Why Condition on the Claim Count?
An aggregate loss is a random sum: the number of claims is random, and so are the amounts paid on those claims. The calculation becomes more manageable when separated into two questions: first, what is the probability of each possible claim count; second, given that count, what is the probability that the sum of severities falls in the required range? Weighting the answers to the second question by the probabilities from the first gives the aggregate distribution.
This approach applies whether the claim count has a finite or infinite range and whether severities are discrete or continuous. Its key assumption is that the claim count is independent of the severities, which are identically distributed. If that assumption fails, the conditional severity model may change with the claim count.
The Mixture Formula
The events \(N=0,N=1,N=2,\ldots\) are mutually exclusive and together cover all possible claim counts. The event \(S\le s\) can therefore be split across those events. For each count, the probability of \(S\le s\) and \(N=n\) is the probability of \(N=n\) multiplied by the conditional probability of \(S\le s\) given \(N=n\). Adding these disjoint contributions gives the aggregate CDF.
If \(N\) can only take values from \(0\) through \(m\), the sum ends at \(m\). If \(N\) has an unbounded range, the expression is an infinite sum of nonnegative terms. In either case, the claim-count probabilities are the mixture weights: they are nonnegative and sum to one.
This equality uses independence of \(N\) and the severities, as well as the identical severity distribution. Conditioning on \(N=n\) tells us how many severities to add; it does not change the distribution of any one severity. The conditional probability is evaluated using the model for the sum of \(n\) severities. For discrete severities, this can be done by listing the combinations of severity values whose total is at most \(s\). For continuous severities, it is the probability over the region \(x_1+\cdots+x_n\le s\), or it can be obtained from a known distribution for the sum.
A useful way to organize a calculation is to make a row for each possible value of \(n\), record its frequency probability, find the corresponding conditional probability, and multiply. Only after all rows are complete should the weighted contributions be added. This keeps the count model separate from the severity-sum calculation.
Use the support of \(N\), including \(0\) when it is possible.
For each \(n\), calculate the probability that the sum of \(n\) severities is at most the specified threshold.
Multiply each conditional probability by \(P(N=n)\).
The result must lie between zero and one; the weights used must sum to one.
Worked Applications
Worked Example: A Discrete Severity Model
A hypothetical policy has claim count \(N\) with \(P(N=0)=0.25\), \(P(N=1)=0.50\), and \(P(N=2)=0.25\). Given a claim, its severity is \(1\) or \(3\) thousand dollars, each with probability \(0.50\). Severities are independent of \(N\) and independent across claims. Find \(P(S\le 3)\).
Method. Condition on \(N\), then enumerate the severity outcomes for each count. The enumeration avoids treating the threshold as a limit on each individual claim rather than on the total.
For \(N=0\), \(S=0\), so \(P(S\le 3\mid N=0)=1\). For \(N=1\), either possible severity is at most \(3\), so \(P(S\le 3\mid N=1)=1\).
For \(N=2\), the four ordered pairs of severities are \((1,1),(1,3),(3,1),(3,3)\), each with probability \(0.50\times0.50=0.25\). Their totals are \(2,4,4,6\), respectively. Only \((1,1)\) has total at most \(3\), giving \(P(S\le 3\mid N=2)=0.25\).
Weight and add the conditional probabilities:
Thus \(P(S\le 3)=0.8125\), or 0.8125 to four decimal places. The weights check: \(0.25+0.50+0.25=1\), and the answer is a valid probability.
Worked Example: Continuous Severities and a Sum Region
For a hypothetical portfolio, \(P(N=n)=0.10,0.20,0.30,0.40\) for \(n=0,1,2,3\), respectively. Given a claim, its severity is uniformly distributed between \(0\) and \(2\) thousand dollars. Claim severities are independent of the count and of one another. Find \(P(S\le 2)\).
Method. Find the conditional probability for each count using the area of the relevant region in the severity space, then use the count probabilities as weights. The threshold applies to the aggregate, so it is not enough to check whether each severity is below \(2\).
For \(n=0\), the aggregate is zero, so the conditional probability is \(1\). For \(n=1\), a uniform severity on \([0,2]\) is at most \(2\) with probability \(1\).
For \(n=2\), the possible pairs fill the square \(0\le x_1\le2,\ 0\le x_2\le2\), whose area is \(2\times2=4\). The condition \(x_1+x_2\le2\) selects a right triangle with legs of length \(2\) and area \(\frac12(2)(2)=2\). Thus the conditional probability is \(2/4=0.50\).
