Worked Case: Deductibles, Policy Limits, and Quota Share

Introduction

A portfolio calculation can go wrong even when each individual formula is familiar. The deductible, the policy limit, the frequency model, the quota-share percentage, and the premium basis must all refer to the same covered payment. This tutorial works through those links in order: define the insurer’s payment per loss, find its probability and moments, and then apply quota share to premium and aggregate loss.

Throughout, $X$ is the gross loss on one claim. The deductible $d$ is the amount the insured bears first, and $u$ is the insurer’s maximum payment per claim. Thus $u$ is a payment limit above the deductible, not a maximum gross loss. All examples are hypothetical.

Definition: For $X\geq 0$, $d\geq 0$, and $u>0$, the insurer’s payment per loss is $$ Y^L=\min\bigl((X-d)_+,u\bigr),\qquad (z)_+=\max(z,0). $$ It is zero when $X\leq d$, equals $X-d$ when $d<X<d+u$, and equals $u$ when $X\geq d+u$. The probability of a positive payment is $P(X>d)=S_X(d)$. If a contract instead states a maximum gross loss $L$ subject to a deductible, the maximum payment is $L-d$, not $L$.

Payment Probabilities and Moments

The payment distribution has a probability mass at zero whenever some losses do not exceed the deductible. If the severity distribution is continuous, it also has a mass at $u$: all losses at or above $d+u$ produce the same capped payment. In particular,

P(Y^L=0)=P(X\leq d),\qquad P(Y^L=u)=P(X\geq d+u).

For a severity distribution with atoms, retain the weak inequality in the cap probability: a loss exactly equal to $d+u$ pays $u$. For a continuous severity, the probability of equality is zero, so the cap probability can also be written $S_X(d+u)$.

A useful way to obtain moments is to use the survival integral for a nonnegative random variable. For $0\leq y<u$, $Y^L>y$ exactly when $X>d+y$. This gives the following expressions, valid for continuous or discrete severity:

E[Y^L]=\int_0^u P(X>d+y)\,dy =\int_0^u S_X(d+y)\,dy, $$ $$ E[(Y^L)^2]=2\int_0^u y\,P(X>d+y)\,dy =2\int_0^u yS_X(d+y)\,dy, \qquad \operatorname{Var}(Y^L)=E[(Y^L)^2]-\{E[Y^L]\}^2.

Derivation: For any nonnegative payment $Y$, write $Y=\int_0^\infty \mathbf{1}_{\{Y>y\}}\,dy$. Taking expectations gives $E[Y]=\int_0^\infty P(Y>y)\,dy$. For the capped payment, $P(Y^L>y)=P(X>d+y)$ for $0\leq y<u$, and it is zero for $y\geq u$. This proves the first moment formula. Similarly, $Y^2=\int_0^Y 2y\,dy$; taking expectations gives $E[Y^2]=2\int_0^\infty yP(Y>y)\,dy$, which proves the second moment formula after substituting the same survival probability.

The moments above are per loss: a loss below the deductible contributes a zero. If a severity is required per positive payment, denote it by $Y^P$, with the conditional distribution of $Y^L$ given $Y^L>0$. Provided $P(X>d)>0$,

E[Y^P]=\frac{E[Y^L]}{P(X>d)},\qquad E[(Y^P)^2]=\frac{E[(Y^L)^2]}{P(X>d)}.

These conditional moments must not be substituted for per-loss moments when the claim count includes every exposure or every underlying loss opportunity. The zero-payment probability is already reflected in $E[Y^L]$.

Worked Example: Exponential Severity with a Deductible and Limit

A hypothetical policy has gross claim severity $X$ exponentially distributed with mean $\theta=1000$ dollars, so $S_X(x)=e^{-x/1000}$ for $x\geq0$. The deductible is $d=200$ dollars and the maximum insurer payment is $u=800$ dollars. Find the positive-payment and capped-payment probabilities, the mean payment per loss, and its variance.

Method and probabilities. The model is continuous, and the payment is $Y^L=\min((X-200)_+,800)$. Therefore

P(Y^L>0)=P(X>200)=e^{-0.2}=0.8187, \qquad P(Y^L=800)=P(X\geq1000)=e^{-1}=0.3679.

The probability of zero payment is $1-0.8187308=0.1812692$. The probability of a positive payment below the cap is $0.8187308-0.3678794=0.4508513$.

Mean. Apply the survival-integral formula and integrate:

E[Y^L]=\int_0^{800}e^{-(200+y)/1000}\,dy =1000\bigl(e^{-0.2}-e^{-1}\bigr) =450.8513\text{ dollars}.

Second moment and variance. Let $a=u/\theta=0.8$. Substituting $z=y/\theta$ in the second-moment integral gives

E[(Y^L)^2] =2\theta^2e^{-d/\theta}\int_0^a ze^{-z}\,dz =2\theta^2e^{-d/\theta}\bigl[1-(1+a)e^{-a}\bigr].

