Second Moments of Capped Aggregate Payments

Why the Second Moment Matters

A stop-loss contract with an attachment and a payment cap pays only for aggregate losses in a particular layer. Its expected payment measures the average amount ceded, but the variance describes how much the payment can fluctuate around that average. Calculating the variance requires both the first and second moments of the capped payment.

The key technique is to integrate the probability that the payment exceeds a threshold. For the second moment, the integrand includes the threshold itself. This produces a tail formula that works for continuous, discrete, and mixed aggregate-loss distributions, with atoms handled correctly.

Definition: Let $S$ be a nonnegative aggregate loss, $d\geq 0$ an attachment, and $L>0$ a payment limit. The capped stop-loss payment is $Z=\min((S-d)_+,L)$, where $(x)_+=\max(x,0)$. Thus $Z=0$ when $S\leq d$, $Z=S-d$ when $d<S<d+L$, and $Z=L$ when $S\geq d+L$.

The payment is the loss falling in the layer from $d$ to $d+L$, not the entire aggregate loss above $d$. All results below concern undiscounted payments. Interest and mortality assumptions are not needed: the model is a probability distribution for $S$, specified separately in each example.

Tail-Integral Formulas

For any nonnegative random variable $Z$, its moments can be written in terms of the probabilities $P(Z>t)$. For the capped payment, $Z\leq L$, so the relevant threshold $t$ ranges only from 0 to $L$.

Theorem: For $Z=\min((S-d)_+,L)$, $$ E[Z]=\int_0^L P(S>d+t)\,dt $$ and $$ E[Z^2]=2\int_0^L t\,P(S>d+t)\,dt. $$ Equivalently, by setting $x=d+t$, $$ E[Z]=\int_d^{d+L}P(S>x)\,dx,\qquad E[Z^2]=2\int_d^{d+L}(x-d)P(S>x)\,dx. $$ Consequently, provided the moments are finite (as they are for finite $L$), $$ \operatorname{Var}(Z)=2\int_0^L t\,P(S>d+t)\,dt- \left(\int_0^L P(S>d+t)\,dt\right)^2. $$

The strict inequality in $P(S>d+t)$ follows from the event $Z>t$. For $0\leq t<L$, the payment exceeds $t$ exactly when $S>d+t$. At $t=L$, the event is empty; the value of an integrand at that single endpoint does not affect an integral. In a discrete model, an atom at the attachment produces no payment, while an atom at or above $d+L$ produces the capped payment. The tail integral accounts for both.

Why the Second-Moment Formula Holds

The derivation comes from expressing a nonnegative quantity as the area under its threshold indicators. For each realized value $z$, the indicator $I(z>t)$ equals 1 for thresholds below $z$ and 0 for thresholds at or above $z$. Therefore the area under that indicator is $z$; weighting the indicator by $2t$ gives $z^2$.

Write the pointwise identity.
For every $z\geq 0$, $z^2=\int_0^z 2t\,dt=\int_0^L 2t\,I(z>t)\,dt$ when $z\leq L$.

Take expectations.
Since the integrand is nonnegative and bounded over the finite interval, interchange expectation and integration: $E[Z^2]=\int_0^L 2t\,E[I(Z>t)]\,dt$.

Translate the event.
For $0\leq t<L$, $Z>t$ if and only if $S>d+t$. Thus $E[I(Z>t)]=P(S>d+t)$, giving the stated second-moment formula. Replacing $2t$ by 1 in the pointwise identity $z=\int_0^L I(z>t)\,dt$ gives the first-moment formula.

The factor $2t$ is essential: it is the derivative of $t^2$, and it makes the integral recover a square rather than the original payment. Once the two moments are found, variance follows from $\operatorname{Var}(Z)=E[Z^2]-E[Z]^2$. The first moment must be squared only after it has been calculated.

