Aggregate Loss With Deterministic Severities

Why Fixed Severity Simplifies Aggregate Loss

Aggregate loss usually depends on both how many claims occur and how large each claim is. When every claim has the same fixed severity, the second source of uncertainty disappears: knowing the claim count determines the aggregate loss exactly. The calculation is then a translation from claim-count probabilities to dollar amounts.

This special case is useful both as a complete model in its own right and as a way to check whether an aggregate-loss calculation makes sense. It is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus. The focus here is the aggregate distribution for binomial, Poisson, and negative binomial claim counts—not models with variable claim amounts.

Definition: Let $N$ be a nonnegative integer-valued claim count, and suppose each claim produces the same fixed, positive payment $c$. The aggregate payment $S$ is $$ S=\sum_{j=1}^{N}c=cN, $$ where the sum is zero when $N=0$. Thus the aggregate payment is a scaled claim count, not a new uncertain quantity that must be calculated by convolving claim amounts.

The payment $c$ can represent a fixed claim severity or a fixed insurer payment per claim after a contract’s terms have been applied. The model assumes that every claim in the period produces exactly $c$, with no additional variable payments or expenses. The count model must also describe the same period and portfolio to which the aggregate payment refers.

From Count Probabilities to Aggregate Probabilities

Since $c>0$, each possible count $k$ maps to exactly one aggregate amount $ck$. Conversely, an aggregate amount has positive probability only if it is a nonnegative integer multiple of $c$. If the count probability mass function is $P(N=k)$, then the aggregate probability at that amount is the same probability:

P(S=ck)=P(N=k),\qquad k=0,1,2,\ldots

For any real threshold $x\geq 0$, the largest integer count whose aggregate payment does not exceed $x$ is $\lfloor x/c\rfloor$, where $\lfloor\cdot\rfloor$ denotes the floor (greatest integer not exceeding its argument). Therefore,

P(S\leq x)=P\!\left(N\leq \left\lfloor\frac{x}{c}\right\rfloor\right).

For $x<0$, $P(S\leq x)=0$, because $S$ cannot be negative. The floor is important: if $x$ is between two possible aggregate amounts, the aggregate distribution does not acquire a partial probability at $x$. Its cumulative probability stays constant until the next multiple of $c$.

The count model supplies $P(N=k)$; the fixed severity supplies the conversion from count $k$ to amount $ck$. The following assumptions and mass functions specify the three frequency models used here.

Count model	Assumption and parameters	Count probability
Binomial	There are $n$ independent opportunities, each producing at most one claim with probability $p$.	$P(N=k)=\binom{n}{k}p^k(1-p)^{n-k}$, for $k=0,\ldots,n$.
Poisson	The count over the specified period has Poisson parameter $\lambda>0$.	$P(N=k)=e^{-\lambda}\lambda^k/k!$, for $k=0,1,\ldots$.
Negative binomial	$N$ is the number of failures before the $r$th success in independent trials, each with success probability $q$. Here $r$ is a positive integer and $0<q<1$.	$P(N=k)=\binom{k+r-1}{k}q^r(1-q)^k$, for $k=0,1,\ldots$.

The negative binomial formula uses a particular convention: $N$ counts failures, not the total number of trials. The stated convention is essential because other presentations may define a negative binomial variable differently. Once the count probabilities are specified, the aggregate distribution follows by replacing each count $k$ with the payment $ck$.

Why the Scaling Rule Holds

The probability transfer follows directly from the definition of aggregate payment. If $N=k$, the sum contains $k$ terms, each equal to $c$; hence $S=ck$. For a positive $c$, different counts cannot map to the same amount: $ck_1=ck_2$ implies $k_1=k_2$. Therefore the event $S=ck$ is exactly the event $N=k$, which proves the mass-function rule.

\{S=ck\}=\{N=k\} \quad\Longrightarrow\quad P(S=ck)=P(N=k).

Likewise, $S\leq x$ is equivalent to $cN\leq x$. Dividing by positive $c$ gives $N\leq x/c$. Since $N$ is an integer, this is exactly $N\leq\lfloor x/c\rfloor$, proving the cumulative-probability rule. This argument does not require claims to be independently sized: their payments are fixed by assumption. The probability calculation comes entirely from the specified count distribution.

Worked Examples

Worked Example: Binomial Count and a Strict Dollar Threshold

A hypothetical equipment warranty portfolio has four independent items. During the year, each item generates at most one covered claim, with probability $0.5$. Each claim produces a fixed payment of $1,200. Find the full aggregate distribution, $P(S\geq \$2{,}400)$, and $P(S<\$2{,}400)$.

