The Survival-Function Formula For Limited Expected Loss

Why the Survival Curve Measures Limited Loss

A policy limit caps the insurer’s payment on an individual loss. To price different limits consistently, an actuary needs the expected payment at each limit, not just the probability that a loss exceeds that limit. The survival function provides a direct way to calculate this expected payment: the expected limited loss is the area under the severity survival curve up to the policy limit.

This representation is useful when the severity distribution is described by probabilities at several loss amounts or by a fitted curve. It also makes the increased limits factor (ILF) a ratio of areas. The derivation and calculations below are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Definition: Let $X$ be a nonnegative ground-up loss, $u\geq 0$ a per-loss policy limit, and $S_X(x)=P(X>x)$ the severity survival function. The limited loss at limit $u$ is $\min(X,u)$, and its expected value is $E[\min(X,u)]$. For a positive base limit $b$ with $E[\min(X,b)]>0$, the increased limits factor at $u$ relative to $b$ is $$ \operatorname{ILF}(u;b)=\frac{E[\min(X,u)]}{E[\min(X,b)]}. $$

The strict inequality in $S_X(x)=P(X>x)$ matters at a point where losses have positive probability. The survival function is right-continuous: if a probability mass occurs at $x=a$, then $S_X(a)$ is already after the drop associated with that mass. A left-limit value such as $S_X(a-)$ can therefore differ from $S_X(a)$. When integrating over a continuous interval, changing the value at a single endpoint does not change the area, but the endpoint must still be stated consistently if a curve is being specified as a probability distribution.

Deriving the Limited Expected Loss Formula

For a fixed realized loss $X$, the amount $\min(X,u)$ is the length of the interval from zero up to the smaller of $X$ and $u$. Equivalently, it is the area under an indicator that equals one while the loss exceeds the point on the horizontal axis:

\min(X,u)=\int_0^u \mathbf{1}_{\{X>x\}}\,dx.

If $X\leq u$, the indicator is one from zero to $X$, so the integral is $X$. If $X>u$, it is one throughout the interval from zero to $u$, so the integral is $u$. Thus the identity holds in both cases, including $X=u$, since the value of the indicator at a single point does not affect the integral.

Taking expectations gives the tail-integral formula. Interchanging expectation and integration is valid because the indicator is nonnegative; equivalently, Tonelli’s theorem applies. Since $u$ is finite, the limited loss is bounded by $u$, so its expectation is finite even if the ground-up loss has an infinite mean.

\begin{aligned} E[\min(X,u)] &=E\left[\int_0^u \mathbf{1}_{\{X>x\}}\,dx\right]\\ &=\int_0^u E[\mathbf{1}_{\{X>x\}}]\,dx\\ &=\int_0^u P(X>x)\,dx\\ &=\int_0^u S_X(x)\,dx. \end{aligned}

The formula assumes $X\geq 0$ and $u\geq 0$, as is standard for ground-up insurance severity and a policy limit. It is not a direct formula for a loss variable that can be negative. The integration variable $x$ and the limit $u$ must also use the same monetary units as $X$.

Theorem: For a nonnegative loss $X$ and finite limit $u\geq 0$, $$ E[\min(X,u)]=\int_0^u S_X(x)\,dx. $$ Consequently, if the base limited expected loss is positive, the ILF is the ratio of the two areas under the same severity survival function: $$ \operatorname{ILF}(u;b)=\frac{\int_0^u S_X(x)\,dx}{\int_0^b S_X(x)\,dx}. $$

Calculating Areas Under a Survival Curve

When the survival function is specified as linear between successive points, each interval contributes a trapezoid. If the survival values at the ends of an interval of width $w$ are $s_1$ and $s_2$, its area is $w(s_1+s_2)/2$. Add the interval areas up to the selected limit. For an ILF, calculate both limited expected losses on the same severity basis before taking their ratio.

Worked Example: A Curve With an Atom at the Limit

A hypothetical liability severity curve is measured in thousands of dollars. It is linear from $(0,0.80)$ to $(20,0.50)$, from $(20,0.50)$ to $(40,0.20)$, and from $(40,0.20)$ to $(50-,0.10)$. At $x=50$, set $S_X(50)=0$, and let $S_X(x)=0$ for $x>50$. Calculate the limited expected losses at limits 20, 40, and 50, all in thousands of dollars, and the ILF at 50 relative to 20.