Using $e^{-0.8}=0.4493290$, the bracket is $1-1.8(0.4493290)=0.1912079$. Hence

E[(Y^L)^2] =2(1000)^2(0.8187308)(0.1912079) =313095.6\text{ dollars}^2, $$ $$ \operatorname{Var}(Y^L)=313095.6-(450.8513)^2 =109828.7\text{ dollars}^2.

Check and trap avoided. The cap applies at gross loss $d+u=1000$, not at gross loss $u=800$. Thus the calculated cap probability is $P(X\geq1000)$, and it cannot exceed the positive-payment probability. The reported probabilities satisfy $0.3679<0.8187$, as required. Values are rounded to four decimal places for probabilities and one decimal place for dollar moments.

From Per-Loss Payments to Aggregate Loss

Let $N$ be the number of claims and let $Y^L_1,Y^L_2,\ldots$ be independent payments with the same per-loss distribution, independent of $N$. The aggregate covered loss before reinsurance is $S_N=\sum_{j=1}^{N}Y^L_j$. Its moments depend on the frequency model:

E[S_N]=E[N]E[Y^L],\qquad \operatorname{Var}(S_N)=E[N]\operatorname{Var}(Y^L)+\operatorname{Var}(N)\{E[Y^L]\}^2.

For a Poisson claim count with mean $\lambda$, $\operatorname{Var}(N)=E[N]=\lambda$, so $\operatorname{Var}(S_N)=\lambda E[(Y^L)^2]$. For a fixed count $N=n$, the variance is instead $n\operatorname{Var}(Y^L)$. The frequency assumption must therefore be stated before using an aggregate variance formula.

Quota-share result: If the reinsurer cedes fraction $c$ of each covered payment, the insurer retains fraction $r=1-c$. Per claim, insurer and reinsurer payments are $rY^L$ and $cY^L$. For any aggregate covered loss $S_N$, the retained aggregate loss is $rS_N$, so its mean is $rE[S_N]$ and its variance is $r^2\operatorname{Var}(S_N)$. In a proportional premium allocation with no ceding commission or other adjustment, the reinsurer receives fraction $c$ of the gross premium and the insurer retains fraction $r$.

Worked Example: Discrete Severity, Quota-Share Premium, and Poisson Aggregate

A hypothetical portfolio has gross severity distribution $P(X=100)=0.20$, $P(X=400)=0.30$, $P(X=1000)=0.30$, and $P(X=2000)=0.20$, with amounts in dollars. Each policy has deductible $d=300$ dollars and maximum payment $u=700$ dollars. The annual claim count is Poisson with mean $\lambda=10$. The gross premium is set at 115% of expected annual covered loss, then allocated 40% to the reinsurer under quota share. Find the payment probability and moments, the premium allocation, and the insurer’s retained aggregate moments.

Step 1: Map gross severities to covered payments. Apply the deductible and then cap the resulting payment at 700 dollars:

\begin{array}{c|rrrr} X&100&400&1000&2000\\ Y^L&0&100&700&700\\ P&0.20&0.30&0.30&0.20 \end{array}

A positive payment occurs for the last three severity values, so $P(Y^L>0)=0.30+0.30+0.20=0.80$. A payment at the cap occurs at $X=1000$ or $X=2000$, so $P(Y^L=700)=0.30+0.20=0.50$.

Step 2: Calculate per-loss moments.

E[Y^L]=0.20(0)+0.30(100)+0.30(700)+0.20(700)=380, $$ $$ E[(Y^L)^2]=0.20(0)+0.30(10000)+0.30(490000)+0.20(490000) =248000, $$ $$ \operatorname{Var}(Y^L)=248000-380^2=103600.

Step 3: Calculate the gross premium and its allocation. The expected annual covered loss is $10(380)=3800$ dollars. The gross premium is $1.15(3800)=4370$ dollars. Thus the reinsurer’s allocated premium is $0.40(4370)=1748$ dollars, and the insurer retains $0.60(4370)=2622$ dollars.

Step 4: Calculate retained aggregate loss moments. For a Poisson count, the pre-reinsurance aggregate mean is $10(380)=3800$ dollars and its variance is $10(248000)=2480000$ dollars squared. The insurer retains $r=0.60$, giving

E[S_{\mathrm{ret}}]=0.60(3800)=2280\text{ dollars}, \qquad \operatorname{Var}(S_{\mathrm{ret}})=0.60^2(2480000) =892800\text{ dollars}^2.

Check and trap avoided. The severity probabilities sum to 1, and the payment is never greater than 700 dollars. A tempting but wrong deductible-only calculation is $E[X]-dP(X>d)=840-300(0.80)=600$ dollars. It incorrectly includes the $100$-dollar losses in $E[X]$, although they generate no payment, and it ignores the cap. The table-based result $380$ dollars correctly excludes those losses and applies the limit. Dollar amounts are shown to the cent where relevant; exact whole-dollar results need no rounding.

Fixed Counts and the Choice of Variance

A quota share scales an aggregate payment by a constant, but it does not change the underlying frequency assumption. The next case uses a fixed claim count to make that distinction explicit. It also checks the capped payment distribution directly using a continuous severity.

