Deriving Payment Moments From Tail Integrals

Why Use the Payment Survival Function?

Coverage modifications can make a payment distribution awkward to describe directly. A deductible creates a probability of no payment; a policy limit can create an atom at the maximum payment; and coinsurance rescales positive payments. In each case, however, the survival function of the payment can often be written as a simple probability about the original loss. Integrating that survival function gives the payment moments without first deriving a full probability distribution.

Let $X$ be a nonnegative loss and $Y^L$ the insurer's payment per loss under a specified coverage. The formulas below concern the unconditional payment, including the possibility that $Y^L=0$. They do not require a continuous severity distribution. Unless stated otherwise, examples use hypothetical losses, undiscounted payments, and no mortality assumption; interest is not involved.

Definition: For a nonnegative random variable $Y$, its survival function is $S_Y(t)=P(Y>t)$, for $t\geq 0$. If the relevant moments are finite, its first and second moments are $$ E[Y]=\int_0^\infty S_Y(t)\,dt,\qquad E[Y^2]=2\int_0^\infty tS_Y(t)\,dt. $$ Consequently, $\operatorname{Var}(Y)=E[Y^2]-(E[Y])^2$. These formulas include any point masses in the distribution.

Why the Tail-Integral Formulas Hold

For each fixed nonnegative value $y$, the area under the indicator $I(t<y)$ from zero to infinity is exactly $y$. Likewise, integrating $2t$ over that interval gives $y^2$. Apply these identities to the random variable $Y$, then take expectations. Since the integrands are nonnegative, Tonelli's theorem permits the order of expectation and integration to be exchanged, even before finiteness is established.

Y=\int_0^\infty I(Y>t)\,dt,\qquad Y^2=\int_0^\infty 2tI(Y>t)\,dt

E[Y]=\int_0^\infty E[I(Y>t)]\,dt =\int_0^\infty P(Y>t)\,dt

E[Y^2]=\int_0^\infty 2tE[I(Y>t)]\,dt =2\int_0^\infty tP(Y>t)\,dt

The key point is that $P(Y>t)$, not the density of $Y$, is being integrated. The result therefore remains valid when the payment has atoms, including an atom at zero or at a policy maximum. A value of the survival function at one endpoint does not affect the integral.

Translating Coverage into a Payment Tail

For an ordinary deductible $d$, the payment is $Y^L=(X-d)_+$, where $(z)_+=\max(z,0)$. For $t\geq0$, the event $Y^L>t$ is exactly the event $X>d+t$. Thus the payment survival function is the original loss survival function shifted by the deductible. If the payment is capped at $M$, then the tail is zero for $t\geq M$. For a proportional payment $cX$, where $c>0$, the event $cX>t$ is $X>t/c$. These event equivalences are the main step in applying the formulas.

Write the payment function.
State how the deductible, coinsurance, and payment maximum transform the loss.

Convert a payment-tail event.
For each relevant $t$, solve $Y^L>t$ as an event involving $X$.

Integrate over the payment support.
Use the first- or second-moment tail formula, setting the tail to zero beyond any payment maximum.

Find variance and check.
Subtract the square of the unconditional mean from the second raw moment, using unrounded expressions until the final rounding.

Worked Example: An Exponential Loss With a Deductible

A hypothetical policy has payment $Y^L=(X-d)_+$, with deductible $d=\$50$. Assume $X$ has an exponential distribution with mean $\theta=\$100$, so $S_X(x)=e^{-x/100}$ for $x\geq0$. Find the unconditional mean, second raw moment, and variance of the payment. Payments are undiscounted; there is no mortality assumption or interest rate.

Method: derive the payment tail and integrate. For $t\geq0$, $Y^L>t$ if and only if $X>50+t$. Therefore

S_{Y^L}(t)=e^{-(50+t)/100}=e^{-0.5}e^{-t/100}.

