Comparing Layer Costs Across Attachment Points

Why Attachment Point Changes Layer Cost

An aggregate stop-loss contract may cover only a band of losses: the reinsurer pays after losses pass an attachment point, but payment stops at a specified upper boundary. Two bands can have the same width and still have different expected costs. The lower band generally costs more because it covers losses that occur more often. The exact difference is determined by how much probability lies in the upper tail across the two bands.

Here, “cost” means the expected aggregate payment, before any expense allowance or risk loading. The calculations apply to a hypothetical aggregate loss \(X\), which is nonnegative. They are aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.

Definition: For attachment point \(a\geq 0\) and layer width \(h>0\), the payment from the layer is \(Y_{a,h}=\min((X-a)_+,h)\), where \(z_+=\max(z,0)\). Equivalently, \(Y_{a,h}=(X-a)_+-(X-a-h)_+\). The expected layer cost is \(C_h(a)=E[Y_{a,h}]\). The layer covers losses between \(a\) and \(a+h\), subject to its cap \(h\).

The layer starts paying only when \(X>a\). It pays the amount of loss above \(a\), but never more than \(h\). Thus a loss above \(a+h\) exhausts the layer, and its payment remains \(h\) even if the aggregate loss grows further.

Stop-Loss Transforms and Layer Costs

Define the stop-loss transform at attachment \(d\) by \(\pi(d)=E[(X-d)_+]\). It is the expected amount by which aggregate loss exceeds \(d\). Subtracting two such expectations isolates the payment in the band between their attachment points.

C_h(a)=\pi(a)-\pi(a+h).

This identity follows directly from the payment definition: the first stop-loss payment includes all loss above \(a\), while the second removes the part above the layer’s upper boundary \(a+h\). The remaining payment is exactly the capped layer payment.

For a nonnegative \(X\), let \(S_X(x)=P(X>x)\). The tail-integral identity for a nonnegative random variable gives \(E[(X-d)_+]=\int_d^\infty S_X(x)\,dx\). Therefore, for finite-mean \(X\),

C_h(a)=\int_a^{a+h} S_X(x)\,dx =\int_0^h S_X(a+t)\,dt.

The integral formula also applies when \(X\) has a discrete distribution or atoms. The value of the survival function at individual jump points does not change the integral. A layer payment is bounded by \(h\), so its expectation remains finite even if \(E[X]\) is not finite; in that case use the finite-interval survival integral rather than subtracting two stop-loss transforms that might both be infinite.

Comparison result: Consider two layers of the same width \(h\), with attachments \(a<b\). Since \(S_X\) is nonincreasing, \(S_X(a+t)\geq S_X(b+t)\) for every \(t\) from \(0\) to \(h\). Consequently, \(C_h(a)\geq C_h(b)\). The inequality is strict if \(S_X(a+t)>S_X(b+t)\) over a set of \(t\)-values with positive length. It can be equality if the two corresponding tail probabilities agree almost everywhere across the bands.

Why the Lower Layer Costs at Least as Much

The comparison is a point-by-point comparison of corresponding positions within each layer, not a claim that every loss in the lower band is less than every loss in the upper band. Pair the point \(a+t\) in the lower band with \(b+t\) in the upper band. Because \(a<b\), the second point is farther into the tail. The probability of exceeding that farther point cannot be larger.

Write both costs over the same range.
\(C_h(a)=\int_0^h S_X(a+t)\,dt\) and \(C_h(b)=\int_0^h S_X(b+t)\,dt\).

Compare corresponding tails.
For each \(t\in[0,h]\), \(a+t<b+t\). Since \(S_X\) is nonincreasing, \(S_X(a+t)\geq S_X(b+t)\).

Integrate the difference.
\(C_h(a)-C_h(b)=\int_0^h [S_X(a+t)-S_X(b+t)]\,dt\geq 0\).

