Approximate Aggregate Tails After Coverage Modifications

Why Approximate an Aggregate Tail?

A covered-payment model may have an atom at zero, a long right tail, or both. Its aggregate loss can therefore be difficult to describe with a single familiar distribution, even when the first few moments are available. An approximation can turn those moments into an estimate of a tail probability or an aggregate stop-loss amount. The estimate is useful only if its assumptions and limitations are kept visible.

Let $N$ be the number of claims in a period, and let $Y_1,Y_2,\ldots$ be independent, identically distributed covered payments, independent of $N$. The aggregate covered loss is $S=\sum_{j=1}^{N}Y_j$, with $S=0$ if $N=0$. The payment variable $Y$ already reflects the applicable coverage modification; it is not the ground-up loss $X$.

Definition: Write $m_r=E[Y^r]$, and let $\mu_N=E[N]$, $\sigma_N^2=\operatorname{Var}(N)$. Under the independence and identical-distribution assumptions above, $$ E[S]=\mu_Nm_1,\qquad \operatorname{Var}(S)=\mu_N(m_2-m_1^2)+\sigma_N^2m_1^2. $$ If $N\sim\operatorname{Poisson}(\lambda)$, these reduce to $E[S]=\lambda m_1$ and $\operatorname{Var}(S)=\lambda m_2$.

For a Poisson count, the aggregate variance uses the second raw moment $m_2$, not the payment variance alone. In general, the aggregate can have a large probability at zero: for compound Poisson losses with positive payments whenever a claim occurs, $P(S=0)=e^{-\lambda}$. Coverage modifications can also make $Y=0$ for some claims, increasing that probability.

Normal Approximation and Stop-Loss Estimates

When the aggregate is a sum of many contributions and no small number of claims dominates its variability, a normal approximation can be a useful first estimate. Use the aggregate mean $\mu=E[S]$ and standard deviation $\sigma=\sqrt{\operatorname{Var}(S)}$, and approximate $S$ by $N(\mu,\sigma^2)$. This approximation is not exact merely because the moments are known. In particular, it can perform poorly for highly skewed aggregates or at thresholds where an atom or a discrete lattice matters.

For a continuous normal approximation, the tail probability above a threshold $d$ is estimated using $z=(d-\mu)/\sigma$. The stop-loss amount at retention $d$ is $E[(S-d)_+]$, where $(x)_+=\max(x,0)$. For $S$ approximated by $N(\mu,\sigma^2)$, it is

E[(S-d)_+]\ \approx\ \sigma\phi(z)+(\mu-d)\bigl[1-\Phi(z)\bigr], \qquad z=\frac{d-\mu}{\sigma},

where $\phi$ and $\Phi$ are the standard normal density and distribution functions. This formula estimates the average amount by which the aggregate exceeds $d$; it is not the tail probability multiplied by a fixed excess.

Justification: For $Z\sim N(\mu,\sigma^2)$, integrate the excess over the region $Z>d$, and substitute $t=(Z-\mu)/\sigma$: $$ E[(Z-d)_+]=\int_z^\infty (\mu+\sigma t-d)\phi(t)\,dt. $$ Since $\int_z^\infty \phi(t)\,dt=1-\Phi(z)$ and $\int_z^\infty t\phi(t)\,dt=\phi(z)$, the stated formula follows.

A lattice correction can improve a normal approximation for a discrete aggregate. If the aggregate lies on a lattice with spacing $h$, and $d$ is a lattice point, the event $S>d$ begins at $d+h$. Approximate its probability with the normal tail above the midpoint $d+h/2$. Do not apply this correction mechanically to continuous aggregates, or to a stop-loss expectation: a continuity correction is a tail-probability device, not a change to the stop-loss definition.

Worked Example: Normal Stop-Loss After a Coverage Modification

A hypothetical policy portfolio has exactly 40 claims. Its covered payment $Y$, after the contract terms have been applied, is zero with probability $0.25$; otherwise it is exponential with mean $400. Estimate the aggregate stop-loss amount at $d=\mu+\sigma$ using a normal approximation. State the resulting tail probability as well.

