Variance of Deductible Payments

Why Deductible-Payment Variance Needs Care

An ordinary deductible reduces the insurer’s payment on each loss, but it also creates a probability of no payment at all. That zero-payment probability affects the payment’s variance. A calculation based only on the distribution of positive payments can therefore give the wrong unconditional moments unless it accounts for the chance that a loss does not exceed the deductible.

This tutorial derives the variance using the payment’s survival function. The method works for continuous and discrete nonnegative loss amounts and makes the role of the zero-payment probability explicit.

Definition: Let $X$ be a nonnegative loss amount and let $d\geq 0$ be an ordinary deductible. The insurer’s payment per loss is $Y^L=(X-d)^+=\max(X-d,0)$. Define the loss survival function by $S_X(x)=P(X>x)$. Then $P(Y^L=0)=P(X\leq d)=1-S_X(d)$, and for $t\geq 0$, $P(Y^L>t)=S_X(d+t)$.

The identity for $P(Y^L>t)$ follows directly from the payment rule: a payment exceeds $t$ exactly when the loss exceeds $d+t$. In particular, the probability of a positive payment is $P(Y^L>0)=S_X(d)$. These statements use the strict-survival convention $S_X(x)=P(X>x)$, so a loss exactly equal to the deductible produces no payment.

Tail Integrals for the First and Second Moments

For any nonnegative random variable $Y$ with a finite $r$th moment, the tail-integral identity is $E[Y^r]=r\int_0^\infty t^{r-1}P(Y>t)\,dt$, for $r>0$. It does not require $Y$ to have a density. It therefore applies whether the payment distribution is continuous, discrete, or a mixture of the two.

Theorem: If $X$ is nonnegative and the relevant moments are finite, the ordinary-deductible payment has $$ E[Y^L]=\int_0^\infty S_X(d+t)\,dt=\int_d^\infty S_X(x)\,dx $$ and $$ E[(Y^L)^2]=2\int_0^\infty tS_X(d+t)\,dt =2\int_d^\infty (x-d)S_X(x)\,dx. $$ Consequently, $$ \operatorname{Var}(Y^L)=2\int_d^\infty (x-d)S_X(x)\,dx- \left(\int_d^\infty S_X(x)\,dx\right)^2. $$

The second form of the second-moment integral follows by setting $x=d+t$. Variance is then the unconditional second moment minus the square of the unconditional mean. Both moments include the probability of zero payment; no separate point-mass adjustment is needed when the unconditional survival function is used.

Why the Second-Moment Formula Holds

For each fixed nonnegative payment value $y$, $y^2=\int_0^y 2t\,dt$. Equivalently, $y^2=\int_0^\infty 2t\,\mathbf{1}_{\{y>t\}}\,dt$, where the indicator is one when $y>t$ and zero otherwise. Taking expectations and interchanging the nonnegative integral and expectation gives the tail formula.

E[(Y^L)^2] =E\left[\int_0^\infty 2t\,\mathbf{1}_{\{Y^L>t\}}\,dt\right] =\int_0^\infty 2tP(Y^L>t)\,dt =2\int_0^\infty tS_X(d+t)\,dt.

The interchange is valid for a nonnegative integrand; if the second moment is infinite, the identity still holds with an infinite value. The variance formula is useful when the moments are finite. For practical calculations, integrate only over the range where $S_X(d+t)$ is positive.

Worked Examples

Worked Example: Exponential Loss with a Deductible

A hypothetical equipment policy has an ordinary deductible of $\$500$. Suppose the loss amount $X$ is exponential with rate $\lambda=1/1000$ per dollar, so $S_X(x)=e^{-x/1000}$ for $x\geq0$. Find the probability of no payment, the mean payment, the second moment, and the variance.

Method and zero-payment probability. The payment is $Y^L=(X-500)^+$. The probability of a positive payment is $P(Y^L>0)=S_X(500)=e^{-0.5}\approx0.6065$. Thus $P(Y^L=0)=1-e^{-0.5}\approx0.3935$, with probabilities rounded to four decimal places.

Mean. Substitute $t$ into the deductible survival function: $P(Y^L>t)=e^{-(500+t)/1000}=e^{-0.5}e^{-t/1000}$. Therefore $E[Y^L]=e^{-0.5}\int_0^\infty e^{-t/1000}\,dt =1000e^{-0.5}\approx \$606.53$.