Method. Use the binomial count model with $n=4$, $p=0.5$, and $c=\$1{,}200$. First calculate each count probability, then map count $k$ to aggregate payment $1{,}200k$. This method avoids treating a dollar threshold as though it were itself a claim count.

For $k=0,1,2,3,4$, the binomial coefficients are $1,4,6,4,1$. Since $p=1-p=0.5$, each count probability is its coefficient times $(0.5)^4=1/16$:

\begin{aligned} P(N=0)&=\frac{1}{16},& P(N=1)&=\frac{4}{16},& P(N=2)&=\frac{6}{16},\\ P(N=3)&=\frac{4}{16},& P(N=4)&=\frac{1}{16}.&& \end{aligned}

The aggregate amounts are $0$, $1,200, $2,400, $3,600, and $4,800, respectively, with those same probabilities. The probabilities sum to $(1+4+6+4+1)/16=1$, checking the distribution.

“At least $2,400” means $S\geq \$2{,}400$, so $1{,}200N\geq2{,}400$, or $N\geq2$. Thus

P(S\geq \$2{,}400)=P(N\geq2) =\frac{6+4+1}{16} =\frac{11}{16} =0.6875.

“Less than $2,400” means $N<2$, or $N\leq1$. Therefore

P(S<\$2{,}400)=P(N\leq1) =\frac{1+4}{16} =\frac{5}{16} =0.3125.

The check is $0.6875+0.3125=1$: the two events partition all possible aggregate outcomes. The strict inequality excludes the $2,400 outcome.

Worked Example: Poisson Count with a Noninteger Threshold Ratio

A hypothetical repair program has a Poisson claim count with parameter $\lambda=1.5$ for the year. Every claim results in a fixed payment of $800. Find $P(S=\$1{,}600)$, $P(S\leq \$2{,}400)$, and $P(S>\$2{,}400)$.

Method. Use $N\sim\operatorname{Poisson}(1.5)$ and $S=800N$. For the exact amount, solve for the corresponding count. For the threshold, divide by the per-claim payment and use the integer count boundary. This avoids evaluating a Poisson probability at a dollar amount.

The exact aggregate amount $1,600 corresponds to $N=1{,}600/800=2$. Hence

\begin{aligned} P(S=\$1{,}600) &=P(N=2)\\ &=e^{-1.5}\frac{1.5^2}{2!}\\ &=e^{-1.5}\frac{2.25}{2}\\ &=1.125e^{-1.5}\\ &\approx 0.2510. \end{aligned}

For $S\leq \$2{,}400$, the count must satisfy $800N\leq2{,}400$, so $N\leq3$. The four probabilities from counts zero through three are

\begin{aligned} P(N=0)&=e^{-1.5},\\ P(N=1)&=1.5e^{-1.5},\\ P(N=2)&=1.125e^{-1.5},\\ P(N=3)&=\frac{1.5^3}{3!}e^{-1.5}=0.5625e^{-1.5}. \end{aligned}

Adding them gives

\begin{aligned} P(S\leq \$2{,}400) &=(1+1.5+1.125+0.5625)e^{-1.5}\\ &=4.1875e^{-1.5}\\ &\approx 0.9344. \end{aligned}

The complementary event is $S>\$2{,}400$, equivalently $N\geq4$. Therefore,

P(S>\$2{,}400)=1-0.9343575\ldots=0.0656.

The displayed probabilities are rounded to four decimal places. As a check, the cumulative probability and its complement sum to one before rounding. The strict upper event starts at four claims, not three.

Worked Example: Negative Binomial Parameterization and Aggregate Amounts

A hypothetical monitoring service records claims until its third successful inspection. Each inspection independently succeeds with probability $q=0.5$, and $N$ counts failures before the third success. Each failure produces a fixed payment of $500. Find $P(S=\$2{,}000)$, $P(S\leq \$1{,}000)$, and $P(S\geq \$1{,}500)$.

Method. Use the negative binomial convention stated in this problem: $r=3$ successes, success probability $q=0.5$, and $N$ failures before the third success. Then convert dollar amounts to counts. The key mistake avoided is confusing failures with total trials.