The mean is

E[Y^L]=\int_0^\infty e^{-0.5}e^{-t/100}\,dt =e^{-0.5}(100)=100e^{-0.5} =\$60.65.

For the second raw moment, use the second-moment formula and the standard integral $\int_0^\infty t e^{-t/100}\,dt=100^2$:

E[(Y^L)^2] =2e^{-0.5}\int_0^\infty t e^{-t/100}\,dt =20{,}000e^{-0.5} =\$12{,}130.61\text{ squared}.

Keep the unrounded mean expression when calculating variance:

\operatorname{Var}(Y^L) =20{,}000e^{-0.5}-(100e^{-0.5})^2 =20{,}000e^{-0.5}-10{,}000e^{-1} =\$8{,}451.82\text{ squared}.

Check. The probability of a positive payment is $P(X>50)=e^{-0.5}$. By the memoryless property, conditional on $X>50$, the excess $X-50$ has exponential mean $\$100$ and second moment $\$20{,}000$. Multiplying these conditional moments by $e^{-0.5}$ reproduces the unconditional moments above.

Payment Limits and Atoms

If the payment is $Y^L=\min(X,M)$, then $S_{Y^L}(t)=S_X(t)$ for $0\leq t<M$, and $S_{Y^L}(t)=0$ for $t\geq M$. The capped payment has an atom at $M$ whenever $P(X\geq M)>0$. The tail integral still accounts for that atom: payment values at the cap contribute to the tail for every threshold below $M$. No separate atom term is needed.

Worked Example: A Limited Exponential Payment

A hypothetical contract pays $Y^L=\min(X,\$200)$. Let $X$ be exponential with mean $\theta=\$100$, so $S_X(x)=e^{-x/100}$. Find the payment mean, second raw moment, and variance. Payments are undiscounted; no mortality assumption or interest rate applies.

For $0\leq t<200$, the event $Y^L>t$ is $X>t$; at and above $200$, the payment cannot exceed the threshold. Hence the tail is $e^{-t/100}$ below $200$, and zero above it. The mean is

E[Y^L]=\int_0^{200}e^{-t/100}\,dt =100(1-e^{-2}) =\$86.47.

For the second raw moment, integration by parts gives $\int_0^M t e^{-t/\theta}\,dt=\theta^2[1-(1+M/\theta)e^{-M/\theta}]$. Using $M/\theta=2$,

E[(Y^L)^2] =2\int_0^{200}t e^{-t/100}\,dt =20{,}000(1-3e^{-2}) =\$11{,}879.88\text{ squared}.

Using the exact mean expression in the variance calculation gives

\operatorname{Var}(Y^L) =20{,}000(1-3e^{-2})-10{,}000(1-e^{-2})^2 =\$4{,}403.43\text{ squared}.

Check. The capped payment is always between zero and $\$200$, and the calculated mean is within that range. Its second raw moment is also below $200E[Y^L]$, as required by $(Y^L)^2\leq200Y^L$. The limit atom is included in the tail integral over thresholds below $\$200$.

Proportional Coverage With a Payment Maximum

The same approach handles multiple payment features. Suppose the payment is $\min(cX,M)$, with $0<c\leq1$. For $0\leq t<M$, its survival function is $S_X(t/c)$. If a deductible is also present, the event must be translated using the full payment function; it is not generally correct to apply the deductible after scaling unless the contract defines that order.

Worked Example: Coinsurance With a Payment Cap

A hypothetical policy pays $Y^L=\min(0.5X,\$100)$. Assume $X$ is exponential with mean $\theta=\$100$, so $S_X(x)=e^{-x/100}$. Calculate the mean, second raw moment, and variance. Payments are undiscounted; no mortality assumption or interest rate applies.

For $0\leq t<100$, $Y^L>t$ is equivalent to $X>2t$. Therefore $S_{Y^L}(t)=e^{-2t/100}=e^{-t/50}$ on this interval, and the tail is zero for $t\geq100$. The mean is

E[Y^L]=\int_0^{100}e^{-t/50}\,dt =50(1-e^{-2}) =\$43.23.

