Limited Expected Value Identities for Deductibles

Why Limited Expected Values Help

An ordinary deductible makes the insurer’s payment depend on how much a loss exceeds a threshold. A useful companion quantity is the part of the loss retained up to that threshold: the limited expected value. It provides a direct way to calculate the expected deductible payment and a check that works for both continuous and discrete loss distributions.

Let $X$ be a nonnegative loss amount, and let $d\geq 0$ be an ordinary deductible. The insurer’s payment per loss is $Y^L=(X-d)^+$, where $z^+=\max(z,0)$. The limited amount is $\min(X,d)$. These two amounts partition the loss: the insured bears the amount up to $d$, and the insurer pays the amount above $d$.

Definition: The limited expected value of $X$ at threshold $d$ is $E[\min(X,d)]$. For an ordinary deductible $d$, the expected payment per loss is $E[(X-d)^+]$. Both are expectations of dollar amounts, but they describe different parts of the same loss.

Unless an example states otherwise, amounts below are hypothetical gross losses in dollars, with no interest adjustment, no mortality contingency, and no policy limit. Thus, discounting and life-contingent assumptions do not enter the calculations. We assume $X$ has a finite mean when using an identity involving $E[X]$.

The Deductible Identity

For every possible nonnegative loss $x$, the amount retained up to the deductible plus the amount above the deductible equals the full loss. This pointwise relationship is the reason the expectation identity holds; it does not require a continuous distribution.

x=\min(x,d)+(x-d)^+

Taking expectations gives the main result. The assumption $X\geq 0$ ensures the two terms represent the intended loss portions, and a finite $E[X]$ ensures the expectations are finite.

Theorem: If $X\geq 0$, $d\geq 0$, and $E[X]<\infty$, then

E[(X-d)^+]=E[X]-E[\min(X,d)].

This identity is useful when the loss mean and limited expected value are easier to find than the payment expectation directly. It also gives an immediate check: the expected loss must equal the sum of the expected retained amount and expected payment.

Survival-Function Formulas

For a nonnegative loss, write $S_X(t)=P(X>t)$. The limited expected value can be found by integrating the survival function only up to $d$; the deductible payment can be found by integrating it beyond $d$, provided the mean is finite.

E[\min(X,d)]=\int_0^d S_X(t)\,dt

E[(X-d)^+]=\int_d^\infty S_X(t)\,dt

To justify the first formula, represent the limited amount as the length of the interval from zero up to the lesser of $X$ and $d$. At each point $t$ in that interval, the indicator that the loss reaches beyond $t$ is one; otherwise it is zero. Integrating these indicators gives the limited amount.

\min(X,d)=\int_0^d \mathbf{1}_{\{X>t\}}\,dt

Taking expectations and interchanging expectation and integration for this nonnegative integrand yields the limited expected value formula, since $E[\mathbf{1}_{\{X>t\}}]=P(X>t)=S_X(t)$. Likewise, $(X-d)^+=\int_d^\infty \mathbf{1}_{\{X>t\}}\,dt$, which gives the payment formula. Strict versus non-strict inequality at a single integration point does not change an integral, so these formulas also apply when $X$ has an atom at $d$.

For a discrete distribution with possible values $x_j$ and probabilities $p_j$, the corresponding direct calculations are

E[\min(X,d)]=\sum_j p_j\min(x_j,d), \qquad E[(X-d)^+]=\sum_j p_j(x_j-d)^+.

A mass exactly at the deductible contributes $d$ times its probability to the limited expected value and zero to the deductible payment. It must not be treated as a payment above the deductible.

Worked Calculations

Worked Example: Exponential Loss and a Deductible

A hypothetical policy covers a loss $X$ with an ordinary deductible of $d=\$500$. Assume $X$ is exponential with rate $\lambda=0.001$ per dollar, so $E[X]=1/\lambda=\$1{,}000$. There is no interest adjustment, mortality contingency, or policy limit. Calculate the limited expected value and the expected insurer payment, then check their sum.

Method. Use the exponential survival function $S_X(t)=e^{-0.001t}$ to calculate the limited expected value, then apply the deductible identity.

Limited expected value. Integrate the survival function from zero to the deductible:

E[\min(X,500)] =\int_0^{500}e^{-0.001t}\,dt =\left[-1000e^{-0.001t}\right]_0^{500} =1000(1-e^{-0.5}).

Using $e^{-0.5}=0.60653066$, this is $1000(1-0.60653066)=\$393.46934$, or \$393.47 rounded to the nearest cent.

Expected payment. The loss mean is $\$1{,}000$, so

E[(X-500)^+] =1000-393.46934 =\$606.53066 \approx \boxed{\$606.53}.

Check. The rounded amounts sum to $\$393.47+\$606.53=\$1{,}000.00$, the given mean. The common mistake this method avoids is subtracting the deductible itself from the mean: $1000-500=\$500$ is not the expected payment, because some losses are below the deductible and pay nothing.

Worked Example: Uniform Loss and Survival Areas

A hypothetical coverage has an ordinary deductible $d=\$3{,}000$. Assume $X$ is uniform on $[0,\$12{,}000]$, with no interest adjustment, mortality contingency, or policy limit. Calculate both the limited expected value and expected payment using survival integrals, and verify the identity.

Method. For $0\leq t\leq 12{,}000$, the survival function is $S_X(t)=(12{,}000-t)/12{,}000$. The mean of this uniform distribution is $\$6{,}000$.

