Deductibles and Claim Frequency

Why a Deductible Changes Claim Frequency

A per-claim deductible affects more than the amount paid on a claim. Claims at or below the deductible produce no positive insurer payment, so the frequency of payable claims is lower than the frequency of ground-up claims. The claims that do receive a payment also have a different average severity: their severity is the average payment conditional on exceeding the deductible, not the average ground-up loss.

Keeping these two changes separate is essential in frequency-severity ratemaking. The insurer’s expected loss per exposure can be calculated either from all ground-up claims and their deductible-adjusted payments, or as payable claim frequency times average payment per payable claim. The two approaches must agree when they use the same assumptions and definitions.

Definition: Let $X$ be the ground-up severity of one claim and let $d\geq 0$ be an ordinary per-claim deductible. The insurer’s payment on the claim is $Y^L=(X-d)_+=\max(X-d,0)$. A claim is payable when $X>d$. If the ground-up claim frequency is $\lambda$ claims per exposure, the expected payable claim frequency is $\lambda P(X>d)$, and the average payment per payable claim is $E[X-d\mid X>d]$, provided $P(X>d)>0$.

The frequency convention here counts a claim only if it produces a positive insurer payment. A claim exactly equal to $d$ has payment zero, so it is not a payable claim. In practice, reported-claim counts can follow different rules; the definition used in a data set or rating calculation must be checked before treating a count as payable frequency.

Frequency, Severity, and Expected Insurer Loss

Assume that ground-up claims have independent, identically distributed severities $X$, independent of the claim count. Let $q_d=P(X>d)$. For each ground-up claim, the probability of a positive payment is $q_d$. Thus, when the ground-up frequency is $\lambda$, the expected number of payable claims per exposure is $\lambda q_d$. This expected-frequency identity does not require the total claim count to be Poisson.

For continuous severity, $q_d=S_X(d)$, where $S_X(x)=P(X>x)$. The conditional average payment per payable claim is

E[Y^P]=E[X-d\mid X>d] =\frac{E[(X-d)_+]}{S_X(d)}.

Here $Y^P$ denotes payment per payable claim. The numerator is expected insurer payment per ground-up claim, including zero payments; dividing by the probability of a positive payment converts it to an average conditional on payment. The expression requires $S_X(d)>0$. For a severity distribution with atoms, use the conditional expectation directly and include only outcomes with $X>d$.

If the number of ground-up claims per exposure has mean $\lambda$, the expected insurer loss per exposure is

\lambda E[(X-d)_+] =\lambda P(X>d)\,E[X-d\mid X>d].

Why the Frequency-Severity Product Must Reconcile

The identity follows by conditioning the payment on whether the claim exceeds the deductible. On $X\leq d$, the insurer pays zero; on $X>d$, the payment is $X-d$. Therefore,

E[(X-d)_+] =E[(X-d)\mathbf{1}_{\{X>d\}}] =P(X>d)E[X-d\mid X>d].

Multiplying by $\lambda$, the expected number of ground-up claims per exposure, gives expected insurer loss per exposure. Equivalently, $\lambda P(X>d)$ is expected payable claims per exposure, and $E[X-d\mid X>d]$ is the average payment for each such claim. This is a reconciliation of expected values; it does not imply that every realized portfolio has exactly this claim count or average payment.

For a continuous nonnegative severity distribution, the expected payment per ground-up claim can also be written using the survival function:

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

To see why, a nonnegative payment $Y^L$ has expectation $\int_0^\infty P(Y^L>y)\,dy$. For $y\geq0$, $Y^L>y$ exactly when $X>d+y$. Substituting $x=d+y$ gives the stated integral. Dividing this integral by $S_X(d)$ gives the conditional average payment per payable claim.

If the ground-up count $N$ is Poisson with mean $\lambda$ per exposure, and severities are independent of $N$, independently retaining each claim with probability $q_d$ gives a Poisson payable count with mean $\lambda q_d$. If $N$ is not Poisson, $\lambda q_d$ remains the expected payable count, but the payable count need not be Poisson.

Worked Example: Exponential Severity and Poisson Claim Counts

Hypothetical model and assumptions. A policy has a ground-up claim count $N\sim\operatorname{Poisson}(0.20)$ per policy-year. Claim severities are independent of $N$ and independent across claims, with $X$ exponentially distributed with mean $\theta=\$1{,}000$. The exponential survival function is $S_X(x)=e^{-x/\theta}$ for $x\geq0$. An ordinary deductible is $d=\$500$. Find the payable frequency, average payment per payable claim, and expected insurer loss per policy-year.

