Expected Payments By Integrating Loss Layers

Why Expected Layer Payments Require Careful Splitting

An insurance payment often depends on which part of a loss falls within the policy’s covered range. A deductible sets the attachment point: losses below it produce no payment. A limit sets an exhaustion point: the insurer does not pay for loss above the covered layer. Coinsurance can further reduce the insurer’s share. To find an expected payment, the calculation must reflect all of these thresholds—not just the probability that a loss exceeds the deductible.

The earlier material on policy limits and coinsurance represents payments as piecewise functions of the loss. Here, the objective is to take the expectation of such a payment efficiently. The key result is that the expected value of a loss layer can be found by integrating the loss survival function across that layer. This works for continuous losses and, with the same formula, for losses with atoms.

Definition: For a nonnegative loss $X$, a layer with attachment $a$ and width $L$, where $a\geq 0$ and $L>0$, pays $Y=\min((X-a)_+,L)$, where $(z)_+=\max(z,0)$. Its payment is zero if $X\leq a$, increases with $X$ between $a$ and $a+L$, and is capped at $L$ if $X\geq a+L$. If the insurer pays a constant proportion $c$ of the layer, with $0\leq c\leq 1$, its payment is $cY$.

The Expected-Payment Formula

Write $S_X(x)=P(X>x)$ for the survival function of $X$. For a layer attached at $a$ and exhausted at $a+L$, its expected payment is

E[\min((X-a)_+,L)]=\int_a^{a+L} S_X(x)\,dx.

If $X$ has a continuous distribution with density $f_X$, the same expectation can be calculated directly from the payment function:

E[\min((X-a)_+,L)] =\int_a^{a+L}(x-a)f_X(x)\,dx +L\int_{a+L}^{\infty}f_X(x)\,dx.

The first integral accounts for losses inside the layer, where the payment is $x-a$. The second accounts for losses that reach or exceed exhaustion, where the payment is fixed at $L$. When there is an atom at a threshold, the direct density formula must be supplemented with the corresponding probability mass; the survival-integral formula already handles atoms without modification.

Theorem: If a layer has coinsurance proportion $c$, its expected insurer payment is $c\int_a^{a+L}S_X(x)\,dx$. For several layers on the same loss, calculate the expected payment of each layer over its own attachment-to-exhaustion interval, multiply by that layer’s coinsurance proportion, and add the results.

Why the Survival Integral Works

For any fixed realized loss $X$, the layer payment is the length of the portion of the interval from $a$ to $a+L$ that lies below $X$. Equivalently, it is the integral over that interval of an indicator that is one when the loss exceeds the integration point:

\min((X-a)_+,L) =\int_a^{a+L}\mathbf{1}_{\{X>x\}}\,dx.

For example, if $X\leq a$, the indicator is zero throughout the interval. If $a<X<a+L$, it is one from $a$ up to $X$, giving an integral of $X-a$. If $X\geq a+L$, it is one across the full interval, giving $L$. Taking expectations and interchanging expectation and integration for this nonnegative integrand gives

E[\min((X-a)_+,L)] =\int_a^{a+L} E[\mathbf{1}_{\{X>x\}}]\,dx =\int_a^{a+L}P(X>x)\,dx.

The equality also holds if $X$ has a point mass at a threshold: changing the indicator at a single integration point does not change the integral’s value. For a layer with coinsurance $c$, multiply the realized payment and its expectation by $c$. For multiple layers, add their payments and use linearity of expectation; independence between the layers is not required when they are all functions of the same loss.

Worked Examples

Worked Example: A Uniform Loss With a Deductible and Limit

A hypothetical policy covers a loss $X$, measured in thousands of dollars, with $X\sim\operatorname{Uniform}(0,100)$. The deductible is 20 (thousand dollars), and the policy limit is 50 (thousand dollars) above the deductible. The insurer pays 80% of the covered layer. Find the expected insurer payment. This is a one-period loss model; no interest rate or mortality assumption is involved.

Method. The attachment is $a=20$, the layer width is $L=50$, and exhaustion occurs at $a+L=70$. The coinsurance proportion is $c=0.8$. For this uniform distribution, the density is $f_X(x)=1/100$ on $0\leq x\leq100$, and its survival function on that interval is $S_X(x)=1-x/100$.

