Compound Negative Binomial Models

Why the Claim-Count Model Matters

An aggregate loss combines a random number of claims with random severities. Even when the severity model is fixed, changing the claim-count distribution changes how much weight is assigned to each possible number of claims. A negative binomial count model is useful when counts are more variable than a Poisson model with the same mean. For an exact aggregate probability, the practical task is to combine each possible count probability with the corresponding conditional severity probability.

This tutorial uses a specific negative binomial convention, since other conventions count successes rather than failures. All insurance amounts and models below are hypothetical. The calculations use independent and identically distributed claim severities that are independent of the claim count.

Definition: Let $N$ be the number of claims. In this tutorial, $N\sim\operatorname{NB}(r,p)$ means that $N$ counts the failures before the $r$th success, where $r$ is a positive integer and $0<p<1$. Thus, for $n=0,1,2,\ldots$, $$ P(N=n)=\binom{n+r-1}{n}p^r(1-p)^n. $$ The aggregate loss is $S=\sum_{j=1}^{N}X_j$, with $S=0$ when $N=0$. The claim amounts $X_1,X_2,\ldots$ are independent and identically distributed, and are independent of $N$.

Under this convention, $P(N=0)=p^r$, $E[N]=r(1-p)/p$, and $\operatorname{Var}(N)=r(1-p)/p^2$. In particular, the variance exceeds the mean for finite $r$, because $$ \operatorname{Var}(N)-E[N]=\frac{r(1-p)^2}{p^2}>0. $$ The parameter $p$ is the success probability in the count definition; it is not the probability of a claim.

Mixing Conditional Severity Probabilities

For a fixed claim count $n$, define $S_n=X_1+\cdots+X_n$, with $S_0=0$. Let $q_n(s)=P(S_n=s)$ for a discrete severity model. The aggregate probability at $s$ is obtained by summing over the count values that can produce $s$:

Theorem: For a discrete aggregate loss, $$ P(S=s)=\sum_{n=0}^{\infty}P(N=n)P(S_n=s) =\sum_{n=0}^{\infty}\binom{n+r-1}{n}p^r(1-p)^n q_n(s). $$ Terms for which $q_n(s)=0$ contribute nothing. For a threshold event, the same conditioning principle gives $P(S\le c)=\sum_{n=0}^{\infty}P(N=n)P(S_n\le c)$.

The justification is a direct partition by claim count. The events $\{N=n\}$ are mutually exclusive and cover all possibilities. On $\{N=n\}$, the aggregate has the distribution of $S_n$, because the severities are independent of $N$. Therefore,

P(S=s)=\sum_{n=0}^{\infty}P(S=s\mid N=n)P(N=n) =\sum_{n=0}^{\infty}P(S_n=s)P(N=n).

This is a mixture, not a single severity calculation: the conditional probability for each count must be weighted by the probability of that count. For lattice severities, the possible aggregate values often quickly rule out most counts. Checking this support before doing arithmetic prevents unnecessary terms and helps avoid missing a contributing count.

Write the count probabilities.
Use the stated negative binomial convention to calculate $P(N=n)$ for relevant $n$, checking the zero-count term separately.

Find possible count values.
Use the smallest and largest possible claim amounts, or the severity support, to decide which counts can produce the target aggregate or event.

Calculate conditional probabilities.
For each remaining count, find $P(S_n=s)$ or $P(S_n\le c)$ from the severity model.

Weight and add.
Multiply each conditional probability by $P(N=n)$, then sum the contributions. Check that no feasible count was omitted.

Worked Count Mixtures

Worked Example: An Exact Aggregate Probability With Two Severity Values

A hypothetical portfolio has $N\sim\operatorname{NB}(2,0.5)$, using the failures-before-successes convention. Each claim amount is independently 1 or 4 units, each with probability $0.5$, and severities are independent of $N$. Find $P(S=5)$.

Method. Condition on $N$, first identifying all counts that can sum to 5. With $n$ claims, the aggregate is between $n$ and $4n$. Counts $n=0$ and $n=1$ cannot produce 5. Counts $n=2$ and $n=5$ can. For $n=3$, the possible sums are 3, 6, 9, and 12; for $n=4$, the minimum is 4 but the possible sums are 4, 7, 10, 13, and 16. Counts $n\ge6$ have minimum aggregate at least 6. Thus only $n=2$ and $n=5$ contribute.

The count probabilities are $$ P(N=2)=\binom{3}{2}(0.5)^2(0.5)^2=3(0.25)(0.25)=0.1875 $$ and $$ P(N=5)=\binom{6}{5}(0.5)^2(0.5)^5=6(0.25)(0.03125)=0.046875. $$

For two claims, a total of 5 requires one amount of 1 and one amount of 4. There are two possible orders, so $$ P(S_2=5)=2(0.5)(0.5)=0.5. $$ For five claims, a total of 5 requires all five amounts to be 1, giving $$ P(S_5=5)=(0.5)^5=0.03125. $$

The weighted contributions are $0.1875(0.5)=0.09375$ and $0.046875(0.03125)=0.00146484375$. Hence $$ P(S=5)=0.09375+0.00146484375=0.09521484375. $$ The method is count conditioning with support checks. The common mistake it avoids is using only the two-claim case: five claims can also sum to 5 when every severity is 1. Rounded to four decimal places, the answer is 0.0952.

Worked Example: Separating the Zero Atom From a Positive Aggregate

A hypothetical model has $N\sim\operatorname{NB}(2,0.5)$. Each claim amount is independently 1 or 2 units with equal probability, and severities are independent of the count. Find $P(S=0)$ and $P(S=2)$.

