Expected Deductible Payments After Severity Trending

Expected Payments After Trending

When claim severities are adjusted for inflation or another multiplicative trend, the deductible must be applied to the trended loss if the goal is to estimate current insurer payments under a current policy. Trending changes the loss amount; it does not automatically change the deductible. The expected payment therefore depends on both the trended severity distribution and the deductible that applies after trending.

This tutorial considers an ordinary deductible and calculates the expected insurer payment per loss. It does not include a policy limit, coinsurance, or a change in claim frequency. The key tools are the survival function of the trended severity and an equivalent calculation on the historical severity scale.

Definition: Let $X$ be a nonnegative historical ground-up loss, and let $g>0$ be a deterministic multiplicative trend factor. The trended loss is $Z=gX$. Under an ordinary deductible $d\geq 0$, the insurer payment per loss is $Y^L=(Z-d)_+=\max(Z-d,0)$. The expected insurer payment per loss is $E[Y^L]$.

Here, $d$ is the deductible in the same price basis as $Z$. If historical losses are being brought to current prices, $d$ is the current deductible. The model assumes that the trend factor applies to the ground-up severity before the coverage term is applied.

Survival-Function Formula

For a nonnegative loss $Z$, write $S_Z(t)=P(Z>t)$ for its survival function. The expected payment under an ordinary deductible can be written as the area under this survival function above the deductible.

E[(Z-d)_+]=\int_d^\infty S_Z(t)\,dt

This formula applies to continuous, discrete, and mixed loss distributions, provided the expectation exists. For a discrete loss, the survival function has jumps, but the integral remains valid. Whether a loss exactly equal to $d$ is described as “at” or “above” the deductible does not affect the payment: that loss produces zero payment.

Express the excess as an area.
For any fixed nonnegative value $z$, $(z-d)_+=\int_0^\infty \mathbf{1}_{\{z>d+u\}}\,du$. The integrand is 1 for the values of $u$ that fall below the excess and 0 otherwise.

Take expectations.
Taking expectations of the indicator gives $P(Z>d+u)$. Thus $E[(Z-d)_+]=\int_0^\infty P(Z>d+u)\,du$.

Change the integration variable.
Set $t=d+u$, so $dt=du$. The lower limit becomes $d$, giving $E[(Z-d)_+]=\int_d^\infty S_Z(t)\,dt$.

Since $Z=gX$ and $g>0$, the trended survival function is $S_Z(t)=P(gX>t)=S_X(t/g)$. Substituting this into the payment formula and setting $x=t/g$ gives a useful calculation on the historical scale:

E[(gX-d)_+] =\int_d^\infty S_X(t/g)\,dt =g\int_{d/g}^\infty S_X(x)\,dx

The historical-scale lower limit is $d/g$, not $d$. A current deductible of $d$ is equivalent, for this calculation, to a historical-scale threshold of $d/g$. The leading factor $g$ then converts the payment amount back to the trended scale.

Worked Calculations

Worked Example: Exponential Severity With a Current Deductible

A hypothetical portfolio has historical claim severity $X$ that is exponential with mean $1,000, so $S_X(x)=e^{-x/1000}$ for $x\geq 0$. Claims are trended by a factor $g=1.2$, and the current ordinary deductible is $d=$$1,200. Find the expected insurer payment per loss.

Method and assumptions. The trend is deterministic and multiplicative, and it is applied to each ground-up loss before the deductible. The trended loss $Z=1.2X$ is exponential with mean $1.2(1000)=$$1,200, so $S_Z(t)=e^{-t/1200}$.

Using the survival-function formula, integrate from the current deductible:

E[(Z-1200)_+] =\int_{1200}^{\infty} e^{-t/1200}\,dt =\left[-1200e^{-t/1200}\right]_{1200}^{\infty} =1200e^{-1}

Since $e^{-1}\approx 0.3678794$, the expected payment is $1200(0.3678794)=441.4553\ldots$. Rounded only at the final step to the nearest cent, this is $441.46 per loss.

Check. On the historical scale, the deductible threshold is $1200/1.2=1000$. The equivalent calculation is $1.2\int_{1000}^{\infty}e^{-x/1000}\,dx=1.2(1000e^{-1})=1200e^{-1}$, agreeing with the result. The method avoids treating the current deductible as a historical deductible.

Worked Example: Uniform Severity and a Fixed Current Deductible

A hypothetical historical severity $X$ is uniform on $[0,4000]$ dollars. All losses are trended by $g=1.25$, and the current ordinary deductible is $2,500. Calculate the expected insurer payment per loss.

Method and assumptions. The distribution has density $1/4000$ on $[0,4000]$, and the trend is applied before coverage. The trended loss $Z=1.25X$ is uniform on $[0,5000]$. Therefore, for $0\leq t\leq 5000$, $S_Z(t)=1-t/5000$, and the survival function is zero above $5,000.