Method and payment moments. The conditional exponential payment has second raw moment $E[Y^2\mid Y>0]=2(400)^2=320{,}000$. Including the zero-payment probability gives $$ m_1=0.75(400)=300,\qquad m_2=0.75(320{,}000)=240{,}000. $$ Thus $\operatorname{Var}(Y)=240{,}000-300^2=150{,}000$.

Aggregate moments. With a fixed count of 40 independent payments, $$ \mu=40(300)=12{,}000,\qquad \operatorname{Var}(S)=40(150{,}000)=6{,}000{,}000. $$ Therefore $\sigma=\sqrt{6{,}000{,}000}=2{,}449.4897$, and $d=12{,}000+2{,}449.4897=14{,}449.4897$.

Tail and stop-loss estimates. Here $z=(d-\mu)/\sigma=1$, so the normal tail estimate is $1-\Phi(1)=0.1587$. Substituting $d=\mu+\sigma$ in the stop-loss formula gives $$ E[(S-d)_+]\ \approx\ \sigma[\phi(1)-(1-\Phi(1))] =2{,}449.4897(0.2419707-0.1586553) \approx \$204.08. $$ The tail probability is reported to four decimals and the stop-loss amount to the nearest cent.

Check and trap avoided. Since $d$ is one standard deviation above the mean, a standard normal tail of about 0.1587 is consistent. The stop-loss amount is much smaller than $\sigma$, as expected at this high retention. A common mistake is to use $m_2$ as the payment variance; the correct payment variance is $m_2-m_1^2=150{,}000$.

Moment-Matched Gamma Approximation

A gamma distribution is positively supported and can provide a right-skewed alternative to the normal. For aggregate mean $\mu>0$ and variance $\sigma^2>0$, a gamma distribution with shape $\alpha$ and scale $\theta$ that matches these two moments has

\alpha=\frac{\mu^2}{\sigma^2},\qquad \theta=\frac{\sigma^2}{\mu}.

This match uses only the first two moments. It does not guarantee that the aggregate has a gamma distribution, and a continuous gamma model does not reproduce an atom at zero. If $G$ has this gamma distribution, its stop-loss estimate at $d\geq 0$ is

E[(G-d)_+] =\mu\,\overline F_{\Gamma(\alpha+1,\theta)}(d) -d\,\overline F_{\Gamma(\alpha,\theta)}(d),

where $\overline F_{\Gamma(a,\theta)}$ is the survival function of a gamma variable with shape $a$ and scale $\theta$. To see why, the integral for $E[G\,1_{\{G>d\}}]$ is $\mu$ times the survival probability for a gamma distribution with shape increased by one; subtracting $dP(G>d)$ gives the excess expectation.

Worked Example: Comparing Normal and Gamma Aggregate Tails

A hypothetical book has $N\sim\operatorname{Poisson}(4)$, and each claim produces a covered payment of exactly $1,000. Estimate $P(S> \$5{,}000)$ using both a continuity-corrected normal approximation and a two-moment gamma approximation. Also estimate the stop-loss amount at $5,000 under each continuous approximation.

Aggregate moments. Here $m_1=1{,}000$, $m_2=1{,}000{,}000$, and $\lambda=4$. Therefore $$ \mu=4(1{,}000)=4{,}000,\qquad \sigma^2=4(1{,}000{,}000)=4{,}000{,}000,\qquad \sigma=2{,}000. $$ The aggregate takes values in increments of $1,000. The event $S>\$5{,}000$ begins at $6,000, so the normal continuity-correction boundary is $5,500.

Normal tail. The corrected standardized threshold is $z=(5{,}500-4{,}000)/2{,}000=0.75$. Hence $$ P(S>\$5{,}000)\ \approx\ 1-\Phi(0.75)=0.2266. $$ For the stop-loss estimate, use the actual retention of $5,000 in the continuous normal formula, not the corrected tail boundary. Here $z=(5{,}000-4{,}000)/2{,}000=0.5$, so $$ E[(S-5{,}000)_+]\ \approx\ 2{,}000\phi(0.5)-1{,}000[1-\Phi(0.5)] =2{,}000(0.352065)-1{,}000(0.308538) \approx \$395.59. $$

