Translate Covered Claim Counts into Frequency Models

Why a Deductible Changes Claim Frequency

A policy may experience many ground-up losses, but a data system that records only claims above a deductible observes fewer events. The recorded count is therefore not generally the same random variable as the underlying claim count. To model it correctly, combine the frequency model with the probability that one ground-up loss exceeds the deductible.

This conversion is an instance of binomial thinning: conditional on the number of underlying claims, each claim is independently retained with a specified probability. The resulting frequency distribution depends on the original count model. A Poisson count remains Poisson, while a negative binomial count remains negative binomial with an adjusted parameter.

Definition: Let $N$ be the number of ground-up claims in a fixed exposure period, and let $X_j$ be the severity of claim $j$. For a deductible $d$, define the recorded claim count as $$ K=\sum_{j=1}^{N} I(X_j>d), $$ where $I(\cdot)$ equals 1 when its condition is true and 0 otherwise. If severities are independent and identically distributed, independent of $N$, then each claim is retained with probability $q=P(X>d)$. For a strict exceedance, $q=S_X(d)=1-F_X(d)$. If a contract or data rule counts losses equal to $d$, use $P(X\ge d)$ instead.

The thinning result requires the same retention probability for every claim and independent retention decisions conditional on $N$. These conditions hold, for example, when claim severities are independent draws from one common severity distribution. A common fixed deductible does not itself make the retention probability equal to one: it is the severity tail probability $q$ that determines which claims are observed.

The count $K$ is a count of claims exceeding the deductible, not a count of all policies with a positive payment after other policy-level rules. If retention depends on policy characteristics, if claims within a policy have dependent severities, or if the deductible varies by claim in a way that changes retention probabilities, a single common-$q$ thinning model may not apply without further conditioning or modification.

Poisson Frequency: The Retained Count Is Poisson

Suppose $N\sim\operatorname{Poisson}(\lambda)$, so $P(N=n)=e^{-\lambda}\lambda^n/n!$, and each claim is retained independently with probability $q$. Conditional on $N=n$, $K$ has a binomial distribution with parameters $n$ and $q$. Summing over the possible underlying counts gives the recorded-count distribution.

\begin{aligned} P(K=k) &=\sum_{n=k}^{\infty}P(K=k\mid N=n)P(N=n)\\ &=\sum_{n=k}^{\infty}\binom{n}{k}q^k(1-q)^{n-k} e^{-\lambda}\frac{\lambda^n}{n!}\\ &=e^{-\lambda}\frac{(\lambda q)^k}{k!} \sum_{m=0}^{\infty}\frac{[\lambda(1-q)]^m}{m!}\\ &=e^{-\lambda q}\frac{(\lambda q)^k}{k!}. \end{aligned}

In the third line, $m=n-k$, and the exponential series gives $\sum_{m=0}^{\infty}[\lambda(1-q)]^m/m!=e^{\lambda(1-q)}$. Thus $K\sim\operatorname{Poisson}(\lambda q)$. Its expected count and variance are both $\lambda q$.

Worked Example: Poisson Claims Above an Exponential-Model Deductible

A hypothetical portfolio has a Poisson number of ground-up claims per policy-year with parameter $\lambda=8$. Claim severities are independent of frequency and have an exponential distribution with mean $\theta=\$500$, so $S_X(x)=e^{-x/500}$ for $x\ge0$. Only claims strictly exceeding a deductible $d=500\ln 4$ dollars are recorded. Find the distribution of the recorded count and the probabilities of zero, one, and at least two recorded claims.

Method and retention probability. For the exponential severity, the probability of exceeding the deductible is $$ q=P(X>d)=e^{-d/500}=e^{-(500\ln 4)/500}=e^{-\ln 4}=\frac14. $$ The deductible is approximately $\$693.15$, but the exact value $500\ln 4$ makes the retention probability exact.

Recorded frequency. Poisson thinning gives $K\sim\operatorname{Poisson}(\lambda q)=\operatorname{Poisson}(8\cdot\tfrac14)=\operatorname{Poisson}(2)$. Therefore $$ P(K=0)=e^{-2}=0.1353,\qquad P(K=1)=e^{-2}\frac{2^1}{1!}=0.2707. $$ Then $P(K\ge2)=1-P(K=0)-P(K=1)=1-0.1353-0.2707=0.5940$. The three probabilities sum to $0.1353+0.2707+0.5940=1.0000$, as required. Probabilities are rounded to four decimal places.

Common mistake avoided. Using $\operatorname{Poisson}(8)$ for the recorded count would treat every ground-up claim as recorded. The correct rate is $8\cdot\tfrac14=2$, because only one quarter of ground-up claims exceed the deductible.

Negative Binomial Frequency: Adjust the Success Probability

Use the following explicit negative binomial convention. $N\sim\operatorname{NB}(r,p)$ means that $N$ counts failures before $r$ successes, with $$ P(N=n)=\binom{n+r-1}{n}p^r(1-p)^n,\qquad n=0,1,2,\ldots, $$ where $r>0$ and $0<p<1$. Under this parameterization, $E[N]=r(1-p)/p$ and $\operatorname{Var}(N)=r(1-p)/p^2$. The probability $p$ is a distribution parameter; it is not the deductible retention probability $q$.

