Worked Chain: Poisson Frequency With Deductible and Limit

From Ground-Up Claims to Aggregate Payments

A deductible and a per-claim payment limit change both the amount paid on a claim and the number of claims that produce a payment. For a Poisson claim count and exponential ground-up severity, the memoryless property makes this calculation especially manageable: positive payments still have a simple capped severity distribution, and their count is Poisson.

We will use $d$ for the deductible and $u$ for the insurer’s maximum payment per claim. Thus, a claim of amount $x$ produces payment $\min((x-d)_+,u)$, where $(z)_+=\max(z,0)$. This convention means that the limit applies to the insurer’s payment, not to the ground-up claim amount.

Model and assumptions: During a fixed period, the number of ground-up claims $N$ has a Poisson distribution with mean $\lambda$. Given $N$, claim amounts $X_1,\ldots,X_N$ are independent and identically distributed exponential random variables with rate $\beta>0$, independent of $N$. A claim payment is $Y^L=\min((X-d)_+,u)$, with $d\geq 0$ and $u>0$. The aggregate insurer payment is $S=\sum_{j=1}^N Y_j^L$. All amounts use the same monetary unit.

For an exponential severity, $P(X>d)=e^{-\beta d}$. A claim therefore produces a positive payment precisely when its ground-up amount exceeds $d$. Let $M$ be the number of positive payments. Poisson thinning gives $M\sim\operatorname{Poisson}(\mu)$, where $\mu=\lambda e^{-\beta d}$. The cap $u$ does not remove any additional positive payments; even a claim that reaches the cap is still one positive payment.

Payment Severity and Expected Aggregate Loss

Conditional on a positive payment, the excess $X-d$ has an exponential distribution with the same rate $\beta$. In fact, for $z\geq 0$,

P(X-d>z\mid X>d) =\frac{P(X>d+z)}{P(X>d)} =\frac{e^{-\beta(d+z)}}{e^{-\beta d}} =e^{-\beta z}.

Thus, conditional on a positive payment, the insurer pays an exponential amount capped at $u$. There is a continuous part below $u$, and a point mass at $u$: a payment reaches the cap when the excess $X-d$ is at least $u$.

Conditional payment distribution: Given $Y^L>0$, the payment is $\min(Z,u)$, where $Z$ is exponential with rate $\beta$. For $0\leq y<u$, $P(Y^L>y\mid Y^L>0)=e^{-\beta y}$; the probability of a payment exactly equal to $u$ is $e^{-\beta u}$. Its conditional mean is $(1-e^{-\beta u})/\beta$.

The mean follows by integrating the conditional survival probability. For a nonnegative random variable $W$, $E[W]=\int_0^\infty P(W>y)\,dy$. Here the conditional payment cannot exceed $u$, so

E[Y^L\mid Y^L>0] =\int_0^u e^{-\beta y}\,dy =\frac{1-e^{-\beta u}}{\beta}.

A ground-up claim has zero payment with probability $1-e^{-\beta d}$, so its unconditional expected insurer payment is $e^{-\beta d}(1-e^{-\beta u})/\beta$. Conditioning on the number of positive payments, or equivalently using the mean of a compound Poisson sum, gives the aggregate expectation.

E[S] =E[M]E[Y^L\mid Y^L>0] =\lambda e^{-\beta d}\frac{1-e^{-\beta u}}{\beta} =\lambda E[Y^L].

Thin the frequency.
Use $P(X>d)=e^{-\beta d}$ to obtain $M\sim\operatorname{Poisson}(\lambda e^{-\beta d})$.

Modify the severity.
Conditional on payment, use the exponential excess $X-d$, capped at $u$.

Calculate the target.
Use the conditional payment mean for expected aggregate loss, or identify the count and payment events for an aggregate probability.

Aggregate Probabilities

Because $S=0$ exactly when there are no positive payments, the probability of no aggregate payment is the zero probability of the thinned Poisson count. Its complement is the probability of at least one payment.

P(S=0)=P(M=0)=e^{-\mu}, \qquad P(S>0)=1-e^{-\mu}, \qquad \mu=\lambda e^{-\beta d}.