Second moment. Using the second-moment tail integral, $E[(Y^L)^2]=2e^{-0.5}\int_0^\infty t e^{-t/1000}\,dt$. For an exponential integral, $\int_0^\infty t e^{-t/1000}\,dt=1000^2=1{,}000{,}000$. Hence $E[(Y^L)^2]=2{,}000{,}000e^{-0.5}\approx \$1{,}213{,}061.32$.

Variance and check. The squared mean is $(1000e^{-0.5})^2=1{,}000{,}000e^{-1}\approx \$367{,}879.44$ squared. Thus $\operatorname{Var}(Y^L)=2{,}000{,}000e^{-0.5}-1{,}000{,}000e^{-1} \approx \$845{,}181.88$ squared. To the nearest cent, the variance is $\$845{,}181.88$ squared. It is positive, as required. The method uses unconditional moments; it avoids treating every loss as though it generated a payment.

Worked Example: Uniform Loss and a Large Zero-Payment Mass

Suppose a hypothetical property loss is uniformly distributed on $[0,\$3{,}000]$, and the ordinary deductible is $\$1{,}000$. Find the mean, second moment, and variance of the payment. Give money amounts to the nearest cent.

Payment survival and zero mass. For $0\leq t<\$2{,}000$, $P(Y^L>t)=P(X>1000+t)=(2000-t)/3000$; for $t\geq \$2{,}000$, it is zero. The probability of a positive payment is $P(X>1000)=2/3$, so the zero-payment probability is $1/3$.

Mean. The tail integral gives $E[Y^L]=\int_0^{2000}(2000-t)/3000\,dt$. Evaluating, $\int_0^{2000}(2000-t)\,dt=[2000t-t^2/2]_0^{2000} =4{,}000{,}000-2{,}000{,}000=2{,}000{,}000$. Therefore $E[Y^L]=2{,}000{,}000/3000=\$666.67$ to the nearest cent; exactly, it is $\$2{,}000/3$.

Second moment. $$ E[(Y^L)^2]=\frac{2}{3000}\int_0^{2000}t(2000-t)\,dt =\frac{2}{3000}\left[1000t^2-\frac{t^3}{3}\right]_0^{2000}. $$ At the upper endpoint the bracket is $4{,}000{,}000{,}000-\frac{8{,}000{,}000{,}000}{3} =\frac{4{,}000{,}000{,}000}{3}$. Thus the second moment is $\frac{8{,}000{,}000{,}000}{9{,}000}=\frac{8{,}000{,}000}{9} \approx \$888{,}888.89$ squared.

Variance and check. The exact squared mean is $4{,}000{,}000/9$. Subtracting, $\operatorname{Var}(Y^L)=8{,}000{,}000/9-4{,}000{,}000/9 =4{,}000{,}000/9\approx \$444{,}444.44$ squared. The positive payment is at most $\$2{,}000$, and the mean is below that bound, consistent with the result.

Worked Example: Discrete Losses and a Stepwise Survival Function

For a hypothetical cargo policy, the loss $X$ is $\$0,\$400,\$1{,}000,$ or $\$1{,}800$, with respective probabilities $0.20,0.30,0.30,$ and $0.20$. The ordinary deductible is $\$600$. Calculate the payment’s mean, second moment, and variance.

Payment values and zero probability. Applying $Y^L=(X-600)^+$ gives payments $\$0,\$0,\$400,$ and $\$1{,}200$ with probabilities $0.20,0.30,0.30,$ and $0.20$. Thus $P(Y^L=0)=0.50$. The unconditional survival function is $0.50$ for $0\leq t<400$, $0.20$ for $400\leq t<1200$, and zero for $t\geq1200$.

Mean from the survival function. $E[Y^L]=\int_0^{400}0.50\,dt+\int_{400}^{1200}0.20\,dt =200+160=\$360$.

Second moment from the survival function. $$ E[(Y^L)^2] =2\left(\int_0^{400}0.50t\,dt+\int_{400}^{1200}0.20t\,dt\right). $$ The first integral inside parentheses is $0.50(400^2/2)=40{,}000$. The second is $0.20(1200^2-400^2)/2=128{,}000$. Hence the second moment is $2(40{,}000+128{,}000)=\$336{,}000$ squared.

