Measure Layer-Cost Sensitivity to Severity Parameters

Why Measure Layer-Cost Sensitivity?

An excess-of-loss treaty’s expected payment depends not only on its attachment and limit, but also on the assumed severity distribution. A small change in a fitted parameter can materially change the probability of reaching a high layer. Sensitivity analysis quantifies that effect: it shows how much expected treaty cost changes for a change in a severity parameter, and which assumptions deserve the closest pricing attention.

The focus here is the expected payment for a single claim in a fixed per-claim layer. Let $X$ be the ground-up loss, let $a$ be the attachment point, and let $u>a$ be the exhaustion point. The treaty payment is zero below $a$, increases with loss between $a$ and $u$, and is capped at $u-a$. All monetary quantities in the examples are stated in thousands of dollars unless otherwise specified.

Definition: For a severity distribution indexed by parameter vector $\theta$, define the expected payment per claim in the layer attaching at $a$ and exhausting at $u$ by $$ L(a,u;\theta)=E[\min((X-a)^+,u-a)] =\int_a^u S_X(x;\theta)\,dx, $$ where $(z)^+=\max(z,0)$ and $S_X(x;\theta)=P(X>x)$. This is a per-claim layer cost, not an aggregate stop-loss cost.

The survival-integral form makes sensitivity analysis direct. Once a severity model is specified, differentiate its survival function with respect to the parameter of interest, integrate over the layer, and interpret the result in the model’s units. A derivative is a local rate of change: it approximates the cost change for a sufficiently small parameter perturbation, but it is not generally the exact change for a large perturbation.

Derivatives of Expected Layer Cost

Suppose the layer endpoints are fixed and the survival function depends on a scalar parameter $\theta$. Under suitable regularity conditions, the derivative is

\frac{\partial L(a,u;\theta)}{\partial\theta} =\int_a^u \frac{\partial S_X(x;\theta)}{\partial\theta}\,dx.

A sufficient condition for this interchange is that, in a neighborhood of the parameter value under consideration, the derivatives $\partial S_X(x;\theta)/\partial\theta$ are bounded in absolute value by an integrable function of $x$ on $[a,u]$. For example, continuity of the parameter derivative on a compact parameter-by-$x$ region is sufficient. Differentiability at each $x$, by itself, does not establish this interchange.

Write the payment and its expectation.
Use the fixed per-claim layer payment $\min((X-a)^+,u-a)$, then express its expectation as an integral of $S_X$ from $a$ to $u$.

Differentiate the survival function.
Hold the layer endpoints fixed and take the parameter derivative of $S_X(x;\theta)$.

Integrate and interpret.
Integrate the derivative over the layer. Check its sign, units, and whether it is a per-claim or annual cost sensitivity.

The formula follows because the payment can also be written as an integral of indicators: $\min((X-a)^+,u-a)=\int_a^u \mathbf{1}_{\{X>x\}}\,dx$, apart from endpoint values that do not affect the integral. Taking expectations and interchanging expectation and integration gives $\int_a^u P(X>x)\,dx$. Differentiating this survival integral under the stated domination condition gives the sensitivity formula.

Worked Example: Pareto Shape Sensitivity in a Fixed Layer

A hypothetical portfolio has Pareto Type I ground-up severity with scale $x_m=2$ thousand dollars and shape $\alpha=3$. Thus $S_X(x)=(x_m/x)^\alpha$ for $x\ge x_m$. A per-claim treaty attaches at $a=4$ and exhausts at $u=8$, both in thousands of dollars. Find the expected payment per claim and its local sensitivity to $\alpha$.

Method and model check. The Pareto survival function applies throughout this layer because $4\le x\le 8$ is above the scale $x_m=2$. The layer cost is

L(4,8;\alpha)=\int_4^8 \left(\frac{2}{x}\right)^\alpha dx =\frac{2^\alpha}{\alpha-1}\left(4^{1-\alpha}-8^{1-\alpha}\right), \qquad \alpha>1.

The condition $\alpha>1$ ensures the displayed closed form is valid; the integral over this finite layer itself is finite for other positive shapes as well. At $\alpha=3$, the integrand is $8x^{-3}$, so

L(4,8;3)=8\int_4^8 x^{-3}\,dx =8\left[-\frac{1}{2x^2}\right]_4^8 =8\left(-\frac{1}{128}+\frac{1}{32}\right) =8\left(\frac{3}{128}\right) =0.1875.

