Ordinary Deductible¶

Definition¶

For a nonnegative ground-up loss $X$ (the full loss before any deductible or limit is applied) and deductible $d\ge0$ $0$ , the insurer's per-loss payment is $Y^L=(X-d)_+=\max(X-d,0)$ $d, 0)$ . The insured retains $\min(X,d)$ . The per-payment distribution is the conditional law of $X-d$ given $X>d$ (written $Y^P$ ), defined only when $\mathbb{P}(X>d)>0$ $0$ .

Interpretation¶

Every loss is included in the per-loss model, with a zero payment below the deductible. The per-payment model describes only cases generating a positive payment.

Exponential loss density with deductible d = 100: the area left of d, probability 0.632, is shaded blue as zero payment and the visible part of the full right-tail probability 0.368 red as payment.

For $X\sim\operatorname{Exp}(0.01)$ and $d=100$ $100$ , the blue area is $\mathbb{P}(Y^L=0)=F_X(d)=1-\mathrm{e}^{-1}\approx0.632$ $0.632$ : these losses stay in the per-loss model with payment zero. The full right-tail probability $\bar F_X(d)=\mathrm{e}^{-1}\approx0.368$ holds the losses that produce a payment $X-d$ $d$ ; the shading shows only losses up to 410, with probability $\mathrm{e}^{-1}-\mathrm{e}^{-4.1}\approx0.35131$ $0.35131$ , and omits probability $\mathrm{e}^{-4.1}\approx0.01657$ $0.01657$ . Rescaling the full tail by $1/\bar F_X(d)$ gives the per-payment law $Y^P$ , which here is again exponential with mean 100. Hence $\mathbb{E}[Y^L]=100\mathrm{e}^{-1}\approx36.79$ $36.79$ , as in the example.

Move the deductible $d$ and the rate $\lambda$ : the blue area $\mathbb P(Y^L=0)=F_X(d)$ grows with $d$ , while the per-payment mean $\mathbb E[Y^P]=1/\lambda$ does not move, because the exponential law is memoryless. The default $d=100$ $100$ , $\lambda=0.01$ gives $\mathbb E[Y^L]=36.79$ $36.79$ .

The survivor function $\bar F_X(d)=\mathbb{P}(X>d)$ gives the probability that the deductible is exceeded. Dividing the shifted tail by this probability normalizes the per-payment law over cases receiving a positive payment.

Key Results¶

$\mathbb{P}(Y^L=0)=F_X(d)$ and $F_{Y^L}(y)=F_X(d+y)$ for $y\ge0$ $0$ .
For $y\ge0$ $0$ , $\mathbb{P}(Y^P>y)=\bar F_X(d+y)/\bar F_X(d)$ $y) / \overset{ˉ}{F}_{X} (d)$ .
$\mathbb{E}[Y^L]=\bar F_X(d)\mathbb{E}[Y^P]$ when finite.
For continuous $X$ , $f_{Y^P}(y)=f_X(d+y)/\bar F_X(d)$ for $y>0$ $0$ .

Examples¶

For $X\sim\operatorname{Exp}(0.01)$ and $d=100$ $100$ , the probability of payment is $\mathrm{e}^{-1}$ . By the memoryless property, conditional payment has mean 100, so mean payment per loss is $100\mathrm{e}^{-1}\approx36.79$ $36.79$ .

Prerequisites¶

Sources¶

Risk Insurance · Lecture 1

Aliases: excess deductible, per-loss deductible payment, straight deductible Keywords: (X - d)+, per-loss variable Y^L, per-payment variable Y^P, ground-up loss, memoryless exponential deductible Symbols: lambda

Last updated: 27 October 2026 11:52