Why Combine Frequency and Severity?
An aggregate loss for a period depends on both how many claims occur and how large those claims are. Empirical frequency data summarize the first component; observed claim amounts summarize the second. Combining them gives a model-based estimate of probabilities for total loss, even when the observed data do not contain every aggregate outcome of interest.
The construction is a plug-in estimate: treat the observed frequency proportions and observed severity proportions as the distributions for a future period. This is useful only after the assumptions behind that construction are stated. In particular, the standard model assumes that claim count is independent of claim severities, and that severities are independent and identically distributed conditional on the count. These assumptions may be unsuitable if, for example, high-count periods systematically have larger claims.
For example, if 8 comparable periods had counts \(0,0,1,1,1,2,2,3\), the empirical frequency probabilities would be \(p_0=2/8\), \(p_1=3/8\), \(p_2=2/8\), and \(p_3=1/8\). If the severity observations were \(2,2,5,5,5\) (in thousands of dollars), the empirical severity distribution would assign probability \(2/5\) to 2 and \(3/5\) to 5. Repeated observations retain their multiplicity: they contribute repeated empirical probability mass.
Conditioning and Discrete Convolution
For a specified aggregate total \(s\), first find the probability of that total conditional on each possible claim count. Then weight each conditional probability by the empirical probability of that count. This is the law of total probability applied to \(N\).
For \(n=0\), the sum of severities is defined as zero, so its conditional distribution places probability one at \(s=0\). For larger \(n\), the conditional aggregate distribution is obtained by repeatedly convolving the severity distribution with itself. If \(q_n(s)=P(X_1+\cdots+X_n=s)\), then \(q_0(0)=1\), \(q_0(s)=0\) for \(s\ne0\), and
The sum over \(x\) is taken over the severity values in the empirical distribution. This recurrence includes all ordered sequences of severities. For two claims, for instance, the outcomes \(2,5\) and \(5,2\) are distinct possible draws, even though they have the same total.
Worked Example: Convolving Two Empirical Distributions
A hypothetical portfolio has four comparable observed periods, with claim counts \(0,1,1,2\). Its empirical severity observations are \(2\) and \(4\), each in thousands of dollars. Under the independent empirical model, find the aggregate-loss distribution and \(P(L>4)\).
Method and assumptions. Each observed period receives equal frequency weight, so \(p_0=1/4\), \(p_1=2/4=1/2\), and \(p_2=1/4\). Each severity draw is independently 2 or 4, each with probability \(1/2\), and is made with replacement. Frequencies and severities are independent.
Zero claims. If \(N=0\), then \(L=0\). Its contribution to \(P(L=0)\) is \(1/4\).
One claim. Given \(N=1\), aggregate loss is 2 or 4, each with probability \(1/2\). Weighting by \(p_1=1/2\), the contributions are \(P(L=2)= (1/2)(1/2)=1/4\) and \(P(L=4)=(1/2)(1/2)=1/4\).
Two claims. The four equally likely ordered pairs are \((2,2),(2,4),(4,2),(4,4)\). Their totals are 4, 6, 6, and 8. Given \(N=2\), the probabilities of totals 4, 6, and 8 are \(1/4,1/2,\) and \(1/4\). Weighting by \(p_2=1/4\), their contributions are \(1/16,1/8,\) and \(1/16\), respectively.
Combine the contributions. Thus \(P(L=0)=1/4\), \(P(L=2)=1/4\), \(P(L=4)=1/4+1/16=5/16\), \(P(L=6)=1/8\), and \(P(L=8)=1/16\). The probabilities sum to \(4/16+4/16+5/16+2/16+1/16=1\), as required. Therefore \(P(L>4)=P(L=6)+P(L=8)=1/8+1/16=3/16=0.1875\).
Check and common mistake. The aggregate mean should equal \(E[N]E[X]\) under independence. Here \(E[N]=0(1/4)+1(1/2)+2(1/4)=1\), and \(E[X]=(2+4)/2=3\) thousand dollars. From the distribution, \(E[L]=0(1/4)+2(1/4)+4(5/16)+6(1/8)+8(1/16)=3\) thousand dollars, which checks. A common error is to report \(3/4\) for \(P(L>4)\) conditional on two claims, without weighting by \(P(N=2)=1/4\), or to forget that the one-claim outcome of 4 is not greater than 4.
Estimating a Probability by Convolution
A probability such as \(P(L\leq c)\) can be found by summing the aggregate probabilities at all totals no greater than \(c\). Alternatively, condition on \(N\) and calculate the conditional probability for each count. The second approach is often shorter when the event has a simple description, but it still requires accounting for every count with positive empirical probability.
Worked Example: A Cumulative Aggregate-Loss Probability
A hypothetical group of four periods has counts \(0,0,1,2\). The observed claim severities are \(1,3,5\), each occurring once; amounts are in thousands of dollars. Under independent sampling with replacement, find \(P(L\leq3)\).
Method and assumptions. The empirical count probabilities are \(p_0=1/2\), \(p_1=1/4\), and \(p_2=1/4\). Each severity is independently 1, 3, or 5 with probability \(1/3\), independent of \(N\).
Condition on the count. For \(N=0\), \(L=0\), so \(P(L\leq3\mid N=0)=1\). For \(N=1\), the qualifying severities are 1 and 3, so \(P(L\leq3\mid N=1)=2/3\). For \(N=2\), the nine ordered pairs are equally likely. Only \((1,1)\) has total at most 3: \((1,3)\) and \((3,1)\) total 4, and every pair involving 5 has total greater than 3. Thus \(P(L\leq3\mid N=2)=1/9\).
Weight and add. The law of total probability gives \(P(L\leq3)=(1/2)(1)+(1/4)(2/3)+(1/4)(1/9)=1/2+1/6+1/36=25/36\). As a decimal to four places, the estimate is \(0.6944\).
