le d<2\), \(2\le d<6\), \(6\le d<8\), and \(d\ge8\). (Answer: \(Y\le_{cx}X\); the means are 4, and \(\Pi_Y(d)\le\Pi_X(d)\) on every interval.)
- Use the aggregate distributions in the worked example with supports \(\{0,4,8\}\) and \(\{0,6\}\). Solve for the attachment in \(4\le d<6\) at which their stop-loss premiums are equal, then state whether either convex-order relationship holds. (Answer: the premiums meet at \(d=24/5=4.8\); they cross, so neither loss is smaller in convex order.)