Risk Modelling and Survival Analysis · Loss distributions, with and without risk sharing

Deductibles, Policy Limits and Excess of Loss Payments

Updated 11 October 2026 · Fact-checked

A deductible d turns a loss X into a payment of max(X − d, 0). A policy limit caps the payment. To find the mean, integrate the survival function of X between d and the upper limit. For variance, find E[Y²] the same way and subtract the squared mean. State clearly whether you are working per loss or per payment.

Understand Deductibles, Policy Limits and Excess of Loss

A loss X is the amount of damage. The payment is what the insurer actually pays. A deductible and a limit change the payment, so its distribution differs from that of X.

With an ordinary (excess) deductible d, the insurer pays only the part of the loss above d. The payment per loss is Y = (X − d)+ = max(X − d, 0). Small losses give a payment of zero, so Y has a probability mass at 0 equal to P(X ≤ d). Think of it as an excess of loss layer that starts at d.

With a policy limit, the insurer pays at most a fixed amount. If the cover stops at loss u, the payment is min(X, u). If the deductible is d and cover stops at loss u, the payment is min(X, u) − min(X, d). The maximum payment is then u − d. Read the question carefully: a limit may be stated as the maximum payment or as the maximum loss covered.

Two viewpoints matter. The per-loss variable counts every loss, including those with zero payment. The per-payment variable only counts losses where a payment is made (X > d). It is the per-loss variable conditioned on Y > 0. Mean per payment = mean per loss ÷ P(X > d).

A franchise deductible is different. If X ≤ d, nothing is paid. If X > d, the full loss X is paid, with no subtraction. So the franchise payment is larger than the ordinary one by exactly d whenever X > d.

Key rules to remember

Ordinary deductible, per loss: Y = (X − d)+ = max(X − d, 0); Payment is zero when X ≤ d. Its distribution has a mass at 0 of F(d).
Limited loss: min(X, u) = X if X ≤ u; u if X > u; Use for a policy limit with no deductible.
Payment with deductible d and loss cover up to u: Y = min(X, u) − min(X, d), for d < u; Maximum payment is u − d.
Expected payment per loss (deductible): E[(X − d)+] = ∫ from d to ∞ of S(x) dx; S(x) = 1 − F(x) is the survival function. Needs the mean of X to be finite.
Expected limited loss: E[min(X, u)] = ∫ from 0 to u of S(x) dx; Also E[X] = E[min(X, d)] + E[(X − d)+].
Expected payment with deductible and limit: E[min(X, u) − min(X, d)] = ∫ from d to u of S(x) dx; Area under the survival curve between d and u.
Second moment of limited loss: E[min(X, u)²] = ∫ from 0 to u of 2x S(x) dx; Use with the mean to get variance: Var = E[Y²] − (E[Y])².
Per payment versus per loss: E[Y per payment] = E[(X − d)+] ÷ S(d), where S(d) = P(X > d); Same relation holds for E[Y²]. Mean excess loss e(d) = E[X − d | X > d].
Exponential with rate λ: E[(X − d)+] = e^(−λd) ÷ λ; E[min(X, u)] = (1 − e^(−λu)) ÷ λ; By the lack of memory, X − d given X > d is again exponential with rate λ.
Pareto with parameters α, λ: S(x) = (λ ÷ (λ + x))^α: X − d given X > d is Pareto(α, λ + d); mean per payment = (λ + d) ÷ (α − 1); Needs α > 1 for the mean and α > 2 for the variance. Variance of Pareto(α, λ) = αλ² ÷ ((α − 1)²(α − 2)).
Franchise deductible: Y = 0 if X ≤ d; Y = X if X > d. E[Y] = E[(X − d)+] + d × S(d); Pays the full loss once the loss exceeds d.

How to solve Deductibles, Policy Limits and Excess of Loss questions

Use this order for any question that applies a deductible, a limit or both to a loss distribution.

1Write the payment as a function of X. Decide if it is an ordinary deductible, a franchise deductible, a limit on payment or a limit on loss, or a mix.
2Decide if the question asks per loss or per payment. Per payment means conditioning on X > d.
3Write the survival function S(x) of X and note the parameter conditions, such as α > 1 or α > 2 for the Pareto.
4For the mean, integrate S(x) over the range where the payment grows: d to ∞ for a deductible only, 0 to u for a limit only, d to u for both.
5For the second moment, use E[Y²] = ∫ 2y × P(Y > y) dy, or use the shifted distribution for exponential and Pareto. Then Var = E[Y²] − (E[Y])².
6If the question wants the distribution, give the mass at 0 (F(d)), the density or survival function for 0 < y < maximum, and any mass at the maximum payment.
7Check: the per-loss mean must be less than E[X], and per-payment mean = per-loss mean ÷ S(d). Write the answer in rupees with units.

Quickest way: Use the shift and the survival-area shortcut

When to use it: When X is exponential or Pareto and the question asks for mean or variance of a payment after a deductible.

Exponential: the per-payment variable is the same exponential. Mean per payment = 1/λ and variance = 1/λ². Multiply by S(d) = e^(−λd) for the per-loss mean.
Pareto(α, λ): the per-payment variable is Pareto(α, λ + d). Use the standard Pareto mean and variance with λ + d in place of λ.
For the per-loss second moment, multiply the per-payment second moment by S(d).
With a limit, take the area under S(x) from d to u. For the exponential this is (e^(−λd) − e^(−λu)) ÷ λ.
Check the limit case: if u = ∞ the result must reduce to the deductible-only result.

