Why Condition on a Positive Payment?
A deductible can turn many reported losses into zero insurer payments. As a result, the average payment across all claims is not the same as the average payment among claims for which the insurer actually pays. These quantities answer different questions: the first averages over every loss, including losses below the deductible; the second describes payment size once a payment has occurred.
Let \(X\) be a nonnegative loss and \(d\geq 0\) an ordinary deductible. The insurer payment per loss is \(Y^L=(X-d)_+\), where \(z_+=\max(z,0)\). A positive payment occurs precisely when \(X>d\). The strict inequality matters: a loss exactly equal to the deductible produces no payment.
Deriving the Conditional Distribution
For \(y\geq 0\), the event \(0<Y^L\leq y\) is exactly \(d<X\leq d+y\). Conditioning this event on \(Y^L>0\), or \(X>d\), gives the conditional cumulative distribution function.
The subtraction of \(F_X(d)\) removes all losses at or below the deductible from the numerator. In particular, an atom at \(X=d\) does not become a positive-payment atom; it produces a zero payment and is removed by the conditioning. Any atom at \(X=d+y\), for \(y>0\), is included in the conditional CDF at that value of \(y\).
The corresponding conditional survival function has a compact form. For \(y\geq 0\), \(Y^L>y\) is equivalent to \(X>d+y\). Since \(Y^L>0\) is already implied by \(Y^L>y\), the conditional probability is a ratio of tail probabilities.
If \(X\) has a density \(f_X\) and no relevant atoms, the conditional payment density for \(y>0\) is \(f_X(d+y)/S_X(d)\). For a mixed or discrete severity, the CDF and survival formulas remain valid, but the conditional payment distribution can retain atoms at positive payment amounts. A density formula alone would not describe those atoms.
Conditional Expected Payment
For any nonnegative random variable, its expectation is the integral of its survival function. Applying this tail-integral identity to the conditional payment distribution gives the conditional mean. The same result follows by dividing the unconditional mean payment by the probability of a positive payment.
To see why the numerator is the tail integral starting at \(d\), write \(E[(X-d)_+]=\int_0^\infty P((X-d)_+>y)\,dy\). The event inside the probability is \(X>d+y\), so changing variables to \(x=d+y\) gives \(\int_d^\infty S_X(x)\,dx\). Also, \(Y^L=0\) when there is no payment, so \(E[Y^L]=E[Y^L\mid Y^L>0]P(Y^L>0)\). Rearranging yields the ratio.
This conditional mean is sometimes called the mean excess above the deductible. It is not generally equal to \(E[X]-d\): conditioning changes which losses are averaged. Nor is it the unconditional expected payment \(E[Y^L]\), which averages in the zero payments.
Calculate \(P(Y^L>0)=P(X>d)=1-F_X(d)\), paying attention to any atom at \(d\).
For a payment threshold \(y\geq 0\), use \(P(d<X\leq d+y)/P(X>d)\), or use the equivalent tail ratio.
Divide the unconditional expected payment by the probability of a positive payment, or integrate the conditional survival function.
Worked Example: Exponential Severity With a Deductible
A hypothetical insurer models a claim loss \(X\) as exponential with mean \(\theta=1{,}000\) dollars, so \(S_X(x)=e^{-x/1000}\) for \(x\geq0\). The ordinary deductible is \(d=500\) dollars. Find the probability that the payment is at most 300 dollars given that it is positive, and find the mean positive payment. No interest or mortality assumption is involved; the specified severity distribution is exponential with rate \(1/1000\) per dollar.
Method and positive-payment probability. A payment is positive when \(X>500\), so \(P(Y^L>0)=S_X(500)=e^{-0.5}=0.6065307\).
Conditional probability. A positive payment at most 300 dollars means \(500<X\leq800\). Using the conditional CDF,
Conditional mean. The exponential tail integral gives \(E[(X-500)_+]=\int_{500}^{\infty}e^{-x/1000}\,dx=1000e^{-0.5}=606.5307\) dollars. Therefore,
The distribution formula also gives \(P(Y^L>y\mid Y^L>0)=e^{-y/1000}\), confirming that the conditional payment is exponential with the same mean. The displayed probability is rounded to four decimals and the payment to the nearest cent. Check: multiplying the conditional mean by the positive-payment probability returns \(1{,}000e^{-0.5}\), the unconditional expected payment. The method avoids confusing that unconditional amount with the mean conditional payment.
Worked Example: A Discrete Severity With an Atom at the Deductible
For a hypothetical product, a loss \(X\) has the following distribution: \(P(X=0)=0.10\), \(P(X=400)=0.20\), \(P(X=600)=0.30\), and \(P(X=1{,}000)=0.40\), with amounts in dollars. The ordinary deductible is \(d=400\). Find the conditional payment CDF and the mean payment given a positive payment. No interest or mortality assumption is involved; this is a discrete severity model with the stated four-point support.
