From Event Frequency and Severity to Annual Loss
An operational loss model separates two sources of uncertainty. The frequency model describes how many loss events occur in a year; the severity model describes the amount of each event. The annual total depends on both. Combining them carefully makes clear which frequency and severity assumptions are needed for an aggregate distribution and its moments.
Throughout, consider a hypothetical portfolio and let \(N\) be the number of loss events during one year. Let \(X_i\) be the amount of the \(i\)th event, and define the annual aggregate loss \(S\) as the sum of the event amounts. If there are no events, the sum is zero.
The independence condition is substantive. If high event counts tend to occur in years with larger event severities, the standard compound formulas need not hold. Testing frequency-severity independence is therefore relevant to the choice of model, not just a technical preliminary. The model also treats event amounts as additive and does not, by itself, describe dependence among event severities, changing exposure during the year, or changes in the loss environment over time.
The Probability Generating Function
For a nonnegative integer-valued random variable \(W\), its probability generating function (PGF) is \(G_W(z)=E[z^W]\), for values of \(z\) where the expectation exists. When both event counts and severities are integer-valued in a chosen unit, the aggregate \(S\) is also integer-valued and has a PGF. For example, a model might measure loss amounts in whole thousands of dollars. That unit choice must be consistent throughout the model.
Write \(G_N(z)=E[z^N]\) for the count PGF and \(G_X(z)=E[z^X]\) for the severity PGF. Condition on \(N=n\). Given \(n\), the aggregate is the sum of \(n\) independent severities, so its PGF is \(G_X(z)^n\). Averaging over the possible counts gives the compound PGF.
This composition is the key distributional result: evaluate the frequency PGF at the severity PGF. It follows from conditioning and the assumed independence; it does not require a particular frequency distribution. The identity applies wherever the PGFs and the composition are defined.
If severities are measured on a continuous scale, an ordinary PGF is not available for \(X\). A corresponding result uses the Laplace transform \(L_X(s)=E[e^{-sX}]\), for nonnegative loss amounts and \(s\geq 0\):
The same conditioning argument proves this formula. It is useful for stating the aggregate distributional relationship, but it does not turn continuous severities into integer-valued PGFs.
Deriving the Aggregate Mean and Variance
Let \(\mu_X=E[X]\), \(\sigma_X^2=\operatorname{Var}(X)\), \(m_N=E[N]\), and \(v_N=\operatorname{Var}(N)\). Assume these moments are finite and the independent compound model holds. Conditional on \(N=n\), the sum has mean \(n\mu_X\) and variance \(n\sigma_X^2\), by independence of the severities. Applying the laws of total expectation and total variance yields:
The variance has two components. The first, \(m_N\sigma_X^2\), is uncertainty in severities across the expected number of events. The second, \(v_N\mu_X^2\), is uncertainty from the event count itself, scaled by squared average severity. These formulas are not valid without the stated independence and finite-moment assumptions.
For a nonnegative integer-valued \(W\), if \(E[W]\) is finite, then \(G_W'(1)=E[W]\). If \(E[W(W-1)]\) is finite, then \(G_W''(1)=E[W(W-1)]\). The second condition matters: existence of a finite first moment alone does not guarantee a finite second derivative or second moment. When the second factorial moment is finite, \(\operatorname{Var}(W)=G_W''(1)+G_W'(1)-[G_W'(1)]^2\).
Differentiating the compound PGF gives another justification for the moment formulas. Put \(\mu_X=G_X'(1)\), and suppose the first moments are finite. Then the chain rule gives \(G_S'(1)=G_N'(1)G_X'(1)=m_N\mu_X\). If the second factorial moments are also finite, a second differentiation gives:
Since \(E[X(X-1)]=\sigma_X^2+\mu_X^2-\mu_X\) and \(G_N''(1)=v_N+m_N^2-m_N\), substituting these into \(G_S''(1)+G_S'(1)-[G_S'(1)]^2\) reduces to \(m_N\sigma_X^2+v_N\mu_X^2\). This agrees with the conditional-variance derivation.