Apply the survival-integral formula across the covered interval:

\begin{aligned} E[Y] &=0.8\int_{20}^{70}\left(1-\frac{x}{100}\right)\,dx\\ &=0.8\left[x-\frac{x^2}{200}\right]_{20}^{70}\\ &=0.8\left[\left(70-\frac{4900}{200}\right) -\left(20-\frac{400}{200}\right)\right]\\ &=0.8[(70-24.5)-(20-2)]\\ &=0.8(45.5-18)\\ &=0.8(27.5)=22 \end{aligned}

The result is 22 thousand dollars, or $22,000.00. As a check, the expected payment before coinsurance is 27.5 thousand dollars, which is less than the maximum layer payment of 50 thousand dollars. Applying 80% gives 22 thousand dollars. A common mistake is to multiply the full layer width by the probability of exceeding the deductible: $50P(X>20)=50(0.8)=40$ thousand dollars, before coinsurance. That incorrectly assumes every loss above the deductible exhausts the layer; the survival integral correctly accounts for partial payments between 20 and 70.

Worked Example: An Exponential Loss With a Capped Payment

A hypothetical one-period loss $X$, measured in thousands of dollars, has an exponential distribution with mean 50 thousand dollars. Thus its rate is $\lambda=1/50=0.02$ per thousand dollars, and $S_X(x)=e^{-0.02x}$ for $x\geq0$. A policy has a deductible of 10 thousand dollars and a limit of 20 thousand dollars above the deductible. The insurer pays 75% of this layer. Find its expected payment. No interest rate or mortality assumption is involved.

Method. The layer attaches at $a=10$ and is exhausted at $a+L=30$, with all amounts in thousands of dollars. The survival-integral formula, including the 75% coinsurance, gives

\begin{aligned} E[Y] &=0.75\int_{10}^{30}e^{-0.02x}\,dx\\ &=0.75\left[-\frac{1}{0.02}e^{-0.02x}\right]_{10}^{30}\\ &=0.75(50)(e^{-0.2}-e^{-0.6}). \end{aligned}

Using $e^{-0.2}=0.8187308$ and $e^{-0.6}=0.5488116$, the difference is $0.2699192$. Therefore,

$$ E[Y]=37.5(0.2699192)=10.12197 $$

The expected insurer payment is 10.12197 thousand dollars, which is $10,121.97 to the nearest cent. As a check, it is positive and below the maximum insurer payment of $0.75(20)=15$ thousand dollars. A frequent error is to integrate to infinity because the exponential distribution has an unbounded support. The policy’s payment stops increasing at 30 thousand dollars of loss, so the layer integral ends at 30; losses beyond that point are already represented by survival at the exhaustion threshold.

Worked Example: Adding Expected Payments From Two Coinsured Layers

A hypothetical loss $X$, measured in thousands of dollars, has density $f_X(x)=2x/10{,}000$ for $0\leq x\leq100$, and density zero elsewhere. The policy pays 50% of the layer from 20 to 60 thousand dollars, then 30% of the layer from 60 to 90 thousand dollars. Find the expected total insurer payment. This one-period model has no interest or mortality assumption.

Method. First check that the specified density integrates to one:

\int_0^{100}\frac{2x}{10{,}000}\,dx =\left[\frac{x^2}{10{,}000}\right]_0^{100}=1.

The survival function on the support is $S_X(x)=1-x^2/10{,}000$. Indeed, integrating the density from $x$ to 100 gives $P(X>x)=1-x^2/10{,}000$. The expected payment is the sum of the two coinsured layer integrals:

\begin{aligned} E[Y] &=0.5\int_{20}^{60}\left(1-\frac{x^2}{10{,}000}\right)dx +0.3\int_{60}^{90}\left(1-\frac{x^2}{10{,}000}\right)dx\\ &=0.5\left[x-\frac{x^3}{30{,}000}\right]_{20}^{60} +0.3\left[x-\frac{x^3}{30{,}000}\right]_{60}^{90}. \end{aligned}

For the first layer, the integral before coinsurance is

\left(60-\frac{216{,}000}{30{,}000}\right) -\left(20-\frac{8{,}000}{30{,}000}\right) =52.8-19.7333333=33.0666667.