Worked Applications

Worked Example: A Uniform Aggregate Loss

A hypothetical pool has aggregate loss $S\sim\operatorname{Uniform}(0,10)$, with density $1/10$ on $[0,10]$. A stop-loss contract attaches at $d=2$ and has limit $L=4$. Find the mean, second moment, and variance of its payment. Treat losses as undiscounted; there is no interest or mortality model.

Method. For $0\leq t\leq4$, the payment exceeds $t$ when $S>2+t$. The uniform survival probability on this interval is $P(S>2+t)=(10-(2+t))/10=(8-t)/10$.

The first moment is $$ E[Z]=\int_0^4\frac{8-t}{10}\,dt =\frac{1}{10}\left[8t-\frac{t^2}{2}\right]_0^4 =\frac{32-8}{10}=2.4. $$

The second moment is $$ E[Z^2]=2\int_0^4t\frac{8-t}{10}\,dt =\frac{1}{5}\left[4t^2-\frac{t^3}{3}\right]_0^4 =\frac{1}{5}\left(64-\frac{64}{3}\right) =\frac{128}{15}=8.5333\ldots. $$

Therefore $$ \operatorname{Var}(Z)=\frac{128}{15}-(2.4)^2 =\frac{208}{75}=2.7733\ldots. $$ To four decimal places, the mean is 2.4000, the second moment is 8.5333, and the variance is 2.7733, in the corresponding loss units and squared loss units.

Check. The payment is $S-2$ for $2<S<6$, equals 4 for $S\geq6$, and is zero below 2. Directly, $E[Z]=0.1\int_0^4t\,dt+4(0.4)=0.8+1.6=2.4$. Likewise, $E[Z^2]=0.1\int_0^4t^2\,dt+16(0.4)=32/15+6.4=128/15$, confirming the tail-integral results. The method avoids treating every loss above the attachment as an uncapped payment.

Worked Example: A Discrete Aggregate-Loss Table

A hypothetical portfolio has aggregate loss $S$ with the following distribution. A contract attaches at $d=5$ and has payment limit $L=4$. Calculate the payment variance. Payments are undiscounted; no interest or mortality assumption is involved.

Aggregate loss $S$	0	5	7	9	12
Probability	0.10	0.20	0.30	0.25	0.15

Method. The payment values at these five aggregate losses are $0,0,2,4,4$. The probability of $S=5$ is not a payment: the loss is exactly at the attachment.

Using the table directly, $$ E[Z]=(0.10)(0)+(0.20)(0)+(0.30)(2)+(0.25)(4)+(0.15)(4) =0.60+1.00+0.60=2.20. $$ The second moment is $$ E[Z^2]=(0.30)(2^2)+(0.25)(4^2)+(0.15)(4^2) =1.20+4.00+2.40=7.60. $$ Thus $\operatorname{Var}(Z)=7.60-(2.20)^2=7.60-4.84=2.76$.

Tail-integral check. For $0\leq t<2$, $P(S>5+t)=0.30+0.25+0.15=0.70$. For $2\leq t<4$, the tail probability is $0.25+0.15=0.40$; endpoint choices at individual thresholds do not alter the integrals. Hence $$ E[Z]=0.70(2)+0.40(2)=2.20, $$ and $$ E[Z^2]=2\left(0.70\int_0^2t\,dt+0.40\int_2^4t\,dt\right) =2\left(0.70(2)+0.40(6)\right)=7.60. $$ The check agrees with the direct calculation. The strict tail event correctly excludes the atom at the attachment and includes the atom at the cap threshold for thresholds below the limit.

Worked Example: An Exponential Aggregate Loss

Suppose a hypothetical aggregate loss has an exponential distribution with mean 4 loss units, so $P(S>x)=e^{-x/4}$ for $x\geq0$. Let $d=2$ and $L=3$. Calculate the payment variance. The payment is undiscounted, and no interest or mortality assumption enters this model.