Worked Example: Poisson Frequency and Two-Point Severity
A hypothetical control-failure model has a Poisson event count with annual mean \(\lambda=2\). Each event causes a loss of either 1 or 3 units, each with probability \(0.5\), independently across events and independently of \(N\). Find the aggregate PGF, mean, variance, and probability of no annual loss.
Method and assumptions. Use the compound PGF and moment formulas. The Poisson count has \(G_N(z)=\exp(2(z-1))\), \(E[N]=2\), and \(\operatorname{Var}(N)=2\). The severity PGF is \(G_X(z)=0.5z+0.5z^3\).
Aggregate PGF. Substitute the severity PGF into the frequency PGF:
Severity moments. The mean is \(E[X]=0.5(1)+0.5(3)=2\) units. The second moment is \(E[X^2]=0.5(1^2)+0.5(3^2)=0.5+4.5=5\) units squared. Thus \(\operatorname{Var}(X)=5-2^2=1\) unit squared.
Aggregate moments. The mean is \(E[S]=2(2)=4\) units. The variance is \(2(1)+2(2^2)=2+8=10\) units squared.
Probability of zero loss. Both possible event severities are positive, so \(S=0\) exactly when \(N=0\). Therefore \(P(S=0)=e^{-2}=0.1353\), to four decimal places. As a check, \(G_S(0)=\exp(2(0-1))=e^{-2}\), the same probability.
Common mistake avoided. The variance is not just \(E[N]\operatorname{Var}(X)=2\). That term accounts for severity variation conditional on frequency; the additional \(2(2^2)=8\) accounts for count variation.
Worked Example: Geometric Frequency and a Compound PGF
Suppose a hypothetical system has a geometric event count with \(P(N=n)=0.5(0.5)^n\) for \(n=0,1,2,\ldots\). Each event loss is 1 or 2 units with equal probability. Find the PGF and aggregate mean and variance.
Method and frequency moments. For this geometric convention, \(G_N(z)=0.5/(1-0.5z)=1/(2-z)\), \(E[N]=(1-p)/p=1\), and \(\operatorname{Var}(N)=(1-p)/p^2=2\), where \(p=0.5\). This convention includes zero events; it is not the geometric distribution supported on counts beginning at one.
Severity PGF and composition. The severity PGF is \(G_X(z)=0.5z+0.5z^2\). Hence:
Severity moments. \(E[X]=0.5(1)+0.5(2)=1.5\) units. Also \(E[X^2]=0.5(1)+0.5(4)=2.5\) units squared, so \(\operatorname{Var}(X)=2.5-1.5^2=0.25\) unit squared.
Aggregate moments. \(E[S]=1(1.5)=1.5\) units. The variance is \(1(0.25)+2(1.5^2)=0.25+4.5=4.75\) units squared.
Check using factorial moments. Here \(G_N'(1)=1\), \(G_N''(1)=2\), \(G_X'(1)=1.5\), and \(G_X''(1)=1\). Thus \(G_S''(1)=2(1.5^2)+1(1)=5.5\), while \(G_S'(1)=1.5\). Therefore \(E[S^2]=5.5+1.5=7\) and \(\operatorname{Var}(S)=7-1.5^2=4.75\), as above.
Common mistake avoided. Do not substitute \(E[X^2]\) for \(\operatorname{Var}(X)\) in the first variance component while leaving the rest unchanged. The correct expression is \(E[N]\operatorname{Var}(X)+\operatorname{Var}(N)(E[X])^2\), which gives \(0.25+4.5=4.75\), not \(E[N]E[X^2]+\operatorname{Var}(N)(E[X])^2=7\).
Worked Example: Binomial Frequency and a Low-Loss Probability
A hypothetical process can produce at most three loss events in a year. Each of three independent opportunities produces an event with probability \(0.4\), so \(N\sim\operatorname{Binomial}(3,0.4)\). Each event is independently a loss of 1 or 2 units with equal probability. Find the aggregate mean and variance, and calculate \(P(S\leq 1)\).
Method and frequency moments. The count PGF is \(G_N(z)=(0.6+0.4z)^3\), with \(E[N]=3(0.4)=1.2\) and \(\operatorname{Var}(N)=3(0.4)(0.6)=0.72\). The severity mean is \(1.5\), and its variance is \(2.5-1.5^2=0.25\).
