Worked Discrete Compound Aggregate And Tail Calculation

Introduction

A compound aggregate model combines uncertainty in the number of claims with uncertainty in each claim's size. Even when both components are simple, the resulting aggregate distribution can have several possible values and non-obvious tail probabilities. Exact calculation is useful for checking approximations and for evaluating discrete aggregate risk measures.

This tutorial works with a binomial claim count and a severity distribution concentrated at two amounts. The calculations are aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus. The emphasis is on connecting the aggregate probability function to its moments, exceedance probabilities, and stop-loss expectation.

Definition: Let $N\sim\operatorname{Bin}(n,q)$ be the number of claims, where $n$ is the number of independent exposure opportunities and $q$ is the probability of a claim on each opportunity. Let $Y_1,Y_2,\ldots$ be independent and identically distributed claim severities, independent of $N$. The aggregate loss is $S=\sum_{j=1}^{N}Y_j$, with $S=0$ when $N=0$. For a two-point severity, $P(Y_j=a)=r$ and $P(Y_j=b)=1-r$, where $0<a<b$.

Independence matters: the binomial model assumes the exposure outcomes are independent, and the severity model assumes claim amounts are independent of the count and of one another. If the number of claims depends on severity, or claims are dependent, the formulas below need not apply. When amounts are stated in thousands of dollars, all aggregate values and stop-loss amounts remain in thousands until converted; probabilities are always dimensionless.

Building the Exact Aggregate Distribution

A convenient method is to treat each exposure as contributing either zero or one claim severity. Its contribution has probability $1-q$ of being zero and probability $qr$ or $q(1-r)$ of being $a$ or $b$, respectively. The probability generating function (pgf) of that contribution is $1-q+q(rz^a+(1-r)z^b)$, when the amounts are integer multiples of a common unit. Independence of the $n$ exposures gives the aggregate pgf:

G_S(z)=E[z^S]=\left(1-q+q\left(rz^a+(1-r)z^b\right)\right)^n.

The coefficient of $z^s$ is $P(S=s)$. If the severity amounts are not integer multiples of a convenient unit, conditioning on $N$ and enumerating severity combinations still gives an exact finite distribution. In either method, probability mass at $S=0$ includes the event $N=0$; if both possible severities are positive, $P(S=0)=(1-q)^n$.

Worked Example: Exact Distribution, Moments, and a Tail Layer

A hypothetical portfolio has $N\sim\operatorname{Bin}(3,1/2)$. Each claim is independently either \$1,000 or \$3,000, each with probability $1/2$. Find the aggregate probability function, mean and variance, $P(S> \$4{,}000)$, and $E[(S-\$4{,}000)_+]$.

Method and units. Measure aggregate loss in thousands of dollars, so the claim severities are $Y=1$ and $Y=3$. For each of the three opportunities, the contribution is zero with probability $1/2$, one with probability $1/4$, and three with probability $1/4$. Thus

G_S(z)=\left(\frac{2+z+z^3}{4}\right)^3 =\frac{8+12z+6z^2+13z^3+12z^4+3z^5+6z^6+3z^7+z^9}{64}.

Therefore the possible aggregate amounts and probabilities, in thousands of dollars, are $0:8/64$, $1:12/64$, $2:6/64$, $3:13/64$, $4:12/64$, $5:3/64$, $6:6/64$, $7:3/64$, and $9:1/64$. The coefficients sum to $8+12+6+13+12+3+6+3+1=64$, so the probabilities sum to one.

Mean and variance. A single severity has $E[Y]=(1+3)/2=2$ and $\operatorname{Var}(Y)=((1-2)^2+(3-2)^2)/2=1$. The frequency has $E[N]=3(1/2)=3/2$ and $\operatorname{Var}(N)=3(1/2)(1/2)=3/4$. Hence

E[S]=E[N]E[Y]=\frac32(2)=3, \qquad \operatorname{Var}(S)=E[N]\operatorname{Var}(Y)+\operatorname{Var}(N)(E[Y])^2 =\frac32(1)+\frac34(4)=\frac92.

The mean is 3 thousand dollars, or \$3,000. The variance is $4.5$ in squared thousands of dollars.

Strict exceedance. $S>4$ includes aggregate amounts 5, 6, 7, and 9, but not 4. Thus $P(S>\$4{,}000)=(3+6+3+1)/64=13/64=0.2031$ to four decimals (20.3125%).

Stop-loss expectation. For attachment $d=\$4{,}000$, the positive excesses in thousands are 1 at $S=5$, 2 at $S=6$, 3 at $S=7$, and 5 at $S=9$. Therefore

E[(S-4)_+]=1\frac{3}{64}+2\frac{6}{64}+3\frac{3}{64}+5\frac{1}{64} =\frac{29}{64}=0.453125\text{ thousand dollars}.