Therefore, the expected payment is $0.1875$ thousand dollars, or $187.50 per claim.

Differentiate the survival function first: $\partial S_X(x;\alpha)/\partial\alpha=(2/x)^\alpha\ln(2/x)$. Hence

\frac{\partial L}{\partial\alpha}\bigg|_{\alpha=3} =\int_4^8 8x^{-3}\ln(2/x)\,dx.

For an explicit evaluation, write $L=2^\alpha B/(\alpha-1)$, where $B=4^{1-\alpha}-8^{1-\alpha}$. At $\alpha=3$, $B=1/16-1/64=3/64$, and $B'=-\ln(4)/16+\ln(8)/64=-5\ln(2)/64$. Thus $B'/B=-5\ln(2)/3$, giving

\frac{\partial L}{\partial\alpha} =L\left(\ln 2-\frac{1}{2}-\frac{5\ln 2}{3}\right) =0.1875\left(-\frac{1}{2}-\frac{2\ln 2}{3}\right) \approx -0.1804.

The derivative is in thousands of dollars per unit of $\alpha$. Its negative sign is sensible: with scale fixed, a higher Pareto shape makes the upper tail lighter and reduces expected payment in this layer. The result is checked against the integral form: $\ln(2/x)<0$ everywhere from 4 to 8, so the derivative must be negative. This method avoids confusing the layer cost with a formula for an unlimited excess payment.

Scale, Shape, and Pricing Interpretation

The direction and size of sensitivity depend on the parameter’s role in the model. A scale increase typically moves losses upward and increases layer cost. A shape change may make the tail heavier or lighter, depending on the model’s parameterization. Always differentiate the actual survival function being used; parameter names alone do not determine the sign.

The absolute derivative is expressed in cost units per unit of parameter, so it can be hard to compare parameters measured on different scales. A useful dimensionless summary is the elasticity:

\mathcal{E}_{\theta} =\frac{\theta}{L}\frac{\partial L}{\partial\theta}.

Locally, a 1% increase in $\theta$ changes $L$ by approximately $\mathcal{E}_{\theta}\%$. This is a first-order approximation; it is not an exact percentage change for a large parameter adjustment.

Worked Example: Compare Pareto Scale and Shape Effects

Use the Pareto model and layer from the preceding calculation: $x_m=2$, $\alpha=3$, $a=4$, and $u=8$, all monetary amounts in thousands of dollars. Compare the local layer-cost effect of a 1% increase in scale with that of a 1% increase in shape.

Method. Since the layer endpoints are both above $x_m$, the scale appears in the survival function as $x_m^\alpha$. Therefore $\partial L/\partial x_m=\alpha L/x_m$. Using $L=0.1875$,

\frac{\partial L}{\partial x_m} =\frac{3(0.1875)}{2} =0.28125 \quad\text{thousand dollars per thousand dollars of scale}.

A 1% increase in $x_m=2$ is an increase of $0.02$ thousand dollars. The first-order cost change is therefore $0.28125(0.02)=0.005625$ thousand dollars, or about $5.63 per claim. The scale elasticity is $(2/0.1875)(0.28125)=3$.

From the shape calculation, $\partial L/\partial\alpha\approx-0.1804$ thousand dollars per unit. A 1% increase in $\alpha=3$ is $0.03$, so the first-order change is approximately $(-0.1804)(0.03)=-0.005412$ thousand dollars, or a decrease of about $5.41 per claim. Its elasticity is approximately $(3/0.1875)(-0.1804)=-2.886$.

Thus, for equal 1% relative changes at this parameter point, scale has a slightly larger absolute local impact than shape. Both impacts are small in dollars per claim, but a portfolio with many expected claims can magnify them. This is a local comparison, not a claim that scale is always more important for every treaty or severity model.

Worked Example: Convert Sensitivity to Annual Treaty Cost

A hypothetical book has an expected 20 claims per year. Assume claim frequency is independent of severity, and each claim has the Pareto severity and the fixed layer in the prior examples. Find the annual expected treaty payment and the local annual sensitivity to the Pareto shape parameter at $\alpha=3$.