Worked Example: Uniform Severity with a Fixed Number of Claims

A hypothetical warranty portfolio has severity $X\sim\operatorname{Uniform}(0,2000)$ dollars on each claim. The deductible is $d=400$ dollars, the maximum payment is $u=600$ dollars, and exactly five claims occur. The reinsurer takes a 30% quota share. Find the positive-payment and cap probabilities, the insurer’s per-loss moments, and the retained aggregate moments.

Step 1: Probabilities. A payment is positive when $X>400$, hence $P(Y^L>0)=1600/2000=0.80$. A payment reaches the cap when $X\geq400+600=1000$, hence $P(Y^L=600)=1000/2000=0.50$.

Step 2: Per-loss moments. Between gross severities 400 and 1000, the payment ranges uniformly from zero to 600 with density $1/2000$; above 1000, the payment is 600. Therefore

E[Y^L]=\frac{1}{2000}\int_0^{600}y\,dy+(0.50)(600) =\frac{600^2}{2(2000)}+300=390\text{ dollars}, $$ $$ E[(Y^L)^2]=\frac{1}{2000}\int_0^{600}y^2\,dy+(0.50)(600^2) =\frac{600^3}{3(2000)}+180000=216000\text{ dollars}^2, $$ $$ \operatorname{Var}(Y^L)=216000-390^2=63900\text{ dollars}^2.

Step 3: Apply quota share and fixed frequency. The insurer retains $r=1-0.30=0.70$. With exactly five independent claims, retained aggregate mean and variance are

E[S_{\mathrm{ret}}]=5(0.70)(390)=1365\text{ dollars}, \qquad \operatorname{Var}(S_{\mathrm{ret}})=5(0.70)^2(63900) =156555\text{ dollars}^2.

Check and trap avoided. The aggregate variance uses the per-claim variance because the count is fixed. Using the Poisson shortcut would instead use $5E[(Y^L)^2]$ before scaling, which is not the variance for five fixed claims. The cap probability $0.50$ is no greater than the positive-payment probability $0.80$, and each retained payment is bounded by $0.70(600)=420$ dollars. Dollar results are exact.

Common Basis Errors

The calculations above rely on matching each quantity to its stated basis. Before combining severity, frequency, treaty, and premium figures, check the following:

Limit basis: Confirm whether the stated limit is a maximum payment or a maximum gross loss. Here $u$ is a maximum payment, so the cap begins at gross loss $d+u$.
Per loss versus per payment: $E[Y^L]$ includes zero payments. A conditional per-payment mean divides by $P(X>d)$ and should be used only with a payment count.
Deductible and cap: Apply the deductible to gross loss first, then apply the stated payment limit. Do not subtract the deductible from losses that do not exceed it.
Frequency and variance: A Poisson aggregate variance is $\lambda E[(Y^L)^2]$; a fixed-count aggregate variance is $n\operatorname{Var}(Y^L)$.
Quota-share basis: State whether the quoted percentage is ceded or retained. A ceded fraction $c$ leaves retention $1-c$, and variances scale by the square of the retained fraction.
Premium basis: A proportional allocation of gross premium is not automatically a new pricing calculation. If there are commissions, expenses, or separate reinsurance pricing terms, they must be specified and handled separately.

A reliable order of operations is to define the payment, calculate its probability and moments, apply the stated frequency model, and only then scale aggregate loss and allocated premium by quota share. This keeps the exposure, claim, payment, and premium bases consistent.

Check Your Understanding

For each hypothetical case, use the stated convention that the policy limit is the maximum payment per claim. Answers are given so that the key quantities can be checked.

A severity is uniform on 0 to 1000 dollars. The deductible is 200 dollars and the maximum payment is 300 dollars. Find the probability of a positive payment, the probability of a payment at the cap, and $E[Y^L]$. (Answer: $0.8000$, $0.5000$, and 195 dollars.)
A severity is exponential with mean 500 dollars. The deductible is 100 dollars and the maximum payment is 400 dollars. Find the positive-payment probability, cap probability, and mean payment per loss. (Answer: $0.8187$, $0.3679$, and 225.43 dollars.)
A severity has $P(X=0)=0.20$, $P(X=500)=0.30$, and $P(X=1000)=0.50$. The deductible is 200 dollars and the maximum payment is 500 dollars. Find the positive-payment probability, cap probability, $E[Y^L]$, and $\operatorname{Var}(Y^L)$. (Answer: $0.8000$, $0.5000$, 340 dollars, and 36400 dollars squared.)
Per-loss covered payments have $E[Y^L]=100$ dollars and $E[(Y^L)^2]=5000$ dollars squared. Claim count is Poisson with mean 4, and 25% is ceded by quota share. Find the insurer’s retained aggregate mean and variance. (Answer: 300 dollars and 11250 dollars squared.)
A severity has $P(X=0)=0.25$, $P(X=200)=0.25$, and $P(X=500)=0.50$. The deductible is 100 dollars and the maximum payment is 300 dollars. Claim count is Poisson with mean 8, and the insurer retains 80%. Find the positive-payment probability, $E[Y^L]$, and retained aggregate mean and variance. (Answer: $0.7500$, 175 dollars, 1120 dollars, and 243200 dollars squared.)

What You'll Learn