Estimating Trend With Time-Series Regression

Why Use Regression to Estimate Trend?

Two observations can define a rate of change, but they cannot show whether the pattern between those observations is steady or unusual. With several historical periods, regression uses all the observations to estimate an overall direction. It also makes the time unit explicit, which matters when periods are not equally spaced.

This tutorial focuses on a log-linear regression of positive period-level loss costs. The estimated slope describes change on a logarithmic scale; converting that slope correctly gives the corresponding compounded trend rate. The estimate is meaningful only if the observations are comparable and the model is a reasonable summary of their pattern.

Definition: Let $C_j$ be the observed loss cost for period $j$, and let $t_j$ be the elapsed time, in years, from a chosen origin to that period. A log-linear trend model is $$ \ln(C_j)=a+bt_j+\varepsilon_j, $$ where $a$ is an intercept, $b$ is the slope per year, and $\varepsilon_j$ is the residual for period $j$. The fitted compounded annual trend rate is $e^b-1$.

The logarithm requires $C_j>0$. The time origin can be any convenient date: changing the origin changes the intercept, but not the fitted slope, provided the same time unit is used. For annual trend, express elapsed time in years. If observations are three years apart, their time coordinates must also be three years apart—not merely one row apart.

Fitting the Regression and Converting the Slope

For observations $(t_j,C_j)$, first set $y_j=\ln(C_j)$. The ordinary least squares slope and intercept are

\hat b= \frac{\sum_j(t_j-\bar t)(y_j-\bar y)} {\sum_j(t_j-\bar t)^2}, \qquad \hat a=\bar y-\hat b\bar t, \qquad \widehat{\ln(C_j)}=\hat a+\hat b t_j.

The slope $\hat b$ is a change in log loss cost per year, not itself the percentage trend. A one-year increase in time changes the fitted log cost by $\hat b$; exponentiating shows that the fitted loss cost is multiplied by $e^{\hat b}$. Thus the annual compounded rate is $e^{\hat b}-1$. For example, a slope of $0.05$ implies a rate of $e^{0.05}-1$, about 5.13%, not exactly 5%.

Interpretation: In the fitted model, a time increase of $h$ years corresponds to a multiplicative change of $e^{\hat b h}$, or a percentage change of $e^{\hat b h}-1$. The intercept gives the fitted log loss cost at $t=0$; exponentiating it gives the fitted loss cost at the chosen time origin.

Worked Example: Fitting an Exact Annual Trend

A hypothetical portfolio has comparable annual loss costs of $100.00, $110.00, $121.00, and $133.10 at elapsed years $t=0,1,2,3$. Fit the log-linear regression and calculate the annual trend. Use unrounded logarithms in the calculations, then report logged values and the slope to nine decimal places and the rate to two decimal places.

Method. Regress $y_j=\ln(C_j)$ on elapsed years $t_j$. The cost sequence is generated by multiplying by 1.10 each year, so it provides a check on the regression result.

The logged costs, shown to nine decimal places, are $4.605170186$, $4.700480366$, $4.795790546$, and $4.891100725$. Using the unrounded logarithms, their mean is $\bar y=4.748135456$; the mean time is $\bar t=1.5$. The centered time values are $-1.5,-0.5,0.5,1.5$. Because the log costs increase by $\ln(1.10)=0.095310180$ per year, their centered values are $-0.142965270,-0.047655090,0.047655090,0.142965270$, to nine decimal places.

The numerator and denominator of the slope are

\begin{aligned} S_{ty} &=(-1.5)(-0.142965270)+(-0.5)(-0.047655090)\\ &\quad +(0.5)(0.047655090)+(1.5)(0.142965270)\\ &\approx 0.476550899,\\ S_{tt}&=(-1.5)^2+(-0.5)^2+(0.5)^2+(1.5)^2=5,\\ \hat b&=\frac{S_{ty}}{S_{tt}}\approx 0.095310180. \end{aligned}

The fitted annual trend is $e^{0.095310180}-1=0.100000$, or 10.00% to two decimal places. The intercept is $\hat a=\bar y-\hat b\bar t\approx4.605170186$, and $e^{\hat a}=100$, as expected at year 0.

