Linear Trend

Study guide: Exam 5

A Linear Trend models a ratemaking statistic as changing by a constant amount per unit of time — a fixed number of dollars of severity or pure premium a year, rather than a fixed percentage. It is fitted by least-squares regression of the statistic itself on time, and projected by adding the slope times the length of the trend period.

$\hat{y}_t = a + b\,t$

$b = \frac{\sum (t - \bar{t})(y_t - \bar{y})}{\sum (t - \bar{t})^2}$

$\text{Projected value} = y + b \times n$

$a$ and $b$ are the intercept and slope of an ordinary least-squares fit, $y$ the historical value being trended, and $n$ the trend period in years — the same period used with an exponential trend (see Loss Trend). Werner & Modlin: with a linear selection the trend is a constant amount, so the projected change is the selected annual amount multiplied by the length of the trend period.
The analysis works on differences between periods rather than percentage changes, and a linear fit's slope is in the units of the statistic (dollars of pure premium per year, claims per exposure per year).
The implied percentage falls over time. For a rising series, the same $b$ is a shrinking share of a growing level, so a linear trend projects a decelerating rate of change.
Its shortcoming. For a falling series, a straight line eventually crosses zero and projects negative frequencies or severities — impossible in insurance. Both models can be reasonable for increasing trends; the exponential model is used far more often because it has neither problem.
Choose between the two by the data's behaviour and the fit statistics ( $R^2$ , residual pattern), not by which gives the more convenient answer; ASOP 13 asks for the choice of model to be reasoned. The same fitting approach serves premium and exposure trend.

Linear Fit to Pure Premiums Accident-year pure premiums: $2020$ : $\$212$ ; $2021$ : $\$219$ ; $2022$ : $\$227$ ; $2023$ : $\$233$ ; $2024$ : $\$241$ . Trend the AY $2022$ pure premium $4.5$ years forward using a linear fit, and compare with an exponential fit.
Answer Centre time at $2022$ ( $t = -2, \dots, 2$ ; $\bar{y} = \$226.40$ ; $\sum t^2 = 10$ ):
 $\begin{align*} \sum t\,(y_t - \bar{y}) &= -2(212) - 219 + 233 + 2(241) \\ &= 72 \\ b &= 72 / 10 = \$7.20 \text{ per year} \\ \text{Projected PP} &= \$227 + \$7.20 \times 4.5 \\ &= \$259.40 \end{align*}$ 
(Since $\sum t = 0$ , subtracting $\bar{y}$ does not change the sum.) The slope is $3.4\%$ of the $2020$ level but only $3.0\%$ of the $2024$ level: a constant amount is a falling percentage.

An exponential fit to the same data gives $3.24\%$ a year and a projection of $\$227 \times e^{4.5 \times 0.03183} = \$261.97$ . The linear projection is $\$2.57$ lower because it does not compound. Over a $4.5$ -year period the two nearly agree; the gap widens with the trend period and the steepness of the trend.

A Falling Frequency Projected Too Far Claim frequency on a line where a safety device is spreading: $0.0800$ , $0.0725$ , $0.0660$ , $0.0590$ , $0.0520$ for five consecutive years ( $t = 0, \dots, 4$ ). Project the frequency $8$ years beyond the last point ( $t = 12$ ) with a linear and an exponential fit.
Answer
 $\begin{align*} b &= \frac{\sum (t - 2)(y_t - \bar{y})}{10} = \frac{-0.0695}{10} \\ &= -0.00695 \text{ per year} \\ a &= \bar{y} - 2b = 0.0659 + 0.0139 = 0.0798 \\ \hat{y}_{12} &= 0.0798 - 0.00695 \times 12 \\ &= -0.0036 \end{align*}$ 
The linear model projects a negative frequency. The exponential fit to the same five points gives an annual trend of $-10.1\%$ and $\hat{y}_{12} = 0.0224$ — still steep, but a possible number.

Even before it crosses zero the linear projection misbehaves: its fixed decline of $0.00695$ is $13\%$ of the last observed value ( $0.0520$ ) but $40\%$ of its own $t = 9$ value ( $0.0173$ ), an accelerating percentage decline that no one intended to select. For a decreasing statistic the exponential form is the defensible default.