Exposure Curves
Study guide: Exam 9
Exposure Curves give, for a class of property risks, the share of expected loss that falls below a retention expressed as a fraction d of each risk's maximum possible loss. An exposure curve is the normalised limited expected value function of the loss-degree distribution, and it is used to split a risk's expected loss between the cedent's retention and a per-risk excess-of-loss layer.
G(d) = \frac{E[X \wedge dM]}{E[X]}
G(d) = \frac{\int_0^d \left(1 - F(y)\right) dy}{\int_0^1 \left(1 - F(y)\right) dy}
E[\text{loss in } (D_1, D_2]] = E[X]\left[G\!\left(\tfrac{D_2}{M}\right) - G\!\left(\tfrac{D_1}{M}\right)\right]
Xis the ground-up loss,Mthe MPL (or sum insured or EML),y = X/M \in [0, 1]the loss degree with CDFF, andd = D/Mthe normalised retention, capped at1. The cedent keepsE[X]\,G(d)and the layer above cedes the rest.- Properties (Bernegger).
G(0) = 0andG(1) = 1, andGis increasing and concave.G'(0) = 1/E[y], so the slope at the origin gives the mean damage ratio.p = G'(1)/G'(0)is the probability of a total loss.F(y) = 1 - G'(y)/G'(0)fory < 1, so the curve and the distribution determine each other.
- The MBBEFD class. Bernegger's loss-distribution-based family of curves has two parameters,
b > 0andg > 1, with separate limiting forms whenb = 1orgb = 1:
G_{b,g}(d) = \frac{\ln\!\left[\dfrac{(g-1)b + (1-gb)\,b^{d}}{1-b}\right]}{\ln(gb)}
- Fitting it. Match the total-loss probability and the mean:
g = 1/pdirectly, then solve forbiteratively from\mu = E[y], which falls asbrises. It can also be fitted to\muand the standard deviation. - The Swiss Re curves. The one-parameter family
b(c) = e^{3.1 - 0.15(1+c)c},g(c) = e^{(0.78 + 0.12c)c}reproduces the Swiss Re curves Y1–Y4 atc = 1.5,2,3,4and a Lloyd's industrial curve atc = 5. Atc = 0every loss is a total loss. Highercgives a more concave curve: small partial losses relative to the MPL, typical of large industrial risks. - Exposure rating. Band the risk profile by MPL, estimate each band's expected loss as
\text{ELR} \times \text{premium}, and applyGband by band. For casualty lines, ILFs play the role ofG(Reinsurance Pricing). The curve prices a per-risk cover and says nothing about catastrophe accumulation.
Building a Curve from a Loss-Degree Distribution For a class of warehouses,
10\%of losses are total, and below a total loss1 - F(y) = 1 - 0.9yfor0 \le y < 1. DeriveG, check the slope properties, and find the share of expected loss between20\%and50\%of MPL.Answer
\begin{align*} E[y] &= \int_0^1 (1 - 0.9y)\,dy \\ &= 0.55 \\[4pt] G(d) &= \frac{d - 0.45d^2}{0.55} \end{align*}
G'(0) = 1/0.55 = 1.818 = 1/E[y], andG'(1) = 0.1/0.55, sop = G'(1)/G'(0) = 0.10, the total-loss share, as it should be.\begin{align*} G(0.2) &= \frac{0.2 - 0.018}{0.55} \\ &= 0.3309 \\[4pt] G(0.5) &= \frac{0.5 - 0.1125}{0.55} \\ &= 0.7045 \\[4pt] \text{Layer share} &= 0.7045 - 0.3309 \\ &= 0.3736 \end{align*}The layer from
20\%to50\%of MPL carries37\%of expected loss. A mean damage ratio of55\%is far heavier than real property experience, which is why fitted curves such as the Swiss Re family are much more concave.
Exposure Rating a Per-Risk Layer A per-risk treaty covers
\$3M xs\$2M, with an ELR of60\%and the Swiss Rec = 3curve, whereG(0.10) = 0.4056,G(0.25) = 0.6002andG(0.40) = 0.7163. Band A has MPL\$5M and\$400{,}000of premium. Band B has MPL\$20M and\$600{,}000of premium. Find the expected layer loss.Answer Band A:
d_1 = 2/5 = 0.40andd_2 = \min(5/5, 1) = 1. Band B:d_1 = 2/20 = 0.10andd_2 = 5/20 = 0.25.\begin{align*} \text{Band A} &= 240{,}000 \times (1 - 0.7163) \\ &= \$68{,}088 \\[4pt] \text{Band B} &= 360{,}000 \times (0.6002 - 0.4056) \\ &= \$70{,}056 \\[4pt] \text{Total} &= \$138{,}144 \end{align*}That is
13.8\%of subject premium, or23.0\%of expected loss. For band A the layer reaches the MPL, so it absorbs the whole top of the curve, total losses included. For band B it is a thin slice low on the curve, but the band's larger expected loss makes the dollars similar. The reinsurer then adds its expense and risk load, and checks the result against experience rating.