CDFs After an Ordinary Deductible

From Ground-Up Loss to Payment CDF

An ordinary deductible changes the amount paid on each loss: the insured bears the first $d$ dollars, and the insurer pays the excess above $d$. To find the distribution of the insurer’s payment, translate each payment threshold back to the corresponding ground-up loss threshold. The boundary at zero matters: a payment of zero occurs whenever the loss does not exceed the deductible, including when it equals the deductible.

Assume throughout that $X$ is a nonnegative ground-up loss with cumulative distribution function $F_X(x)=P(X\le x)$, and that the deductible $d$ is nonnegative. The insurer’s payment per loss is $Y=(X-d)^+=\max(X-d,0)$. No continuity or density assumption is needed for the CDF formula.

Definition: The cumulative distribution function of a random variable $Y$ is $F_Y(y)=P(Y\le y)$. For the ordinary-deductible payment $Y=(X-d)^+$, where $X\ge 0$ and $d\ge 0$, $$ F_Y(y)= \begin{cases} 0, & y<0,\\ F_X(d+y), & y\ge 0. \end{cases} $$ In particular, $F_Y(0)=F_X(d)=P(X\le d)$. The inequality is inclusive: a loss exactly equal to $d$ results in zero payment.

The formula applies whether the ground-up loss distribution is discrete, continuous, or a mixture. A CDF is defined using a “less than or equal to” event, so the formula also handles jumps in $F_X$. At positive $y$, the relevant ground-up threshold is $d+y$, not $y$ alone.

Why the Formula Holds

For negative $y$, $Y$ cannot be negative, so $P(Y\le y)=0$. For $y\ge 0$, split the event according to whether the loss is at or below the deductible. If $X\le d$, then $Y=0\le y$. If $X>d$, then $Y=X-d$, and $Y\le y$ is equivalent to $X\le d+y$. Together these cases give exactly the event $X\le d+y$.

\begin{aligned} \{Y\le y\} &=\{X\le d\}\ \cup\ \{d<X\le d+y\}\\ &=\{X\le d+y\}, \qquad y\ge 0,\\ F_Y(y) &=P(X\le d+y)=F_X(d+y). \end{aligned}

At $y=0$, the second event in the union is empty, leaving $\{X\le d\}$. Thus the CDF’s value at zero follows from the same event argument; it is not an extra rule. If the ground-up distribution has an atom at $d$, that loss is included in $F_X(d)$, because it produces zero payment.

Useful consequence: For $y\ge 0$, $P(Y>y)=P(X>d+y)=1-F_X(d+y)$. If the survival function is defined by $S_X(x)=1-F_X(x)$, then $P(Y>y)=S_X(d+y)$. The strict inequality in this tail probability is important when the ground-up distribution has an atom at $d+y$.

Reading the CDF Across Its Support

The CDF of $Y$ is flat at zero for every negative argument, then begins at $F_X(d)$ when $y=0$. For nonnegative arguments it follows the ground-up CDF shifted horizontally by $d$. This description is a translation of thresholds, not a change to the probability scale.

Payment threshold	Equivalent ground-up event	Probability
$y<0$	$Y\le y$ is impossible	$F_Y(y)=0$
$y=0$	$X\le d$	$F_Y(0)=F_X(d)$
$y>0$	$X\le d+y$	$F_Y(y)=F_X(d+y)$
$y\ge 0$, upper tail	$X>d+y$	$P(Y>y)=S_X(d+y)$

For a continuous ground-up distribution, one may sometimes express the CDF using an integral of its density. But the CDF identity above is more general and is the safer starting point: it remains valid when the loss distribution has discrete outcomes or a mixture of discrete and continuous parts.

Worked Example: A Discrete Ground-Up Loss Model

A hypothetical repair policy has deductible $d=\$1{,}000$. The ground-up loss $X$ takes values $\$0,\$500,\$1{,}000,\$1{,}800,\$2{,}500$, with respective probabilities $0.10,0.20,0.30,0.25,0.15$. Find $F_Y(0)$, $F_Y(800)$, and $P(Y>800)$.

Method. First accumulate the ground-up probabilities up to each required threshold. The probabilities sum to $0.10+0.20+0.30+0.25+0.15=1.00$, as required.

At zero payment, the equivalent ground-up event is $X\le d=\$1{,}000$. Therefore, $F_Y(0)=F_X(1{,}000)=0.10+0.20+0.30=0.60$. The loss equal to the deductible is included.

For the payment threshold $\$800$, use the ground-up threshold $d+800=\$1{,}800$: $F_Y(800)=F_X(1{,}800)=0.10+0.20+0.30+0.25=0.85$.

The upper-tail probability is the complement: $P(Y>800)=1-F_Y(800)=1-0.85=0.15$. This also follows directly from $P(X>1{,}800)$, which is the probability of the $\$2{,}500$ loss, $0.15$.

Check. $F_Y(800)+P(Y>800)=0.85+0.15=1$. The common mistake avoided here is using $F_X(800)$ rather than $F_X(1{,}800)$ for the payment CDF.

Continuous Model Applications

For a continuous severity model, the same threshold shift applies. The model supplies $F_X$ or $S_X$; the deductible formula determines which ground-up value to evaluate. State the model parameters and support before substituting, because a formula outside its stated support may not apply.

Worked Example: Pareto Type I Payment CDF

Suppose a hypothetical ground-up loss follows a Pareto Type I distribution with scale $\theta=\$1{,}000$ and shape $\alpha=2$, so that $S_X(x)=(\theta/(\theta+x))^\alpha$ for $x\ge0$. The policy has ordinary deductible $d=\$300$. Find $F_Y(0)$, $F_Y(500)$, and the probability $0<Y\le500$. Give probabilities to four decimals.

