Expected Payments and Moments Under Deductibles

Why Deductible Payment Moments Matter

An ordinary deductible changes more than the probability that an insurer makes a payment. It also changes the size and variability of payments on the claims that exceed the deductible. Pricing and risk analysis therefore require moments of the payment amount itself, not just moments of the ground-up loss.

Let $X$ be a nonnegative ground-up loss and $d\geq 0$ an ordinary deductible. The insurer pays the part of the loss above $d$, if any. In this tutorial, $Y^L$ denotes payment per loss; there is one such payment amount for each modeled loss, including a zero payment when the loss does not exceed the deductible. No interest or mortality assumptions are needed for these severity calculations.

Definition: Under an ordinary deductible $d$, the insurer’s payment per loss is $Y^L=(X-d)^+=\max(X-d,0)$. Its mean is $E[Y^L]$, its second raw moment is $E[(Y^L)^2]$, and its variance is $\operatorname{Var}(Y^L)=E[(Y^L)^2]-(E[Y^L])^2$.

The point mass at zero matters: any loss at or below $d$ produces no insurer payment. Consequently, an expected payment is not generally the same as the expected excess loss conditional on the loss exceeding $d$. The payment moments below are unconditional, so they include the possibility of no payment.

Moment Formulas From a Density or Survival Function

Suppose first that $X$ has a density $f_X$. For a positive payment, the ground-up loss must exceed $d$, and the payment is $x-d$. Integrating that payment, or its square, over the density gives the moments directly.

E[Y^L]=\int_d^\infty (x-d)f_X(x)\,dx,\qquad E[(Y^L)^2]=\int_d^\infty (x-d)^2f_X(x)\,dx.

These density integrals apply when the loss distribution is absolutely continuous. If $X$ has discrete probability masses or a mixture of continuous and discrete parts, the general definition is an expectation with respect to the entire distribution; summation terms for the masses must be included. A survival-function representation works for continuous, discrete, and mixed nonnegative losses.

Theorem: For any nonnegative loss $X$, let $S_X(x)=P(X>x)$. Provided the relevant moments are finite, $$ E[Y^L]=\int_0^\infty S_X(d+t)\,dt =\int_d^\infty S_X(x)\,dx, $$ $$ E[(Y^L)^2]=2\int_0^\infty tS_X(d+t)\,dt =2\int_d^\infty (x-d)S_X(x)\,dx. $$ The payment variance is the second expression minus the square of the mean.

The argument of the survival function is $d+t$, not just $t$: a payment greater than $t$ occurs precisely when the ground-up loss exceeds $d+t$. This shift is central to using the formulas correctly.

Why the Survival-Function Formulas Hold

For any nonnegative random variable $Z$, write its value as the area under an indicator: $Z=\int_0^\infty \mathbf{1}_{\{Z>t\}}\,dt$. Taking expectations gives its tail-integral formula. For the second moment, first write $Z^2=\int_0^Z 2t\,dt$. Taking expectations again yields a weighted tail integral.

E[Z]=\int_0^\infty P(Z>t)\,dt,\qquad E[Z^2]=2\int_0^\infty tP(Z>t)\,dt.

Set $Z=Y^L$. For $t\geq 0$, the event $Y^L>t$ is equivalent to $X>d+t$. Therefore $P(Y^L>t)=S_X(d+t)$. Substitution into the two tail-integral formulas gives the stated results. The use of strict exceedance does not cause a boundary problem: changing the integrand at isolated threshold points does not change these integrals.

Identify the payment.
Write $Y^L=(X-d)^+$, and keep the deductible inside the positive-part expression.

Choose a moment method.
Use density integrals when a density is available; use the survival form when the tail is simpler or the distribution may contain masses.

Find the raw moments first.
Calculate $E[Y^L]$ and $E[(Y^L)^2]$ before subtracting the squared mean to obtain the variance.

Worked Calculations

Worked Example: Exponential Loss With a Deductible Equal to Its Mean

A hypothetical insurer models an individual claim amount $X$ in dollars as exponential with rate $\lambda=1/500$ per dollar, so $E[X]=500$ dollars and $S_X(x)=e^{-x/500}$ for $x\geq0$. The ordinary deductible is $d=\$500$. Find the mean, second raw moment, and variance of the insurer’s payment per loss. Give monetary results to the nearest cent.

Method. Use the survival-function formulas. For $t\geq0$, $S_X(d+t)=e^{-(500+t)/500}=e^{-1}e^{-t/500}$.

Mean. Pull out the constant factor and integrate the exponential tail:

E[Y^L]=\int_0^\infty e^{-1}e^{-t/500}\,dt =e^{-1}(500) =\frac{500}{e} \approx \$183.94.

Second raw moment. For $a>0$, $\int_0^\infty te^{-t/a}\,dt=a^2$; here $a=500$. Thus

E[(Y^L)^2] =2e^{-1}\int_0^\infty te^{-t/500}\,dt =2e^{-1}(500^2) =\frac{500{,}000}{e} \approx \$183{,}939.72.

Variance. The squared mean is $(500/e)^2=250{,}000/e^2\approx \$33{,}833.82$ in squared-dollar units. Therefore

\operatorname{Var}(Y^L) =\frac{500{,}000}{e}-\frac{250{,}000}{e^2} \approx 150{,}105.90\ \text{dollars}^2.

As a check, the mean payment is positive but below the ground-up mean of $\$500$, as expected with a positive deductible. The variance is also positive. Common mistake avoided: subtracting the deductible from the ground-up mean would give $E[X]-d=0$, which is not the mean insurer payment; losses above the deductible still produce positive payments.