Step 1: Probability of a payable claim. The probability a ground-up claim exceeds the deductible is

q_d=P(X>500)=e^{-500/1000}=e^{-0.5}=0.6065307.

Step 2: Payable claim frequency. Thinning the Poisson claim count gives a Poisson payable count with mean

$$ 0.20(0.6065307)=0.1213061 $$

payable claims per policy-year.

Step 3: Average payment per payable claim. By the memoryless property of the exponential distribution, conditional on $X>500$, the excess $X-500$ has an exponential distribution with mean $\$1{,}000$. Hence $E[Y^P]=\$1{,}000$.

Step 4: Reconcile expected loss. Frequency times conditional severity gives $0.1213061(\$1{,}000)=\$121.3061$ per policy-year. Directly, the expected payment per ground-up claim is $\theta e^{-d/\theta}=\$1{,}000(0.6065307)=\$606.5307$, so the ground-up frequency calculation gives $0.20(\$606.5307)=\$121.3061$. Both methods agree. Reported to the nearest cent, expected insurer loss is \$121.31 per policy-year. The equality of ground-up and conditional average severity would not generally hold for a non-exponential model.

Calculating From a Discrete Severity Distribution

For a discrete or empirical severity distribution, identify which outcomes exceed the deductible, subtract the deductible from each such loss, and retain their original probabilities when calculating expected payment per ground-up claim. Divide by the probability of exceeding the deductible only when calculating the conditional average payment per payable claim.

Worked Example: A Discrete Model and a Common Frequency Error

Hypothetical model and assumptions. A book has ground-up frequency $\lambda=0.10$ claims per policy-year. Claim severities are independent of the count and take the values $\$200,\$600,\$1{,}400$ with probabilities $0.20,0.30,0.50$, respectively. Each claim is subject to an ordinary deductible of $d=\$500$. Calculate payable frequency, average payment per payable claim, and expected insurer loss per policy-year.

Step 1: Identify positive payments. The $\$200$ loss produces no payment. The $\$600$ and $\$1{,}400$ losses produce payments of $\$100$ and $\$900$. Thus,

$$ q_d=0.30+0.50=0.80. $$

Step 2: Calculate payable frequency. The expected number of payable claims per policy-year is $0.10(0.80)=0.08$.

Step 3: Calculate payment severity. Expected insurer payment per ground-up claim is

0.20(0)+0.30(\$100)+0.50(\$900) =\$30+\$450 =\$480.

Dividing by the payable probability gives average payment per payable claim: $\$480/0.80=\$600$.

Step 4: Reconcile. Payable frequency times average payment is $0.08(\$600)=\$48$ per policy-year. Directly, ground-up frequency times expected payment per ground-up claim is $0.10(\$480)=\$48$ per policy-year. The calculations agree.

Common mistake. Multiplying the payable frequency $0.08$ by the ground-up mean severity, $0.20(\$200)+0.30(\$600)+0.50(\$1{,}400)=\$920$, gives $\$73.60$, which is incorrect. The $\$920$ includes losses below the deductible and does not represent the payment severity of payable claims. Use either $\$0.08$ times the conditional payment mean of $\$600$, or $\$0.10$ times expected payment per ground-up claim of $\$480$.

Using Empirical Claim Data

An empirical severity distribution assigns equal probability to each observed claim when no other weights are specified. At a selected deductible, the fraction of observations above the deductible estimates $P(X>d)$. The mean of the positive deductible-adjusted payments estimates $E[Y^P]$. Their product, multiplied by ground-up claim frequency, should equal the empirical average payment per ground-up claim multiplied by that frequency.

Worked Example: Reconciling an Empirical Severity Sample

Hypothetical data and assumptions. A sample contains ten ground-up claims with severities, in dollars, of $\$100,\$250,\$400,\$550,\$700,\$900,\$1{,}200,\$1{,}600,\$2{,}000,\$3{,}000$. Treat these observations as equally weighted representatives of the severity distribution. The book has 2,000 policy-years of exposure and 300 ground-up claims, so estimated ground-up frequency is $300/2{,}000=0.15$ claims per policy-year. For $d=\$400$, estimate payable frequency, conditional average payment, and expected insurer loss per policy-year.

Step 1: Count payable sample claims. Only losses strictly greater than $\$400$ produce positive payments. There are seven such observations, so the estimated payable probability is $7/10=0.70$.

