Tail-Integral Methods for Modified Expected Losses

Why Tail Integrals Help

A coverage modification changes the insurer’s payment without changing the underlying ground-up loss. For a loss $X$, an ordinary deductible removes the first part of the loss, a payment limit caps the insurer’s amount, and coinsurance can scale that amount. The resulting payment may have a complicated distribution, especially when the ground-up loss is discrete or has both continuous and discrete components. Its expectation can still be calculated directly from the survival function of $X$.

The central idea is to add up the probability that the payment exceeds each possible dollar level. For a deductible and payment limit, the relevant probabilities are simply survival probabilities of the ground-up loss at shifted thresholds. This often avoids constructing the full distribution of the insurer’s payment.

Definition: Let $X$ be a nonnegative ground-up loss with finite expectation, and define the survival function by $S_X(x)=P(X>x)$. For deductible $d\geq 0$, the insurer’s payment before any limit is $(X-d)_+=\max(X-d,0)$. With a payment limit $u>0$, the payment is $Y=\min((X-d)_+,u)$. A proportional coinsurance share $c$, where $0\leq c\leq 1$, changes the payment to $cY$.

Here $u$ is the maximum amount the insurer pays on a loss after applying the deductible. Thus, a payment limit $u$ corresponds to a ground-up loss threshold of $d+u$. This convention is important: a limit on the insurer’s payment is not generally the same as a limit on the ground-up loss.

The Tail-Integral Identity

For any nonnegative random variable $Z$, its expectation equals the integral of its survival probabilities. The identity works for continuous, discrete, and mixed distributions, provided the expectation exists. It does not require a probability density.

Theorem: If $Z\geq 0$, then

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

To see why, write each realized value $Z$ as the total length of the interval from zero up to $Z$. The indicator $1_{\{t<Z\}}$ equals one exactly for those levels $t$ that lie below $Z$. Integrating those indicators gives $Z$. Taking expectations and interchanging the order of integration and expectation for a nonnegative integrand gives the identity.

Z=\int_0^\infty 1_{\{t<Z\}}\,dt \quad\Longrightarrow\quad E[Z] =\int_0^\infty E[1_{\{t<Z\}}]\,dt =\int_0^\infty P(Z>t)\,dt.

For the limited deductible payment $Y=\min((X-d)_+,u)$, the event $Y>t$, for $0\leq t<u$, is exactly the event $X>d+t$. For $t\geq u$, the payment cannot exceed $t$. Applying the identity therefore gives the main coverage formula.

Key formula: For a nonnegative loss $X$, deductible $d\geq 0$, and payment limit $u>0$,

E[\min((X-d)_+,u)] =\int_0^u S_X(d+t)\,dt =\int_d^{d+u} S_X(x)\,dx.

The same argument gives useful special cases. With a deductible but no payment limit, the integration continues to infinity. With a payment limit and no deductible, it starts at zero. A proportional coinsurance share multiplies the expectation by $c$, because $E[cY]=cE[Y]$.

E[(X-d)_+]=\int_d^\infty S_X(x)\,dx, \qquad E[\min(X,u)]=\int_0^u S_X(x)\,dx, \qquad E[cY]=cE[Y].

Stop-Loss Differences for Combined Coverage

Define the stop-loss expectation at threshold $a\geq 0$ as $\Pi_X(a)=E[(X-a)_+]$. This is the expected amount of loss above the threshold $a$, before applying a cap. A deductible-limit payment can be expressed as the difference between two such expectations.

\begin{aligned} \Pi_X(d)-\Pi_X(d+u) &=\int_d^\infty S_X(x)\,dx-\int_{d+u}^\infty S_X(x)\,dx\\ &=\int_d^{d+u} S_X(x)\,dx\\ &=E[\min((X-d)_+,u)]. \end{aligned}

The subtraction removes the part of the insurer’s uncapped payment above the payment limit. This relation is useful when stop-loss expectations are available directly, even if evaluating the limited-payment integral from scratch would be inconvenient. The integral formula itself remains valid for discrete or mixed losses; individual point masses do not require a density because isolated points have zero width in the integration variable.

Specify the payment.
For a deductible $d$ and payment limit $u$, write $Y=\min((X-d)_+,u)$. Confirm that $u$ is a cap on the insurer’s payment.