When each underlying claim is independently retained with probability $q$, the retained count is also negative binomial with the same $r$, but a new success probability. Write $$ p'=\frac{p}{p+(1-p)q}. $$ The denominator is positive for the stated parameter ranges. The complementary probability is $1-p'=(1-p)q/[p+(1-p)q]$.

Theorem: If $N\sim\operatorname{NB}(r,p)$ under the failures-before-$r$-successes convention and claims are independently retained with common probability $q$, then $$ K\sim\operatorname{NB}\left(r,\frac{p}{p+(1-p)q}\right). $$ In particular, $E[K]=q\,r(1-p)/p$. The number of retained claims is not, in general, negative binomial with the original parameter $p$.

The parameter adjustment can be verified directly from the conditional binomial probabilities. For $k\ge0$, sum over $n=k,k+1,\ldots$, and set $m=n-k$. The identity $\binom{n}{k}\binom{n+r-1}{n}=\binom{k+r-1}{k}\binom{m+r+k-1}{m}$ then gives

\begin{aligned} P(K=k) &=\binom{k+r-1}{k}p^r[(1-p)q]^k \sum_{m=0}^{\infty}\binom{m+r+k-1}{m}[(1-p)(1-q)]^m\\ &=\binom{k+r-1}{k} \left(\frac{p}{D}\right)^r \left(\frac{(1-p)q}{D}\right)^k, \qquad D=p+(1-p)q. \end{aligned}

For the second equality, use the standard series identity $\sum_{m=0}^{\infty}\binom{m+a-1}{m}z^m=(1-z)^{-a}$ for $|z|<1$, with $a=r+k$. Here $1-(1-p)(1-q)=D$. The displayed probability is exactly the negative binomial probability with $p'=p/D$, since $1-p'=(1-p)q/D$. This proves the result.

Worked Example: Correcting a Negative Binomial Count After Thinning

For a hypothetical group of policies, the ground-up claim count per year is $N\sim\operatorname{NB}(2,0.5)$. Claim severities are independent of frequency, and the deductible retains each claim with probability $q=0.25$. Find the distribution of the recorded count, $P(K=0)$, $P(K=1)$, and $P(K\ge2)$.

Method. First adjust the negative binomial parameter: $$ p'=\frac{p}{p+(1-p)q} =\frac{0.5}{0.5+(0.5)(0.25)} =\frac{0.5}{0.625}=0.8. $$ Thus $K\sim\operatorname{NB}(2,0.8)$. With this convention, $P(K=0)=(p')^r$, so $$ P(K=0)=(0.8)^2=0.64. $$ For one recorded claim, $$ P(K=1)=\binom{1+2-1}{1}(0.8)^2(1-0.8)^1 =2(0.64)(0.2)=0.256. $$ Therefore $P(K\ge2)=1-0.64-0.256=0.104$. The probabilities for zero, one, and at least two sum to $0.64+0.256+0.104=1$.

Check and common mistake avoided. The original expected count is $2(1-0.5)/0.5=2$, so the retained expected count should be $2(0.25)=0.5$. The adjusted model gives $E[K]=2(1-0.8)/0.8=0.5$, confirming the parameter adjustment. Using the original $p=0.5$ for the retained count would give $P(K=0)=0.25$, not $0.64$. Probabilities are shown exactly here; to four decimal places, $P(K=0)=0.6400$, $P(K=1)=0.2560$, and $P(K\ge2)=0.1040$.

Moments and a Second Negative Binomial Calculation

The mean of the retained count provides a useful check across frequency models. Conditional on $N$, $E[K\mid N]=Nq$. Hence $E[K]=E[E[K\mid N]]=qE[N]$. Also, conditional on $N$, $K$ is binomial, so $\operatorname{Var}(K\mid N)=Nq(1-q)$. Applying the variance decomposition gives $$ \operatorname{Var}(K)=E[\operatorname{Var}(K\mid N)] +\operatorname{Var}(E[K\mid N)] =q(1-q)E[N]+q^2\operatorname{Var}(N). $$ For a Poisson count, this reduces to $\lambda q$. For $N\sim\operatorname{NB}(r,p)$, it equals $r(1-p)q[p+(1-p)q]/p^2$, which is also the variance of $\operatorname{NB}(r,p')$.

Worked Example: Mean, Variance, and Zero-Count Probability

A hypothetical policy group has yearly ground-up counts $N\sim\operatorname{NB}(3,0.6)$. Independent claim severities follow an exponential distribution with mean $\$1{,}000$. A deductible of $1{,}000\ln 2$ dollars is used, and claims are recorded only when they strictly exceed it. Find the distribution, mean, variance, and probability of at least one recorded claim.

Retention probability. The exponential survival probability at the deductible is $$ q=e^{-d/1000}=e^{-(1000\ln 2)/1000}=\frac12. $$ The original mean count is $3(1-0.6)/0.6=2$, so the retained mean must be $2(0.5)=1$.