This is \$453.125, or \$453.13 to the nearest cent. The check is dimensional: $13/64$ is a probability, whereas $29/64$ thousand dollars is an expected payment.

Common mistake avoided. The exceedance probability cannot be converted into dollars. The correct results are a dimensionless probability of $13/64$ and an expected excess of $29/64$ thousand dollars.

Why the Aggregate Moment Formulas Hold

The aggregate mean and variance can be obtained without expanding the pgf. Conditional on $N=k$, the sum contains $k$ independent severities, so $E[S\mid N=k]=kE[Y]$ and $\operatorname{Var}(S\mid N=k)=k\operatorname{Var}(Y)$. Applying the laws of total expectation and total variance gives:

E[S]=E[N]E[Y], \qquad \operatorname{Var}(S)=E[N]\operatorname{Var}(Y)+\operatorname{Var}(N)(E[Y])^2.

For the variance, the law of total variance says $\operatorname{Var}(S)=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])$. Substituting the conditional expressions gives $E[N\operatorname{Var}(Y)]+\operatorname{Var}(NE[Y])$, which equals the displayed formula because $E[Y]$ is constant. This reasoning requires the independent, identically distributed severities and their independence from $N$.

Worked Example: Two Possible Claim Amounts with Unequal Probabilities

A hypothetical book has $N\sim\operatorname{Bin}(2,1/3)$. Each claim is \$2,000 with probability $3/4$ and \$5,000 with probability $1/4$. Find the aggregate distribution, its mean and variance, $P(S>\$5{,}000)$, and the expected excess above \$5,000.

Method. Work in thousands of dollars. One exposure contributes zero with probability $2/3$, 2 with probability $(1/3)(3/4)=1/4$, and 5 with probability $(1/3)(1/4)=1/12$. Squaring its pgf gives the aggregate distribution:

G_S(z)=\left(\frac23+\frac14z^2+\frac1{12}z^5\right)^2 =\frac49+\frac13z^2+\frac1{16}z^4+\frac19z^5+\frac1{24}z^7+\frac1{144}z^{10}.

The masses sum to $64/144+48/144+9/144+16/144+6/144+1/144=1$. The severity has $E[Y]=(3/4)(2)+(1/4)(5)=11/4$, and $E[Y^2]=(3/4)(4)+(1/4)(25)=37/4$. Thus $\operatorname{Var}(Y)=37/4-(11/4)^2=27/16$. Also $E[N]=2/3$ and $\operatorname{Var}(N)=2(1/3)(2/3)=4/9$. It follows that

E[S]=\frac23\frac{11}{4}=\frac{11}{6}\text{ thousand dollars}, \qquad \operatorname{Var}(S)=\frac23\frac{27}{16}+\frac49\left(\frac{11}{4}\right)^2 =\frac{323}{72}\text{ (thousand dollars)}^2.

For the strict event $S>5$, only amounts 7 and 10 qualify, so $P(S>5)=6/144+1/144=7/144=0.0486$ to four decimals. The excesses above 5 are 2 and 5, respectively; hence $E[(S-5)_+]=2(6/144)+5(1/144)=17/144$ thousand dollars, or \$118.06 to the nearest cent. Here $17/144$ thousand dollars equals $17{,}000/144=\$118.055\ldots$, which rounds to \$118.06.

Common mistake avoided. The atom at exactly 5 is excluded from $P(S>5)$, but contributes zero—not a negative amount—to the stop-loss expectation. The calculation uses the strict tail for the probability and positive excesses only for the payment.

Reading the Tail and Calculating Stop-Loss Values

For any discrete aggregate distribution, an exceedance probability is a sum of the probabilities of all aggregate outcomes satisfying the specified inequality. In particular, $P(S>d)$ excludes an atom at $d$, while $P(S\ge d)$ includes it. The stop-loss expectation at attachment $d$ is $E[(S-d)_+]$, where $(x)_+=\max(x,0)$. If $S$ takes values $s_j$ with probabilities $p_j$, then

E[(S-d)_+]=\sum_j \max(s_j-d,0)p_j.

This follows directly by evaluating the random variable $(S-d)_+$ at each possible outcome: it is zero when $s_j\le d$, and $s_j-d$ when $s_j>d$. The stop-loss expectation is an amount of money, not a probability. The formula applies whether or not there is probability mass at the attachment; a mass exactly at $d$ contributes zero.