Method. Integrate the specified survival curve interval by interval. The value $0.10$ is the left limit at 50; the drop to $S_X(50)=0$ represents probability mass at exactly 50. This is a valid right-continuous survival function.

From zero to 20, the trapezoid area is $$ 20\left(\frac{0.80+0.50}{2}\right)=20(0.65)=13. $$ Therefore, $E[\min(X,20)]=13$ thousand dollars.

From 20 to 40, the added area is $$ 20\left(\frac{0.50+0.20}{2}\right)=20(0.35)=7. $$ Thus $E[\min(X,40)]=13+7=20$ thousand dollars.

From 40 to 50, the area uses the left-limit value at 50: $$ 10\left(\frac{0.20+0.10}{2}\right)=10(0.15)=1.5. $$ The point-value drop at 50 has zero width and adds no area. Hence $E[\min(X,50)]=20+1.5=21.5$ thousand dollars.

The requested ILF is $$ \operatorname{ILF}(50;20)=\frac{21.5}{13}=1.653846\ldots\approx 1.6538. $$ Check and common mistake avoided: The limited expected loss increases with the limit, as $21.5>13$. Using $S_X(50)=0$ as the right endpoint of the final trapezoid would incorrectly omit the probability mass at 50 from the left-limit area. The correct endpoint for the interval approaching 50 is $S_X(50-)=0.10$.

Worked Example: ILF From a Piecewise-Linear Rating Curve

A hypothetical property severity curve, in thousands of dollars, is linear through $(0,1.00)$, $(10,0.70)$, $(30,0.30)$, and $(60,0)$, and is zero above 60. Find the limited expected losses at 30 and 60, then calculate the ILF at 60 relative to 30.

Method. Sum the trapezoid areas to each limit, then divide the limited expected loss at the higher limit by that at the base limit.

For the interval from 0 to 10, the area is $$ 10\left(\frac{1.00+0.70}{2}\right)=10(0.85)=8.5. $$ For the interval from 10 to 30, the area is $$ 20\left(\frac{0.70+0.30}{2}\right)=20(0.50)=10. $$ Therefore, $$ E[\min(X,30)]=8.5+10=18.5 $$ thousand dollars.

The area from 30 to 60 is $$ 30\left(\frac{0.30+0}{2}\right)=30(0.15)=4.5. $$ Thus $$ E[\min(X,60)]=18.5+4.5=23 $$ thousand dollars, and $$ \operatorname{ILF}(60;30)=\frac{23}{18.5}=\frac{46}{37}\approx 1.2432. $$ Check and common mistake avoided: The incremental area from 30 to 60 is 4.5, not the area of a rectangle at the starting survival value. The survival probability declines across the interval, so the trapezoid average is $(0.30+0)/2=0.15$. The total at 60, 23, is larger than the total at 30, 18.5.

Worked Example: Interpolating an ILF From a Coarse Curve

A hypothetical commercial severity curve is expressed in thousands of dollars. It is linear from $(0,0.95)$ to $(25,0.65)$, from $(25,0.65)$ to $(75,0.25)$, and from $(75,0.25)$ to $(125-,0.05)$. At 125, $S_X(125)=0$, and the survival function remains zero above 125. Calculate the limited expected losses at 25, 75, and 125, and the ILF at 75 relative to 25.

Method. Apply the tail-integral formula to the piecewise-linear curve. The endpoint description makes the final drop a mass at 125 and keeps the survival function right-continuous.

At limit 25, the area is $$ 25\left(\frac{0.95+0.65}{2}\right)=25(0.80)=20 $$ thousand dollars.

From 25 to 75, the added area is $$ 50\left(\frac{0.65+0.25}{2}\right)=50(0.45)=22.5. $$ Therefore, $E[\min(X,75)]=20+22.5=42.5$ thousand dollars.

From 75 to 125, use the left-limit survival value at 125: $$ 50\left(\frac{0.25+0.05}{2}\right)=50(0.15)=7.5. $$ It follows that $E[\min(X,125)]=42.5+7.5=50$ thousand dollars. The requested factor is $$ \operatorname{ILF}(75;25)=\frac{42.5}{20}=2.1250. $$ Check and common mistake avoided: The limited expected loss at 125 is 50 thousand dollars, which exceeds the value at 75 by the positive area 7.5. Using the right-continuous value $S_X(125)=0$ in place of $S_X(125-)=0.05$ for the entire final interval would understate that increment.

