Ordinary Deductible Calculation Workshop

From Loss Amount to Insurer Payment

An ordinary deductible is applied separately to each covered loss. For a single loss $X$ and a fixed deductible $d\geq 0$, the insurer pays only the amount by which the loss exceeds $d$. Thus the deductible calculation transforms the loss distribution into a payment distribution, including a possible point mass at zero.

The calculations below concern one claim, with no policy limit, coinsurance, or other payment adjustment. Unless an example explicitly says otherwise, losses are nonnegative and have a finite mean. Interest discounting and mortality assumptions do not apply to these single-claim severity calculations.

Definition: The insurer's payment per loss under an ordinary deductible $d$ is $Y^L=(X-d)_+=\max(X-d,0)$. A payment occurs exactly when $X>d$, so $P(Y^L>0)=P(X>d)=S_X(d)$. The payment conditional on a positive payment is $Y^P=Y^L\mid(Y^L>0)$, defined only when $P(X>d)>0$.

The strict inequality matters: a loss exactly equal to the deductible produces a zero payment. This distinction is consequential when $X$ has a point mass at $d$, though for a continuous loss distribution $P(X=d)=0$.

Distribution of the Payment

For $y<0$, $Y^L$ cannot be at most $y$. For $y\geq0$, the event $Y^L\leq y$ is exactly the event $X\leq d+y$. This gives the distribution function, whether or not $X$ is continuous:

F_{Y^L}(y)= \begin{cases} 0, & y<0,\\ F_X(d+y), & y\geq0. \end{cases}

In particular, $P(Y^L=0)=F_X(d)$. The probability of a positive payment is $1-F_X(d)$, not $1-F_X(d-)$. If $X$ has an atom at $d$, that atom belongs to the zero-payment probability. For $y>0$, the payment distribution inherits the loss distribution above the deductible, shifted down by $d$.

When a payment-conditioned distribution is needed, normalize the positive-payment part by $P(X>d)$. For $y\geq0$ and $P(X>d)>0$,

F_{Y^P}(y) = P(Y^L\leq y\mid Y^L>0) = \frac{F_X(d+y)-F_X(d)}{1-F_X(d)}.

This describes the size of a payment given that one occurs; it is not the distribution of payment per loss. The latter still includes the probability mass at zero.

Worked Example: Exponential Loss With a Deductible

A hypothetical equipment policy has one loss $X$ with an exponential distribution of mean $\theta=\$10{,}000$, so $S_X(x)=e^{-x/10{,}000}$ for $x\geq0$. The ordinary deductible is $d=\$4{,}000$. Find the probability of an insurer payment, the payment distribution, and the expected payment. There is no interest discounting or mortality assumption.

Method. Use the survival function at the deductible for the payment probability, the shifted loss distribution for the payment CDF, and the integrated survival function for the expected payment.

Payment probability.

P(Y^L>0)=P(X>4{,}000) =e^{-4{,}000/10{,}000} =e^{-0.4} =0.6703.

Therefore $P(Y^L=0)=1-0.670320\ldots=0.3297$. For $y\geq0$, measured in dollars,

F_{Y^L}(y)=F_X(4{,}000+y) =1-e^{-(4{,}000+y)/10{,}000}.

At $y=0$, this is $1-e^{-0.4}=0.329679\ldots$, confirming the zero-payment probability. Conditional on a positive payment, the memoryless property gives an exponential payment with mean $\$10{,}000$; its CDF is $1-e^{-y/10{,}000}$ for $y\geq0$.

Expected payment. Integrating the survival function above the deductible gives

E[Y^L] =\int_{4{,}000}^{\infty}e^{-x/10{,}000}\,dx =10{,}000e^{-0.4} =\$6{,}703.20.

As a check, the conditional mean payment is $\$10{,}000$, and $\$10{,}000(0.670320\ldots)=\$6{,}703.20$, matching the unconditional expectation. Results are rounded only at the final step: probabilities to four decimals and dollars to the cent.

Expected Payment and Why the Tail Integral Works

For a nonnegative loss with finite mean, the expected ordinary-deductible payment can be calculated directly from the survival function:

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

This identity applies to continuous and discrete loss distributions, including mixtures. An atom at the deductible does not create a payment, and atoms above it are included in the expectation. For a distribution with a density, the same expectation can also be written as $\int_d^\infty (x-d)f_X(x)\,dx$.

To justify the tail-integral formula, write a nonnegative payment as the length of the interval of levels $y$ below that payment. For each realized loss,

(X-d)_+=\int_0^\infty \mathbf{1}_{\{X>d+y\}}\,dy.

Taking expectations and interchanging the nonnegative integral and expectation gives

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

where the last equality uses $x=d+y$. This explains why the expected payment depends on the whole tail above $d$, not simply on the mean loss minus the deductible.

When the probability of payment is positive, a useful check is

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

The unconditional expectation includes zero-payment claims. The conditional mean $E[Y^P]$ does not, so it must be multiplied by the payment probability to recover $E[Y^L]$.

Worked Example: Discrete Losses and an Atom at the Deductible

For a hypothetical single-claim warranty, the loss $X$, measured in thousands of dollars, is $0, 5, 12,$ or $20$ with respective probabilities $0.10, 0.25, 0.40,$ and $0.25$. The deductible is $d=8$ thousand dollars. Find the payment distribution, payment probability, and expected payment. No interest discounting or mortality assumption applies.

Method. Apply $Y^L=(X-d)_+$ to each possible loss, then combine the probabilities for equal payment amounts.

Loss $X$ (thousands)	Probability	Payment $Y^L$ (thousands)
0	0.10	0
5	0.25	0
12	0.40	4
20	0.25	12

The payment probability is $0.40+0.25=0.65$, because only losses strictly greater than $8$ produce a payment. The distribution of $Y^L$ has probability $0.35$ at zero, $0.40$ at $4$ thousand, and $0.25$ at $12$ thousand.