Common mistakes in Deductibles, Policy Limits and Excess of Loss

Using E[X] − d as the expected payment.
It looks natural to subtract d from every loss, but small losses give zero, not a negative payment.
Fix: Use E[(X − d)+] = ∫ from d to ∞ of S(x) dx. It is larger than E[X] − d whenever P(X < d) > 0.
Mixing up per-loss and per-payment means.
Both are called the expected claim payment in everyday language.
Fix: Read the wording. Per loss includes zero payments. Per payment divides by S(d). Write the label next to each answer.
Taking the maximum payment as u when the limit is a loss limit with a deductible.
The same word limit is used for both ideas.
Fix: If cover stops at loss u, the maximum payment is u − d. If the maximum payment is m, the loss limit is d + m.
Using Pareto(α, λ − d) or leaving λ unchanged after a deductible.
Students confuse the Pareto shift with the lack of memory of the exponential.
Fix: For X > d, the excess X − d is Pareto(α, λ + d). Only the exponential keeps its parameter unchanged.
Confusing a franchise deductible with an ordinary one.
Both pay nothing for small losses, so the difference is easy to miss.
Fix: Ordinary pays X − d above d. Franchise pays the full X above d. Franchise mean = ordinary mean + d × S(d).
Applying the mean formula when the mean does not exist, such as a Pareto with α ≤ 1.
The formula (λ + d) ÷ (α − 1) gives a number or a negative value that looks plausible.
Fix: Check α > 1 for the mean and α > 2 for the variance before you use the formulas.

Worked examples

Example 1

Losses X are exponential with mean ₹50,000. An insurer pays the loss in excess of a deductible of ₹10,000, up to a maximum payment of ₹40,000 per loss. Find the probability that a loss produces a payment, the probability that the maximum payment is made, and the expected payment per loss.

Show the solution

Here λ = 1/50,000 = 0.00002 and S(x) = e^(−x/50,000). The loss limit is u = 10,000 + 40,000 = ₹50,000.
P(payment > 0) = P(X > 10,000) = e^(−0.2) = 0.8187.
P(maximum payment) = P(X ≥ 50,000) = e^(−1) = 0.3679.
Expected payment per loss = ∫ from 10,000 to 50,000 of S(x) dx = 50,000 × (e^(−0.2) − e^(−1)).
e^(−0.2) = 0.818731 and e^(−1) = 0.367879, so the difference is 0.450852.
Expected payment = 50,000 × 0.450852 = ₹22,542.6, about ₹22,543.

Answer: P(payment) = 0.8187; P(maximum payment) = 0.3679; expected payment per loss ≈ ₹22,543.

Example 2

Losses X follow a Pareto distribution with α = 3 and λ = ₹20,000, so S(x) = (20,000 ÷ (20,000 + x))³. A policy has an ordinary deductible of ₹10,000 and no limit. Find the mean and variance of the payment per payment and per loss.

Show the solution

P(X > 10,000) = S(10,000) = (20,000 ÷ 30,000)³ = (2/3)³ = 8/27 = 0.2963.
Per payment: X − 10,000 given X > 10,000 is Pareto(3, 30,000).
Mean per payment = 30,000 ÷ (3 − 1) = ₹15,000.
Variance per payment = αλ² ÷ ((α − 1)²(α − 2)) = 3 × 900,000,000 ÷ (4 × 1) = 675,000,000. The standard deviation is about ₹25,981.
Second moment per payment = 675,000,000 + 15,000² = 675,000,000 + 225,000,000 = 900,000,000. Check: 2λ² ÷ ((α − 1)(α − 2)) = 2 × 900,000,000 ÷ 2 = 900,000,000.
Per loss mean = 15,000 × 8/27 = ₹4,444.44.
Per loss second moment = 900,000,000 × 8/27 = 266,666,667.
Per loss variance = 266,666,667 − 4,444.44² = 266,666,667 − 19,753,086 = 246,913,581.

Answer: Per payment: mean ₹15,000, variance 675,000,000 (₹²). Per loss: mean ₹4,444.44, variance about 246,913,581 (₹²).

Exam tips

Write the payment variable Y in terms of X as your first line. Examiners award marks for a correct definition of Y, and it prevents per-loss and per-payment errors.
For exponential and Pareto questions, use the shift results instead of integrating. State the result you are using, then quote the standard mean and variance.
If a question asks for the distribution of the payment, give the probability mass at 0 and, if there is a limit, the mass at the maximum payment. Many students omit these.
State the conditions for existence of moments (α > 1, α > 2) when you use Pareto formulas.
In computer-based paper work, check the result by simulation or numerical integration of S(x), and report the formula and the answer together.

Practice questions from Loss distributions, with and without risk sharing

Deductibles, Policy Limits and Excess of Loss: frequently asked questions

How do I calculate the expected claim payment with a deductible?

Find E[(X − d)+], which equals the integral of the survival function S(x) from d to infinity. This is the expected payment per loss. To get the expected payment per payment, divide by P(X > d).

What is the difference between a franchise deductible and an ordinary deductible?

An ordinary deductible subtracts d from every loss above d, so the payment is X − d. A franchise deductible pays the whole loss X once X exceeds d, and nothing otherwise. The franchise mean is the ordinary mean plus d × P(X > d).

What happens to an exponential loss after a deductible?

The excess X − d, given X > d, is again exponential with the same rate λ, because of the lack of memory property. The per-payment mean stays 1/λ. The per-loss mean is e^(−λd) ÷ λ.

What happens to a Pareto loss after a deductible?

The excess X − d, given X > d, is Pareto with the same α and with λ replaced by λ + d. Its mean per payment is (λ + d) ÷ (α − 1) for α > 1. The mean excess loss rises with d.

Is a policy limit the same as excess of loss?

No. A policy limit caps what the insurer pays. Excess of loss is a layer that begins at a retention and covers losses above it, often up to a limit. A deductible with a limit on the same policy forms exactly such a layer.