Deductibles and the Frequency of Positive Payments

Why a Deductible Changes the Claim Count

A policy may have a positive number of claims but no payments if every loss is at or below the deductible. To count paid claims correctly, separate two questions: how many claims occur, and what is the probability that one claim exceeds the deductible? Once that probability is known, the positive-payment count can be calculated from the original claim count.

This tutorial considers an ordinary deductible applied separately to each claim. It focuses on whether a payment is positive, not on the size of the payment. No interest or mortality assumption is needed for this model.

Definition: Let $X$ be the loss amount on one claim and let $d\geq 0$ be an ordinary per-claim deductible. The insurer's payment per loss is $Y^L=(X-d)_+=\max(X-d,0)$. Define $q=P(Y^L>0)=P(X>d)$. If $N$ is the number of claims and $K$ is the number of claims producing positive payments, then $K=\sum_{j=1}^{N} I(X_j>d)$, where $I(\cdot)$ equals 1 when its condition holds and 0 otherwise.

The strict inequality matters: a loss exactly equal to $d$ produces payment zero. With $F_X(x)=P(X\leq x)$, the positive-payment probability is $q=1-F_X(d)$. Equivalently, under the convention $S_X(x)=P(X>x)$, it is $q=S_X(d)$. For a continuous severity distribution, the probability of equality at one point is zero; for a discrete or mixed distribution, it may not be.

Conditional Binomial Thinning

Assume the claim count $N$ is a nonnegative integer-valued random variable. Assume claim losses $X_1,X_2,\ldots$ are independent and identically distributed, and independent of $N$. Each claim then independently produces a positive payment with the same probability $q$. Given $N=n$, the number of positive payments is binomial.

Theorem: Under these assumptions, $K\mid N=n\sim\operatorname{Binomial}(n,q)$, so $P(K=k\mid N=n)=\binom{n}{k}q^k(1-q)^{n-k}$ for $k=0,\ldots,n$. If $N$ is random, the unconditional probability is $P(K=k)=\sum_{n=k}^{\infty}P(N=n)\binom{n}{k}q^k(1-q)^{n-k}$.

The conditional statement follows because, once $n$ claims have occurred, each has an independent indicator of positive payment with probability $q$. The unconditional formula averages those conditional probabilities over the possible values of $N$. In particular, $P(K=0)=E[(1-q)^N]$: conditional on $N=n$, all $n$ claims fail to exceed the deductible with probability $(1-q)^n$.

$$ E[K]=qE[N]. $$

To justify the expectation, condition on the number of claims. A binomial random variable with parameters $n$ and $q$ has mean $nq$, so $E[K\mid N]=Nq$. Taking expectations of both sides gives $E[K]=E[E[K\mid N]]=qE[N]$, provided $E[N]$ is finite. This identity does not require $N$ itself to be binomial; it does require that each claim has the same positive-payment probability and that severity indicators behave independently conditional on the count.

A useful equivalent calculation uses probability generating functions. Define $G_N(s)=E[s^N]$ and $G_K(s)=E[s^K]$. Conditional on $N=n$, the generating function of $K$ is $(1-q+qs)^n$, because each claim contributes a factor $s$ with probability $q$ and a factor 1 otherwise. Averaging over $N$ gives:

$$ G_K(s)=G_N(1-q+qs). $$

Setting $s=0$ recovers $P(K=0)=G_N(1-q)$. Differentiating at $s=1$, when the mean exists, gives $E[K]=qE[N]$. These are useful checks on a calculation made by mixing conditional binomial probabilities.

Worked Examples

Worked Example: Continuous Severity and a Fixed Number of Claims

A hypothetical portfolio has exactly four claims, so $N=4$. Claim severities are independent exponential random variables with rate $\lambda=0.001$ per dollar, and the ordinary deductible is $d=\$1{,}000$. Find the probability of a positive payment on one claim, the distribution of $K$, and $P(K=0)$. There is no discounting or mortality component.