This derivation explains both the direction and the possible equality. The lower layer has no smaller tail probability at any corresponding point, so its expected payment cannot be less. However, the comparison need not be strict: if the aggregate-loss distribution places no probability in the regions that distinguish the two layers, the costs may coincide.

Worked Example: Comparing Two Layers for a Discrete Aggregate Loss

A hypothetical portfolio has aggregate loss \(X\) with the following distribution. Compare the expected payments for width-40 layers attached at 20 and 80.

\(P(X=0)=0.2,\ P(X=40)=0.3,\ P(X=100)=0.3,\) and \(P(X=160)=0.2\). The loss amounts and layer payments are in the same monetary units.

Method: calculate the payment at each possible loss. For the layer attached at 20, the payment is \(\min((X-20)_+,40)\). At \(X=0,40,100,160\), respectively, payments are \(0,20,40,40\). Thus

C_{40}(20)=0.2(0)+0.3(20)+0.3(40)+0.2(40) =0+6+12+8=26.

For the layer attached at 80, payments at the same four loss values are \(0,0,20,40\). Therefore

C_{40}(80)=0.2(0)+0.3(0)+0.3(20)+0.2(40) =0+0+6+8=14.

The lower layer costs \(26-14=12\) more in expected payment. As a check using stop-loss transforms, \(\pi(20)=58\), \(\pi(60)=32\), \(\pi(80)=22\), and \(\pi(120)=8\), found by taking the probability-weighted excess above each threshold. Hence \(C_{40}(20)=58-32=26\), and \(C_{40}(80)=22-8=14\), matching the direct calculation. The common mistake avoided here is to assume that the higher layer costs more because it covers larger losses; its lower probability of being reached also matters.

Worked Example: Uniform Aggregate Loss

Suppose, hypothetically, that aggregate loss is uniform on the interval from 0 to 120 monetary units. Compare width-30 layers attached at 30 and 60. State the model as \(X\sim\operatorname{Uniform}(0,120)\); no interest or mortality assumption is involved.

For \(0\leq x\leq120\), the survival function is \(S_X(x)=(120-x)/120\). The first layer covers losses from 30 to 60, so its expected payment is

C_{30}(30)=\int_{30}^{60}\frac{120-x}{120}\,dx =\frac{1}{120}\left[120x-\frac{x^2}{2}\right]_{30}^{60} =\frac{(7200-1800)-(3600-450)}{120} =\frac{2250}{120}=18.75.

The second layer covers losses from 60 to 90:

C_{30}(60)=\int_{60}^{90}\frac{120-x}{120}\,dx =\frac{1}{120}\left[120x-\frac{x^2}{2}\right]_{60}^{90} =\frac{(10800-4050)-(7200-1800)}{120} =\frac{1350}{120}=11.25.

Thus the lower layer has expected payment \(18.75-11.25=7.50\) units greater. As a stop-loss check, for this uniform model \(\pi(d)=(120-d)^2/240\) when \(0\leq d\leq120\). Therefore \(\pi(30)=33.75\), \(\pi(60)=15\), and \(\pi(90)=3.75\). The layer costs are \(33.75-15=18.75\) and \(15-3.75=11.25\), as before. This example avoids the trap of comparing only the attachment points: both layers have the same width, so the full interval of tail probabilities determines the difference.

Worked Example: Exponential Aggregate Loss and Tail Decay

A hypothetical aggregate loss has an exponential distribution with mean 100 monetary units. Equivalently, its rate parameter is \(0.01\) per unit, and \(S_X(x)=e^{-x/100}\) for \(x\geq0\). Compare width-50 layers attached at 0 and 100.

