Expected Payment With a Deductible

Why the Survival Function Matters

An ordinary deductible changes how much an insurer pays for each loss. If a loss is below the deductible, the insurer pays nothing; if it exceeds the deductible, the insurer pays only the amount above it. To calculate the average payment, it is therefore not enough to know the average ground-up loss. We need to account for how much probability lies at and above the deductible, and how the severity distribution behaves beyond that point.

This tutorial develops the expected payment directly from the ground-up survival function. The central calculation is useful for continuous, discrete and mixed severity distributions, provided the expected payment is finite. It is aligned with the SOA Exam ASTAM (Advanced Short-Term Actuarial Mathematics) syllabus.

Definition: Let $X$ be a nonnegative ground-up loss and let $d\geq 0$ be an ordinary deductible. The insurer’s payment per loss is $Y^L=(X-d)_+=\max(X-d,0)$. Define the survival function by $S_X(x)=P(X>x)$. The probability of a positive payment is $P(Y^L>0)=P(X>d)=S_X(d)$.

The strict inequality matters when $X$ has a probability mass at $d$: a loss equal to the deductible generates no payment. For a continuous distribution, $P(X=d)=0$, so $P(X>d)$ and $P(X\geq d)$ agree. Expected payment, however, is an amount, not a probability; knowing $S_X(d)$ alone generally does not determine it.

Expected Payment From the Survival Function

For a nonnegative random variable, its expectation can be found by adding up the probabilities that it exceeds each possible payment amount. Applied to the deductible payment, this gives a direct integral of the ground-up survival function starting at the deductible.

Theorem: If $X\geq 0$ and the integral is finite, the expected insurer payment under an ordinary deductible $d$ is

E[Y^L]=E[(X-d)_+]=\int_0^\infty P(Y^L>t)\,dt =\int_d^\infty S_X(x)\,dx.

The tail probability at the deductible also provides a useful conditional representation. When $P(X>d)>0$,

E[Y^L]=P(X>d)\,E[X-d\mid X>d].

This representation separates the chance of a payment from its average size when a payment occurs. It is not a substitute for the survival integral if the conditional expectation is not already known.

Why the Integral Starts at the Deductible

Set $Z=(X-d)_+$, which is nonnegative. For any nonnegative random variable $Z$, the identity $E[Z]=\int_0^\infty P(Z>t)\,dt$ follows by writing each realized value $Z$ as the area under an indicator: $Z=\int_0^\infty \mathbf{1}_{\{Z>t\}}\,dt$. Taking expectations gives the integral of the probabilities $P(Z>t)$.

For this payment variable, $Z>t$ precisely when $X>d+t$. Therefore,

E[(X-d)_+] =\int_0^\infty P(X>d+t)\,dt =\int_d^\infty P(X>x)\,dx =\int_d^\infty S_X(x)\,dx,

where the second equality uses the change of variable $x=d+t$. The survival function is integrated over every possible payment layer above the deductible. The value $S_X(d)$ is the height of the tail at the starting point, not the entire area under that tail.

Worked Example: Exponential Loss With a Deductible

A hypothetical equipment policy has a ground-up loss $X$ that is exponential with mean $\theta=\$1{,}000$, so its rate is $\lambda=1/1{,}000$ per dollar. The ordinary deductible is $d=\$400$. Find the probability of a positive payment and the expected insurer payment per loss. No interest discounting or mortality assumption is involved; this is a per-loss severity calculation.

Method. For an exponential loss, $S_X(x)=e^{-\lambda x}$ for $x\geq 0$. The tail probability at the deductible is

P(X>d)=S_X(400)=e^{-400/1000}=e^{-0.4}=0.670320\ldots.

Thus the probability of a positive payment is $0.6703$, to four decimal places. The expected payment is the area under the survival function from $400$ onward:

\begin{aligned} E[Y^L] &=\int_{400}^\infty e^{-x/1000}\,dx\\ &=\left[-1000e^{-x/1000}\right]_{400}^\infty\\ &=1000e^{-0.4}\\ &=670.320\ldots\text{ dollars}. \end{aligned}

Rounded to the nearest cent, the expected insurer payment is $670.32 per loss. As a check, the exponential memoryless property gives an expected amount above the deductible of $\$1{,}000$ conditional on exceeding it. Multiplying by the payment probability gives $0.670320\ldots\times \$1{,}000=\$670.32$. The common mistake this method avoids is reporting the tail probability $0.6703$ as though it were the expected dollar payment.

Discrete Losses and a Mass at the Deductible

The survival-integral formula is not limited to continuous distributions. A discrete loss distribution produces a step-shaped survival function, and its integral is the sum of the areas of the steps. In particular, a probability mass exactly at the deductible contributes no payment.

Worked Example: A Discrete Severity With a Deductible-Size Loss

A hypothetical coverage has ground-up loss $X$ with the following distribution. The ordinary deductible is $d=\$400$. Find the probability of a positive payment, the expected payment per loss and the average payment conditional on a positive payment. No interest discounting or mortality assumption is involved.

\begin{array}{c|cccc} x & 0 & 400 & 600 & 1000\\ \hline P(X=x) & 0.15 & 0.25 & 0.35 & 0.25 \end{array}

Method. The probabilities sum to $0.15+0.25+0.35+0.25=1$. A positive payment occurs only for losses of $600$ or $1{,}000$, so

$$ P(X>400)=0.35+0.25=0.60. $$

The payment amounts at the four possible losses are $0,0,200,$ and $600$ dollars, respectively. Directly weighting those amounts gives

E[Y^L] =0.15(0)+0.25(0)+0.35(200)+0.25(600) =70+150 =\$220.