Step 1: Write the ground-up CDF. From $F_X(x)=1-S_X(x)$, $F_X(x)=1-(1000/(1000+x))^2$ for $x\ge0$.

Step 2: Evaluate at zero. The loss threshold is the deductible, $\$300$: $F_Y(0)=F_X(300)=1-(1000/1300)^2=1-(10/13)^2=1-100/169=69/169=0.4083$.

Step 3: Evaluate at payment $\$500$. The ground-up threshold is $\$300+\$500=\$800$: $F_Y(500)=F_X(800)=1-(1000/1800)^2=1-(5/9)^2=1-25/81=56/81=0.6914$.

Step 4: Find the probability of a positive payment no greater than $\$500$. This event is $0<Y\le500$, so subtract the CDF at zero from the CDF at $\$500$: $P(0<Y\le500)=F_Y(500)-F_Y(0)=56/81-69/169=(9464-5589)/13689=3875/13689=0.2831$.

Check. The result is positive and less than $F_Y(500)=0.6914$. The common mistake avoided is subtracting $F_X(300)$ from $F_X(500)$: the correct upper ground-up threshold is $\$800$, not $\$500$.

Worked Example: Exponential Severity and a Tail Probability

A hypothetical loss amount $X$ is exponentially distributed with rate $\lambda=0.001$ per dollar, so $S_X(x)=e^{-0.001x}$ for $x\ge0$. An ordinary deductible is $d=\$400$. Find $F_Y(0)$, $F_Y(600)$, and $P(Y>600)$. Give probabilities to four decimals.

Step 1: Find the zero-payment CDF value. Since $F_X(x)=1-e^{-0.001x}$, $F_Y(0)=F_X(400)=1-e^{-0.4}=1-0.670320=0.329680$, or $0.3297$.

Step 2: Shift the positive payment threshold. A payment of at most $\$600$ corresponds to a loss of at most $\$400+\$600=\$1{,}000$: $F_Y(600)=F_X(1{,}000)=1-e^{-1}=1-0.367879=0.632121$, or $0.6321$.

Step 3: Compute the tail. $P(Y>600)=S_X(1{,}000)=e^{-1}=0.367879$, or $0.3679$.

Check. $0.6321+0.3679=1.0000$ after rounding. The common mistake avoided is using $e^{-0.6}$ for the tail, which would omit the $\$400$ ground-up loss amount absorbed by the deductible.

Common CDF Traps

Forgetting the deductible shift. For $y\ge0$, evaluate $F_X$ at $d+y$, not at $y$.
Using a strict inequality at the deductible. $F_Y(0)=P(X\le d)$. Losses exactly equal to $d$ produce zero payment and belong in this event.
Giving a nonzero CDF below zero. Because $Y\ge0$, $F_Y(y)=0$ for $y<0$.
Confusing a CDF with a positive-payment probability. $F_Y(y)$ includes all payments from zero through $y$. For $y\ge0$, $P(0<Y\le y)=F_X(d+y)-F_X(d)$.
Taking a complement with the wrong boundary. $P(Y>y)=P(X>d+y)=1-F_X(d+y)$; when there is an atom at $d+y$, that atom is not in the strict upper tail.

These translations matter whenever a problem asks for probabilities of insurer payments rather than ground-up losses. A useful routine is to write the payment event first, convert it to an event involving $X$, and only then substitute into the ground-up CDF or survival function. This makes both the deductible shift and the boundary convention explicit.

Key takeaway: For an ordinary deductible, the payment CDF is zero below zero and equals the ground-up CDF shifted by the deductible at every nonnegative threshold: $F_Y(y)=F_X(d+y)$. In particular, the CDF begins at $F_X(d)$, not at zero, because all losses at or below the deductible produce zero payment.

Check Your Understanding

For each problem, translate the payment threshold to a ground-up threshold before calculating.

A hypothetical loss is $\$0,\$400,\$900,\$1{,}500,\$2{,}400$ with respective probabilities $0.10,0.20,0.25,0.30,0.15$. The deductible is $\$900$. Find $F_Y(0)$, $F_Y(600)$, and $P(0<Y\le600)$. (Answer: $0.55,\ 0.85,\ 0.30$)
$X$ is uniform on $[0,2000]$ dollars and $d=\$500$. Find $F_Y(0)$, $F_Y(300)$, and $P(Y>300)$. (Answer: $0.2500,\ 0.4000,\ 0.6000$)
$X$ is exponential with rate $0.002$ per dollar and $d=\$250$. Find $F_Y(0)$, $F_Y(250)$, and $P(0<Y\le250)$, to four decimals. (Answer: $0.3935,\ 0.6321,\ 0.2387$)
$X$ is Pareto Type I with $\theta=\$1{,}000$, $\alpha=2$, and $d=\$300$. Find $F_Y(0)$, $F_Y(500)$, and $P(0<Y\le500)$, to four decimals. (Answer: $0.4083,\ 0.6914,\ 0.2831$)
A nonnegative ground-up loss has $F_X(400)=0.10$, $F_X(1000)=0.45$, and $F_X(1600)=0.80$. With deductible $d=\$400$, find $F_Y(0)$, $F_Y(600)$, and $P(Y>600)$. (Answer: $0.10,\ 0.80,\ 0.20$)

Payment threshold	Equivalent ground-up event	Probability
\(y<0\)	\(Y\le y\) is impossible	\(F_Y(y)=0\)
\(y=0\)	\(X\le d\)	\(F_Y(0)=F_X(d)\)
\(y>0\)	\(X\le d+y\)	\(F_Y(y)=F_X(d+y)\)
\(y\ge 0\), upper tail	\(X>d+y\)	\(P(Y>y)=S_X(d+y)\)

What You'll Learn