Method: calculate the exceedance probability, then apply the conditional binomial model. For an exponential severity, $S_X(x)=e^{-\lambda x}$ for $x\geq0$. Therefore:

q=P(X>1{,}000)=e^{-(0.001)(1{,}000)}=e^{-1}\approx0.3679.

Since $N=4$ is fixed and the four severity draws are independent, $K\sim\operatorname{Binomial}(4,e^{-1})$. Thus $P(K=k)=\binom{4}{k}(e^{-1})^k(1-e^{-1})^{4-k}$, for $k=0,1,2,3,4$. In particular:

P(K=0)=(1-e^{-1})^4\approx(0.6321206)^4\approx0.1597.

The mean count is $E[K]=4e^{-1}\approx1.4715$. As a check, the probability of no positive payment is less than one, and the expected number of positive payments is between zero and four. The common mistake avoided here is using the probability $P(X\geq d)$ without checking the severity distribution; for this continuous model it equals $P(X>d)$, but that equality need not hold when there is an atom at the deductible.

Worked Example: A Severity Atom at the Deductible

A hypothetical insurer observes independent claim severities with the following distribution: $P(X=\$200)=0.15$, $P(X=\$500)=0.25$, $P(X=\$900)=0.35$, and $P(X=\$1{,}500)=0.25$. The deductible is $d=\$500$. The claim count has $P(N=0)=0.20$, $P(N=2)=0.50$, and $P(N=3)=0.30$, independently of severities. Find the distribution of $K$ and its mean. There is no discounting or mortality component.

Method: determine the strict exceedance probability before mixing the conditional binomials. Only losses of \$900 and \$1,500 exceed the deductible. A loss of exactly \$500 produces no payment. Therefore $q=0.35+0.25=0.60$, and $1-q=0.40$.

For $N=0$, $K=0$ with probability 1. For $N=2$, the probabilities for $K=0,1,2$ are $0.40^2=0.16$, $2(0.60)(0.40)=0.48$, and $0.60^2=0.36$. For $N=3$, the probabilities for $K=0,1,2,3$ are $0.40^3=0.064$, $3(0.60)(0.40^2)=0.288$, $3(0.60^2)(0.40)=0.432$, and $0.60^3=0.216$.

Weight each conditional probability by the corresponding claim-count probability:

\begin{aligned} P(K=0)&=0.20+0.50(0.16)+0.30(0.064)=0.2992,\\ P(K=1)&=0.50(0.48)+0.30(0.288)=0.3264,\\ P(K=2)&=0.50(0.36)+0.30(0.432)=0.3096,\\ P(K=3)&=0.30(0.216)=0.0648. \end{aligned}

The probabilities sum to $0.2992+0.3264+0.3096+0.0648=1.0000$. Also, $E[N]=0(0.20)+2(0.50)+3(0.30)=1.90$, so $E[K]=0.60(1.90)=1.14$. Checking from the distribution gives $0(0.2992)+1(0.3264)+2(0.3096)+3(0.0648)=1.1400$. The central trap is counting $X=d$ as a positive payment; doing so would incorrectly make $q=0.85$.

Worked Example: A Mixed Claim Count and a Zero-Payment Probability

A hypothetical annual portfolio has $P(N=1)=0.40$ and $P(N=4)=0.60$. Independent claim losses exceed the ordinary deductible with probability $q=0.50$. Find $E[K]$, $P(K=0)$, and $P(K\geq2)$. The severity assumption is identical independent exceedance indicators across claims; there is no discounting or mortality component.

Method: use conditional binomial probabilities and then average over the two possible claim counts. First calculate the original expected claim count:

$$ E[N]=1(0.40)+4(0.60)=0.40+2.40=2.80. $$

Hence $E[K]=qE[N]=0.50(2.80)=1.40$. For zero positive payments, if $N=1$ the probability is $0.50$; if $N=4$ it is $0.50^4=0.0625$. Thus:

$$ P(K=0)=0.40(0.50)+0.60(0.0625)=0.2000+0.0375=0.2375. $$

For $K\geq2$, the $N=1$ case contributes zero. When $N=4$, the binomial probability is $P(K\geq2\mid N=4)=1-P(K=0\mid N=4)-P(K=1\mid N=4)$. The two subtracted probabilities are $0.50^4=0.0625$ and $\binom41(0.50)(0.50^3)=4(0.50)(0.125)=0.25$. Therefore:

P(K\geq2)=0.40(0)+0.60(1-0.0625-0.25)=0.60(0.6875)=0.4125.