For \(n=3\), the region \(x_1+x_2+x_3\le2\), with all three severities nonnegative, is a three-dimensional simplex. Its volume is \(2^3/3!=8/6=4/3\). The full cube has volume \(2^3=8\), so the conditional probability is \((4/3)/8=1/6\).
The mixture is therefore
Rounded only at the final step, \(P(S\le2)=\mathbf{0.5167}\) to four decimal places. The claim-count weights sum to \(1.00\); the conditional probabilities are all between zero and one.
Worked Example: A Threshold That Includes One Claim but Not Every Two-Claim Outcome
A hypothetical portfolio has \(P(N=0)=0.10\), \(P(N=1)=0.40\), and \(P(N=2)=0.50\). Each claim severity is \(2\) thousand dollars with probability \(0.70\), or \(5\) thousand dollars with probability \(0.30\). Severities are independent of the count and of one another. Find \(P(S\le6)\).
Method. Calculate the aggregate-threshold probability separately for each claim count. This avoids the trap of multiplying the probability that one severity meets a condition without checking the total when there are multiple claims.
For \(N=0\), the aggregate is zero and the conditional probability is \(1\). For \(N=1\), both possible severities, \(2\) and \(5\), are at most \(6\), so the conditional probability is also \(1\).
For \(N=2\), the possible ordered pairs and their probabilities are: \((2,2)\) with probability \((0.70)(0.70)=0.49\); \((2,5)\) with probability \(0.21\); \((5,2)\) with probability \(0.21\); and \((5,5)\) with probability \(0.09\). Their totals are \(4,7,7,10\). Only the first total is at most \(6\), so \(P(S\le6\mid N=2)=0.49\).
Apply the count weights:
The answer is 0.7450 to four decimal places. The count probabilities sum to \(1\), and the result reflects that two claims can exceed the threshold even though either single claim does not.
Assumptions, Interpretation, and Exam Checks
The formula is an application of the law of total probability. Its main benefit is conceptual as well as computational: it makes clear which part of the model supplies each input. The distribution of \(N\) supplies the weights; the severity model supplies the conditional probabilities for the sums.
The standard calculation above assumes that the severities are identically distributed and independent of the claim count. If claim sizes depend on the number of claims—for example, if high-frequency periods also tend to have larger claims—then \(P(S\le s\mid N=n)\) must be calculated using the severity distribution conditional on \(N=n\). It cannot automatically be replaced by the sum of \(n\) severities from one common, count-independent distribution.
For a threshold below zero, a nonnegative aggregate satisfies \(P(S\le s)=0\). For a nonnegative threshold, be careful with equality: the event is \(S\le s\), not \(S<s\). In a discrete model, an aggregate exactly equal to \(s\) is included. Also distinguish a conditional probability from its weighted contribution: \(P(S\le s\mid N=n)\) is not itself the probability contribution for count \(n\); that contribution is \(P(N=n)P(S\le s\mid N=n)\).
Check Your Understanding
For each problem, condition on the claim count before combining the conditional probabilities.
- A hypothetical portfolio has \(P(N=0)=0.20\), \(P(N=1)=0.50\), and \(P(N=2)=0.30\). Each severity is \(1\) or \(4\) thousand dollars with equal probability, independently of the count and other severities. Find \(P(S\le2.5)\). (Answer: 0.5250)
- Claim count probabilities are \(P(N=0)=0.25\), \(P(N=1)=0.50\), and \(P(N=2)=0.25\). Severities are independent Uniform\((0,1)\) amounts, also independent of \(N\). Find \(P(S\le1)\). (Answer: 0.8750)
- A portfolio has \(P(N=0)=0.10\), \(P(N=1)=0.20\), \(P(N=2)=0.30\), and \(P(N=3)=0.40\). Severities are independent Uniform\((0,1)\) amounts and independent of \(N\). Find \(P(S\le1.5)\). (Answer: 0.7625)
- A hypothetical count model has \(P(N=1)=0.60\) and \(P(N=2)=0.40\). Each independent severity is \(2\) thousand dollars with probability \(0.80\), or \(6\) thousand dollars with probability \(0.20\); severities are independent of \(N\). Find \(P(S\le5)\). (Answer: 0.7360)