Expected Loss Components Under Limits and Deductibles

Why Limits and Deductibles Change Expected Loss

A gross loss is not necessarily the amount the insurer pays. A deductible leaves an initial portion of each loss with the insured, while a limit can leave the portion above the coverage layer with the insured as well. Expected losses used for rating or experience analysis therefore depend on the covered layer, not just on the expected gross severity.

The key is to define exactly what “limit” means. In this tutorial, an insurer-payment limit is the maximum indemnity paid per loss. Some policy descriptions instead give a limit on the total loss amount eligible for coverage. With a deductible in place, these conventions imply different layer widths unless the numbers are translated carefully. All calculations below are per loss unless a claim frequency is explicitly included. They concern expected loss only, not premium, expenses, or profit.

Definition: Let $X\geq 0$ be a gross loss before insurance terms. With per-loss deductible $d\geq 0$ and maximum insurer payment $u\geq 0$, the insurer payment is $Y=\min((X-d)_+,u)$, where $(z)_+=\max(z,0)$. The insured-retained loss is $R=X-Y$. The gross loss is $X$; it is not the same as either $Y$ or $R$.

If the policy instead states a maximum covered gross-loss amount $L$, then, when $L>d$, the insurer pays $Y=\min((X-d)_+,L-d)$. The covered layer attaches at $d$ and exhausts at $L$, so its width is $L-d$. If $L\leq d$, there is no positive covered layer under this convention. In contrast, an insurer-payment limit of $u$ gives a layer from $d$ to $d+u$.

Expected Loss in a Covered Layer

Write $S_X(t)=P(X>t)$ for the survival function of gross loss. For a layer attaching at $a$ and having width $w$, the payment is $Y_{a,w}=\min((X-a)_+,w)$. Its expected value can be calculated by accumulating the probability that the loss reaches each point in the layer.

E[Y_{a,w}] = E[\min((X-a)_+,w)] = \int_a^{a+w} S_X(t)\,dt

This formula applies to nonnegative losses and a per-loss layer. It is valid for discrete as well as continuous loss distributions: the integral of the survival function accounts for probability masses as well as density between loss values. For a layer with no upper limit, take $w$ to infinity if the expected excess is finite.

Set the attachment and exhaustion points.
A deductible $d$ is the attachment. If $u$ is the maximum insurer payment, the layer exhausts at $d+u$. If $L$ is a total-loss limit, it exhausts at $L$.

Find the expected insurer payment.
Use $E[Y]=\int_{\text{attachment}}^{\text{exhaustion}}S_X(t)\,dt$, or compute each possible payment and probability for a discrete severity.

Reconcile retained and gross expected losses.
Since $R=X-Y$, $E[R]=E[X]-E[Y]$. Check that expected gross loss equals expected insurer payment plus expected retained loss.

Why the Layer Formula Holds

For a fixed loss value $X=x$, the layer payment is the length of the part of the interval from $a$ to $a+w$ that lies below $x$. In particular, the payment is zero if $x\leq a$, is $x-a$ if $a<x<a+w$, and is $w$ if $x\geq a+w$. These three cases can be written together as

\min((x-a)_+,w) = \int_a^{a+w} \mathbf{1}_{\{x>t\}}\,dt

Taking expectations on both sides replaces the indicator at each point $t$ by its probability, $P(X>t)=S_X(t)$. Interchanging expectation and integration is justified because the integrand is nonnegative. Thus

E[\min((X-a)_+,w)] = \int_a^{a+w} E[\mathbf{1}_{\{X>t\}}]\,dt = \int_a^{a+w} S_X(t)\,dt

For a deductible $d$ and insurer-payment limit $u$, the same result also gives a useful difference of excess-loss expectations:

$$ E[Y] = E[(X-d)_+]-E[(X-d-u)_+] $$

The second term removes the amount of loss excess above the layer’s exhaustion point. This difference is useful when an excess-loss expectation is already known. It does not mean that the insurer pays the full excess above the deductible; the upper layer has been removed by the limit.

Worked Calculations

Worked Example: Uniform Severity With a Deductible and Payment Limit

A hypothetical insurer covers individual losses $X\sim\operatorname{Uniform}(0,1000)$, in dollars. The policy has a per-loss deductible of $d=\$100$ and a maximum insurer payment of $u=\$300$ per loss. Find the expected insurer payment and expected retained loss. There is no discounting, so an interest rate is not applicable; mortality is not part of this loss-severity model.