Step 2: Calculate positive payments. The payments are $\$150,\$300,\$500,\$800,\$1{,}200,\$1{,}600,\$2{,}600$. Their total is

150+300+500+800+1{,}200+1{,}600+2{,}600=\$7{,}150.

The average payment conditional on a payable claim is $\$7{,}150/7=\$1{,}021.4286$.

Step 3: Calculate payable frequency. Estimated frequency is $0.15(0.70)=0.105$ payable claims per policy-year.

Step 4: Reconcile expected loss. The frequency-severity calculation gives $0.105(\$1{,}021.4286)=\$107.25$ per policy-year. Alternatively, expected payment per ground-up claim under the empirical distribution is $\$7{,}150/10=\$715$, so $0.15(\$715)=\$107.25$ per policy-year. Across 2,000 policy-years, this corresponds to $2{,}000(\$107.25)=\$214{,}500$ of estimated aggregate insurer loss. The sample loss exactly equal to the deductible contributes zero payment and is not counted as payable.

Interpretation and Exam Traps

A deductible generally reduces payable frequency and expected payment per ground-up claim. The conditional average payment per payable claim, however, need not decrease when the deductible rises: higher deductibles remove smaller claims, so the remaining payable claims can have a larger average payment. The expected insurer loss per ground-up claim is $E[(X-d)_+]$; it is not the same quantity as conditional payment severity $E[Y^P]$.

Mixing frequency bases. Do not multiply ground-up claim frequency by average payment per payable claim. Pair ground-up frequency with expected payment per ground-up claim, or payable frequency with conditional payment severity.
Counting claims at the threshold. Under an ordinary deductible, a loss exactly equal to $d$ has zero insurer payment. For payable frequency, use $P(X>d)$, not $P(X\geq d)$, unless the contract or data definition explicitly specifies otherwise.
Assuming all claim counts are Poisson. Expected payable frequency equals $\lambda P(X>d)$ under the stated independence assumptions, but a Poisson payable-count distribution follows only when the ground-up count is Poisson.
Forgetting the exposure basis. If $\lambda$ is claims per policy-year, the resulting payable frequency and expected loss are also per policy-year. If the exposure unit changes, the frequency unit must change with it.

These calculations matter when rating plans need a consistent view of both how often an insurer pays and how much it pays when it does. A frequency-severity estimate is internally consistent only when the frequency definition, conditional severity, deductible treatment, and exposure basis agree.

Check Your Understanding

For each problem, distinguish payable frequency from payment severity and check the expected loss by both methods.

A hypothetical book has ground-up frequency $0.12$ claims per policy-year. Severity is $\$300,\$700,\$1{,}500$ with probabilities $0.25,0.50,0.25$, and the deductible is $\$500$. Find the payable probability, payable frequency, average payment per payable claim, and expected insurer loss per policy-year. (Answer: $0.75$; $0.09$ payable claims per policy-year; $\$466.67$; $\$42.00$ per policy-year.)
Ground-up frequency is $0.25$ per policy-year and severity is exponential with mean $\$800$. For $d=\$400$, find the payable frequency, conditional average payment, and expected insurer loss per policy-year. (Answer: $0.1516$ payable claims per policy-year; $\$800$; $\$121.31$ per policy-year.)
The ground-up claim count is Poisson with mean $2$ per policy-year. Severity is $\$100,\$600,\$1{,}600$ with probabilities $0.20,0.50,0.30$, independently of the count. For $d=\$500$, find the mean payable count, conditional average payment, expected insurer loss, and probability of no payable claims in a policy-year. (Answer: $1.6$; $\$475$; $\$760$ per policy-year; $e^{-1.6}=0.2019$.)
A hypothetical severity sample has equally weighted losses of $\$200,\$500,\$800,\$1{,}100,\$1{,}700$. Ground-up frequency is $0.20$ claims per exposure-year and $d=\$500$. Find the empirical payable probability, payable frequency, conditional average payment, and expected insurer loss per exposure-year. (Answer: $0.60$; $0.12$ payable claims per exposure-year; $\$700$; $\$84.00$ per exposure-year.)
Ground-up frequency is $0.40$ per policy-year, and severity is $\$100,\$500,\$1{,}200$ with probabilities $0.20,0.50,0.30$. Compare payable frequency and expected insurer loss at $d=\$200$ and $d=\$600$. (Answer: At $d=\$200$, payable frequency $0.32$ and expected loss $\$180$ per policy-year; at $d=\$600$, payable frequency $0.12$ and expected loss $\$72$ per policy-year.)

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