Variance and check. The squared mean is $360^2=\$129{,}600$ squared, so $\operatorname{Var}(Y^L)=336{,}000-129{,}600=\$206{,}400$ squared. As a direct check, the payment values give $0.30(400^2)+0.20(1200^2)=48{,}000+288{,}000=336{,}000$, matching the survival-integral result. The stepwise method correctly includes the zero-payment probability.

Conditioning on a Positive Payment: A Common Trap

If $p=S_X(d)>0$, the payment conditional on being positive has survival function $P(Y^L>t\mid Y^L>0)=S_X(d+t)/p$. The unconditional survival function is $p$ times this conditional survival function. Therefore, for any positive moment order $r$, $E[(Y^L)^r]=pE[(Y^L)^r\mid Y^L>0]$. Omitting the factor $p$ calculates a conditional moment, not the required unconditional moment.

Common exam trap: In the exponential example, the positive payment conditional on exceeding the deductible is exponential with mean $\$1{,}000$, so its conditional second moment is $\$2{,}000{,}000$ squared. Using those conditional values directly would incorrectly report $E[Y^L]=\$1{,}000$ and $E[(Y^L)^2]=\$2{,}000{,}000$ squared. The correction is to multiply each by $p=e^{-0.5}$: the unconditional mean is $1000e^{-0.5}$, and the unconditional second moment is $2{,}000{,}000e^{-0.5}$. Then subtract the square of the unconditional mean to obtain variance.

A second error is to subtract the square of the conditional mean from the unconditional second moment. That mixes two different distributions and does not give either conditional or unconditional variance. Keep the calculation consistent: either calculate both unconditional moments and use $\operatorname{Var}(Y^L)=E[(Y^L)^2]-E[Y^L]^2$, or calculate conditional moments and then account for the zero-payment mass when obtaining unconditional moments.

When the Method Applies

The survival-integral formulas apply to a nonnegative loss with an ordinary deductible and no other payment modification. The tail-integral identity itself does not depend on a density, so it accommodates atoms in the loss distribution, as the discrete example demonstrates. For finite variance, the first and second moments in the variance calculation must be finite.

The probability mass at zero should not be confused with a positive payment at the deductible amount: a loss equal to $d$ produces payment zero. If a contract also caps or otherwise changes the payment, its payment survival function must reflect those terms; the formulas must be applied to that actual payment random variable rather than assuming the ordinary-deductible survival function still holds.

Key takeaway: For $Y^L=(X-d)^+$, use $P(Y^L>t)=S_X(d+t)$. Integrating this unconditional survival function gives the moments with the zero-payment probability already included; variance is the second moment minus the square of the mean.

Check Your Understanding

For each problem, calculate the requested unconditional payment moments. Round monetary answers to the nearest cent where appropriate.

A hypothetical loss is exponential with mean $\$500$, and the ordinary deductible is $\$500$. Find the zero-payment probability, mean payment, second moment, and variance. (Answer: $P(Y^L=0)=0.6321$; $E[Y^L]=\$183.94$; $E[(Y^L)^2]=\$183{,}939.72$ squared; $\operatorname{Var}(Y^L)=\$150{,}105.90$ squared.)
A loss is uniform on $[0,\$5{,}000]$ with an ordinary deductible of $\$2{,}000$. Use the survival integral to find the mean payment, second moment, and variance. (Answer: $E[Y^L]=\$900$; $E[(Y^L)^2]=\$1{,}800{,}000$ squared; $\operatorname{Var}(Y^L)=\$990{,}000$ squared.)
A hypothetical loss takes values $\$0,\$500,\$900,\$1{,}500$ with probabilities $0.10,0.20,0.40,0.30$, respectively. With a $\$700$ ordinary deductible, find the zero-payment probability and payment variance. (Answer: $P(Y^L=0)=0.30$; $\operatorname{Var}(Y^L)=\$96{,}000$ squared.)
A loss is a mixture: with probability $0.5$ it is exponential with mean $\$1{,}000$, and with probability $0.5$ it is exponential with mean $\$2{,}000$. Apply a $\$1{,}000$ ordinary deductible. Find the unconditional mean and variance of the payment. (Answer: $E[Y^L]=500e^{-1}+1000e^{-0.5}\approx \$790.47$; $\operatorname{Var}(Y^L)=500{,}000e^{-1}+4{,}000{,}000e^{-0.5}-(500e^{-1}+1000e^{-0.5})^2\approx \$2{,}169{,}158.66$ squared.)

What You'll Learn