Method. For a compound claim total with expected claim count $E[N]=20$ and identically distributed claim layer payments, linearity of expectation gives annual expected payment $20L$. This result does not require the layer payments to be independent of one another, although the stated frequency-severity model assumes frequency is independent of severity.

E[\text{annual layer payment}] =20(0.1875)=3.75 \quad\text{thousand dollars}.

The count basis is held fixed when measuring severity sensitivity. Therefore the annual derivative is 20 times the per-claim derivative:

\frac{\partial E[\text{annual layer payment}]}{\partial\alpha} =20(-0.1804) \approx -3.608 \quad\text{thousand dollars per unit of }\alpha.

A small increase of $0.03$ in $\alpha$ gives an approximate annual change of $(-3.608)(0.03)=-0.1082$ thousand dollars, or a decrease of about $108.24 per year. The check is that this equals 20 times the per-claim approximation of a $5.41 decrease. If frequency itself depends on the parameter, its derivative must also be included; it is held fixed here.

Numerical Perturbations and Common Traps

When an analytic derivative is inconvenient, a centered finite difference can estimate local sensitivity:

\frac{\partial L}{\partial\theta} \approx \frac{L(a,u;\theta+h)-L(a,u;\theta-h)}{2h}.

Choose $h$ small enough to approximate a local slope, but not so small that rounding or numerical integration error dominates. Compare results using more than one step size when practical. A one-sided difference may be necessary at a parameter boundary, but it has greater approximation error under ordinary smoothness conditions. Recalculate the layer cost at each perturbed parameter using the same fixed treaty terms and the same severity model.

A frequent error is to differentiate a formula for unlimited excess loss when the treaty has a finite limit. The correct finite-layer cost integrates survival only from attachment to exhaustion. Another is to report a per-claim derivative as an annual price sensitivity without multiplying by the expected claim count. Unit checks catch both: a per-claim cost derivative and an annual cost derivative have different cost bases.

Keep the treaty terms fixed. A sensitivity to severity parameters is not a sensitivity to attachment or limit. If a boundary also varies, its effect must be added.
Check the derivative’s sign. For a parameter that increases survival throughout the layer, the derivative should be positive; for one that decreases survival throughout, it should be negative.
State the perturbation scale. A derivative per unit of a shape parameter is not itself a percent change. Use an elasticity for relative comparisons.
Do not apply a per-claim layer formula to an aggregate treaty. Aggregate stop-loss depends on the distribution of total loss, not just the single-claim survival function.
Keep units consistent. If $L$ is in thousands of dollars, multiplying by expected claims gives thousands of dollars per period; convert to dollars only at the end.

Key takeaway: For a fixed per-claim layer, expected payment is the area under the severity survival curve between attachment and exhaustion. Its parameter sensitivity is the corresponding area under the survival-function derivative, provided differentiation under the integral is justified. Interpret the derivative using its units, sign, and claim-count basis.

Check Your Understanding

For each problem, state whether the requested result is per claim or annual, and keep the monetary units specified.

A hypothetical exponential severity has mean $\theta=5$ thousand dollars. A per-claim layer attaches at 5 and exhausts at 10 thousand dollars. Find its expected payment per claim. (Answer: $5(e^{-1}-e^{-2})\approx1.163$ thousand dollars.)
A hypothetical Pareto Type I severity has scale $x_m=1$ thousand dollars and shape $\alpha=2$. A layer attaches at 2 and exhausts at 5 thousand dollars. Find its expected payment per claim. (Answer: $1(1/2-1/5)=0.300$ thousand dollars.)
For exponential severity with mean $\theta=4$ thousand dollars and layer $[4,8]$, find the derivative of expected layer payment with respect to $\theta$ at 4. (Answer: $2e^{-1}-3e^{-2}\approx0.3298$ thousand dollars per thousand dollars of mean.)
A portfolio has expected claim count 12 per year and per-claim expected layer payment $0.300$ thousand dollars. Find annual expected layer payment. (Answer: $12(0.300)=3.600$ thousand dollars per year.)
For Pareto Type I severity with $x_m=2$, $\alpha=3$, and layer $[4,8]$, the per-claim layer cost is $0.1875$ thousand dollars. Using $\partial L/\partial\alpha=L(-1/2-2\ln(2)/3)$, find the local shape sensitivity to four decimals. (Answer: approximately $-0.1804$ thousand dollars per unit of $\alpha$.)

What You'll Learn