Its expected insurer payment is $0.5(33.0666667)=16.5333333$ thousand dollars. For the second layer, the integral before coinsurance is

\left(90-\frac{729{,}000}{30{,}000}\right) -\left(60-\frac{216{,}000}{30{,}000}\right) =65.7-52.8=12.9.

Its expected insurer payment is $0.3(12.9)=3.87$ thousand dollars. Thus the total expected payment is $16.5333333+3.87=20.4033333$ thousand dollars, or $20,403.33 to the nearest cent. As a check, a loss of at least 90 thousand dollars produces a maximum total payment of $0.5(40)+0.3(30)=29$ thousand dollars, so the calculated expectation is below the maximum. The common mistake to avoid is applying one coinsurance percentage to the entire interval: each layer has its own percentage and must be integrated separately.

Practical Checks and Common Traps

A reliable calculation starts by translating the contract into attachment and exhaustion points. For a deductible $d$ and a limit $u$ that caps the insurer’s payment above the deductible, the covered interval is from $d$ to $d+u$, not from $d$ to $u$. With constant coinsurance $c$, the expected payment is

E[Y^L]=c\int_d^{d+u}S_X(x)\,dx.

Here $Y^L$ denotes payment per loss. The formula assumes the payment is a constant proportion of the covered layer. If the coinsurance proportion changes across layers, split the integral at every point where it changes, as in the third example. If payments depend on additional random quantities, such as a separate random coinsurance rate, those quantities require their own treatment rather than being replaced automatically by a fixed $c$.

Do not confuse a payment limit with a loss threshold. A limit $u$ above a deductible $d$ gives exhaustion at $d+u$.
Do not multiply a layer width by the deductible-exceedance probability. Some losses exceed the deductible but do not exhaust the layer, so their payments are smaller than the full width.
Use consistent units. If losses and thresholds are in thousands of dollars, the integral is also in thousands of dollars. Convert to dollars only after computing the expectation.
Respect each layer’s coinsurance. When the percentage changes, compute each interval’s expected payment separately before adding.
Check the range. A nonnegative payment bounded above by a maximum cannot have an expectation outside those bounds.

These methods are aligned with the SOA Exam P (Probability) syllabus because they combine a piecewise insurance payment with distribution-based expectation. In an exam problem, identify the payment bands first, then select either direct integration against $f_X$ or integration of $S_X$ across each covered layer. The survival method is especially useful when the survival function has a simpler form than the density-based payment integral.

Check Your Understanding

For each problem, identify the layer thresholds before calculating. Amounts described as being in thousands of dollars should remain in those units until the final answer.

A hypothetical loss is uniform on 0 to 80 thousand dollars. A policy pays 100% above a 10-thousand-dollar deductible, up to a limit of 30 thousand dollars above the deductible. Find the expected payment. (Answer: $20,625.00)
A hypothetical exponential loss has mean 25 thousand dollars. A policy has a 5-thousand-dollar deductible and a 20-thousand-dollar limit above it, with 100% coinsurance. Find the expected payment to the nearest cent. (Answer: $11,271.28)
A hypothetical loss has density $f_X(x)=2x/14{,}400$ for $0\leq x\leq120$, with $x$ in thousands of dollars, and density zero otherwise. A policy pays 25% of the layer from 20 to 70 thousand dollars. Find the expected payment to the nearest cent. (Answer: $10,561.34)
A hypothetical exponential loss has mean 40 thousand dollars. A policy pays 100% of the layer from 10 to 30 thousand dollars and 50% of the layer from 30 to 60 thousand dollars. Find the total expected payment to the nearest cent. (Answer: $17,242.10)
A hypothetical loss takes values 0, 10, 40, and 100 thousand dollars with respective probabilities 0.2, 0.3, 0.3, and 0.2. A policy has a 10-thousand-dollar deductible and a 50-thousand-dollar limit above it. Find the expected payment. (Answer: $19,000.00)

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