Expected Annual Losses Under Frequency-Severity Models

Why Combine Frequency and Severity?

A deductible changes the amount allocated to each claim, but an annual estimate also depends on how many claims occur. Frequency describes the number of claims in a policy period; severity describes the amount of an individual claim. A frequency-severity model combines these components to estimate annual insured reimbursements and insurer net claim payments.

The calculations in this tutorial concern a per-claim deductible applied to each covered loss. Assume losses are nonnegative, the deductible is fixed, and there is no policy limit, annual aggregate deductible, reimbursement cap, or other coverage adjustment. These qualifications matter: a policy term that acts on the annual total generally cannot be handled by simply applying a per-claim formula to each loss.

Definition: Let $N$ be the number of claims during a policy period, and let $X_j$ be the gross amount of claim $j$, for $j=1,\ldots,N$. Under a per-claim deductible $d$, the insured's reimbursement for claim $j$ is $R_j=\min(X_j,d)$, and the insurer's ultimate net claim payment is $I_j=(X_j-d)^+$, where $z^+=\max(z,0)$. Annual reimbursements are $R=\sum_{j=1}^{N}R_j$, and annual insurer net losses are $I=\sum_{j=1}^{N}I_j$.

The insurer may initially pay the gross claim and later collect the reimbursement from the insured. In this tutorial, “insurer net loss” means the ultimate claim payment after that reimbursement, not the insurer's initial cash payment. For every individual claim, the allocation satisfies $X_j=R_j+I_j$. Therefore, aggregate gross losses also reconcile as the sum of annual reimbursements and annual insurer net losses.

Expected Annual Loss Under Frequency-Severity Assumptions

Suppose the claim count $N$ has finite mean $E[N]$, and claim severities are identically distributed with the same distribution as $X$. Assume the severities are independent of $N$. For the basic formula below, claims need not be Poisson: only the expected claim count and the expected per-claim amounts are needed. Independence between claim severities is a conventional modeling assumption, though linearity of expectation does not require severities to be mutually independent when their individual expectations are the same.

E[R]=E[N]\,E[\min(X,d)], \qquad E[I]=E[N]\,E[(X-d)^+].

These expectations can be evaluated from a discrete severity table by multiplying each possible payment by its probability. For a nonnegative continuous severity with distribution function $F_X$ and survival function $S_X(x)=P(X>x)$, useful equivalent expressions are:

E[\min(X,d)]=\int_0^d S_X(x)\,dx, \qquad E[(X-d)^+]=\int_d^\infty S_X(x)\,dx.

The survival-function expressions are tail-integral formulas. For any nonnegative random variable $Z$, $E[Z]=\int_0^\infty P(Z>x)\,dx$. For $Z=\min(X,d)$, the event $Z>x$ has probability $S_X(x)$ when $0\le x<d$ and probability zero for $x\ge d$. For $Z=(X-d)^+$, the event $Z>x$ is $X>d+x$. Integrating these probabilities gives the two expressions above.

Why the Frequency-Severity Formula Holds

Condition on the claim count. If $N=n$, the annual reimbursement is the sum of $n$ per-claim reimbursements. Since each has expected value $E[\min(X,d)]$, linearity of expectation gives a conditional expected annual reimbursement of $nE[\min(X,d)]$. Taking the expectation over $N$ gives the annual formula:

E[R] =E\bigl[E[R\mid N]\bigr] =E\bigl[N E[\min(X,d)]\bigr] =E[N]E[\min(X,d)].

The same reasoning applies to insurer net payments, replacing the per-claim reimbursement by $(X-d)^+$. The independence assumption between $N$ and severity makes the per-claim expected amount the same regardless of the number of claims. If severity changes with the claim count or other shared risk conditions, the product of the two unconditional means need not be correct.

Key identity: Since $\min(X,d)+(X-d)^+=X$ for every $X\ge 0$, the expected per-claim reimbursement and insurer net payment sum to $E[X]$. Under the stated assumptions, $E[R]+E[I]=E[N]E[X]$, the expected annual gross loss.

Worked Calculations

Worked Example: Discrete Severity and a Poisson Claim Count

A hypothetical commercial policy has a per-claim deductible of $\$4{,}000$. The annual claim count is $N\sim\operatorname{Poisson}(8)$. Each claim has gross severity $\$1{,}000$ with probability $0.25$, $\$5{,}000$ with probability $0.50$, and $\$11{,}000$ with probability $0.25$, independently of $N$. Find expected annual insured reimbursements and expected annual insurer net claim payments.