The same result follows from the survival function. For $400\leq x<600$, $S_X(x)=0.60$; for $600\leq x<1{,}000$, $S_X(x)=0.25$; and for $x\geq 1{,}000$, $S_X(x)=0$, apart from endpoint conventions that do not change the integral. Hence

\int_{400}^\infty S_X(x)\,dx =(200)(0.60)+(400)(0.25) =120+100 =\$220.

Conditional on a positive payment, the average payment is $\$220/0.60=\$366.67$, rounded to the nearest cent. The loss of exactly $\$400$ has probability $0.25$, but its payment is zero. Using $P(X\geq 400)=0.85$ as the probability of a positive payment would therefore be incorrect.

Mixtures: Integrate the Whole Tail

For a finite mixture, the ground-up survival function is the weighted sum of the component survival functions. The expected payment can consequently be calculated by integrating that mixture survival function, or equivalently by weighting the component expected payments. The deductible applies to the loss after the component generating the loss has been selected.

Worked Example: A Two-Component Exponential Mixture

Suppose a hypothetical loss is generated by selecting an exponential severity with probability $0.70$ and mean $\$500$, or an exponential severity with probability $0.30$ and mean $\$2{,}000$. The ordinary deductible is $d=\$600$. Find the probability of a positive payment and the expected payment per loss. No interest discounting or mortality assumption is involved.

Method. The two component rates are $\lambda_1=1/500$ and $\lambda_2=1/2{,}000$, per dollar. Thus the mixture survival function is

S_X(x)=0.70e^{-x/500}+0.30e^{-x/2000}.

Evaluate it at the deductible to obtain the positive-payment probability:

\begin{aligned} P(X>600) &=0.70e^{-600/500}+0.30e^{-600/2000}\\ &=0.70e^{-1.2}+0.30e^{-0.3}\\ &=0.70(0.3011942)+0.30(0.7408182)\\ &=0.4330814\ldots\\ &\approx 0.4331. \end{aligned}

Integrate each component’s survival function from $600$ to infinity:

\begin{aligned} E[Y^L] &=0.70\int_{600}^{\infty}e^{-x/500}\,dx +0.30\int_{600}^{\infty}e^{-x/2000}\,dx\\ &=0.70(500)e^{-600/500} +0.30(2000)e^{-600/2000}\\ &=350(0.3011942)+600(0.7408182)\\ &=105.4180+444.4909\\ &=\$549.9089\ldots. \end{aligned}

Rounded to the nearest cent, the expected payment is $549.91 per loss. As a check, the result is a weighted average of the component expected payments: $0.70(\$500e^{-1.2})+0.30(\$2{,}000e^{-0.3})$, the same calculation as above. A common error is to replace the mixture with a single exponential having the mixture’s mean; that changes the tail shape and does not preserve this deductible payment.

Common Traps and Scope

The result is useful whenever the ground-up survival function is available, whether it comes from a named severity model, a mixture or a discrete distribution. Before calculating, identify the payment definition and confirm that the contract has only the ordinary deductible being considered. A policy limit, coinsurance or a different deductible arrangement changes the payment function and cannot be silently added to this formula.

Integrating from zero. $\int_0^\infty S_X(x)\,dx=E[X]$, the ground-up mean. With a deductible, start at $d$: $\int_d^\infty S_X(x)\,dx$.
Confusing probability with amount. $S_X(d)=P(X>d)$ is dimensionless. Expected payment is an amount and requires the area under the tail beyond $d$.
Including losses equal to the deductible. For an ordinary deductible, $X=d$ produces zero payment. The positive-payment probability is $P(X>d)$, not $P(X\geq d)$ when there is an atom at $d$.
Using the wrong tail. The integrand is the survival function of the ground-up loss $X$, evaluated from the deductible upward, not the distribution function $F_X$.

The survival integral must be finite for the expected payment to be finite. If the ground-up loss has an infinite mean, a positive deductible does not necessarily make the expected payment finite; the tail beyond the deductible still needs to be checked. Under the stated model, the calculation provides a direct per-loss expected insurer amount, not a premium by itself.

Key takeaway: For an ordinary deductible, integrate the ground-up survival function above the deductible. Use $S_X(d)$ to find the probability of a positive payment, but use $\int_d^\infty S_X(x)\,dx$ to find the expected payment.

Check Your Understanding

For each problem, distinguish the positive-payment probability from the expected payment. Amounts are in dollars.

A hypothetical loss is exponential with mean $\$800$, and the ordinary deductible is $\$200$. Find the probability of a positive payment and the expected payment. (Answer: probability $0.7788$; expected payment $\$623.04$.)
A ground-up loss is $100, 300,$ or $700$, with probabilities $0.20, 0.30,$ and $0.50$, respectively. The deductible is $\$300$. Find the positive-payment probability and expected payment. (Answer: probability $0.5000$; expected payment $\$200.00$.)
A hypothetical mixture selects an exponential loss with probability $0.40$ and mean $\$400$, or one with probability $0.60$ and mean $\$1{,}000$. For a $\$500$ deductible, find the positive-payment probability and expected payment, rounding money to cents. (Answer: probability $0.4785$; expected payment $\$409.76$.)
A loss is uniform on $[0,\$1{,}200]$, with a $\$300$ ordinary deductible. Find the probability of a positive payment and the expected payment. (Answer: probability $0.7500$; expected payment $\$337.50$.)
A loss takes values $0,250,500,$ and $900$, with probabilities $0.10,0.20,0.40,$ and $0.30$. With a $\$250$ deductible, find the positive-payment probability, expected payment, and conditional average payment given a positive payment. (Answer: probability $0.7000$; expected payment $\$295.00$; conditional average payment $\$421.43$.)

What You'll Learn