The capped severity also creates positive point probabilities in the aggregate distribution. For example, $S=u$ occurs when there is exactly one positive payment and that payment reaches the cap. If there are two or more positive payments, their sum can equal exactly $u$ through continuous payment amounts only with probability zero. Therefore,

P(S=u)=P(M=1)e^{-\beta u} =e^{-\mu}\mu e^{-\beta u}.

More generally, for an integer $k\geq 1$, the point mass at $ku$ is $e^{-\mu}\mu^k e^{-k\beta u}/k!$: it occurs when there are exactly $k$ positive payments and each reaches the cap. Continuous combinations of uncapped payments do not add point mass at that exact total. These formulas identify particular atoms; they do not provide the full aggregate distribution or a general formula for $P(S\leq s)$.

Worked Example: Positive-Payment Frequency and Expected Loss

A hypothetical one-year portfolio has $N\sim\operatorname{Poisson}(\lambda)$ ground-up claims, with $\lambda=e^{0.5}$. Each claim amount is exponential with rate $\beta=0.01$ per dollar. The policy has a deductible $d=\$50$ and a maximum insurer payment $u=\$100$ per claim. Find the expected number of positive payments and expected aggregate insurer payment.

Method. Thin the original Poisson count by the probability a claim exceeds the deductible, then multiply the positive-payment mean count by the conditional mean payment.

The probability of a positive payment is

P(X>50)=e^{-(0.01)(50)}=e^{-0.5}=0.6065307.

The mean number of positive payments is therefore

\mu=e^{0.5}(0.6065307)=e^{0.5}e^{-0.5}=1.

Given a positive payment, the excess is exponential with rate $0.01$, capped at $\$100$. Its mean is

E[Y^L\mid Y^L>0] =\frac{1-e^{-(0.01)(100)}}{0.01} =\frac{1-e^{-1}}{0.01} =63.2121\text{ dollars}.

Thus the expected aggregate payment is

E[S]=1(63.2121)=\$63.21.

The expected number of positive payments is $1$, and the expected aggregate payment is $\$63.21$, rounded to the nearest cent. As a check, the expected payment per original claim is $e^{-0.5}(63.2121)$; multiplying by $\lambda=e^{0.5}$ again gives $\$63.2121$.

Common mistake avoided: The limit does not change the positive-payment probability. It caps payment size; the deductible determines whether payment is positive.

Worked Example: Point Probabilities in the Aggregate

Use the model in the preceding example. Calculate $P(S=0)$, $P(S=\$100)$, and $P(S=\$200)$, and check that the first probability is consistent with the positive-payment frequency.

Method. The thinned count has mean $\mu=1$. For zero aggregate payment, there must be zero positive claims. For the aggregate to equal one or two times the per-claim cap, the corresponding number of positive claims must all reach the cap.

First,

P(S=0)=e^{-1}=0.3679.

A positive payment reaches $\$100$ with conditional probability $e^{-(0.01)(100)}=e^{-1}$. Exactly one positive payment has probability $e^{-1}(1)=e^{-1}$. Hence

P(S=\$100)=e^{-1}(1)e^{-1}=e^{-2}=0.1353.

For an aggregate of $\$200$, there must be exactly two positive payments, both capped. The Poisson probability of two positive payments is $e^{-1}/2$, and the probability both are capped is $e^{-2}$. Thus

P(S=\$200)=\frac{e^{-1}}{2}e^{-2} =\frac{e^{-3}}{2} =0.0249.

Finally, $P(S>0)=1-e^{-1}=0.6321$, the complement of $P(S=0)=0.3679$, as required. These three point probabilities do not need to sum to one: other aggregate payment amounts are possible.

Common mistake avoided: $P(S=\$100)$ is not merely the probability that a single claim reaches the cap. The portfolio must also have exactly one positive payment and no other positive payments.

Worked Example: Checking the Attachment and Cap Thresholds

A hypothetical portfolio has a Poisson mean of $3$ ground-up claims for the period. Claim amounts are exponential with rate $\beta=0.02$ per dollar. The deductible is $d=\$25$ and the insurer’s per-claim payment limit is $u=\$50$. Find the mean positive-payment count, expected aggregate payment, probability of no aggregate payment, and probability that the aggregate payment is exactly $\$50$.

Method. Apply the deductible to identify positive payments, then use the cap on the excess above the deductible. A payment reaches $\$50$ when the ground-up claim is at least $\$25+\$50=\$75$.

