Inferring Claim Count From Aggregate Loss

Why an Aggregate Loss Can Reveal Claim Count

An observed total claim amount does not usually identify how many claims produced it. A total of six dollars, for example, could arise from two claims of three dollars or three claims of two dollars. To infer the claim count, combine two ingredients: how plausible each count was before observing the total, and how likely that count is to produce the observed total.

This is a direct application of Bayes’ theorem. The frequency model supplies the prior distribution of the claim count $N$; the severity model supplies the likelihood of the observed aggregate for each possible value of $N$. The resulting conditional probabilities describe the count distribution after the aggregate has been observed.

Definition: Let $N$ be the number of claims and let $X_1,X_2,\ldots$ denote their severities. The aggregate loss is $L=\sum_{j=1}^{N}X_j$, with $L=0$ when $N=0$. For a discrete severity model, define $g_n(\ell)=P(L=\ell\mid N=n)$. Then, whenever $P(L=\ell)>0$, $$ P(N=n\mid L=\ell)= \frac{P(N=n)g_n(\ell)} {\displaystyle\sum_{k}P(N=k)g_k(\ell)}. $$

The factor $P(N=n)$ is the prior probability of count $n$, and $g_n(\ell)$ is the likelihood of observing aggregate $\ell$ if the count were $n$. Their product is an unnormalized posterior weight. Dividing by the sum of those weights makes the posterior probabilities add to one.

Building the Aggregate Likelihood

Suppose claim severities are independent and identically distributed, independent of $N$, and have a discrete probability mass function $p_x=P(X=x)$. Given $N=n$, the aggregate is the sum of $n$ severities. Thus $g_n(\ell)$ is the probability that those $n$ severities sum to $\ell$. It can be found by listing the combinations that add to $\ell$, including all possible claim orderings.

For $n=0$, the aggregate is zero with probability one, so $g_0(0)=1$ and $g_0(\ell)=0$ for $\ell\ne0$. For $n>0$, any count whose smallest and largest possible totals exclude $\ell$ has likelihood zero and can be removed immediately. Counts with nonzero likelihood still need to be weighted by their prior probabilities.

The formula follows by applying Bayes’ theorem to the events $N=n$ and $L=\ell$. Their joint probability is the prior probability of the count multiplied by the conditional probability of the aggregate:

P(N=n,L=\ell)=P(N=n)P(L=\ell\mid N=n).

The events $\{N=k\}$ partition all possible claim counts. Therefore the probability of observing $\ell$ is the sum of these joint probabilities over all counts $k$. Dividing the joint probability for count $n$ by that total gives the stated posterior formula. This also explains why a count with a larger prior probability need not have a larger posterior probability: its likelihood may be smaller.

List feasible counts.
Use the possible severity values to identify every count that can produce the observed aggregate. Counts with zero likelihood receive posterior probability zero.

Calculate each likelihood.
For each feasible count, find the probability that the severities sum to the observed amount. Include distinct orderings when they produce the same total.

Apply the count prior.
Multiply each likelihood by the prior probability of that count under the stated frequency model.

Normalize and check.
Divide each weight by the sum of all weights. Confirm that the posterior probabilities sum to one.

Worked Posterior Calculations

Worked Example: A Poisson Count with Two Possible Severities

A hypothetical portfolio has $N\sim\operatorname{Poisson}(1)$. Given $N$, claim severities are independent, with $P(X=1)=P(X=2)=0.5$, and independent of $N$. An aggregate loss of $3$ is observed. Find the conditional distribution of $N$.

Method. Find the feasible counts, calculate each aggregate likelihood, and weight it by the Poisson prior. This avoids treating the likelihood alone as the posterior.

Each claim is between $1$ and $2$, so an aggregate of $3$ requires $2\leq N\leq3$. If $N=2$, the severities must be $1$ and $2$, in either order:

$$ g_2(3)=2(0.5)(0.5)=0.5. $$

If $N=3$, all three claims must have severity $1$:

$$ g_3(3)=(0.5)^3=0.125. $$

The Poisson prior probabilities are $P(N=2)=e^{-1}/2$ and $P(N=3)=e^{-1}/6$. The unnormalized posterior weights are therefore

w_2=\frac{e^{-1}}{2}(0.5)=\frac{e^{-1}}{4}, \qquad w_3=\frac{e^{-1}}{6}(0.125)=\frac{e^{-1}}{48}.

Their sum is $13e^{-1}/48$. Dividing each weight by that sum gives

P(N=2\mid L=3)=\frac{12}{13}, \qquad P(N=3\mid L=3)=\frac{1}{13}.

All other counts have zero likelihood, and the two posterior probabilities sum to one. As a further check, the conditional mean count is $2(12/13)+3(1/13)=27/13$. The common mistake avoided here is forgetting the two orderings of the severities when $N=2$; omitting one would understate that count’s likelihood.

Worked Example: A Larger Poisson Mean and Competing Count Explanations

A hypothetical portfolio has $N\sim\operatorname{Poisson}(2)$. Claim severities are independent of $N$ and of one another, with $P(X=1)=0.6$ and $P(X=2)=0.4$. Given an observed aggregate loss of $2$, find the posterior probabilities of the possible claim counts.

Method. Enumerate counts capable of producing the observed total, then compare their prior-weighted likelihoods. This prevents the plausible but incorrect shortcut of assuming that an aggregate of $2$ means one claim of $2$.

There are two possible explanations. With $N=1$, the sole claim must be $2$, so $g_1(2)=0.4$. With $N=2$, both claims must be $1$, so $g_2(2)=(0.6)^2=0.36$. A count of zero cannot produce a positive aggregate, and a count of three or more has a minimum aggregate of $3$.

