Severity Distribution

Study guides: Exam ASTAM, Exam MAS-I, Exam 8, Exam 9

A severity distribution is the probability distribution of the size $X$ of a single loss or claim, given that one happens. It is described by its CDF $F(x) = P(X \le x)$ or its density $f(x)$ . Paired with a claim-count distribution, it prices anything that depends on how large claims are: deductibles, limits, excess layers and exposure curves.

$E[X \wedge u] = \int_0^u \big[1 - F(x)\big]\,dx$

$G(d) = \frac{E[X \wedge d]}{E[X]}$

$E[X \wedge u]$ is the limited expected value, the expected loss capped at $u$ . A layer's expected cost is the difference of two of these, so this integral is how the severity distribution turns into expected losses by layer and increased limits factors.
Families. The MAS-I reading's four are the exponential, gamma, Weibull and Pareto. The first three have every moment, so they are light-tailed by the moment criterion, though a Weibull with shape below 1 has a decreasing hazard rate, the mark of a heavier tail. The Pareto has moments only of order below its shape $\alpha$ and is heavy-tailed. The lognormal has every moment but no moment generating function, which makes it heavy-tailed under the more common MGF criterion. A finite mixture of exponentials has every moment and an MGF, so it is light-tailed. A layer's cost depends only on $1 - F$ inside the layer, so for excess layers the tail decides the answer. Parameters are fitted by maximum likelihood. The fit must allow for truncation by deductibles and censoring at policy limits.
Exposure curves (Exam 9). Property losses grow with the size of the risk, so they are expressed as a fraction $X \in [0,1]$ of the risk's maximum possible loss or sum insured. $G(d)$ is then the share of expected loss kept below a deductible of $d$ . It rises from $G(0) = 0$ to $G(1) = 1$ and is concave. The curve and the distribution determine each other: $E[X] = 1/G'(0)$ , and $1 - F(d) = G'(d)/G'(0)$ for $d < 1$ .
The MBBEFD class. Bernegger's MBBEFD distributions give such curves in closed form. The one-parameter subfamily with $c = 1.5, 2, 3, 4$ coincides very well with the Swiss Re Y1–Y4 curves, and $c = 5$ with a Lloyd's curve used for rating industrial risks.
Inflation multiplies the scale parameter: if losses rise by $r$ , then $\theta$ becomes $\theta(1+r)$ . With the limits held fixed, the effect is leveraged (Inflation): under positive inflation, losses capped at a basic limit grow no faster than $r$ , and the losses excess of it, with no cap above, at least as fast. A layer with both an attachment and a cap can grow more slowly than $r$ : for an exponential with mean $1{,}000$ , $5\%$ inflation raises the expected cost per payment in a $500$ xs $100$ layer by only $1.1\%$ .

Fitting Severity to Deductible Data Four claims are reported under a $\$1{,}000$ deductible, with ground-up sizes $\$1{,}500$ , $\$2{,}500$ , $\$4{,}000$ and $\$7{,}000$ . Losses below $\$1{,}000$ are never reported.

Fit an exponential by maximum likelihood.
Answer Each observation is left-truncated at $1{,}000$ , so its likelihood is $f(x_i) / [1 - F(1000)]$ :
 $\begin{align*} L(\theta) &= \prod_{i=1}^{4} \frac{\theta^{-1} e^{-x_i/\theta}}{e^{-1000/\theta}} \\ &= \theta^{-4}\,e^{-(500 + 1500 + 3000 + 6000)/\theta} \\ &= \theta^{-4}\,e^{-11{,}000/\theta} \end{align*}$ 
Setting $\frac{d}{d\theta}\ln L = -4/\theta + 11{,}000/\theta^2$ to zero gives $\hat\theta = 11{,}000/4 = 2{,}750$ , the mean excess over the deductible.

Ignoring the truncation gives $\hat\theta = 15{,}000/4 = 3{,}750$ , which overstates severity by $36\%$ . The fitted curve also says the four reports stand for $4 / e^{-1000/2750} = 5.75$ ground-up losses. Frequency needs the same correction as severity.

An Exposure Curve from a Destruction-Rate Distribution For a class of property risks, the loss as a fraction of MPL has survival function $1 - F(x) = (1 - x)^3$ on $[0, 1]$ . One risk has an MPL of $\$10{,}000{,}000$ , premium of $\$50{,}000$ and an expected loss ratio of $60\%$ .

Derive the exposure curve and find the expected loss in a $\$3$ M excess of $\$2$ M per-risk layer.
Answer
 $\begin{align*} E[X \wedge d] &= \int_0^d (1 - x)^3\,dx \\ &= \frac{1 - (1 - d)^4}{4} \\[4pt] G(d) &= 1 - (1 - d)^4 \end{align*}$ 
The layer runs from $d = 0.2$ to $d = 0.5$ of MPL. Expected loss is $0.60 \times 50{,}000 = \$30{,}000$ .
 $\begin{align*} G(0.5) - G(0.2) &= 0.9375 - 0.5904 \\ &= 0.3471 \\[4pt] \text{Layer loss} &= 0.3471 \times 30{,}000 \\ &= \$10{,}413 \end{align*}$ 
The cedant keeps $0.5904 \times 30{,}000 = \$17{,}712$ below $\$2$ M, and $\$1{,}875$ lies above $\$5$ M. Recovering the distribution from the curve: $G'(d) = 4(1 - d)^3$ , so $E[X] = 1/G'(0) = 0.25$ and $G'(d)/G'(0) = (1 - d)^3$ . That matches the survival function we started from.