The positive-payment probability and mean count are

P(X>25)=e^{-(0.02)(25)}=e^{-0.5}=0.6065307, \qquad \mu=3e^{-0.5}=1.819592.

The conditional mean payment is

E[Y^L\mid Y^L>0] =\frac{1-e^{-(0.02)(50)}}{0.02} =\frac{1-e^{-1}}{0.02} =31.6060\text{ dollars}.

Consequently,

E[S]=1.819592(31.6060)=\$57.51.

The no-payment probability is

P(S=0)=e^{-1.819592}=0.1621.

For $S=\$50$, there must be exactly one positive payment, and that payment must be capped. The conditional probability of reaching the cap is $e^{-(0.02)(50)}=e^{-1}$. Therefore,

P(S=\$50)=e^{-1.819592}(1.819592)e^{-1}=0.1085.

The results are consistent with the model: $P(S>0)=1-0.1621=0.8379$, and the probability of a capped payment is evaluated among positive payments, not among all ground-up claims.

Common mistake avoided: Using $P(X\geq \$50)$ as the cap probability would apply the limit to the ground-up claim rather than to the insurer’s payment. The correct ground-up threshold for a capped payment is $d+u=\$75$.

Common Errors and Practical Checks

The chain is short, but each parameter has a different role. The deductible controls thinning, the limit controls the positive-payment severity, and the original Poisson mean controls the starting claim count. Keeping those roles separate prevents several frequent errors.

Thinning at the wrong threshold. Use $P(X>d)$ for positive payments. The positive-payment frequency is not $\lambda P(X>d+u)$; that latter tail describes claims reaching the payment cap, not all claims producing payment.
Using the wrong capped severity. Conditional payment is $\min(X-d,u)$ given $X>d$, not $\min(X,u)$. The payment cap $u$ is measured above the deductible.
Omitting the probability of no other payments. An aggregate point event such as $S=u$ includes a condition on the positive-payment count as well as a capped payment.
Applying memorylessness outside its model. The unchanged exponential rate for $X-d\mid X>d$ depends on exponential ground-up severity. It does not hold for an arbitrary severity distribution.

This calculation is aligned with the SOA Exam FAM (Fundamentals of Actuarial Mathematics) syllabus because it combines frequency thinning, coverage-modified severity, expected aggregate loss, and exact probabilities associated with a compound model. For a non-exponential severity, the same modeling order still applies, but the conditional excess distribution and its capped mean must be calculated from that severity’s own survival function.

Key takeaway: For Poisson frequency with exponential ground-up severity, deductible $d$, and per-claim payment limit $u$, the positive-payment count has mean $\mu=\lambda e^{-\beta d}$, and $E[S]=\mu(1-e^{-\beta u})/\beta$. Keep the deductible threshold and payment-cap threshold distinct.

Check Your Understanding

For each hypothetical model, identify the positive-payment frequency before calculating the requested quantity. Amounts are in dollars; probabilities are rounded to four decimal places.

A portfolio has Poisson ground-up frequency $\lambda=2$, exponential claim rate $\beta=0.01$, deductible $d=(\ln 2)/0.01$, and payment limit $u=100$. Find the mean positive-payment count and expected aggregate payment. (Answer: mean count $1$; expected aggregate payment $\$63.21$)
A portfolio has $\lambda=5$, exponential rate $\beta=0.05$, and deductible $d=20$. Find the probability of no positive payments. (Answer: $e^{-5e^{-1}}=0.1589$)
A portfolio has $\lambda=e^{0.4}$, $\beta=0.02$, deductible $d=20$, and payment limit $u=50$. Find the probability that the aggregate payment is exactly $\$50$. (Answer: $e^{-2}=0.1353$)
A portfolio has $\lambda=4$, $\beta=0.01$, and both deductible and payment limit equal to $(\ln 2)/0.01$. Find the mean positive-payment count and expected aggregate payment. (Answer: mean count $2$; expected aggregate payment $\$100.00$)
A portfolio has $\lambda=2.5$, $\beta=0.02$, deductible $d=0$, and payment limit $u=50$. Find $P(S=0)$ and $P(S=50)$. (Answer: $P(S=0)=e^{-2.5}=0.0821$; $P(S=50)=2.5e^{-3.5}=0.0755$)

What You'll Learn