Zero aggregate. The claim amounts are strictly positive, so $S=0$ occurs exactly when there are no claims. Therefore, $$ P(S=0)=P(N=0)=\binom{1}{0}(0.5)^2(0.5)^0=0.25. $$

Aggregate of 2. One claim produces 2 only if its severity is 2. The count probability is $P(N=1)=\binom{2}{1}(0.5)^2(0.5)=0.25$, so this case contributes $0.25(0.5)=0.125$. With two claims, a total of 2 requires both severities to be 1, with conditional probability $(0.5)^2=0.25$. Since $P(N=2)=\binom{3}{2}(0.5)^2(0.5)^2=0.1875$, that case contributes $0.1875(0.25)=0.046875$. Three or more claims have aggregate at least 3, and zero claims cannot produce 2.

Adding the feasible contributions gives $P(S=2)=0.125+0.046875=0.171875$. Rounded to four decimal places, $P(S=0)=\mathbf{0.2500}$ and $P(S=2)=\mathbf{0.1719}$. The method distinguishes the zero-count atom from positive-count contributions; it avoids assuming that a positive aggregate can arise only from one claim.

Worked Example: A Threshold Probability and a Count Cutoff

A hypothetical portfolio has $N\sim\operatorname{NB}(1,0.5)$, so the count probabilities follow the specified negative binomial convention with $r=1$. Each claim amount is independently 1 or 2 units with equal probability. Find $P(S\le2)$.

For $r=1$, $P(N=n)=(0.5)^{n+1}$. Thus $P(N=0)=0.5$, $P(N=1)=0.25$, and $P(N=2)=0.125$. For $n\ge3$, the aggregate is at least 3, so those counts cannot contribute to $S\le2$.

When $N=0$, the aggregate is 0 and the conditional event probability is 1. When $N=1$, either severity is at most 2, so the conditional probability is also 1. When $N=2$, the sum is at most 2 only if both severities equal 1, so the conditional probability is $(0.5)^2=0.25$. Therefore, $$ P(S\le2)=0.5(1)+0.25(1)+0.125(0.25) =0.5+0.25+0.03125=0.78125. $$ Rounded to four decimal places, $P(S\le2)=\mathbf{0.7813}$. The method is a threshold mixture with a support-based cutoff; it avoids summing count probabilities for $n\ge3$, none of which can satisfy the event.

Zero Probability and Overdispersion

When claim amounts are strictly positive, the aggregate is zero if and only if the count is zero. The negative binomial model therefore gives $P(S=0)=p^r$. If the severity model allows a zero claim amount, this equality no longer holds: a positive count can also yield a zero aggregate. In that case, if $P(X_j=0)=q$, independence gives $$ P(S=0)=\sum_{n=0}^{\infty}P(N=n)q^n. $$ The $n=0$ term is $P(N=0)$; the other terms account for positive counts whose severities are all zero.

The negative binomial count distribution places more variability in claim counts than a Poisson distribution with the same mean. For example, with $N\sim\operatorname{NB}(2,0.5)$, the mean count is 2, the variance is 4, and the zero-count probability is 0.25. A Poisson count with mean 2 has zero-count probability $e^{-2}$, approximately 0.1353. This comparison illustrates that the model choice affects the mixture weights, including the weight on no claims. It does not imply that every aggregate probability is larger under the negative binomial model; individual probabilities depend on the severity distribution and the aggregate event.

Key takeaway: A compound negative binomial probability is a count-weighted sum of conditional severity probabilities. State the negative binomial convention, include the zero-count case when relevant, and check the severity support before deciding which counts contribute.

Common Exam Traps

Mixing parameter conventions. Under the convention here, $N$ counts failures before $r$ successes, so $P(N=0)=p^r$. Do not substitute a formula from a convention that counts successes or uses a different probability parameter.
Ignoring a feasible count. A target aggregate can sometimes be produced by a larger count when all severities are small. Check every possible count, rather than stopping at the most obvious one.
Adding conditional probabilities without weights. The mixture contribution is $P(N=n)P(S_n=s)$, not just $P(S_n=s)$.
Assuming only zero claims produce zero aggregate. That is true only when each claim amount is strictly positive. If zero-valued severities are allowed, positive claim counts can also produce $S=0$.
Confusing count overdispersion with a universal increase in aggregate probabilities. Greater count variability changes the mixture weights; it does not make every particular aggregate event more likely.

Check Your Understanding

For each hypothetical model, use the failures-before-successes negative binomial convention and assume independent severities that are independent of the claim count.

$N\sim\operatorname{NB}(1,0.5)$, and every claim amount is exactly 2 units. Find $P(S=4)$. (Answer: 0.1250)
$N\sim\operatorname{NB}(2,0.5)$, and each claim amount is 1 or 3 units with equal probability. Find $P(S=3)$. (Answer: 0.1406)
$N\sim\operatorname{NB}(2,0.5)$, and each claim amount is 1 or 2 units with equal probability. Find $P(S=3)$. (Answer: 0.1094)
$N\sim\operatorname{NB}(1,0.5)$, and each claim amount is 1 or 2 units with equal probability. Find $P(S\le3)$. (Answer: 0.8516)
Claim amounts are strictly positive and $N\sim\operatorname{NB}(2,0.5)$. Find the aggregate zero probability and compare it with that for a Poisson claim count of mean 2. (Answer: Negative binomial $P(S=0)=0.2500$; Poisson $P(S=0)=e^{-2}\approx0.1353$.)

What You'll Learn