Method. The insurer’s layer attaches at $\$100$ and exhausts at $100+300=\$400$. For $0\leq t\leq1000$, $S_X(t)=1-t/1000$. Therefore,

E[Y] = \int_{100}^{400}\left(1-\frac{t}{1000}\right)dt = \left[t-\frac{t^2}{2000}\right]_{100}^{400}

At $t=400$, the bracketed expression is $400-400^2/2000=400-80=320$. At $t=100$, it is $100-100^2/2000=100-5=95$. The expected insurer payment is therefore $320-95=\$225$.

The gross expected loss for a uniform distribution on $[0,1000]$ is $E[X]=(0+1000)/2=\$500$. Hence,

E[R]=E[X]-E[Y]=500-225=\$275

Check. The expected deductible portion is $E[\min(X,100)]=100-100^2/2000=\$95$. The expected loss above $\$400$ is $E[(X-400)_+]=(1000-400)^2/(2\cdot1000)=\$180$. Together these retained components give $95+180=\$275$, matching the result. The common mistake is to treat the $\$300$ payment limit as an upper bound on gross loss; it is an upper bound on the insurer’s payment, so the layer ends at $\$400$, not $\$300$.

Worked Example: Discrete Severity and a Per-Loss Payment Cap

A hypothetical equipment policy has gross severity $X$ with this distribution: $P(X=0)=0.20$, $P(X=200)=0.30$, $P(X=600)=0.30$, and $P(X=1200)=0.20$, with amounts in dollars. It has a deductible of $\$200$ and a maximum insurer payment of $\$500$ per loss. Find the expected insurer payment and the expected retained loss. There is no discounting, and mortality is not applicable.

Method. Apply $Y=\min((X-200)_+,500)$ to each possible gross loss. For $X=0$, the payment is $\min(0,500)=0$. For $X=200$, it is also $\min(0,500)=0$. For $X=600$, it is $\min(400,500)=400$. For $X=1200$, it is $\min(1000,500)=500$. Weighting these payments by their probabilities gives

E[Y]=(0.20)(0)+(0.30)(0)+(0.30)(400)+(0.20)(500) =120+100=\$220

The expected gross loss is

E[X]=(0.20)(0)+(0.30)(200)+(0.30)(600)+(0.20)(1200) =60+180+240=\$480

Therefore $E[R]=480-220=\$260$. As a direct check, the retained amounts at the four gross-loss values are $0,200,200,700$. Their expectation is $(0.20)(0)+(0.30)(200)+(0.30)(200)+(0.20)(700)=60+60+140=\$260$.

Check. Expected insurer payment plus expected retention is $220+260=\$480$, equal to expected gross loss. The common mistake is to pay $\$500$ on the $\$600$ loss. After the $\$200$ deductible, that loss leaves only $\$400$ for the insurer; the limit is a maximum, not a fixed payment.

Worked Example: Exponential Severity With a Total-Loss Limit

For a hypothetical product, gross loss has an exponential distribution with mean $\$1{,}000$, so its rate is $0.001$ per dollar and $S_X(t)=e^{-t/1000}$. The deductible is $\$500$, and the policy limit is $\$1{,}500$ on the total loss amount, not on the insurer’s payment. Find expected insurer payment and expected retention. No interest discounting or mortality assumption is involved.

Method. The layer attaches at $\$500$ and exhausts at the total-loss limit of $\$1{,}500$. Its width is $1500-500=\$1{,}000$. Using the survival integral,

E[Y] = \int_{500}^{1500}e^{-t/1000}\,dt = 1000\left(e^{-0.5}-e^{-1.5}\right)

Using $e^{-0.5}=0.6065307$ and $e^{-1.5}=0.2231302$, the expected payment is $1000(0.3834005)=\$383.40$, rounded to the nearest cent.

The exponential mean is $\$1{,}000$, so expected retention is $1000-383.4005=\$616.60$, rounded to the nearest cent. The check is $\$383.40+\$616.60=\$1{,}000.00$, the expected gross loss.