Check and mistake avoided. Each observed cost equals the previous one multiplied by 1.10, so a 10.00% estimated annual trend is consistent with the data. The calculation uses unrounded logarithms; small discrepancies can result if rounded displayed values are used to recompute the numerator. The slope is a log change, so it must be exponentiated before being reported as a percentage trend.

Why the Slope Gives a Compounded Rate

The conversion follows directly from the fitted log equation. At time $t$, the fitted log cost is $\hat a+\hat b t$. One year later, it is $\hat a+\hat b(t+1)$. Subtracting the first fitted log cost from the second gives $\hat b$. Exponentiating that difference gives the ratio of the fitted costs:

\frac{\exp(\hat a+\hat b(t+1))} {\exp(\hat a+\hat b t)} =e^{\hat b}.

Therefore, the fitted cost increases by the multiplicative factor $e^{\hat b}$ per year, and the corresponding percentage increase is $e^{\hat b}-1$. Over $h$ years the ratio is $e^{\hat b h}$, which is equivalent to compounding the annual factor $e^{\hat b}$ for $h$ years.

Worked Example: Regression With Uneven Observations

A hypothetical equipment-insurance portfolio has period-level loss costs recorded at elapsed years 0, 1, 2, and 3: $100.00, $120.00, $120.00, and $144.00. Estimate the annual trend by log-linear regression. Report the slope to six decimal places and the trend to two decimal places.

Method. Take natural logarithms and regress them on elapsed years. The costs do not follow a perfectly smooth year-to-year pattern, so this example illustrates how the regression uses all observations rather than only the first and last.

Let $c=\ln(1.20)=0.182321557$. The logged costs are $\ln(100)$, $\ln(100)+c$, $\ln(100)+c$, and $\ln(100)+2c$. Since the time coordinates are $0,1,2,3$, $\bar t=1.5$ and $S_{tt}=5$. The constant $\ln(100)$ contributes zero to the centered cross-product. The numerator is

S_{ty}=(-1.5)(0)+(-0.5)c+(0.5)c+(1.5)(2c)=3c=0.546964670.

Thus $\hat b=0.546964670/5=0.109392934$, or $0.109393$ to six decimal places. The annual compounded trend is

e^{0.109392934}-1\approx0.1156=11.56\%.

Check and mistake avoided. The endpoints alone imply a two-interval multiplicative change of $144/100=1.44$, whose annualized rate is 20%; that is not the regression result because the two middle observations also affect the fitted slope. The regression gives an annual trend of 11.56%. Do not substitute an endpoint-only calculation when the requested method is regression.

Time Coordinates, Data Comparability, and Assumptions

Regression cannot repair inconsistent inputs. Period-level loss costs should be prepared on a comparable basis: the measure, coverage, and exposure basis should be consistent across periods, and changes in the portfolio that affect average loss cost should not be mistaken for trend. The earlier material on aligning period dates and choosing loss measures is relevant when constructing the observations. The regression itself estimates a pattern in the supplied values; it does not establish that all observed change is underlying trend.

The log-linear model assumes that the expected log loss cost changes approximately linearly with elapsed time. In practical terms, that represents a roughly constant multiplicative trend over the fitted interval. A sustained shift in the trend, an unusual period, or a curved pattern can make one slope a poor summary. Inspecting the observations and residuals helps assess whether a single trend is reasonable, but does not prove the assumptions.

For ordinary least squares, a standard statistical interpretation also assumes that the log-scale residuals have mean zero conditional on time. For conventional uncertainty calculations, residuals are commonly assumed independent with constant variance. These assumptions concern the statistical behavior around the fitted line; they are not guaranteed by the act of computing the regression.

The fitted value after exponentiating, $e^{\hat a+\hat b t}$, is a fitted value on the loss-cost scale associated with the fitted log cost. If errors are normally distributed on the log scale, exponentiating the mean log cost gives the conditional median on the original scale, not the arithmetic mean. For a lognormal model with residual variance $\sigma^2$, the arithmetic mean includes the factor $e^{\sigma^2/2}$. If that variance is stable over time, the factor cancels in a ratio between periods; if it changes, the log-scale slope alone need not describe the change in arithmetic mean loss cost.

