Actuarial Notes

Stop-Loss Reinsurance

Study guides: Exam FAM, Exam ASTAM

Stop-Loss Reinsurance covers an insurer's aggregate losses rather than its individual claims. Over a stated period, the reinsurer pays the amount by which the total losses S exceed a retention d, and the expected value of that payment, E[(S - d)_+], is the net stop-loss premium. It is the aggregate counterpart of excess of loss cover, which applies a retention to each loss, and its treaty form with an aggregate limit is aggregate excess of loss.

I_d = (S - d)_+ = \max(S - d,\ 0)

E[(S - d)_+] = E[S] - E[S \wedge d]

E[S \wedge d] = \sum_{s < d} s\,f_S(s) + d\,\Pr(S \ge d)

E\big[(S - (j+1)h)_+\big] = E\big[(S - jh)_+\big] - h\big[1 - F_S(jh)\big]

E[(S - d)_+] = \frac{b - d}{b - a}\,E[(S - a)_+] + \frac{d - a}{b - a}\,E[(S - b)_+]

  • Symbols: Following Klugman, Panjer & Willmot (Loss Models, chapter 9), S = X_1 + \cdots + X_N is the aggregate loss of the aggregate loss model, with E[S] = E[N]\,E[X]. I_d is the reinsurer's payment, and the insurer keeps S \wedge d, so the retention caps the insurer's total cost for the period. The third block is for a discrete S, and the fourth holds when S takes values only at multiples of a span h.
  • Interpolation: The last block holds for any a \le d \le b with \Pr(a < S < b) = 0. Between two possible values of S the distribution function is flat, so the net stop-loss premium is linear in d there. An arithmetic S, built from a discrete or discretized severity by Panjer recursion or convolution, has such gaps between every pair of span points.
  • Why the whole distribution matters: The net stop-loss premium depends on the distribution of S and not only on its mean. Retentions are usually set above E[S], where (E[S] - d)_+ is 0, but the reinsurer still pays in every year whose losses exceed d.
  • Premium and protection: The net stop-loss premium is a pure premium, and the reinsurer's charge adds expense and risk loadings to it. A per-claim retention cannot protect a year with many moderate claims, and a stop-loss retention can, because it applies to their total.

Net Stop-Loss Premium for a Compound Poisson Portfolio The number of claims on a portfolio in a year is Poisson with mean 1.5. Each claim is 1 or 2, in units of $10,000, with probabilities 0.6 and 0.4. A stop-loss treaty pays aggregate claims in excess of 3 units. Find the net stop-loss premium.

Answer The mean claim is 1(0.6) + 2(0.4) = 1.4, so E[S] = 1.5(1.4) = 2.1. Aggregate losses of 2 arise from one claim of 2 or from two claims of 1.

\begin{align*}
f_S(0) &= e^{-1.5} \\
&= 0.22313 \\
f_S(1) &= 1.5e^{-1.5}(0.6) \\
&= 0.20082 \\
f_S(2) &= 1.5e^{-1.5}(0.4) + \tfrac{1.5^2}{2}e^{-1.5}(0.6^2) \\
&= 0.13388 + 0.09037 \\
&= 0.22425
\end{align*}

These give F_S(2) = 0.64820, so \Pr(S \ge 3) = 0.35180.

\begin{align*}
E[S \wedge 3] &= 1(0.20082) + 2(0.22425) + 3(0.35180) \\
&= 0.20082 + 0.44850 + 1.05540 \\
&= 1.70472 \\
E[(S - 3)_+] &= 2.1 - 1.70472 \\
&= 0.39528
\end{align*}

The net stop-loss premium is about $3,953 a year. Comparing the mean with the retention would suggest that nothing is owed, because E[S] = 2.1 is below 3, but S exceeds 3 with probability 0.35.

A Retention Between Span Points For the same portfolio, the insurer considers a retention of 2.4 units ($24,000). Find the net stop-loss premium by stepping the recursion from d = 0 and interpolating.

Answer S is arithmetic with span h = 1, and E[(S - 0)_+] = E[S] = 2.1. Each step subtracts 1 - F_S(j), using F_S(0) = 0.22313, F_S(1) = 0.42395 and F_S(2) = 0.64820.

\begin{align*}
E[(S - 1)_+] &= 2.1 - 0.77687 \\
&= 1.32313 \\
E[(S - 2)_+] &= 1.32313 - 0.57605 \\
&= 0.74708 \\
E[(S - 3)_+] &= 0.74708 - 0.35180 \\
&= 0.39528
\end{align*}

No value of S lies strictly between 2 and 3, so the premium at 2.4 is the weighted average with weights 0.6 and 0.4.

\begin{align*}
E[(S - 2.4)_+] &= 0.6(0.74708) + 0.4(0.39528) \\
&= 0.44825 + 0.15811 \\
&= 0.60636
\end{align*}

Lowering the retention from 3 to 2.4 units raises the net premium from about $3,953 to about $6,064. The recursion reproduces the first example's answer at d = 3, which checks the arithmetic.

How it appears on the exam

  • ASTAM lists stop loss reinsurance among the coverage modifications whose effects candidates must evaluate. A question gives a frequency distribution and a discrete severity, or the aggregate probabilities directly, and asks for the net stop-loss premium at a stated retention.
  • Aggregate probabilities first: When only the frequency and severity are given, the aggregate probabilities have to be built by convolution or by recursion before the stop-loss premium can be found, so the question joins this page to the aggregate models objective.
  • FAM has the same calculation in its aggregate models objective, which asks candidates to calculate the expected payment for stop-loss insurance.
  • Retention between span points: When the retention falls between two possible values of S, the intended route is the linear interpolation rather than a sum over the whole distribution.

Exam traps

  • Mean against retention: (E[S] - d)_+ is not the net stop-loss premium. It is 0 in the first example, where the true value is 0.395, because it ignores every year whose losses exceed d.
  • The last term of the limited mean: E[S \wedge d] includes d\,\Pr(S \ge d). Summing only s\,f_S(s) for s < d leaves out the years in which the insurer pays the full retention, and makes the stop-loss premium far too large.
  • Which tail probability: The recursion subtracts h\big[1 - F_S(jh)\big] = h\Pr(S > jh). Using \Pr(S \ge jh) subtracts the mass at jh as well and understates each later premium.
  • Aggregate or per claim: A stop-loss retention applies to the total for the period. Applying d to each claim computes an excess of loss cost, E[N]\,E[(X - d)_+], which answers a different question.
  • Where interpolation fails: Linear interpolation needs \Pr(a < S < b) = 0. It does not hold for a continuous S, or across a point where S has positive probability.