Under the stated Poisson model,

P(N=1)=2e^{-2}, \qquad P(N=2)=\frac{2^2e^{-2}}{2!}=2e^{-2}.

The unnormalized weights are $w_1=(2e^{-2})(0.4)=0.8e^{-2}$ and $w_2=(2e^{-2})(0.36)=0.72e^{-2}$. Their sum is $1.52e^{-2}$, so

P(N=1\mid L=2)=\frac{0.8}{1.52}=\frac{10}{19}, \qquad P(N=2\mid L=2)=\frac{0.72}{1.52}=\frac{9}{19}.

The probabilities sum to one. The conditional mean is $1(10/19)+2(9/19)=28/19$. Although the prior probabilities of one and two claims are equal here, the posterior probabilities differ because their aggregate likelihoods differ.

Worked Example: A Binomial Prior with Three Feasible Counts

A hypothetical portfolio has $N\sim\operatorname{Binomial}(4,0.5)$. Each claim independently has severity $2$ or $4$, each with probability $0.5$, independently of $N$. An aggregate loss of $6$ is observed. Determine the posterior distribution of $N$.

Method. Combine the binomial prior with the sum probabilities for each feasible count. The key trap is overlooking that both two-claim and three-claim outcomes can total $6$.

A single claim cannot total $6$, while four claims have a minimum total of $8$. With two claims, one must be $2$ and the other $4$, giving $g_2(6)=2(0.5)(0.5)=0.5$. With three claims, all must be $2$, giving $g_3(6)=(0.5)^3=0.125$.

The prior probabilities are $P(N=2)=\binom{4}{2}(0.5)^4=6/16$ and $P(N=3)=\binom{4}{3}(0.5)^4=4/16$. Thus

w_2=\frac{6}{16}\frac{1}{2}=\frac{3}{16}, \qquad w_3=\frac{4}{16}\frac{1}{8}=\frac{1}{32}.

The total weight is $3/16+1/32=7/32$. Normalization gives

P(N=2\mid L=6)=\frac{6}{7}, \qquad P(N=3\mid L=6)=\frac{1}{7}.

The posterior probabilities add to one, and the conditional mean count is $2(6/7)+3(1/7)=15/7$. Using only the most likely prior count would miss the update from the observed aggregate; Bayes’ theorem requires both the prior and the likelihood.

Interpreting the Result and Avoiding Traps

A posterior distribution answers a conditional question: given the observed aggregate and the specified frequency and severity models, how are probabilities allocated across claim counts? It does not establish which count actually occurred. Even a posterior probability of $12/13$ leaves positive probability on another feasible count.

An exact aggregate observation must be interpreted according to the observation model. The examples use discrete severities and an exact lattice value, so the relevant likelihood is a probability such as $P(L=3\mid N=2)$. If the recorded information is instead that the aggregate lies in an interval, use the probability of that interval conditional on each count. Do not substitute a point probability for an interval event. For continuous aggregate models, exact observations require a density-based likelihood rather than point probabilities; the discrete calculations above should not be applied unchanged.

The independence assumptions also matter. The convolution-style likelihood calculation assumes that claim severities are independent and identically distributed and independent of the count. If severities depend on the claim count, the Bayes formula still applies, but $P(L=\ell\mid N=n)$ must be calculated under that dependence rather than from an independent, identical severity model.

Do not confuse prior and posterior. $P(N=n)$ is the count probability before observing the aggregate; $P(N=n\mid L=\ell)$ is updated using the likelihood.
Do not omit arrangements. If different orders of claim severities produce the same total, include all of them in the likelihood.
Do not retain impossible counts. A zero likelihood produces a zero posterior probability, regardless of how large the prior probability is.
Check the normalization. The posterior probabilities over all feasible counts must sum to one, and the denominator must be positive for the observed value to be possible under the model.

Inferring claim count is useful whenever an aggregate is observed but the underlying number of claims is not. The essential discipline is to keep the two model components separate until the final update: first calculate what each count predicts for the observed total, then weight that prediction by the count prior.

Check Your Understanding

For each problem, use the stated count prior and severity model to find the requested conditional count probabilities.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(1)$, with independent severities $P(X=1)=P(X=2)=0.5$, independent of $N$. Given $L=2$, find the posterior probabilities for each possible count. (Answer: $P(N=1\mid L=2)=4/5,\;P(N=2\mid L=2)=1/5$.)
A hypothetical portfolio has $N\sim\operatorname{Binomial}(3,0.4)$. Each independent claim has severity $1$ with probability $0.75$ and severity $2$ with probability $0.25$, independently of $N$. Given $L=3$, find the posterior probabilities for the possible counts. (Answer: $P(N=2\mid L=3)=4/5,\;P(N=3\mid L=3)=1/5$.)
A hypothetical portfolio has $N\sim\operatorname{Poisson}(1)$, with independent severities $2$ and $4$ each having probability $0.5$, independently of $N$. Given $L=8$, find the posterior probabilities for all possible counts. (Answer: $P(N=2\mid L=8)=48/73,\;P(N=3\mid L=8)=24/73,\;P(N=4\mid L=8)=1/73$.)
A hypothetical portfolio has $N\sim\operatorname{Binomial}(4,0.5)$. Each independent claim has severity $1$ or $3$, each with probability $0.5$, independently of $N$. Given $L=5$, find the posterior probabilities for all possible counts. (Answer: $P(N=3\mid L=5)=6/7,\;P(N=4\mid L=5)=1/7$.)

What You'll Learn