Worked Example: Using Actual Elapsed Years

A hypothetical property portfolio has loss costs at elapsed years $t=0,2,5$ of $200.00$, $220.50$, and $255.2563125$. These values were generated as $200(1.05)^t$. Fit the log-linear regression using the actual elapsed years, estimate the annual trend, and project the fitted loss cost to year 6. Report the rate to two decimal places and the projection to the nearest cent.

Method. Use $y=\ln(C)$ and the time coordinates $0,2,5$. Write $c=\ln(1.05)$. The logged costs are exactly $\ln(200)+tc$, so the values lie on a straight line in log cost against elapsed years.

The mean time is $\bar t=(0+2+5)/3=7/3$, and the sum of squared centered times is

S_{tt}=(0-7/3)^2+(2-7/3)^2+(5-7/3)^2 =\frac{49+1+64}{9}=\frac{38}{3}.

Because $y=\ln(200)+tc$, the centered log costs equal $c(t-\bar t)$. Therefore $S_{ty}=cS_{tt}$, and $\hat b=S_{ty}/S_{tt}=c=\ln(1.05)\approx0.048790164$. The annual trend is $e^{\hat b}-1=0.05$, or 5.00%.

At year 6, the fitted loss cost is $200(1.05)^6=268.019128125$, which rounds to $268.02. The year-6 projection is one year beyond the last observation, at year 5.

Check and mistake avoided. The fitted line recovers the generating 5% annual rate and gives the year-6 value by applying one additional year of growth. Treating the three observations as equally spaced row numbers would ignore the actual gaps of two and three years and would not estimate an annual trend correctly.

Common Exam Traps and Practical Interpretation

A regression estimate is a summary of the data and model, not automatic evidence that a constant trend will continue. A defensible interpretation names the time unit, the loss-cost measure, and the period over which the estimate was fitted. It also distinguishes a model-based rate from a rate selected after considering data quality and other relevant evidence.

Reporting $\hat b$ as the percentage trend. If $\hat b=0.095310$, reporting 9.531% is incorrect. The correction is $e^{0.095310}-1$, which is approximately 10.00%.
Using the observation number instead of elapsed time. When dates are unevenly spaced, row numbers change the scale of the slope. Use time in years (or another clearly stated unit).
Taking logarithms of nonpositive values. The log-linear model requires positive loss costs. Zero or negative observations need a different treatment; they cannot simply be logged.
Assuming the intercept is the portfolio's observed starting value. The exponentiated intercept is a fitted value at $t=0$, which may differ from the first observed value, especially when the observations do not fall exactly on the fitted line.
Interpreting a fitted log cost as an arithmetic mean without qualification. With lognormal errors, exponentiating the fitted log mean gives a median; an arithmetic mean requires a variance adjustment.

Key takeaway: Regress the natural logarithm of positive, comparable period-level loss costs on actual elapsed time. Convert the estimated log slope $\hat b$ to a compounded annual rate with $e^{\hat b}-1$, and assess whether a constant log-linear trend is credible for the data and intended use. This treatment is aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus.

Check Your Understanding

For each problem, use the stated elapsed times and report an annual compounded rate unless the question asks for something else.

A hypothetical portfolio has loss costs of $50, $55, and $60 at years 0, 1, and 2. Fit the log-linear regression using all three observations. What is the annual trend, to three decimal places? (Answer: 9.545%)
A hypothetical program records loss costs of $80, $88, and $96 at years 0, 2, and 4. Fit a log-linear regression using actual elapsed years. What is the annual trend, to three decimal places? (Answer: 4.663%)
At years 0, 1, 2, and 3, a hypothetical book has loss costs of $100, $110, $100, and $110. Calculate the OLS log slope and convert it to an annual trend. What is the trend, to three decimal places? (Answer: 1.925%)
A fitted regression has a slope of $0.07$ per year. Calculate its implied annual compounded trend rate to three decimal places. (Answer: 7.251%)
A hypothetical portfolio's loss cost is $100 at year 0 and $115.7625 at year 3. The two observations lie exactly on a constant compounded trend pattern. What annual trend do they imply, and what would the fitted loss cost be at year 4? (Answer: 5.000% annual trend; $121.55 at year 4)

What You'll Learn