Worked Example: A Discrete Tail with a Mass at the Attachment

A hypothetical benefit pool has $N\sim\operatorname{Bin}(2,1/2)$. Each claim is independently either \$1,000 or \$4,000, with equal probability. Calculate $P(S\ge \$4{,}000)$, $P(S>\$4{,}000)$, and $E[(S-\$3{,}000)_+]$.

Method. In thousands of dollars, one exposure contributes 0 with probability $1/2$, 1 with probability $1/4$, and 4 with probability $1/4$. Squaring the pgf gives

G_S(z)=\left(\frac12+\frac14z+\frac14z^4\right)^2 =\frac14+\frac14z+\frac1{16}z^2+\frac14z^4+\frac18z^5+\frac1{16}z^8.

The probabilities add to $4/16+4/16+1/16+4/16+2/16+1/16=1$. At least 4 includes the outcomes 4, 5, and 8, so $P(S\ge4)=4/16+2/16+1/16=7/16=0.4375$. Strictly above 4 excludes the mass at 4, giving $P(S>4)=2/16+1/16=3/16=0.1875$.

For attachment 3, the positive excesses are 1 at aggregate 4, 2 at 5, and 5 at 8. Thus $E[(S-3)_+]=1(4/16)+2(2/16)+5(1/16)=13/16$ thousand dollars, or \$812.50. The result checks against the tail calculation: the payment is positive at all three outcomes above 3, including the outcome equal to 4, so it is not calculated from $P(S>4)$.

Common mistake avoided. Replacing $\ge$ by $>$ would omit the probability $1/4$ at 4. Also, the stop-loss attachment here is 3, not 4; using the exceedance threshold as the attachment would answer a different question.

Checks That Catch Errors

Exact compound calculations are manageable, but small interpretation errors can change the result materially. Use these checks before accepting an answer:

Probability check: all aggregate probabilities must be nonnegative and sum to one, including the zero-claim outcome.
Support check: possible aggregate amounts must be sums of the allowed severities for at most $n$ claims.
Threshold check: mark whether the question asks for $>$, $\ge$, or an excess above an attachment. These are not interchangeable when there is an atom at the threshold.
Moment check: the mean has the same units as severity, but variance has squared units. The aggregate variance includes both severity variation and count variation.
Unit check: if losses are tabulated in thousands, a stop-loss expectation in those units must be multiplied by 1,000 to report dollars. A probability is never multiplied by a currency conversion factor.

The pgf is especially helpful when the count is binomial because each of the $n$ independent opportunities has the same zero-or-claim contribution. For small models, expansion or direct enumeration is transparent; for larger models, exact distributions can be obtained by convolution or recursion. Regardless of the computational method, the interpretation checks remain the same.

Key takeaway: For a binomial count and independent two-point severities, construct the aggregate distribution from the per-exposure contribution, then compute tail probabilities by summing qualifying masses and stop-loss expectations by weighting each mass by its positive excess. Keep probability, monetary amount, and monetary units distinct.

Check Your Understanding

For each problem, work in the units stated and distinguish strict from inclusive tail events.

A hypothetical count is $N\sim\operatorname{Bin}(2,1/4)$, and every claim is \$4,000. Give the aggregate probability function, mean, and variance. (Answer: $P(S=0)=9/16,\ P(S=4{,}000)=3/8,\ P(S=8{,}000)=1/16;\ E[S]=\$2{,}000;\ \operatorname{Var}(S)=\$6{,}000{,}000$.)
Let $N\sim\operatorname{Bin}(3,1/2)$, with every claim equal to \$2,000. Find $P(S=\$4{,}000)$, $P(S>\$4{,}000)$, and $E[(S-\$4{,}000)_+]$. (Answer: $3/8,\ 1/8,\ \$250$.)
Let $N\sim\operatorname{Bin}(2,1/2)$, with severity \$1,000 or \$3,000, each with probability $1/2$. Find $P(S\ge \$4{,}000)$ and $E[(S-\$2{,}000)_+]$. (Answer: $3/16,\ \$750$.)
Let $N\sim\operatorname{Bin}(2,1/3)$, with severity \$2,000 or \$6,000, each with probability $1/2$. Find $P(S>\$6{,}000)$ and $E[(S-\$6{,}000)_+]$. (Answer: $1/12,\ \$277.78$ to the nearest cent.)
Let $N\sim\operatorname{Bin}(3,1/4)$, and let each claim be \$1,000 or \$5,000, each with probability $1/2$, independently. Find $E[S]$, $\operatorname{Var}(S)$, and $P(S=0)$. (Answer: $E[S]=\$2{,}250;\ \operatorname{Var}(S)=8.0625$ squared thousands of dollars, equivalently $\$8{,}062{,}500;\ P(S=0)=27/64$.)

What You'll Learn