The exact amount $2,000 corresponds to $N=2{,}000/500=4$. Thus

\begin{aligned} P(S=\$2{,}000) &=P(N=4)\\ &=\binom{4+3-1}{4}(0.5)^3(1-0.5)^4\\ &=\binom{6}{4}(0.5)^7\\ &=15\left(\frac{1}{128}\right)\\ &=\frac{15}{128}\\ &\approx 0.1172. \end{aligned}

For $S\leq \$1,000$, $500N\leq1{,}000$, so $N\leq2$. The probabilities for counts zero, one, and two are

\begin{aligned} P(N=0)&=\binom{2}{0}(0.5)^3(0.5)^0=\frac{1}{8},\\ P(N=1)&=\binom{3}{1}(0.5)^3(0.5)^1=\frac{3}{16},\\ P(N=2)&=\binom{4}{2}(0.5)^3(0.5)^2=\frac{6}{32}=\frac{3}{16}. \end{aligned}

Therefore,

P(S\leq \$1{,}000) =\frac{1}{8}+\frac{3}{16}+\frac{3}{16} =\frac{1}{2} =0.5000.

For $S\geq \$1,500$, the count must be at least three. This is the complement of $N\leq2$, so

P(S\geq \$1{,}500)=1-\frac{1}{2}=0.5000.

The two threshold events $S\leq \$1,000$ and $S\geq \$1,500$ cover all possible payments because aggregate payments occur in $500 increments. Their probabilities sum to one, as required.

Common Exam Trap: Thresholds and Integer Boundaries

A frequent error is to mishandle a strict inequality at an attainable aggregate amount. In the binomial example, a tempting but incorrect calculation for $P(S<\$2{,}400)$ is to include counts through two:

\text{Incorrect: }\quad P(S<\$2{,}400)=P(N\leq2)=\frac{1+4+6}{16}=\frac{11}{16}.

That calculation includes $N=2$, which produces exactly $2,400 and therefore does not satisfy the strict inequality. The correct conversion is $1{,}200N<2{,}400$, giving $N<2$, or $N\leq1$:

\text{Correct: }\quad P(S<\$2{,}400)=P(N\leq1)=\frac{1+4}{16}=\frac{5}{16}.

For any threshold, write the inequality involving $cN$ first, divide by $c$, and only then identify the admissible integer counts. This sequence makes both strict and inclusive boundaries explicit.

Reading the Result and Its Limits

The aggregate distribution has support only at amounts $0,c,2c,\ldots$, with an upper endpoint $nc$ for a binomial count with $n$ opportunities. For Poisson and negative binomial counts, the support continues over all nonnegative multiples of $c$. If $c>0$, the probability of no aggregate payment is exactly the probability of no claims: $P(S=0)=P(N=0)$.

A useful check is to verify that aggregate probabilities match the corresponding count probabilities and that the probabilities over all possible counts sum to one. For a threshold calculation, also check that the count cutoff is an integer boundary, not an unrounded dollar ratio. These checks catch many arithmetic and interpretation errors.

The simplification depends on deterministic severity. If claim payments can vary, a count alone does not determine the aggregate amount; additional severity information is then required. Likewise, if claim payments include deductibles, limits, or other adjustments that make the payment differ by claim, do not use $S=cN$ unless the resulting payment really is constant for every claim in the model.

Check Your Understanding

For each problem, convert the aggregate event into a count event before calculating.

A hypothetical device program has $N\sim\operatorname{Binomial}(3,0.2)$, with a fixed payment of $1,000 per claim. Find $P(S=\$2{,}000)$. (Answer: 0.0960)
A hypothetical inspection portfolio has $N\sim\operatorname{Poisson}(2)$, with a fixed payment of $500 per claim. Find $P(S\leq \$1{,}000)$, to four decimal places. (Answer: 0.6767)
Let $N$ count failures before the third success, with success probability $q=0.5$. Each failure generates a fixed $400 payment. Find $P(S=\$800)$. (Answer: 0.1875)
A hypothetical warranty count is $N\sim\operatorname{Binomial}(5,0.4)$, and each claim produces $250. Find $P(S>\$750)$, to four decimal places. (Answer: 0.0870)
A hypothetical service plan has $N\sim\operatorname{Poisson}(1)$ and a fixed payment of $750 per claim. Find $P(\$750\leq S\leq \$2{,}250)$, to four decimal places. (Answer: 0.6131)

Count model	Assumption and parameters	Count probability
Binomial	There are \(n\) independent opportunities, each producing at most one claim with probability \(p\).	\(P(N=k)=\binom{n}{k}p^k(1-p)^{n-k}\), for \(k=0,\ldots,n\).
Poisson	The count over the specified period has Poisson parameter \(\lambda>0\).	\(P(N=k)=e^{-\lambda}\lambda^k/k!\), for \(k=0,1,\ldots\).
Negative binomial	\(N\) is the number of failures before the \(r\)th success in independent trials, each with success probability \(q\). Here \(r\) is a positive integer and \(0<q<1\).	\(P(N=k)=\binom{k+r-1}{k}q^r(1-q)^k\), for \(k=0,1,\ldots\).

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