Method and assumptions. Use $E[N]=8$, then compute the expected reimbursement and insurer payment for one claim. The deductible is applied separately to each claim.

For severities $\$1{,}000,\$5{,}000,\$11{,}000$, the insured reimbursements are respectively $\$1{,}000,\$4{,}000,\$4{,}000$. Thus:

E[\min(X,d)] =0.25(1{,}000)+0.50(4{,}000)+0.25(4{,}000) =250+2{,}000+1{,}000 =\$3{,}250.

The insurer net payments for those same severities are $\$0,\$1{,}000,\$7{,}000$. Therefore:

E[(X-d)^+] =0.25(0)+0.50(1{,}000)+0.25(7{,}000) =0+500+1{,}750 =\$2{,}250.

Multiplying each per-claim expectation by the expected claim count gives:

E[R]=8(\$3{,}250)=\$26{,}000,\qquad E[I]=8(\$2{,}250)=\$18{,}000.

Check. Expected gross severity is $0.25(1{,}000)+0.50(5{,}000)+0.25(11{,}000)=\$5{,}500$, so expected annual gross loss is $8(\$5{,}500)=\$44{,}000$. The allocation reconciles: $\$26{,}000+\$18{,}000=\$44{,}000$. The calculation avoids the mistake of treating the $\$4{,}000$ deductible as one annual amount.

Worked Example: A Binomial Claim Count and a Discrete Severity Mix

A hypothetical fleet policy has $N\sim\operatorname{Binomial}(6,0.5)$. Each claim is $\$1{,}000$ with probability $0.20$, $\$4{,}000$ with probability $0.50$, or $\$9{,}000$ with probability $0.30$, independently of the claim count. The per-claim deductible is $\$3{,}000$. Calculate expected annual reimbursements and insurer net payments.

Method and assumptions. A binomial count has mean $E[N]=6(0.5)=3$. The frequency-severity expectation formula uses that mean; it does not require the claim count to be Poisson.

For one claim, the insured pays $\$1{,}000$, $\$3{,}000$, or $\$3{,}000$, respectively. Hence:

E[\min(X,d)] =0.20(1{,}000)+0.50(3{,}000)+0.30(3{,}000) =200+1{,}500+900 =\$2{,}600.

The insurer pays $\$0$, $\$1{,}000$, or $\$6{,}000$, respectively. Therefore:

E[(X-d)^+] =0.20(0)+0.50(1{,}000)+0.30(6{,}000) =0+500+1{,}800 =\$2{,}300.

The annual expectations are:

E[R]=3(\$2{,}600)=\$7{,}800,\qquad E[I]=3(\$2{,}300)=\$6{,}900.

Check. Mean gross severity is $0.20(1{,}000)+0.50(4{,}000)+0.30(9{,}000)=\$4{,}900$. Expected annual gross loss is $3(\$4{,}900)=\$14{,}700$, and $\$7{,}800+\$6{,}900=\$14{,}700$. The calculation avoids assuming that a Poisson model is necessary just because the frequency and severity components are being combined.

Worked Example: Exponential Severity and a Poisson Claim Count

A hypothetical equipment policy has $N\sim\operatorname{Poisson}(4)$. Individual claim severity $X$ follows an exponential distribution with mean $\$10{,}000$, so its rate is $1/10{,}000$ per dollar and $S_X(x)=e^{-x/10{,}000}$ for $x\ge0$. The per-claim deductible is $\$4{,}000$. Find both expected annual amounts, rounded to cents.

Method and assumptions. Apply the survival-integral expressions to the exponential severity, then multiply by $E[N]=4$. The continuous model assumes nonnegative severity and no upper policy limit.

For the insurer's per-claim payment:

E[(X-d)^+] =\int_{4{,}000}^{\infty}e^{-x/10{,}000}\,dx =10{,}000e^{-4{,}000/10{,}000} =10{,}000e^{-0.4} \approx \$6{,}703.2005.

The exponential distribution has $E[X]=\$10{,}000$. By the per-claim allocation identity, the expected reimbursement is:

E[\min(X,d)] =E[X]-E[(X-d)^+] =10{,}000-6{,}703.2005 \approx \$3{,}296.7995.