Integrate the trended survival function from the current deductible to the maximum trended loss:

\begin{aligned} E[(Z-2500)_+] &=\int_{2500}^{5000}\left(1-\frac{t}{5000}\right)\,dt\\ &=\left[t-\frac{t^2}{10000}\right]_{2500}^{5000}\\ &=(5000-2500)-(2500-625)\\ &=625 \end{aligned}

The expected insurer payment is $625.00 per loss. As a check on the integration, the survival curve falls linearly from $0.5$ at the deductible to zero at $5,000. The area is a triangle with base $2500$ and height $0.5$, giving $\tfrac12(2500)(0.5)=625$.

Common mistake. Applying the $2,500 deductible to historical losses and then trending the resulting payment gives $1.25E[(X-2500)_+]$, which is not the stated coverage. Here that incorrect calculation is $1.25\left(\tfrac12(1500)(0.375)\right)=351.5625$. The correction is to apply the deductible after trending, or equivalently to use the historical threshold $2500/1.25=2000$ and multiply the historical-scale excess expectation by $1.25$.

Worked Example: Discrete Severity With an Atom Above the Deductible

A hypothetical historical loss $X$ has the following distribution: $P(X=0)=0.20$, $P(X=1000)=0.50$, and $P(X=3000)=0.30$. Losses are trended by $g=1.2$, and the current ordinary deductible is $1,500. Find the expected insurer payment per loss.

Method and assumptions. Trend each possible ground-up loss, then apply the deductible to each trended amount. The possible trended losses are $0$, $1.2(1000)=1200$, and $1.2(3000)=3600$ dollars, with probabilities unchanged.

The corresponding insurer payments are $(0-1500)_+=0$, $(1200-1500)_+=0$, and $(3600-1500)_+=2100$. Weight each payment by its probability:

$$ E[Y^L]=(0.20)(0)+(0.50)(0)+(0.30)(2100)=630 $$

Thus the expected payment is $630.00 per loss. For a survival-function check, the historical survival probability is $0.30$ from just above $1,000 up to $3,000. The equivalent historical threshold is $1500/1.2=1250$, so $1.2\int_{1250}^{3000}0.30\,dx=1.2(525)=630$. The method handles the atom at $3,000: it contributes its probability-weighted excess payment, not its full loss amount.

Choosing the Right Scale

The two equivalent calculations are useful in different situations. If a trended distribution is already available, integrate its survival function above the current deductible. If the historical distribution is easier to describe, divide the current deductible by the trend factor, integrate the historical survival function above that threshold, and multiply the result by the trend factor.

Available information	Calculation	Key point
Trended severity $Z=gX$	$\int_d^\infty S_Z(t)\,dt$	Use the current deductible $d$
Historical severity $X$	$g\int_{d/g}^\infty S_X(x)\,dx$	Use the equivalent threshold $d/g$, then multiply by $g$

The formula gives a payment per loss, conditional on using the specified severity model for a loss. It is not, by itself, an expected annual or portfolio payment. Such totals would also require a claim count model and assumptions about how claim frequency relates to severity.

A useful check is to compare the deductible with the range of possible trended losses. If every possible trended loss is at or below $d$, the expected payment is zero. If some losses exceed $d$, the survival integral includes only the tail above $d$; it does not include the ground-up loss below the deductible. For a nonnegative severity with finite mean, the expected payment also cannot exceed $E[Z]=gE[X]$.

Key takeaway: For a deterministic trend $g$ followed by an ordinary deductible $d$, calculate $E[(gX-d)_+]$. Use the trended survival function above $d$, or equivalently calculate $g\int_{d/g}^{\infty}S_X(x)\,dx$. Applying the deductible on the wrong price scale changes the coverage being valued.

Check Your Understanding

For each problem, identify the current deductible, account for the trend before applying coverage, and report the expected payment per loss.

A hypothetical historical severity is exponential with mean $2,000. It is trended by $g=1.1$, and the current deductible is $1,100. What is the expected insurer payment, rounded to the nearest cent? (Answer: $1,334.37)
A hypothetical historical loss is uniform on $[0,6000]$. After a trend factor of $1.2$, the current ordinary deductible is $3,600. Find the expected payment per loss. (Answer: $900.00)
A historical loss takes values $500, $1,500, and $4,000 with probabilities $0.20$, $0.50$, and $0.30$, respectively. Apply a trend factor of $1.1$ and a current deductible of $3,000. Find the expected payment. (Answer: $420.00)
A hypothetical historical severity has survival function $S_X(x)=(1000/(1000+x))^2$ for $x\geq0$. Losses are trended by $1.5$ and have a current deductible of $3,000. Use the historical-scale integral to find the expected payment. (Answer: $500.00)
A historical severity is uniform on $[0,10000]$. It is trended by $1.1$ in each of two successive years, and the current deductible is $6,050. Find the expected payment per loss. (Answer: $1,512.50)

Available information	Calculation	Key point
Trended severity \(Z=gX\)	\(\int_d^\infty S_Z(t)\,dt\)	Use the current deductible \(d\)
Historical severity \(X\)	\(g\int_{d/g}^\infty S_X(x)\,dx\)	Use the equivalent threshold \(d/g\), then multiply by \(g\)

What You'll Learn