Growth Curve

Study guide: Exam GI-301

Growth curve $G(x \mid \omega, \theta)$ is a smooth parametric curve, in the form of a cumulative distribution function, giving the expected cumulative proportion of ultimate loss reported (or paid) at age $x$ . In Clark (2003) it replaces a set of selected development factors: the age-to-ultimate factor is its reciprocal.

$\text{LDF}_x = \frac{1}{G(x)}$

$\text{Loglogistic: } G(x) = \frac{x^{\omega}}{x^{\omega} + \theta^{\omega}}$

$\text{Weibull: } G(x) = 1 - \exp\!\left[-\left(\frac{x}{\theta}\right)^{\omega}\right]$

Parameters and age. $\omega$ is the shape ("warp") and $\theta$ the scale. The age $x$ runs from the average accident date to the evaluation date, so an accident year evaluated $t \ge 12$ months after it starts has $x = t - 6$ . For the loglogistic, $\theta$ is the median, so $\text{LDF}_\theta = 2.000$ , and $\text{LDF}_x = 1 + \theta^{\omega}x^{-\omega}$ is Sherman's inverse power curve applied to age-to-ultimate factors. For the Weibull, $\theta$ is about the 63.2nd percentile, so $\text{LDF}_\theta \approx 1.582$ .
Expected emergence. The expected increment between ages $x$ and $y$ is $\text{ULT}_{AY}[G(y) - G(x)]$ in Clark's LDF method, or $\text{Premium}_{AY} \times \text{ELR} \times [G(y) - G(x)]$ in his Cape Cod method. $\omega$ , $\theta$ and the ultimates or ELR are fitted by maximum likelihood on over-dispersed Poisson increments, and the fitted ultimates are the familiar chain ladder and Cape Cod estimates.
Why a curve. Clark gives three reasons. Only two parameters describe the whole pattern. The data need not sit on evenly spaced evaluation dates, so a latest diagonal nine months after the one before, or only the last few diagonals, can be used. And the pattern is smooth instead of following every random movement in the age-to-age factors.
The tail, and truncation. The curve extrapolates to infinite age, so it produces its own tail factor. The loglogistic's is heavy: in Clark's example only $77.24\%$ has emerged after ten years, a tail of $1.2946$ against the Weibull's $1.0525$ . To avoid relying on a mechanical extrapolation, Clark truncates at a selected age $x_T$ (240 months), using $G(x_T)/G(x)$ in place of $1/G(x)$ .
Limits. Emergence must rise strictly from $0\%$ to $100\%$ , so a line with real expected negative development, such as significant salvage recoveries, needs a different model. Every accident year shares one pattern. For a period still being earned, Clark scales the curve by the exposed fraction: $G_{AY}(t) = \text{Expos}(t) \times G^{*}\big(\text{AvgAge}(t)\big)$ , with separate formulas for accident and policy years.

Loglogistic and Weibull LDFs, With and Without Truncation Clark's fits to Mack's triangle were loglogistic $\omega = 1.434294$ , $\theta = 48.6249$ and Weibull $\omega = 1.296906$ , $\theta = 48.88453$ .

An accident year is $48$ months old at year-end, with $3{,}000$ reported ($000s). Estimate its reserve (a) from the loglogistic, (b) from the loglogistic truncated at $240$ months, and (c) from the Weibull.
Answer The average age is $x = 48 - 6 = 42$ months, and the truncation point is $240 - 6 = 234$ .

(a) Loglogistic. With $(x/\theta)^{\omega} = (42/48.6249)^{1.434294} = 0.8105$ :
 $\begin{align*} G(42) &= \frac{0.8105}{1 + 0.8105} \\ &= 0.4477 \\ \text{LDF} &= 1/0.4477 \\ &= 2.2338 \\ \text{Reserve} &= 3{,}000 \times (2.2338 - 1) \\ &= 3{,}701 \end{align*}$ 
(b) Truncated at 240 months. With $(234/48.6249)^{1.434294} = 9.5214$ :
 $\begin{align*} G(234) &= \frac{9.5214}{10.5214} \\ &= 0.90496 \\ \text{LDF}_{42 \to 234} &= 0.90496/0.44767 \\ &= 2.0215 \\ \text{Reserve} &= 3{,}000 \times (2.0215 - 1) \\ &= 3{,}064 \end{align*}$ 
(c) Weibull. With $(42/48.88453)^{1.296906} = 0.8213$ :
 $\begin{align*} G(42) &= 1 - e^{-0.8213} \\ &= 0.5601 \\ \text{LDF} &= 1/0.5601 \\ &= 1.7853 \\ \text{Reserve} &= 3{,}000 \times (1.7853 - 1) \\ &= 2{,}356 \end{align*}$ 
The two curves fit the same ten years of data, yet their reserves differ by $1{,}345$ . About $94\%$ of the gap is development beyond those ten years (average age $114$ ), where the loglogistic's tail factor is $1/0.7724 = 1.2946$ and the Weibull's is $1.0525$ . Truncation takes back $637$ of the loglogistic's tail. The curve form is itself an assumption, and the tail is where it shows.

Growth Function for a Partial Accident Year and a Policy Year A loglogistic pattern has $\omega = 1.5$ and $\theta = 40$ months, defined on the average accident date of the earned exposure. Using Clark's exposure adjustments, find the cumulative percent of ultimate for (a) an accident year $9$ months after it starts, and (b) a policy year of annual policies $18$ months after it starts. Compare (b) with an accident year at the same age.
Answer (a) Accident year at 9 months. Three quarters of the year is exposed, and the earned part has an average age of $4.5$ months. With $(4.5/40)^{1.5} = 0.03773$ , $G^{*}(4.5) = 0.03773/1.03773 = 0.03636$ :
 $\begin{align*} G_{AY}(9) &= \tfrac{9}{12} \times 0.03636 \\ &= 0.02727 \\ \text{LDF} &= 1/0.02727 \\ &= 36.67 \end{align*}$ 
The factor $1/G^{*}(4.5) = 27.50$ alone would develop the $9$ months already exposed. The extra $12/9$ adds the quarter of the year not yet exposed.

(b) Policy year at 18 months. Policies are written evenly over the first $12$ months, and each is exposed for $12$ months:
 $\begin{align*} \text{Expos}(18) &= 1 - \tfrac{1}{2}\left(2 - \tfrac{18}{12}\right)^2 \\ &= 0.875 \\ \text{AvgAge}(18) &= \frac{(18 - 12) + \tfrac{1}{3}(24 - 18)(1 - 0.875)}{0.875} \\ &= 7.143 \end{align*}$ 
With $(7.143/40)^{1.5} = 0.07546$ , $G^{*}(7.143) = 0.07546/1.07546 = 0.07017$ :
 $\begin{align*} G_{PY}(18) &= 0.875 \times 0.07017 \\ &= 0.06140 \\ \text{LDF} &= 16.29 \end{align*}$ 
An accident year at $18$ months is fully exposed with average age $12$ , so $G(12) = 0.1643/1.1643 = 0.1411$ and its LDF is $7.09$ . The policy year is less than half as developed at the same age, because its accidents happen later: some of its exposure is not yet earned, and what is earned is younger.