Interpreting and Using the Result

An ILF compares expected limited losses at two per-loss limits on the same severity basis. If the frequency assumption is unchanged and expected loss per exposure is frequency times expected severity payment, that common frequency factor cancels in the ratio. The ILF is therefore not, by itself, a complete premium indication: expenses, risk loads, coverage differences, or other rating adjustments are not included in the definition above.

For $u\geq b$, the integrand $S_X(x)$ is nonnegative, so $$ E[\min(X,u)]-E[\min(X,b)]=\int_b^u S_X(x)\,dx\geq 0. $$ Thus the limited expected loss cannot decrease as the limit increases. If the base limited expected loss is positive, this also implies $\operatorname{ILF}(u;b)\geq 1$ when $u\geq b$. The increase is zero over any interval where the survival function is zero.

A useful calculation sequence is:

Confirm the loss basis and units.
Use a nonnegative ground-up severity and express the curve coordinates and policy limits in the same monetary units.

Read survival values with the correct endpoint convention.
At an atom, distinguish the left limit $S_X(a-)$ from the right-continuous value $S_X(a)$.

Integrate to each limit.
For a piecewise-linear curve, calculate each trapezoid area and add the areas up to the selected limit.

Form and check the ratio.
Divide the higher-limit limited expected loss by the base-limit value, then check that the result is dimensionless and has the expected direction.

A frequent exam trap is to treat the survival probability at the policy limit as the expected limited loss. They are different quantities: $S_X(u)$ is a probability, while $E[\min(X,u)]$ is a monetary amount obtained by accumulating survival probabilities over all amounts from zero through $u$. Another trap is to multiply an interval width by only its starting survival value; that is a rectangle, not the correct trapezoid area when the curve changes linearly.

For example, if survival decreases linearly from 0.50 to 0.10 over an interval of width 60, the interval area is $$ 60\left(\frac{0.50+0.10}{2}\right)=18, $$ not $60(0.50)=30$. The rectangle overstates the area by 12 because it treats the starting survival probability as constant throughout the interval. Endpoint handling and area checks are therefore essential, especially when the curve includes an atom at its maximum loss.

Key takeaway: Limited expected loss is the area under $S_X(x)=P(X>x)$ from zero to the policy limit. Calculate ILFs as ratios of these areas, preserve the survival function’s endpoint convention, and verify that a higher limit does not produce a lower limited expected loss.

Check Your Understanding

For each piecewise-linear curve, calculate the requested areas first and then form any ILF. Amounts are in thousands of dollars unless stated otherwise.

A hypothetical severity curve decreases linearly from $(0,0.90)$ to $(20,0.50)$. Calculate $E[\min(X,20)]$. (Answer: 14 thousand dollars.)
A curve is linear through $(0,1.00)$, $(10,0.60)$, and $(30,0.20)$. Calculate the limited expected losses at 10 and 30 and the ILF at 30 relative to 10. (Answer: 8 thousand dollars; 16 thousand dollars; ILF $=2.0000$.)
A curve is linear through $(0,0.80)$, $(40,0.40)$, and $(80-,0.20)$, with $S_X(80)=0$ and zero survival above 80. Calculate the limited expected losses at 50 and 80, then the ILF at 80 relative to 50. (Answer: 27.75 thousand dollars; 36 thousand dollars; ILF $\approx1.2973$.)
A hypothetical curve is linear through $(0,0.95)$, $(20,0.75)$, $(50,0.45)$, and $(90,0.15)$, and is zero above 90. Calculate the limited expected losses at 50 and 90 and the ILF at 90 relative to 50. (Answer: 35 thousand dollars; 47 thousand dollars; ILF $\approx1.3429$.)
A curve is linear through $(0,0.70)$, $(10,0.50)$, $(30,0.20)$, and $(40-,0.10)$, with $S_X(40)=0$ and zero survival above 40. Calculate the limited expected losses at 30 and 40 and the ILF at 40 relative to 30. (Answer: 13 thousand dollars; 14.5 thousand dollars; ILF $\approx1.1154$.)

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