For an exponential loss, integrating the survival function gives \(\pi(d)=\int_d^\infty e^{-x/100}\,dx=100e^{-d/100}\). Therefore

C_{50}(0)=\pi(0)-\pi(50) =100(1-e^{-0.5}) =100(1-0.6065307) =39.34693\ldots

The higher layer cost is

C_{50}(100)=\pi(100)-\pi(150) =100(e^{-1}-e^{-1.5}) =100(0.3678794-0.2231302) =14.47492\ldots

Rounded to the nearest cent, the expected payments are 39.35 and 14.47 units. Their difference, calculated before rounding, is \(39.34693\ldots-14.47492\ldots=24.87201\ldots\), or 24.87 units to the nearest cent. The result is consistent with the survival comparison: at every matching position, the layer starting at 0 has the larger exceedance probability. Do not round the two layer costs before subtracting if a more precise difference is required.

Common Exam Trap: Reversing the Comparison

A tempting but incorrect argument is that the higher attachment must have the greater cost because the losses it covers are larger. That confuses the size of a loss conditional on reaching the layer with the probability of reaching it. Expected payment accounts for both. For equal widths, the higher layer is reached less often at every corresponding point in its interval.

For example, if \(a<b\), the incorrect comparison would write \(S_X(a+t)\leq S_X(b+t)\) and conclude that the lower layer costs no more. This reverses the monotonicity of the survival function. The correction is

a+t<b+t \quad\Longrightarrow\quad S_X(a+t)\geq S_X(b+t) \quad\Longrightarrow\quad C_h(a)\geq C_h(b).

Another error is to subtract stop-loss transforms in the wrong order. The layer attached at \(a\) costs \(\pi(a)-\pi(a+h)\), not \(\pi(a+h)-\pi(a)\). The latter is nonpositive because the stop-loss transform decreases as its attachment increases, while a layer payment cannot be negative.

Interpreting the Cost Difference

The comparison formula gives more than a ranking. It expresses the difference in expected cost as an area between two shifted survival curves:

C_h(a)-C_h(b) =\int_0^h\left[S_X(a+t)-S_X(b+t)\right]dt.

If the survival probabilities are close throughout the corresponding bands, the cost difference is small. If the lower band has much greater exceedance probabilities across a substantial part of its width, the difference is larger. This is why attachment comparisons should use the aggregate distribution, rather than just the two attachment values or the maximum possible layer payment.

The result assumes equal layer widths and compares expected payments under the same aggregate-loss distribution. If widths differ, the two integrals have different ranges and the simple ordering need not follow. Likewise, expected payment is not automatically a quoted premium: expenses, profit provisions, capital costs, or other loadings can alter prices. Keep the layer’s expected payment separate from any pricing adjustment.

Key takeaway: For equal-width layers under the same aggregate loss \(X\), the lower attachment has no smaller expected cost. Compute each cost as \(\pi(a)-\pi(a+h)\), or integrate \(S_X\) across the layer. The cost difference is the integral of the difference in corresponding upper-tail probabilities.

Check Your Understanding

For each hypothetical model, compare the stated equal-width layers. Answers are rounded only where indicated.

A discrete aggregate loss has probabilities \(0.1,0.4,0.3,0.2\) at losses \(0,50,100,150\), respectively. Find the expected payments for width-50 layers at attachments 25 and 75, and their difference. (Answer: 35 and 17.5; the lower layer costs 17.5 more.)
Let \(X\) be uniform on \([0,200]\). Compare width-40 layers at attachments 40 and 100 using the survival integral. (Answer: 28 and 16; the lower layer costs 12 more.)
A discrete aggregate loss has probability \(0.25\) at each of \(0,40,80,120\). Find the costs of width-40 layers attached at 20 and 60, then state whether the comparison is strict. (Answer: 25 and 15; the lower layer costs 10 more, so the comparison is strict.)
Let \(X\) be exponential with mean 80 monetary units. Compare width-20 layers at attachments 0 and 60. Give costs and the difference to the nearest cent. (Answer: 17.70 and 8.36; the lower layer costs 9.34 more.)
For a hypothetical aggregate loss, suppose \(\pi(50)=42\), \(\pi(90)=24\), and \(\pi(130)=13\). Compare width-40 layers attached at 50 and 90. (Answer: 18 and 11; the lower layer costs 7 more.)

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