Translate payment levels into loss thresholds.
For $0\leq t<u$, $Y>t$ exactly when $X>d+t$, so $P(Y>t)=S_X(d+t)$.

Integrate over possible payment levels.
Use $E[Y]=\int_0^u S_X(d+t)\,dt$, or substitute $x=d+t$ to integrate from $d$ to $d+u$.

Apply coinsurance, if present.
If the insurer pays share $c$ of the covered amount, multiply the expectation by $c$.

Worked Applications

Worked Example: Exponential Loss With a Deductible and Limit

A hypothetical insurer covers a nonnegative loss $X$ with an exponential distribution of rate $\lambda=0.01$ per dollar, so $S_X(x)=e^{-0.01x}$ for $x\geq 0$. The contract has a deductible of $d=\$100$ and a payment limit of $u=\$100$. Find the expected insurer payment.

Method. Use the tail-integral formula for a deductible and payment limit. The covered payment is $\min((X-100)_+,100)$, so the relevant ground-up loss thresholds run from $100$ to $200$.

E[Y]=\int_{100}^{200}e^{-0.01x}\,dx =\frac{1}{0.01}\left(e^{-1}-e^{-2}\right).

The endpoint exponents are $0.01(100)=1$ and $0.01(200)=2$. Using $e^{-1}=0.3678794$ and $e^{-2}=0.1353353$,

E[Y]=100(0.3678794-0.1353353) =100(0.2325441) =\$23.2544.

Rounded to the nearest cent, the expected payment is $\$23.25$. As a check using stop-loss differences, $\Pi_X(100)=100e^{-1}=\$36.7879$ and $\Pi_X(200)=100e^{-2}=\$13.5335$; their difference is $\$23.2544$. A common mistake is to use the limit as the upper ground-up loss threshold and integrate only to $\$100$. Since the limit caps the insurer’s payment after the deductible, the correct upper threshold is $d+u=\$200$.

Worked Example: Discrete Losses With a Deductible and Limit

For a hypothetical portfolio, the ground-up loss $X$ has the following distribution: $P(X=0)=0.20$, $P(X=100)=0.30$, $P(X=300)=0.30$, and $P(X=600)=0.20$. A policy has deductible $d=\$150$ and payment limit $u=\$250$. Find the expected payment using the survival function, then check it from the possible payments.

Method. The tail-integral range is $150\leq x\leq 400$. Between $150$ and $300$, the losses of $300$ and $600$ exceed the threshold, giving survival probability $0.30+0.20=0.50$. Between $300$ and $400$, only the loss of $600$ exceeds the threshold, giving survival probability $0.20$. Values at the interval endpoints do not affect the integral.

E[Y] =\int_{150}^{300}0.50\,dx+\int_{300}^{400}0.20\,dx =(150)(0.50)+(100)(0.20) =75+20 =\$95.

For a direct check, the payment at $X=0$ is $\$0$, and at $X=100$ it is also $\$0$. At $X=300$, the payment is $\min(300-150,250)=\$150$. At $X=600$, the payment is capped at $\$250$. Therefore,

E[Y]=(0.20)(0)+(0.30)(0)+(0.30)(150)+(0.20)(250) =45+50 =\$95.

The tail-integral calculation and the direct probability-weighted calculation agree. The survival function is particularly useful here because the loss distribution has point masses and no density is needed.

Worked Example: Mixed Loss, Payment Limit, and Coinsurance

A hypothetical loss is zero with probability $0.25$. With probability $0.75$, it has an exponential distribution with rate $0.005$ per dollar. A policy applies a deductible of $\$100$, a payment limit of $\$300$, and 80% coinsurance on the covered amount. Find the expected insurer payment.

Method. For thresholds $x>0$, the loss survival function is $S_X(x)=0.75e^{-0.005x}$: the zero-loss component contributes no survival probability, while the exponential component contributes its conditional survival multiplied by $0.75$. First calculate the expected payment before coinsurance over ground-up thresholds from $\$100$ to $\$400$.

E[Y] =\int_{100}^{400}0.75e^{-0.005x}\,dx =\frac{0.75}{0.005}\left(e^{-0.5}-e^{-2}\right) =150(0.6065307-0.1353353).