As a further check, the expected amount retained below the deductible is $E[\min(X,500)]=1000-E[(X-500)_+]=1000-1000e^{-0.5}=\$393.47$. The expected loss above the total-loss limit is $E[(X-1500)_+]=1000e^{-1.5}=\$223.13$. Their sum is $\$616.60$. The common mistake is to use a $\$1{,}500$ insurer-payment limit, which would exhaust the layer at $\$2{,}000$; the stated limit instead exhausts the layer at a gross loss of $\$1{,}500$.

Reading the Policy Wording Correctly

Before calculating an expected loss, translate the policy wording into an attachment point and exhaustion point. The following comparison assumes $L>d$, and that the deductible applies per loss.

Policy term	Insurer payment	Layer attachment	Layer exhaustion
Deductible $d$, no upper limit	$(X-d)_+$	$d$	No finite exhaustion point
Deductible $d$, insurer-payment cap $u$	$\min((X-d)_+,u)$	$d$	$d+u$
Deductible $d$, total-loss limit $L$	$\min((X-d)_+,L-d)$	$d$	$L$

The retained loss can also be separated into its below-deductible and above-limit portions. For a deductible $d$ and insurer-payment cap $u$,

X=\min(X,d)+\min((X-d)_+,u)+(X-d-u)_+

The three terms are, respectively, the insured’s loss within the deductible, the insurer’s payment, and the insured’s loss above the layer. Taking expectations gives a reconciliation of gross expected loss into those three components. This decomposition is useful for checking calculations, but it does not make the deductible amount constant: it is $\min(X,d)$ for each loss.

From Per-Loss Components to Expected Aggregate Loss

When a policy has a random number $N$ of claims, let each claim have gross severity $X_j$ and insurer payment $Y_j$, with the same deductible and limit applied to each claim. If claim count is independent of the severities and $E[N]$ is finite, the expected aggregate insurer payment is $E[N]E[Y]$. This follows by conditioning on $N=n$: the expected sum of $n$ payments is $nE[Y]$, and then averaging over $N$.

The per-claim expected payment must be calculated on the covered layer before multiplying by expected claim count. Applying the deductible or limit once to the annual aggregate instead would describe a different contract, such as an aggregate deductible or aggregate limit. Per-loss terms and aggregate terms should not be interchanged.

Key takeaway: Define whether the limit caps insurer payment or total covered loss. Then identify the layer’s attachment and exhaustion points, calculate its expected payment, and reconcile that amount to gross and retained expected loss.

Check Your Understanding

For each problem, identify the layer before calculating. Unless a question says otherwise, amounts are per loss, and no discounting or mortality assumption applies.

A hypothetical severity is uniform from $\$0$ to $\$800$. The deductible is $\$100$, and the insurer-payment limit is $\$200$. Find expected insurer payment and expected retained loss. (Answer: expected payment $\$150$; expected retention $\$250$)
Gross severity is $\$0,\$100,\$400,\$900$, with respective probabilities $0.25,0.25,0.30,0.20$. A deductible of $\$100$ and an insurer-payment cap of $\$500$ apply. Find expected payment and retention. (Answer: expected payment $\$190$; expected retention $\$135$)
Gross severity is exponential with mean $\$2{,}000$. The deductible is $\$500$, and the total-loss limit is $\$2{,}500$. Find expected insurer payment and retention, rounded to the nearest cent. (Answer: expected payment $\$984.59$; expected retention $\$1{,}015.41$)
For $X\sim\operatorname{Uniform}(0,1000)$, a deductible of $\$100$ applies. Compare expected insurer payment under (a) a total-loss limit of $\$500$ and (b) an insurer-payment limit of $\$500$. (Answer: (a) $\$280$; (b) $\$325$)
A hypothetical policy has a Poisson claim count with mean $2.5$ per policy-year. Independent claim severities are exponential with mean $\$1{,}000$. Each claim has a $\$200$ deductible and an $\$800$ insurer-payment cap. Find expected aggregate insurer payment per policy-year, rounded to the nearest cent. (Answer: $\$1{,}127.13$)

Policy term	Insurer payment	Layer attachment	Layer exhaustion
Deductible \(d\), no upper limit	\((X-d)_+\)	\(d\)	No finite exhaustion point
Deductible \(d\), insurer-payment cap \(u\)	\(\min((X-d)_+,u)\)	\(d\)	\(d+u\)
Deductible \(d\), total-loss limit \(L\)	\(\min((X-d)_+,L-d)\)	\(d\)	\(L\)

What You'll Learn