Gamma tail and stop loss. The moment-matched gamma parameters are $\alpha=4{,}000^2/4{,}000{,}000=4$ and $\theta=4{,}000{,}000/4{,}000=1{,}000$. For integer shape 4, the gamma survival function at $x$ scale units is $e^{-x}(1+x+x^2/2+x^3/6)$. Using the corrected tail boundary, $x=5.5$, giving $$ P(S>\$5{,}000)\ \approx\ e^{-5.5}\left(1+5.5+\frac{5.5^2}{2}+\frac{5.5^3}{6}\right) \approx 0.2017. $$ For the stop loss, use $d=5{,}000$, or $x=5$. The shape-4 and shape-5 gamma survival probabilities at this point are $39.333333e^{-5}=0.265026$ and $65.375e^{-5}=0.440493$, respectively. Thus $$ E[(G-5{,}000)_+]\approx 4{,}000(0.440493)-5{,}000(0.265026) =\$436.84. $$

Check and trap avoided. The two tail estimates, 0.2266 and 0.2017, are in the same range but need not agree. The exact probability is $P(N\geq6)=1-e^{-4}\sum_{k=0}^{5}4^k/k!\approx0.2149$, providing a check on their scale. A common mistake is to use the $5,500 continuity-correction boundary in the stop-loss formula; the stop loss remains defined at the stated $5,000 retention.

Skewness and the Translated-Gamma Approximation

A useful diagnostic is the aggregate skewness $\gamma_1=E[(S-\mu)^3]/\sigma^3$. For a compound Poisson aggregate with finite third payment moment, the first three cumulants are $\lambda m_1,\lambda m_2,\lambda m_3$, so

\gamma_1=\frac{\lambda m_3}{(\lambda m_2)^{3/2}}.

Positive skewness indicates a longer right tail than a symmetric normal model suggests. A translated-gamma approximation can match the first three aggregate cumulants when $\gamma_1>0$. Write $T=c+G$, where $G$ is gamma with shape $\alpha$ and scale $\theta$. Matching mean, variance, and skewness gives

\alpha=\frac{4}{\gamma_1^2},\qquad \theta=\frac{\sigma\gamma_1}{2},\qquad c=\mu-\frac{2\sigma}{\gamma_1}.

The shift allows the translated gamma to match skewness as well as the first two moments, but can give it a negative lower endpoint. It still does not reproduce a discrete aggregate or its atom at zero. It is not a sound substitute for checking the aggregate’s structure, especially when the event of interest lies far into the tail.

Worked Example: Skewness, a Translated Gamma, and a Tail Warning

A hypothetical portfolio has $N\sim\operatorname{Poisson}(1)$ and a covered payment of exactly $1,000 for each claim. Compare normal and translated-gamma estimates of $P(S>\$2{,}000)$ and the stop-loss amount at $2,000. Use the exact compound-Poisson result as a check.

Moments and skewness. The aggregate mean is $1,000, the variance is $1,000,000, and the standard deviation is $1,000. Its third cumulant is $1(1{,}000^3)=1{,}000{,}000{,}000$, so $\gamma_1=1$. The translated-gamma parameters are $$ \alpha=4,\qquad \theta=500,\qquad c=1{,}000-\frac{2(1{,}000)}{1}=-1{,}000. $$ The approximation is $T=G-1{,}000$, with $G$ gamma of shape 4 and scale $500.

Tail estimates. The normal estimate uses $z=(2{,}000-1{,}000)/1{,}000=1$, giving $1-\Phi(1)=0.1587$. For the translated gamma, $T>2{,}000$ means $G>3{,}000$, or six scale units. Therefore $$ P(T>2{,}000)=e^{-6}\left(1+6+\frac{6^2}{2}+\frac{6^3}{6}\right) =61e^{-6}\approx0.1512. $$ The exact probability is $P(N\geq3)=1-e^{-1}(1+1+1/2)\approx0.0803$. Both approximations overstate this tail substantially.