The difference in parentheses is $0.4711954$, so the expected payment before coinsurance is $150(0.4711954)=\$70.6793$. Applying the 80% share gives

E[0.8Y]=0.8(70.6793)=\$56.5434.

Rounded to the nearest cent, the expected insurer payment is $\$56.54$. The point mass at zero is included through the factor $0.75$; omitting that factor would incorrectly treat every loss as coming from the exponential component. Coinsurance is applied after calculating the covered payment, not by changing the ground-up loss distribution.

Common Exam Trap: A Tail Difference Is Not a Payment Expectation

For a deductible $d$ and payment limit $u$, the difference $S_X(d)-S_X(d+u)$ is generally a probability, not an expected dollar amount. It describes the probability that the loss exceeds the deductible threshold but does not exceed the threshold at which the payment is capped, subject to endpoint conventions. The expected payment instead integrates the survival function across all covered payment levels.

Worked Example: Distinguishing a Probability From an Expected Payment

Use the discrete loss model from the previous example: losses of $\$0,\$100,\$300,\$600$ have probabilities $0.20,0.30,0.30,0.20$. With deductible $d=\$150$ and payment limit $u=\$250$, a candidate writes

E[Y]\ \text{(incorrect)}=S_X(150)-S_X(400)=0.50-0.20=0.30.

This calculation has produced a probability, not an amount in dollars. In this model, $0.30$ is the probability of the $\$300$ loss, which produces a payment between zero and the cap. The $\$600$ loss produces the capped payment and is not counted in that difference. Neither payment amount has been weighted and summed by this probability calculation.

The correct expectation is the integral of the survival function over the payment layer:

E[Y]=\int_{150}^{300}0.50\,dx+\int_{300}^{400}0.20\,dx =75+20=\$95.

The correction is to check the units and the question being answered: a survival probability is dimensionless, while an expected payment is measured in dollars. Integrating survival probabilities over dollar thresholds produces the dollar expectation.

Practical Checks

Before using a tail-integral result, confirm that the loss is nonnegative, the deductible and limit are stated in the right units, and the limit applies to the insurer’s payment or to the ground-up loss as specified by the contract. For combined deductible-limit coverage with payment limit $u$, the integration interval is $[d,d+u]$. If the contract instead caps the ground-up loss at a specified amount, the upper threshold must be read from that contract rather than assumed to be $d+u$.

Check the limiting cases. Setting $d=0$ recovers limited ground-up payment; letting $u$ increase without bound recovers payment above the deductible.
Check the payment range. A payment with limit $u$ must lie between $0$ and $u$, so its expectation must also lie in that interval.
Check the survival function. If the loss distribution has an atom, use its actual survival probabilities; do not assume a density exists.
Check coinsurance. A share $c$ scales the expected covered payment to $cE[Y]$, provided the share applies proportionally to every covered payment.

Key takeaway: Expected modified losses can be calculated by integrating the ground-up loss survival function over the thresholds that correspond to possible insurer payments. For deductible-limit coverage, this is $\int_d^{d+u}S_X(x)\,dx$, equivalently $\Pi_X(d)-\Pi_X(d+u)$.

Check Your Understanding

For each hypothetical contract, calculate the expected insurer payment. Answers are rounded to the nearest cent where applicable.

A loss is uniform on $[0,500]$ dollars. The policy has a deductible of $\$100$ and no payment limit. Find the expected payment using a tail integral. (Answer: $\$160.00$)
A loss is exponential with rate $0.005$ per dollar. The deductible is $\$100$, and the payment limit is $\$200$. Find the expected payment. (Answer: $\$76.68$)
A loss takes values $\$0,\$100,\$250,\$500$ with respective probabilities $0.10,0.20,0.40,0.30$. The deductible is $\$100$, and the payment limit is $\$200$. Find the expected payment. (Answer: $\$120.00$)
A loss is uniform on $[0,1000]$ dollars. The policy has a $\$250$ deductible, a $\$250$ payment limit, and 80% coinsurance. Find the expected insurer payment. (Answer: $\$125.00$)
A loss is zero with probability $0.40$; with probability $0.60$, it is uniform on $[0,1000]$ dollars. The policy has a $\$200$ deductible, a $\$300$ payment limit, and 50% coinsurance. Find the expected insurer payment. (Answer: $\$58.50$)

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