The second raw moment is

E[(Y^L)^2] =2\int_0^{100}t e^{-t/50}\,dt =2(50)^2[1-(1+2)e^{-2}] =5{,}000(1-3e^{-2}) =\$2{,}969.97\text{ squared}.

Subtract the square of the exact mean:

\operatorname{Var}(Y^L) =5{,}000(1-3e^{-2})-[50(1-e^{-2})]^2 =2{,}500(1-4e^{-2}-e^{-4}) =\$1{,}100.86\text{ squared}.

Check. The payment is bounded above by $\$100$, and $E[(Y^L)^2]\leq100E[Y^L]$. Numerically, $\$2{,}969.97$ squared is less than $100(\$43.23)$ in the corresponding dollar-squared units, consistent with the bound.

Common Trap: Conditioning Twice

The tail-integral formulas return unconditional moments. A common error is to calculate a moment conditional on a positive payment and then multiply by the probability of a positive payment more than once. In the deductible example, $P(Y^L>0)=e^{-0.5}$, while $E[Y^L\mid Y^L>0]=\$100$. Thus

E[Y^L]=P(Y^L>0)E[Y^L\mid Y^L>0] =e^{-0.5}(100)=\$60.65.

The incorrect calculation $e^{-0.5}\times e^{-0.5}\times100=\$36.79$ counts the probability of a positive payment twice. A related mistake is to use a conditional payment survival function in the tail integral and treat the resulting conditional mean as the unconditional mean. Always check whether the tail being integrated is $P(Y^L>t)$ or $P(Y^L>t\mid Y^L>0)$.

Another arithmetic trap is to round $E[Y^L]$ before squaring it in the variance formula. For example, an exact mean of $100(1-e^{-2})$ should be squared as $10{,}000(1-e^{-2})^2$, not as the square of its rounded dollar display. Keep exact expressions through the subtraction and round only the final result.

Key takeaway: For a nonnegative payment, integrate its unconditional survival function to obtain the mean, and integrate twice the threshold times that survival function to obtain the second raw moment. Coverage terms are handled by translating the event $Y^L>t$ into an event about $X$; atoms do not invalidate the method.

Check Your Understanding

For each problem, write the payment survival function first. Answers are rounded only where a numerical value is requested.

A hypothetical loss $X$ is exponential with mean $\$80$, and the payment is $Y^L=(X-\$20)_+$. Find $E[Y^L]$. (Answer: $80e^{-0.25}=\$62.30$)
A hypothetical contract pays $Y^L=\min(X,\$120)$, where $X$ is exponential with mean $\$60$. Find the first and second raw moments. (Answer: $60(1-e^{-2})=\$51.88$; $7{,}200(1-3e^{-2})=\$4{,}276.76$ dollars squared)
A hypothetical policy pays $Y^L=\min(0.4X,\$80)$, where $X$ is exponential with mean $\$100$. Find the mean and second raw moment. (Answer: $40(1-e^{-2})=\$34.59$; $3{,}200(1-3e^{-2})=\$1{,}900.78$ dollars squared)
A hypothetical policy pays $Y^L=\min((X-\$50)_+,\$150)$, where $X$ is exponential with mean $\$100$. Find the payment mean. (Answer: $100e^{-0.5}(1-e^{-1.5})=\$47.12$)
A hypothetical loss takes values $\$0,\$40,\$100$ with probabilities $0.2,0.5,0.3$, respectively. The payment is $Y^L=\min((X-\$30)_+,\$50)$. Find the mean, second raw moment, and variance. (Answer: $E[Y^L]=\$20$, $E[(Y^L)^2]=\$800$ dollars squared, and $\operatorname{Var}(Y^L)=\$400$ dollars squared)

What You'll Learn