Stop-loss estimates. The normal estimate at $d=\$2{,}000$ is $$ 1{,}000[\phi(1)-(1-\Phi(1))] =1{,}000(0.2419707-0.1586553) \approx \$83.32. $$ For the translated gamma, the stop loss equals $E[(G-3{,}000)_+]$. The shape-4 and shape-5 gamma survival probabilities at six scale units are $61e^{-6}\approx0.151204$ and $115e^{-6}\approx0.285057$. Thus $$ E[(T-2{,}000)_+] \approx 2{,}000(0.285057)-3{,}000(0.151204) =\$116.50. $$ The exact value is $1{,}000E[(N-2)_+]$. Using $E[N1_{\{N\geq3\}}]=1-2e^{-1}$ and $P(N\geq3)=1-2.5e^{-1}$, this is $$ 1{,}000[(1-2e^{-1})-2(1-2.5e^{-1})] \approx \$103.64. $$

Check and trap avoided. The exact tail is far below either fitted tail, despite the translated gamma matching three cumulants. The Poisson count is small and the aggregate has large jumps and an atom at zero. Matching moments does not eliminate those features, so neither estimate should be treated as an exact tail calculation.

Choosing and Reporting an Approximation

The approximation should reflect both the calculation being requested and the aggregate’s structure. The normal model is convenient for a central range and admits a direct stop-loss formula. A gamma fit preserves positive support and accommodates right skew, while a translated gamma can additionally match the third cumulant when skewness is positive. Neither fitted distribution necessarily captures a mass at zero, a lattice, or unusually influential large claims.

Calculate aggregate moments from the covered-payment variable, not from the ground-up loss if coverage changes payment amounts.
Use the appropriate frequency model. For a Poisson count, $\operatorname{Var}(S)=\lambda E[Y^2]$; for fixed $N=n$, $\operatorname{Var}(S)=n\operatorname{Var}(Y)$.
For a discrete tail event, identify the lattice and apply a continuity correction only when appropriate. Keep the stated retention unchanged in a stop-loss calculation.
Compare the estimated tail with skewness, the probability of zero, and any available exact or recursive benchmark. A large disagreement is a warning to investigate, not a reason to select the most convenient answer.

Key takeaway: A moment-based approximation is an estimate, not a replacement for the aggregate model. State which moments were matched, handle lattice thresholds deliberately, and be especially cautious when the count is small or the aggregate is strongly skewed.

Check Your Understanding

For each problem, use the stated approximation and keep aggregate moments distinct from individual-payment moments.

A hypothetical fixed-count portfolio has $N=20$. Each covered payment is $500 with probability $0.4$, and zero otherwise. Find the aggregate mean and variance, then estimate $P(S>\mu+\sigma)$ by a normal approximation without a lattice correction at this non-lattice threshold. (Answer: $\mu=\$4{,}000$, $\operatorname{Var}(S)=\$1{,}200{,}000$, $\sigma=\$1{,}095.45$, tail $\approx0.1587$.)
A hypothetical aggregate has $N\sim\operatorname{Poisson}(9)$, with a covered payment of $300 per claim. Estimate $P(S>\$3{,}600)$ using a normal approximation with a lattice correction. (Answer: $\mu=\$2{,}700$, $\sigma=\$900$; use boundary $3,750, so $z=1.1667$ and the tail is approximately $0.1217$.)
Approximate the stop-loss amount at $d=\$7{,}000$ when $S$ is modeled as $N(5{,}000,4{,}000{,}000)$. (Answer: $z=1$; $2{,}000[\phi(1)-(1-\Phi(1))]\approx \$166.63$.)
A hypothetical compound Poisson aggregate has $\lambda=4$. Its covered payment is $1,000 with probability $0.5$ and zero otherwise. Fit a two-moment gamma distribution and estimate the stop-loss amount at $3,000. (Answer: $\mu=\$2{,}000$, $\sigma^2=2{,}000{,}000$, $\alpha=2$, $\theta=\$1{,}000$; $2{,}000(0.423190)-3{,}000(0.199148)\approx \$248.94$.)
A hypothetical aggregate has $N\sim\operatorname{Poisson}(1)$, with a covered payment of $1,000 per claim. Find the translated-gamma parameters matching the first three cumulants, then estimate $P(S>\$2{,}000)$ and compare it with the exact probability. (Answer: $\gamma_1=1$, $\alpha=4$, $\theta=\$500$, $c=-\$1{,}000$; translated-gamma tail $61e^{-6}\approx0.1512$, compared with exact $1-2.5e^{-1}\approx0.0803$.)

What You'll Learn