Check and common mistake. The three weighted contributions are \(18/36\), \(6/36\), and \(1/36\), so their sum is \(25/36\). The two-claim conditional probability is \(1/9\), not \(1/3\): the relevant outcomes are ordered pairs, and only one of the nine has a total at most 3. Also, the zero-claim probability contributes because zero is within the requested threshold.
Simulation as an Alternative
When the number of possible severities or claim counts makes exact convolution cumbersome, simulation provides a direct estimate. For each simulated period, draw a count from the empirical frequency distribution; then draw that many severities independently from the empirical severity distribution. The simulated aggregate is their sum. The fraction of simulated aggregates satisfying the event of interest estimates its probability.
Record the frequency probabilities and the severity values with their empirical probabilities. State the independence and with-replacement assumptions.
Draw \(N\), make \(N\) independent severity draws, and add them. If \(N=0\), the aggregate is zero.
For an event such as \(L>c\), divide the number of simulated periods satisfying the event by the total number simulated. More repetitions generally reduce simulation noise, but do not remove model bias or data limitations.
Worked Example: A Finite Simulation Estimate
Use the empirical model with \(P(N=0)=1/2\), \(P(N=1)=1/4\), \(P(N=2)=1/4\), and independent severity draws of 1, 2, or 4 thousand dollars, each with probability \(1/3\). Twelve independent simulated periods produce the outcomes shown below. Estimate \(P(L>2)\).
| Period | Simulated \(N\) | Severity draws (thousands) | Aggregate \(L\) (thousands) | \(L>2\)? |
|---|---|---|---|---|
| 1 | 0 | — | 0 | No |
| 2 | 0 | — | 0 | No |
| 3 | 1 | 1 | 1 | No |
| 4 | 1 | 2 | 2 | No |
| 5 | 2 | 1, 2 | 3 | Yes |
| 6 | 2 | 2, 2 | 4 | Yes |
| 7 | 0 | — | 0 | No |
| 8 | 1 | 2 | 2 | No |
| 9 | 2 | 1, 4 | 5 | Yes |
| 10 | 0 | — | 0 | No |
| 11 | 1 | 1 | 1 | No |
| 12 | 2 | 2, 2 | 4 | Yes |
Calculation. Four of the twelve simulated aggregates exceed 2 thousand dollars: periods 5, 6, 9, and 12. The simulation estimate is therefore \(4/12=1/3=0.3333\), to four decimal places.
Exact check and common mistake. For this model, \(P(L>2\mid N=0)=0\). For \(N=1\), only severity 4 exceeds 2, giving conditional probability \(1/3\). For \(N=2\), all nine ordered pairs qualify except \((1,1)\), so the conditional probability is \(8/9\). The exact probability is \((1/2)(0)+(1/4)(1/3)+(1/4)(8/9)=1/12+2/9=11/36=0.3056\). The simulated estimate differs because it uses only twelve trials. A common mistake is to treat the simulated fraction as the exact model probability; it is an estimate subject to simulation variation.
Assumptions, Interpretation, and Exam Traps
Empirical convolution estimates a future aggregate distribution by combining two empirical distributions. It does not establish that the frequency and severity processes are independent in the underlying population. Independence is an assumption of the model, and the calculation can be misleading if it fails. For example, a portfolio with more claims in certain periods may also have systematically higher severities. In that case, a single severity distribution applied independently to every claim can lose important dependence.
The empirical severity distribution also uses observed claims as its support. A very large severity not present in the sample receives zero probability under the literal empirical distribution. This is a feature of the plug-in estimate, not proof that such a loss is impossible. The result can be particularly sensitive to a small severity sample when estimating tail probabilities.
- Use the observed proportions as probabilities. A count observed twice among eight periods has empirical probability \(2/8\), not probability one.
- Sample severities with replacement under the standard model. The empirical distribution represents a severity law; two claims can independently take the same observed value.
- Keep ordered draws in convolution. For two claims, \((a,b)\) and \((b,a)\) are separate outcomes when \(a\ne b\).
- Weight conditional probabilities. A probability calculated given \(N=n\) must be multiplied by \(p_n\) before adding it to the aggregate probability.
- Track units and the event boundary. If amounts are in thousands, keep thresholds in thousands too; \(L>c\) does not include equality.
A useful diagnostic under the independence model is \(E[L]=E[N]E[X]\), provided the means are finite. This checks an aggregate distribution's mean, but it does not verify the independence assumption or ensure that the estimated tail probabilities are reliable. State the model, show the conditioning or simulation procedure, and interpret the result as an empirical estimate. These are central skills aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus.
Check Your Understanding
For each problem, use independent severity draws with replacement and treat observed proportions as the empirical probabilities. Amounts are in thousands of dollars unless otherwise stated.
- A frequency sample has counts \(1,1,2,2\). The empirical severities are 2 and 5, equally likely. Find \(P(L=5)\). (Answer: \(0.2500\))
- A frequency sample gives \(P(N=0)=1/2\) and \(P(N=2)=1/2\). Severities are 2 and 6, equally likely. Find \(P(L=8)\). (Answer: \(0.2500\))
- Three comparable periods have counts \(0,1,2\). Severities are 3 and 6, equally likely. Find \(P(L=6)\). (Answer: \(1/4=0.2500\))
- Four periods have counts \(0,0,1,2\). Severities are 2 and 4, equally likely. Find \(P(L>4)\). (Answer: \(3/16=0.1875\))
- The empirical count probabilities are \(P(N=0)=1/4\), \(P(N=1)=1/2\), and \(P(N=2)=1/4\). Severities are 1, 3, and 5, equally likely. Find \(P(L\leq3)\). (Answer: \(11/18=0.6111\))