Adjusted model and probabilities. The adjusted parameter is $$ p'=\frac{0.6}{0.6+(0.4)(0.5)}=\frac{0.6}{0.8}=0.75, $$ so $K\sim\operatorname{NB}(3,0.75)$. Its mean is $3(0.25)/0.75=1$, agreeing with the conditional-mean check. Its variance is $$ \operatorname{Var}(K)=\frac{3(1-0.75)}{(0.75)^2} =\frac{0.75}{0.5625}=\frac43=1.3333. $$ Finally, $P(K=0)=(0.75)^3=0.421875$, so $P(K\ge1)=1-0.421875=0.578125$, or $0.5781$ to four decimal places. The mean and variance are rounded to four decimal places; the probability is rounded to four decimal places.

Check. The original negative binomial variance is $3(0.4)/(0.6)^2=3.3333$. The general thinning variance formula gives $(0.5)(0.5)(2)+(0.5)^2(3.3333)=0.5+0.8333=1.3333$, matching the adjusted-distribution result.

Choosing and Checking the Model

The key modeling decision is the distribution of the underlying count before applying the deductible. The severity model supplies $q$; it does not by itself determine the retained frequency family. Once $q$ is calculated, the appropriate thinning result follows from the original count model and its assumptions.

Underlying count model	Recorded count model	Parameter change
$N\sim\operatorname{Poisson}(\lambda)$	$K\sim\operatorname{Poisson}(\lambda q)$	Multiply the rate by $q$
$N\sim\operatorname{NB}(r,p)$, failures-before-$r$-successes convention	$K\sim\operatorname{NB}(r,p')$	$p'=p/[p+(1-p)q]$

Several checks help catch errors in an exam calculation. First, $q$ must lie between 0 and 1; a larger deductible should not increase the probability of retention. Second, $E[K]=qE[N]$, so the recorded mean cannot exceed the original mean when $0\le q\le1$. Third, when $q=1$, every claim is retained and the original model must be recovered. When $q=0$, no claims are retained and $K=0$ with probability 1.

A further trap is substituting a severity distribution conditional on exceeding the deductible for the retention probability. For example, the distribution of $X\mid X>d$ describes the sizes of retained claims. It does not say what fraction of the original claims are retained; that fraction is $P(X>d)$. The frequency conversion uses this unconditional tail probability.

Key takeaway: For an independent, common-probability deductible screen, calculate $q=P(X>d)$, then thin the underlying count. Poisson frequency has its rate multiplied by $q$; negative binomial frequency retains its shape $r$ and uses $p'=p/[p+(1-p)q]$ under the stated parameterization.

Check Your Understanding

For each problem, identify the thinning probability and use the stated negative binomial convention where applicable.

A hypothetical portfolio has $N\sim\operatorname{Poisson}(10)$ ground-up claims per year. A deductible retains each claim independently with probability $0.3$. Give the recorded-count distribution, its mean, and the probability of no recorded claims. (Answer: $K\sim\operatorname{Poisson}(3)$, $E[K]=3$, $P(K=0)=e^{-3}=0.0498$.)
A hypothetical risk has $N\sim\operatorname{Poisson}(15)$, and the severity model gives $P(X>d)=0.2$. Find the probability of exactly two recorded claims. (Answer: $K\sim\operatorname{Poisson}(3)$, $P(K=2)=e^{-3}3^2/2=0.2240$.)
Suppose $N\sim\operatorname{NB}(4,0.5)$, and an independent deductible screen retains claims with probability $0.2$. Give the adjusted distribution, its mean, and its zero-count probability. (Answer: $p'=0.5/[0.5+(0.5)(0.2)]=5/6$; $K\sim\operatorname{NB}(4,5/6)$, $E[K]=0.8$, and $P(K=0)=(5/6)^4=0.4823$.)
Suppose $N\sim\operatorname{NB}(2,0.4)$, and $q=0.75$. Calculate the adjusted parameter, the recorded-count mean, and $P(K=0)$. (Answer: $p'=0.4/[0.4+(0.6)(0.75)]=8/17$; $E[K]=2(9/17)/(8/17)=2.25$; $P(K=0)=(8/17)^2=64/289=0.2215$.)
A hypothetical group has $N\sim\operatorname{NB}(3,0.6)$. A deductible retains each claim with probability $0.5$. Find the probability of at least one recorded claim and verify the recorded mean against the original mean. (Answer: $p'=0.6/[0.6+(0.4)(0.5)]=0.75$; $P(K\ge1)=1-(0.75)^3=0.5781$; $E[N]=2$, so $E[K]=0.5(2)=1$, also $3(1-0.75)/0.75=1$.)

Underlying count model	Recorded count model	Parameter change
\(N\sim\operatorname{Poisson}(\lambda)\)	\(K\sim\operatorname{Poisson}(\lambda q)\)	Multiply the rate by \(q\)
\(N\sim\operatorname{NB}(r,p)\), failures-before-\(r\)-successes convention	\(K\sim\operatorname{NB}(r,p')\)	\(p'=p/[p+(1-p)q]\)

What You'll Learn