Actuarial Notes

Mean Excess Loss

Study guide: Exam ASTAM

The Mean Excess Loss e_X(d) is the expected amount by which a loss exceeds d, given that it does exceed d. In insurance terms, it is the expected payment per payment under an ordinary deductible of d. As a function of d, it is one of the measures used to compare the tails of two severity distributions.

e_X(d) = E[X - d \mid X > d]

e_X(d) = \frac{\int_d^{\infty} S_X(x)\,dx}{S_X(d)}

e_X(d) = \frac{E[X] - E[X \wedge d]}{S_X(d)}

  • Symbols: S_X = 1 - F_X is the survival function and E[X \wedge d] is the limited expected value. For a positive loss, e_X(0) = E[X]. The cost per loss is E[(X - d)_+] = e_X(d)\,S_X(d), which is the per-loss counterpart of this per-payment quantity.
  • Standard shapes: The exponential has a constant mean excess, e_X(d) = \theta, because it is memoryless. The Pareto has e_X(d) = (d + \theta)/(\alpha - 1) for \alpha > 1, which increases linearly, and a uniform loss on (0, \omega) has e_X(d) = (\omega - d)/2, which decreases to zero.
  • Tail comparison: An increasing mean excess function signals a heavy tail and a decreasing one a light tail. It moves opposite to the hazard rate: a decreasing hazard rate gives an increasing mean excess, and an increasing hazard rate a decreasing one. When the limits exist, L'Hôpital's rule gives \lim_{d \to \infty} e_X(d) = \lim_{d \to \infty} 1/h_X(d).
  • Extreme values: The generalized Pareto distribution has a mean excess function that is a straight line. That is why the empirical mean excess function, the average of the observations above x_{(j)} minus x_{(j)}, is plotted to choose the threshold for a peaks-over-threshold fit.

Mean Excess for a Bounded Loss A loss has survival function S_X(x) = (1 - x/10{,}000)^2 for 0 \le x \le 10{,}000. Find the mean excess loss at d = 1{,}000 by the limited expected value formula, then find e_X(d) in general and describe the tail.

Answer Integrating the survival function gives E[X \wedge d] = \frac{10{,}000}{3}\left[1 - (1 - d/10{,}000)^3\right], so E[X] = 3{,}333.33.

\begin{align*}
E[X \wedge 1{,}000] &= 3{,}333.33\,(1 - 0.9^3) \\
&= 903.33 \\
e_X(1{,}000) &= \frac{3{,}333.33 - 903.33}{0.81} \\
&= 3{,}000
\end{align*}

In general, \int_d^{10{,}000} S_X(x)\,dx = \frac{10{,}000}{3}(1 - d/10{,}000)^3, so e_X(d) = (10{,}000 - d)/3. With a deductible of 1,000, the average payment is 3,000, and with a deductible of 4,000, it falls to 2,000. The mean excess falls to zero at the maximum loss, and the hazard rate 2/(10{,}000 - x) increases, so this tail is light.

Mixed Exponentials Versus a Pareto Model A is an equal mixture of exponentials with means 500 and 5,000. Model B is Pareto with \alpha = 3 and \theta = 5{,}500. Both have mean 2,750. Compare their mean excess losses at 1,000 and at 20,000, and decide which tail is heavier.

Answer For Model A, E[X \wedge 1{,}000] = 0.5(500)(1 - e^{-2}) + 0.5(5{,}000)(1 - e^{-0.2}) = 669.34 and S(1{,}000) = 0.5e^{-2} + 0.5e^{-0.2} = 0.477033.

\begin{align*}
e_A(1{,}000) &= \frac{2{,}750 - 669.34}{0.477033} \\
&= 4{,}361.67 \\
e_B(1{,}000) &= \frac{1{,}000 + 5{,}500}{2} \\
&= 3{,}250
\end{align*}

At 20,000, the short-mean component of Model A has all but vanished, so e_A(20{,}000) = 5{,}000, while e_B(20{,}000) = 12{,}750. The two survival probabilities there are close, 0.0092 for A and 0.0100 for B. Model A's mean excess rises and then levels off at 5,000, while Model B's grows without limit, so Model B has the heavier tail even though Model A's mean excess is larger at 1,000. A layer high above 20,000 costs far more under Model B.

How it appears on the exam

  • ASTAM asks candidates to compare two distributions based on characteristics of their tails, including moments, the limiting ratio of tails, hazard rate functions and mean excess functions. A question gives two severity models and asks for each measure and for which tail is heavier.
  • FAM uses the mean excess loss as the expected payment per payment under an ordinary deductible, so it appears in coverage-modification questions through E[X] - E[X \wedge d] divided by S_X(d).
  • Choosing a threshold: In ASTAM's extreme value reading, a question gives the largest observations and asks for the empirical mean excess function at a point, or asks where its plot becomes linear enough to fit a generalized Pareto distribution.

Exam traps

  • Per payment, not per loss: e_X(d) averages over losses above d only. The expected cost per loss is e_X(d)\,S_X(d), which is smaller.
  • The excess, not the loss: e_X(d) is the amount above d, so the average size of a loss that exceeds d is d + e_X(d). The Expected Shortfall uses that larger quantity at d = \text{VaR}.
  • Directions: A heavy tail has an increasing mean excess and a decreasing hazard rate. Swapping the two directions reverses the conclusion.
  • Increasing is not enough: A mixture of exponentials has an increasing mean excess function, yet every moment exists and it stays light-tailed. The limit of e_X(d), not its direction over a short range, is what separates a Pareto-type tail.