The expected payment is

E[Y^L]=(0)(0.35)+(4{,}000)(0.40)+(12{,}000)(0.25) =\$1{,}600+\$3{,}000 =\$4{,}600.

Given that a payment occurs, the conditional probabilities are $0.40/0.65=8/13$ for a $\$4{,}000$ payment and $0.25/0.65=5/13$ for a $\$12{,}000$ payment. The conditional mean is $(4{,}000)(8/13)+(12{,}000)(5/13)=\$7{,}076.92$ to the cent. Multiplying by $0.65$ gives $\$4{,}600$ before rounding, a check on the unconditional calculation.

Mixtures, Endpoint Checks, and Common Traps

A mixture distribution is handled component by component, but the deductible is applied to the loss in every component. The following example combines a point mass exactly at the deductible with a continuous component, making the strict payment condition visible.

Worked Example: A Mixed Loss With Mass at the Deductible

A hypothetical claim has loss $X=\$5{,}000$ with probability $0.25$; with probability $0.75$, $X$ is uniform on $[0,\$15{,}000]$. The ordinary deductible is $\$5{,}000$. Find the payment probability, payment CDF, and expected payment. There is no interest discounting or mortality assumption.

Method. Separate the point-mass component from the uniform component, apply the payment rule to each, and then weight by the mixture probabilities.

The point mass at $\$5{,}000$ produces no payment. In the uniform component, a payment occurs when $X>\$5{,}000$, which has conditional probability $10{,}000/15{,}000=2/3$. Thus

P(Y^L>0)=0.25(0)+0.75\left(\frac{10{,}000}{15{,}000}\right)=0.50.

The zero-payment probability is also $0.25+0.75(5{,}000/15{,}000)=0.50$. For $0\leq y\leq\$10{,}000$, the CDF is

F_{Y^L}(y) =0.25+0.75\left(\frac{5{,}000+y}{15{,}000}\right) =0.50+\frac{y}{20{,}000}.

It equals $0.50$ at $y=0$ and reaches $1$ at $y=\$10{,}000$; for $y\geq\$10{,}000$ it remains $1$. Conditional on payment, the payment is uniform on $(0,\$10{,}000]$, with CDF $y/10{,}000$ over that range.

Only the uniform component contributes to the expected payment. Its deductible payment expectation, before applying the mixture weight, is

\int_{5{,}000}^{15{,}000} (x-5{,}000)\frac{1}{15{,}000}\,dx =\frac{(10{,}000)^2}{2(15{,}000)} =\$3{,}333.33\ldots.

Therefore $E[Y^L]=0.75(\$3{,}333.33\ldots)=\$2{,}500$. The conditional mean payment is $\$5{,}000$, and $0.50(\$5{,}000)=\$2{,}500$, confirming the result.

Common exam trap: counting equality as a payment. In the mixed example, using $P(X\geq\$5{,}000)$ would count the $0.25$ point mass at exactly the deductible and incorrectly give a payment probability of $1.00$. The correct condition is $X>d$, giving $0.50$. Another incorrect shortcut is $E[X]-d$: here $E[X]=0.25(\$5{,}000)+0.75(\$7{,}500)=\$6{,}875$, so that shortcut gives $\$1{,}875$, not the correct $\$2{,}500$. The deductible is subtracted only from losses exceeding it; losses below or equal to it generate zero, not negative payments.

Key takeaway: For an ordinary deductible, find the payment probability from $P(X>d)$, retain $F_X(d)$ as the mass at zero, and calculate the expected payment from $\int_d^\infty S_X(x)\,dx$. For a conditional payment distribution or mean, divide by the payment probability only after isolating the positive-payment outcomes.

Check Your Understanding

For each hypothetical single-claim severity, calculate the requested quantity using the ordinary-deductible payment rule. Report monetary results in the stated units.

A loss is uniform on $[0,\$20{,}000]$, and the deductible is $\$5{,}000$. Find the payment probability, the CDF of $Y^L$ on its nonconstant range, and $E[Y^L]$. (Answer: $P(Y^L>0)=0.7500$; $F_{Y^L}(y)=0.25+y/20{,}000$ for $0\leq y\leq\$15{,}000$; $E[Y^L]=\$5{,}625$.)
A loss is exponential with mean $\$6{,}000$, and the deductible is $\$3{,}000$. Find the payment probability and expected payment. (Answer: $P(Y^L>0)=0.6065$; $E[Y^L]=\$3{,}639.18$.)
A loss in thousands of dollars is $2,6,10,$ or $18$ with probabilities $0.10,0.30,0.40,$ or $0.20$, respectively. The deductible is $6$ thousand dollars. Give the probability distribution of $Y^L$ and its expected value. (Answer: $P(Y^L=0)=0.40$, $P(Y^L=4{,}000)=0.40$, $P(Y^L=12{,}000)=0.20$; $E[Y^L]=\$4{,}000$.)
With probability $0.30$, a loss is exactly $\$4{,}000$; otherwise it is exponential with mean $\$8{,}000$. The deductible is $\$4{,}000$. Find the payment probability and expected payment. (Answer: $P(Y^L>0)=0.4246$; $E[Y^L]=\$3{,}396.57$.)
A Pareto type I loss has survival function $S_X(x)=(\$10{,}000/x)^2$ for $x\geq\$10{,}000$, and a deductible of $\$20{,}000$. Find the payment probability, expected payment, and conditional mean payment. (Answer: $P(Y^L>0)=0.2500$; $E[Y^L]=\$5{,}000$; $E[Y^P]=\$20{,}000$.)

What You'll Learn