Multiplying by the expected claim count and rounding only at the final step to cents:

E[I]=4(6{,}703.2005)\approx \$26{,}812.80,\qquad E[R]=4(3{,}296.7995)\approx \$13{,}187.20.

Check. The expected annual gross loss is $4(\$10{,}000)=\$40{,}000$, and the rounded amounts sum to $\$26{,}812.80+\$13{,}187.20=\$40{,}000.00$. This calculation avoids treating the deductible as a fixed amount subtracted from the mean claim severity in every case; claims below the deductible produce no insurer payment.

Assumptions, Scope, and Common Traps

The basic calculation is powerful because expected values add across claims, but it answers a specific question: the expected annual amount allocated under an uncapped per-claim deductible. Before applying it, confirm that the coverage terms match that structure and that frequency and severity assumptions are appropriate.

Do not deduct once from the annual total. With a per-claim deductible, apply the deductible separately to every claim. An annual aggregate deductible is a different contract structure.
Do not confuse reimbursement with insurer payment. The insured amount is $\min(X,d)$; the insurer amount is $(X-d)^+$. They are complementary parts of the gross claim.
Do not assume $E[N]$ is the frequency parameter in every model. For a Poisson count with parameter $\lambda$, $E[N]=\lambda$; for a binomial count with parameters $n,p$, $E[N]=np$. Use the actual mean of the specified count model.
Check for limits and caps. A payment limit changes the insurer's per-claim layer, and a cap on insured reimbursements changes the reimbursement formula. An annual aggregate term acts on the annual outcome and can require a different calculation.
Check dependence and heterogeneity. If claim severity depends on the number of claims, or different claims have different expected severities, $E[N]E[\text{per-claim payment}]$ may not describe the expected total without additional conditioning or summation.

These expected amounts are loss estimates, not premiums: they do not include expenses, profit provisions, taxes, risk margins, or the timing and collectability of reimbursements. The frequency-severity framework is aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus because it connects claim-count and claim-size assumptions to the expected allocation of losses under a large deductible.

Check Your Understanding

For each problem, calculate expected annual insured reimbursements and insurer net claim payments unless the question asks otherwise. Round dollar amounts to cents when needed.

A hypothetical policy has $N\sim\operatorname{Poisson}(7)$. Each claim is $\$2{,}000$ with probability $0.40$ or $\$7{,}000$ with probability $0.60$. The per-claim deductible is $\$2{,}500$. (Answer: expected reimbursements $\$16{,}100$; expected insurer net payments $\$18{,}900$.)
A hypothetical policy has $N\sim\operatorname{Binomial}(10,0.30)$, with severities $\$1{,}000$ and $\$6{,}000$, each with probability $0.50$, independently of $N$. The per-claim deductible is $\$2{,}000$. Find both expected annual allocations. (Answer: expected reimbursements $\$4{,}500$; expected insurer net payments $\$6{,}000$.)
A hypothetical policy has $N\sim\operatorname{Poisson}(2.5)$, and each severity is exponential with mean $\$8{,}000$. With a per-claim deductible of $\$3{,}000$, calculate both expected annual allocations to cents. (Answer: expected reimbursements $\$6{,}254.21$; expected insurer net payments $\$13{,}745.79$.)
A hypothetical policy has expected claim count $E[N]=4$. Each claim is $\$2{,}000$ with probability $0.20$, $\$5{,}000$ with probability $0.30$, or $\$10{,}000$ with probability $0.50$. For a $\$4{,}000$ per-claim deductible, find expected annual gross loss and its two allocations. (Answer: expected gross loss $\$27{,}600$; expected reimbursements $\$14{,}400$; expected insurer net payments $\$13{,}200$.)
The annual claim count has probabilities $P(N=0)=0.20$, $P(N=2)=0.50$, and $P(N=4)=0.30$. Each claim is $\$3{,}000$ or $\$8{,}000$, each with probability $0.50$, independently of $N$. For a $\$2{,}000$ per-claim deductible, calculate the expected annual reimbursement and insurer payment, showing why the frequency distribution's mean is sufficient. (Answer: $E[N]=2.2$; expected reimbursements $\$4{,}400$; expected insurer net payments $\$7{,}700$.)

What You'll Learn