As a check, the mean must be $0.50$ times the original mean count $2.80$, giving $1.40$, not $1.30$. A frequent error is to multiply $q$ by an incorrectly calculated $E[N]$; here the $N=4$ outcome contributes $4(0.60)=2.40$, not $4(0.50)=2.00$.

Common Traps and Model Limits

Including losses equal to the deductible. Under $Y^L=(X-d)_+$, $X=d$ gives $Y^L=0$. Use $P(X>d)$, especially for discrete or mixed severities.
Using the wrong claim-count weights. For random $N$, conditional binomial probabilities must be multiplied by $P(N=n)$ and summed over $n$. A binomial formula with a single fixed $n$ is not the unconditional distribution.
Confusing expected counts. The expected positive-payment count is $qE[N]$, not $q$ times a mistaken or rounded claim-count mean. Compute $E[N]$ from its full distribution first.
Assuming thinning when indicators are dependent. If claim severities are dependent, have different exceedance probabilities, or depend on $N$, the conditional count need not be binomial. The stated independence and common severity assumptions are essential to this result.
Using a payment amount where only a count is requested. The probability $q$ depends on whether the loss exceeds the deductible. The positive payment's dollar amount is not needed to find $K$.

Key takeaway: For an ordinary deductible, a claim is counted as a positive payment precisely when $X>d$. Under independent, identically distributed severities independent of the claim count, set $q=P(X>d)$; then $K\mid N=n$ is binomial with parameters $n$ and $q$, and $E[K]=qE[N]$.

Check Your Understanding

For each problem, distinguish the probability of exceeding the deductible from the original claim-count distribution.

A policy has exactly 6 claims. For each claim, $P(X=\$100)=0.20$, $P(X=\$400)=0.30$, and $P(X=\$900)=0.50$, independently. The deductible is \$400. Find $E[K]$ and $P(K=0)$. (Answer: $E[K]=3$; $P(K=0)=0.0156$.)
The count is $N\sim\operatorname{Binomial}(3,0.40)$. Claim severities are independent exponential with rate $0.002$ per dollar, independently of $N$, and $d=\$500$. Find $P(K=0)$ and $E[K]$. (Answer: $P(K=0)=(1-0.40e^{-1})^3\approx0.6203$; $E[K]=1.20e^{-1}\approx0.4415$.)
The count probabilities are $P(N=0)=0.10$, $P(N=2)=0.60$, and $P(N=5)=0.30$. Each independent claim exceeds the deductible with probability $0.20$. Find $P(K=0)$ and $E[K]$. (Answer: $P(K=0)=0.10+0.60(0.8^2)+0.30(0.8^5)=0.5823$; $E[K]=0.20(2.7)=0.54$.)
The claim count has $P(N=1)=0.25$, $P(N=3)=0.50$, and $P(N=4)=0.25$. Independent claims exceed the deductible with probability $0.70$. Find $E[K]$ and $P(K\geq1)$. (Answer: $E[K]=0.70(2.75)=1.925$; $P(K\geq1)=1-[0.25(0.30)+0.50(0.30^3)+0.25(0.30^4)]=0.9095$.)
The count has $P(N=2)=0.40$ and $P(N=5)=0.60$, and the independent per-claim probability of exceeding the deductible is $q=0.40$. Find $P(K=1)$ and $E[K]$. (Answer: $P(K=1)=0.40[2(0.40)(0.60)]+0.60[5(0.40)(0.60^4)]=0.3475$; $E[K]=0.40[0.40(2)+0.60(5)]=1.52$.)

What You'll Learn