Method. The tail probability in the layer is $P(S>2+t)=e^{-(2+t)/4}=e^{-1/2}e^{-t/4}$. Thus $$ E[Z]=e^{-1/2}\int_0^3e^{-t/4}\,dt =4\left(e^{-1/2}-e^{-5/4}\right) \approx1.2801. $$

For the second moment, integration by parts gives $\int_0^L t e^{-t/4}\,dt=16[1-e^{-L/4}(1+L/4)]$. Substituting $L=3$, $$ E[Z^2]=32e^{-1/2}\left[1-e^{-3/4}\left(1+\frac{3}{4}\right)\right] \approx3.3647. $$

Therefore $$ \operatorname{Var}(Z)\approx3.3647-(1.2801)^2 \approx3.3647-1.6387=1.7260. $$ Rounded to four decimal places, the variance is 1.7260 squared loss units.

Check. The payment equals $S-2$ between 2 and 5 and equals 3 above 5. Direct integration of the payment density gives $E[Z]=\int_2^5(s-2)(e^{-s/4}/4)\,ds+3e^{-5/4}$, which simplifies to $4(e^{-1/2}-e^{-5/4})$. For the second moment, replacing $s-2$ by $(s-2)^2$ in that integral and adding $9e^{-5/4}$ gives the same second-moment expression above. This verifies that the survival integral includes both the uncapped layer and the exhausted cap.

Common Mistakes and Practical Checks

A frequent error is to use the first-moment tail integral as though it were the second moment. In the uniform example, that mistake would report $\int_0^4P(S>2+t)\,dt=2.4$ as $E[Z^2]$. It is actually $E[Z]$. The correct second-moment calculation weights the tail probability by $2t$, giving $8.5333$. Squaring the mean also does not produce the second moment: it gives $E[Z]^2$, which is used only in the final variance subtraction.

Before finalizing a calculation, check that the integration interval is the payment limit $0\leq t\leq L$, not the full range of aggregate losses. Also check that the threshold is $d+t$, that the payment is capped at $L$, and that the variance is nonnegative. For a discrete model, direct calculation from the payment values provides a useful independent check of the tail integral.

Key takeaway: For a limited stop-loss payment, $E[Z]$ integrates the layer survival probability, while $E[Z^2]$ integrates the same probability weighted by $2t$. Calculate both moments first, then use $\operatorname{Var}(Z)=E[Z^2]-E[Z]^2$.

Check Your Understanding

For each hypothetical model, find the requested payment moments. Losses are undiscounted, and no interest or mortality assumption is required.

$S\sim\operatorname{Uniform}(0,12)$, $d=3$, and $L=5$. Find $E[Z]$, $E[Z^2]$, and $\operatorname{Var}(Z)$. (Answer: $E[Z]=65/24$, $E[Z^2]=425/36$, and $\operatorname{Var}(Z)=2575/576\approx4.4705$.)
Aggregate loss $S$ is 0, 4, 6, or 10 with respective probabilities 0.20, 0.25, 0.35, and 0.20. For $d=4$ and $L=3$, find the payment mean and variance. (Answer: $E[Z]=1.30$ and $\operatorname{Var}(Z)=1.51$.)
Aggregate loss $S$ is 2, 7, or 12 with respective probabilities 0.30, 0.40, and 0.30. For $d=4$ and $L=5$, find the payment mean, second moment, and variance. (Answer: $E[Z]=2.70$, $E[Z^2]=11.10$, and $\operatorname{Var}(Z)=3.81$.)
On the layer $2\leq x\leq6$, suppose $P(S>x)=0.6-0.05(x-2)$. For $d=2$ and $L=4$, find the payment mean, second moment, and variance. (Answer: $E[Z]=2.00$, $E[Z^2]=112/15$, and $\operatorname{Var}(Z)=52/15\approx3.4667$.)
Two independent hypothetical losses are each 4 with probability 0.5 and 0 otherwise; let $S$ be their sum. For $d=3$ and $L=3$, find the payment mean and variance. (Answer: $E[Z]=1.25$ and $\operatorname{Var}(Z)=1.1875$.)

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