Limited expected value. Since the deductible is within the support,

\begin{aligned} E[\min(X,3000)] &=\int_0^{3000}\frac{12000-t}{12000}\,dt\\ &=\left[t-\frac{t^2}{24000}\right]_0^{3000}\\ &=3000-\frac{9{,}000{,}000}{24{,}000}\\ &=3000-375 =\$2{,}625. \end{aligned}

Expected payment. Apply the identity using the mean:

E[(X-3000)^+]=6000-2625=\boxed{\$3{,}375}.

Check by direct integration. Integrating the survival function from the deductible to the largest possible loss gives

\begin{aligned} \int_{3000}^{12000}\frac{12000-t}{12000}\,dt &=\left[t-\frac{t^2}{24000}\right]_{3000}^{12000}\\ &=6000-2625 =\$3{,}375. \end{aligned}

The two expected amounts sum to $\$2{,}625+\$3{,}375=\$6{,}000$, as required. A common error is to report $E[X]-d=\$3{,}000$ as the expected payment. The deductible is not subtracted from every loss: losses below $d$ produce no payment, while losses above $d$ pay only their excess.

Worked Example: A Discrete Loss With an Atom at the Deductible

A hypothetical loss has the following distribution and an ordinary deductible of $d=\$700$. There is no interest adjustment, mortality contingency, or policy limit. Find the limited expected value and expected payment from the probability mass function, and check their sum.

\begin{array}{c|rrrr} x_j & 0 & 400 & 700 & 2500\\ \hline p_j & 0.20 & 0.30 & 0.20 & 0.30 \end{array}

Method. Multiply each possible amount by its probability, using $\min(x_j,700)$ for the limited amount and $(x_j-700)^+$ for the payment.

Loss mean. First calculate the mean needed for the check:

E[X]=0.20(0)+0.30(400)+0.20(700)+0.30(2500) =0+120+140+750 =\$1{,}010.

Limited expected value. The four limited amounts are $0,400,700,700$, respectively. Therefore,

E[\min(X,700)] =0.20(0)+0.30(400)+0.20(700)+0.30(700) =0+120+140+210 =\$470.

Expected payment. The four payments are $0,0,0,1800$, respectively. In particular, the loss equal to the deductible generates no payment.

E[(X-700)^+] =0.20(0)+0.30(0)+0.20(0)+0.30(1800) =\boxed{\$540}.

Check. The amounts add to $\$470+\$540=\$1{,}010=E[X]$. The key trap is to count the $\$700$ loss as exceeding the deductible. It equals the deductible, so its payment is $(700-700)^+=0$; its full amount is included in the limited expected value.

Applying the Identity Reliably

A reliable calculation starts by identifying whether the question asks for the limited part of the loss or the insurer’s payment. For a deductible, these are complementary amounts, not interchangeable names for one expected value. The following sequence keeps the distinction visible.

State the payment rule.
For an ordinary deductible $d$, the payment per loss is $(X-d)^+$. Confirm that there is no additional limit or modification in the stated contract.

Choose a calculation route.
Use $E[\min(X,d)]$ if the limited amount is convenient, or integrate the survival function above $d$ if the tail is convenient. For a discrete loss, sum over its possible values.

Apply the identity and check.
Subtract the limited expected value from $E[X]$, or compare the result with the direct payment calculation. Verify that the limited expected value plus expected payment equals the loss mean.

The identity requires an ordinary deductible applied to a nonnegative loss. It does not by itself account for a policy limit, a franchise deductible, or a proportional payment modification; those features change the payment rule and must be included before calculating an expectation. If $E[X]$ is infinite, the finite-mean subtraction form is not available, though the tail-integral expressions may still be considered as nonnegative extended expectations.

For exam work, the decomposition is more than a formula to memorize. It supplies both a derivation and a diagnostic: if a calculated limited expected value exceeds the mean, or if the proposed expected payment is negative, at least one step is inconsistent with the nonnegative-loss model. Likewise, a payment larger than $E[X]$ cannot be correct under an ordinary deductible, because $(X-d)^+\leq X$ for every $X\geq0$.

Key takeaway: For a nonnegative loss with finite mean, the ordinary-deductible payment and the limited amount partition the loss exactly. Calculate whichever expectation is easier, then use $E[(X-d)^+]=E[X]-E[\min(X,d)]$ and check that the two parts add back to $E[X]$.

Check Your Understanding

For each hypothetical loss model, calculate the requested quantities. Assume gross losses, no interest adjustment, no mortality contingency, and no policy limit.

$X$ is uniform on $[0,\$8{,}000]$, and the ordinary deductible is $\$2{,}000$. Find the limited expected value and expected payment, and check their sum. (Answer: limited expected value $\$1{,}750$; expected payment $\$2{,}250$; sum $\$4{,}000$.)
$X$ is exponential with mean $\$2{,}000$, and the ordinary deductible is $\$1{,}000$. Find both expectations, rounded to the nearest cent. (Answer: limited expected value $\$786.94$; expected payment $\$1{,}213.06$.)
$X$ takes values $\$0,\$500,\$1{,}500$ with probabilities $0.25,0.50,0.25$, respectively, and $d=\$1{,}000$. Calculate $E[X]$, the limited expected value, and the expected payment. (Answer: $E[X]=\$625$; limited expected value $\$500$; expected payment $\$125$.)
$X$ is uniform on $[0,\$10{,}000]$, and $d=\$4{,}000$. Calculate the limited expected value from the survival integral and then find the expected payment. (Answer: limited expected value $\$3{,}200$; expected payment $\$1{,}800$.)
$X$ is exponential with rate $1/1500$ per dollar and $d=\$750$. Find the expected payment and limited expected value, rounded to the nearest cent. (Answer: expected payment $\$909.80$; limited expected value $\$590.20$.)

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