Aggregate PGF. The severity PGF is \(G_X(z)=0.5z+0.5z^2\), giving:
Aggregate moments. The mean is \(1.2(1.5)=1.8\) units. The variance is \(1.2(0.25)+0.72(1.5^2)=0.30+1.62=1.92\) units squared.
Probability of at most one unit. Every event causes at least one unit of loss. Thus \(S=0\) requires no events, with probability \(0.6^3=0.216\). For \(S=1\), there must be exactly one event and its severity must be 1. The probability is \(\binom{3}{1}(0.4)(0.6)^2(0.5)=0.432(0.5)=0.216\). Therefore \(P(S\leq1)=0.216+0.216=0.432\).
Check and common mistake avoided. The coefficient of \(z\) in the aggregate PGF is \(3(0.6)^2(0.2)=0.216\), confirming \(P(S=1)\). Counting every one-event year as a one-unit loss would be wrong: half of those event severities are 2 units.
Using the Results Carefully
The PGF and moment formulas provide compact checks on an aggregate model. The PGF describes the distribution when losses use integer units; its value at zero gives the probability of no aggregate loss. The mean and variance summarize the impact of event count and severity uncertainty, but they do not determine the full aggregate distribution. Different frequency-severity combinations can have the same first two aggregate moments.
A particularly common variance error is to use the second severity moment in place of the severity variance. The correct formula can also be written as \(E[N]E[X^2]+(\operatorname{Var}(N)-E[N])(E[X])^2\); omitting the negative \(E[N](E[X])^2\) adjustment overstates the variance. Another trap is to apply \(G_W''(1)=E[W(W-1)]\) without checking that the second factorial moment is finite. The first derivative requires a finite first moment, while the second derivative identity requires the stronger second-moment condition.
Finally, the formulas assume a common severity distribution independent of the annual count. If event severities vary systematically with the count, or if severities depend on one another, conditioning on \(N\) does not give the simple conditional mean and variance used in the derivation. The analyst should not use the independent compound formulas merely because separate frequency and severity estimates are available. Correct specification of the model boundary and dependence assumptions is essential to interpret the result. This treatment is aligned with the CAS Exam 9 (Risk Management for Actuaries) syllabus.
Check Your Understanding
For each problem, use the stated frequency-severity assumptions and keep the units consistent.
- A hypothetical loss process has \(N\sim\operatorname{Poisson}(3)\). Each event loss is 2 units with probability \(0.75\) and 6 units with probability \(0.25\), independently of the count. Give \(G_S(z)\), \(E[S]\), and \(\operatorname{Var}(S)\). (Answer: \(G_S(z)=\exp(3(0.75z^2+0.25z^6-1))\); mean \(=9\) units; variance \(=36\) units squared.)
- A system has \(N\sim\operatorname{Bernoulli}(0.2)\). If an event occurs, its loss is 1 or 4 units with equal probability. Find the aggregate mean and variance. (Answer: mean \(=0.5\) units; variance \(=1.45\) units squared.)
- For a hypothetical count \(P(N=n)=0.25(0.75)^n\), \(n=0,1,2,\ldots\), each event causes a deterministic loss of 2 units. Find the aggregate PGF, mean, and variance. (Answer: \(G_S(z)=0.25/(1-0.75z^2)\); mean \(=6\) units; variance \(=48\) units squared.)
- A Poisson count has mean \(1.2\); independent event losses are 1 or 3 units with equal probability. Find the aggregate mean and variance, then the probability of no annual loss. (Answer: mean \(=2.4\) units; variance \(=6\) units squared; \(P(S=0)=e^{-1.2}=0.3010\).)
- In a dependent model, \(N=0,1,2\) with probabilities \(0.5,0.25,0.25\), respectively. If \(N=1\), the event loss is 2 units; if \(N=2\), both losses are 10 units each. Calculate the aggregate mean and variance directly, and explain why the standard independent compound formula cannot be applied from a single severity distribution independent of \(N\). (Answer: mean \(=5.5\) units; variance \(=70.75\) units squared; severity depends on \(N\), violating frequency-severity independence.)