Worked Example: Uniform Loss and a Zero-Payment Probability

A hypothetical portfolio has individual ground-up loss $X$ uniformly distributed from $\$0$ to $\$2{,}000$, with density $f_X(x)=1/2000$ on that interval. Each loss has an ordinary deductible of $d=\$500$. Find the mean and variance of the insurer’s payment per loss, to the nearest cent.

The payment ranges from zero to $\$1{,}500$. Losses at or below $\$500$, a probability of $500/2000=0.25$, produce zero payment. For $0<t<1500$, the survival probability of the payment is $P(Y^L>t)=P(X>500+t)=(1500-t)/2000$; above $1500$, it is zero.

Mean by the survival method:

E[Y^L]=\int_0^{1500}\frac{1500-t}{2000}\,dt =\frac{1500^2}{2(2000)} =\$562.50.

Second raw moment:

E[(Y^L)^2] =2\int_0^{1500}t\frac{1500-t}{2000}\,dt =\frac{2}{2000}\left(\frac{1500^3}{2}-\frac{1500^3}{3}\right) =\frac{1500^3}{3(2000)} =\$562{,}500.00.

The squared mean is $(562.50)^2=\$316{,}406.25$ in squared-dollar units. Hence

\operatorname{Var}(Y^L)=562{,}500-316{,}406.25 =246{,}093.75\ \text{dollars}^2.

The calculations retain the 25% chance of no payment; no separate conditional adjustment is needed. Common mistake avoided: treating the payment as uniform from zero to $\$1{,}500$ conditional on a payment and then using those conditional moments as unconditional moments would omit the probability of a zero payment.

Worked Example: Heavy-Tailed Loss With Finite Payment Variance

A hypothetical severity model has survival function $S_X(x)=(1000/(1000+x))^3$ for $x\geq0$. The ground-up loss $X$ is measured in dollars, and the ordinary deductible is $d=\$1{,}000$. Find the mean, second raw moment, and variance of the insurer’s payment. Give monetary results to the nearest cent.

At $d+t=1000+t$, the survival function becomes $S_X(d+t)=1000^3/(2000+t)^3$. The mean is

E[Y^L] =1000^3\int_0^\infty\frac{1}{(2000+t)^3}\,dt =1000^3\left(\frac{1}{2(2000)^2}\right) =\$125.00.

For the second raw moment, use $\int_0^\infty t/(a+t)^3\,dt=1/(2a)$, obtained by substituting $u=a+t$ and integrating $u^{-2}-au^{-3}$. With $a=2000$,

E[(Y^L)^2] =2(1000^3)\int_0^\infty\frac{t}{(2000+t)^3}\,dt =2(1000^3)\left(\frac{1}{2(2000)}\right) =\$500{,}000.00.

The squared mean is $125^2=\$15{,}625.00$ in squared-dollar units, so

\operatorname{Var}(Y^L)=500{,}000-15{,}625 =484{,}375.00\ \text{dollars}^2.

Both moments are finite: the integrals converge, including the integral for the second moment. The deductible shifts the tail used in the payment calculation but does not by itself guarantee finite moments for every possible severity model. Common mistake avoided: assuming that a deductible makes a sufficiently heavy tail harmless; the tail integrals must still converge.

Interpretation and Common Traps

The variance here describes variability in an insurer’s payment per loss. It is not the variance of the number of claims or of an annual aggregate of payments. Those quantities also depend on claim frequency and, for an aggregate, on how individual claims combine.

Keep the deductible shift in place. In the survival formulas, use $S_X(d+t)$, because a payment above $t$ requires a ground-up loss above $d+t$.
Do not confuse raw and central moments. The second raw moment is $E[(Y^L)^2]$; variance requires subtracting $(E[Y^L])^2$.
Include zero payments in unconditional moments. The formulas are for payment per loss, not payment conditional on the loss exceeding the deductible.
Check moment existence. A finite mean does not ensure a finite second moment. Verify convergence of the second-moment integral before reporting a finite variance.

Key takeaway: Under an ordinary deductible, integrate the payment function $(X-d)^+$ or use the shifted tail $S_X(d+t)$. Calculate the first and second raw moments on the same unconditional basis, then subtract the squared mean to obtain the insurer-payment variance.

Check Your Understanding

For each problem, calculate unconditional moments of payment per loss; round monetary answers to the nearest cent where needed.

A hypothetical exponential loss has rate $0.01$ per dollar and an ordinary deductible of $\$100$. Find the mean payment and variance. (Answer: mean $\$36.79$; variance $6{,}004.24$ dollars².)
A hypothetical ground-up loss is uniform on $[0,1200]$ dollars and has an ordinary deductible of $\$300$. Find the mean payment and variance. (Answer: mean $\$337.50$; variance $88{,}593.75$ dollars².)
A hypothetical loss has survival function $S_X(x)=(800/(800+x))^3$ for $x\geq0$, and the deductible is $\$800$. Find the mean payment, second raw moment, and variance. (Answer: $\$100.00$, $\$320{,}000.00$ dollars², and $310{,}000.00$ dollars², respectively.)
A hypothetical loss is $\$100$, $\$500$, or $\$1{,}000$, with probabilities $0.25$, $0.50$, and $0.25$, respectively. The deductible is $\$400$. Find the mean payment and variance, including the zero-payment outcome. (Answer: mean $\$200.00$; variance $55{,}000.00$ dollars².)

What You'll Learn