Method and conditioning event. The loss of 400 dollars is exactly the deductible and produces zero payment. Thus \(P(Y^L>0)=P(X>400)=0.30+0.40=0.70\). Conditional on positive payment, the 600-dollar loss produces a 200-dollar payment, and the 1,000-dollar loss produces a 600-dollar payment. Their conditional probabilities are \(0.30/0.70=3/7\) and \(0.40/0.70=4/7\), respectively.
The conditional CDF is therefore zero for \(y<200\), equals \(3/7\) for \(200\leq y<600\), and equals 1 for \(y\geq600\). At \(y=200\), the CDF includes the atom at the 200-dollar payment.
Conditional mean. Average the two positive payment amounts using their conditional probabilities:
Check and trap avoided. The unconditional expected payment is \(200(0.30)+600(0.40)=300\) dollars; dividing by \(0.70\) gives \(428.57\) dollars, as above. Reporting 300 dollars as the answer would average in the zero payments and answer a different question. The atom at the deductible must also be excluded from the denominator.
Worked Example: Uniform Severity and a Conditional Payment Probability
Suppose a hypothetical loss \(X\) is uniform on \([0,1{,}200]\) dollars, and the ordinary deductible is \(d=300\) dollars. Find the conditional probability that the insurer payment is at most 250 dollars and the mean payment given a positive payment. No interest or mortality assumption is involved; the severity is uniform with density \(1/1{,}200\) on the stated interval.
Method and conditional distribution. The probability of a positive payment is \(P(X>300)=900/1{,}200=0.75\). For \(0\leq y\leq900\),
Consequently, \(P(Y^L\leq250\mid Y^L>0)=250/900=0.2778\). The conditional payment is uniform from 0 to 900 dollars, so its mean is \(900/2=450\) dollars.
Check using the unconditional payment. The expected payment is the area under the payment function over the severity interval, multiplied by the uniform density:
Then \(337.50/0.75=450\) dollars, matching the conditional-distribution calculation. Both the probability and payment are rounded to four decimals and the nearest cent, respectively. The method avoids treating the conditional support as \([300,1{,}200]\): after subtracting the deductible, the payment support is \([0,900]\).
Common Mistakes and Practical Interpretation
- Using the unconditional mean payment. \(E[Y^L]\) includes the zero-payment claims. For the mean among positive payments, divide \(E[Y^L]\) by \(P(X>d)\).
- Including losses equal to the deductible. Under an ordinary deductible, \(X=d\) gives \(Y^L=0\). Use \(P(X>d)\), not \(P(X\geq d)\), when an atom at \(d\) is possible.
- Forgetting to translate the threshold. A payment threshold \(y\) corresponds to a loss threshold \(d+y\), not \(y\). Use \(P(d<X\leq d+y)\) for the conditional CDF.
- Assuming the conditional payment has a density. If the original severity has atoms above the deductible, those atoms remain as positive payment atoms, shifted down by \(d\).
- Conditioning when no positive payment is possible. If \(P(X>d)=0\), the conditional distribution and conditional mean given a positive payment are undefined.
The conditional distribution is useful whenever the question concerns payment size after a claim penetrates the deductible, rather than cost averaged over every loss. The key is to define the conditioning event first, then shift the loss threshold by the deductible and normalize by the probability of a positive payment.
Check Your Understanding
For each problem, find the requested quantities conditional on a positive payment. Treat each deductible as ordinary, so a loss equal to the deductible produces zero payment.
- A hypothetical severity is exponential with mean 800 dollars and deductible 200 dollars. Find the conditional mean payment and \(P(Y^L>400\mid Y^L>0)\). (Answer: mean \(800.00\) dollars; probability \(e^{-0.5}=0.6065\).)
- A discrete loss is 100, 300, 500, or 900 dollars with respective probabilities 0.20, 0.30, 0.30, and 0.20. With deductible 300 dollars, find the conditional mean payment and \(P(Y^L\leq250\mid Y^L>0)\). (Answer: mean \(360.00\) dollars; probability \(0.6000\).)
- A loss is uniform on \([0,1{,}500]\) dollars, with deductible 600 dollars. Find the conditional mean payment and \(P(Y^L\leq300\mid Y^L>0)\). (Answer: mean \(450.00\) dollars; probability \(0.3333\).)
- A hypothetical severity has survival function \(S_X(x)=(1+x/1{,}000)^{-2}\) for \(x\geq0\), and the deductible is 1,000 dollars. Find \(P(Y^L>1{,}000\mid Y^L>0)\) and the conditional mean payment. (Answer: probability \(4/9=0.4444\); mean \(2{,}000.00\) dollars.)