Why Regress Log Frequency on Time?
Claim frequency may rise or fall by a roughly constant percentage each year, even when its dollar or count increase is not constant. For instance, a frequency that grows by 10% annually adds fewer claims in an early year than in a later year because the later increase is applied to a larger base. A straight-line trend in log frequency represents this kind of multiplicative change.
This tutorial estimates a constant annual frequency trend by ordinary least squares (OLS) regression of log frequency on time. The frequencies used below are already stated on a consistent basis, such as claims per 1,000 exposure units. The focus is on the regression and interpretation of its slope, not on how the frequencies or exposure measures are constructed. This method is aligned with the CAS Exam 5 (Basic Ratemaking and Estimated Claim Liabilities) syllabus.
The OLS calculation treats the listed periods equally unless a different regression specification is explicitly given. Its usual unbiasedness interpretation requires the conditional mean of the residual to be zero at each observed time, \(E[\varepsilon_j\mid t_j]=0\). For conventional standard errors and tests, additional assumptions—such as independent residuals with a common variance—are usually imposed. A fitted trend is not automatically evidence that the trend will continue; it summarizes the historical pattern under the stated model.
The log transformation requires strictly positive frequencies. If a period has zero claims, its frequency is zero and its logarithm is undefined. Adding an arbitrary constant to make the logarithm calculable changes the method and can substantially affect the estimated trend; it is not a routine correction to this OLS calculation.
The OLS Slope and Its Meaning
For \(n\) observations, define \(\bar t\) as the average of the times and \(\overline{\ln f}\) as the average of the logged frequencies. The OLS slope and intercept are
The fitted log frequency at time \(t\) is \(\widehat{\alpha}+\widehat{\beta}t\). Exponentiating it gives the fitted frequency on the original scale:
Thus the estimated one-year multiplicative factor is \(\exp(\widehat{\beta})\). The corresponding annual percentage change is
A positive slope means an increasing fitted frequency; a negative slope means a decreasing fitted frequency. The slope itself is a change in log frequency, not an exact percentage. For a small slope, \(\widehat{\beta}\) is approximately the annual proportional change in decimal form, but the exponentiation gives the exact percentage implied by the fitted model.
Why Exponentiating the Slope Gives the Annual Factor
The relationship follows directly from the fitted line. Advancing time by one year changes the fitted log frequency by
Exponentiating both fitted log frequencies and taking their ratio gives
The intercept cancels, so the same multiplicative factor applies at every time under the constant-slope model. Subtracting 1 from that factor converts it to a proportional change; multiplying by 100 expresses the change as a percentage.
Worked Examples
Worked Example: Estimating an Increasing Trend
A hypothetical portfolio has observed frequencies of 4, 5, 5, and 8 claims per 1,000 exposure units at times \(t=0,1,2,3\), respectively. Estimate the OLS log-linear slope and annual percentage trend. Use the stated frequencies as the regression observations.
Method. Regress the natural logarithms of the four frequencies on time. The mean time is \(\bar t=(0+1+2+3)/4=1.5\). The centered time values are \(-1.5,-0.5,0.5,1.5\), and their squared values sum to \(2.25+0.25+0.25+2.25=5\).
The frequency logs are \(\ln4,\ln5,\ln5,\ln8\). Their mean is \[ \overline{\ln f} =\frac{\ln4+2\ln5+\ln8}{4} =\frac{6.684612}{4} =1.671153. \] The numerator of the slope is \[ \begin{aligned} &(-1.5)(\ln4-1.671153) +(-0.5)(\ln5-1.671153)\\ &\quad +(0.5)(\ln5-1.671153) +(1.5)(\ln8-1.671153)\\ &=1.5\ln(8/4)=1.5\ln2=1.039721. \end{aligned} \] Therefore, \[ \widehat{\beta}=\frac{1.039721}{5}=0.207944. \]
The annual factor is \(\exp(0.207944)=2^{0.3}=1.231144\). The estimated annual percentage change is \((1.231144-1)\times100\%=23.1144\%\), or approximately a 23.11% increase per year. As a check, the fitted factor over three years is \(\exp(3\widehat{\beta})=2^{0.9}\), approximately 1.866. OLS fits all four log frequencies; it does not require the fitted three-year factor to equal the observed endpoint ratio \(8/4=2\).
Common mistake avoided: dividing the change between the endpoint log frequencies by three would give an endpoint-based rate, not the OLS slope using all four observations. The OLS numerator and denominator must use all centered time and log-frequency values.
Worked Example: Interpreting a Decreasing Trend
In a hypothetical book, frequencies at \(t=0,1,2\) are 16, 12, and 9 claims per 1,000 exposure units. Estimate the annual trend by OLS and state its direction and magnitude.
The mean time is 1, so the centered times are \(-1,0,1\), and the denominator is \((-1)^2+0^2+1^2=2\). The three frequencies form a geometric sequence: \(12/16=9/12=3/4\). Consequently, their logs lie exactly on a straight line with a one-year log change of \(\ln(3/4)\). Directly applying the OLS formula gives \[ \widehat{\beta} =\frac{(-1)\ln16+(1)\ln9}{2} =\frac{\ln(9/16)}{2} =\ln(3/4) =-0.287682. \]
The annual factor is \(\exp(-0.287682)=0.75\). Therefore, the annual percentage change is \((0.75-1)\times100\%=-25\%\): the fitted frequency decreases by 25% per year. The model’s two-year factor is \(0.75^2=0.5625\), which matches the observed ratio \(9/16=0.5625\). This endpoint check is appropriate here because all three log observations lie exactly on the fitted line.
Common mistake avoided: reporting \(28.77\%\) as the decrease by treating the absolute slope as a percentage gives the wrong answer. The exact annual decrease is \(1-\exp(\widehat{\beta})=25\%\).
Worked Example: Using Unequally Spaced Times
A hypothetical program has frequencies of 4, 6, and 9 claims per 1,000 exposure units at \(t=0,1,3\). Estimate the OLS slope and annual trend. The last observation is three years after the first, not two.
The mean time is \(\bar t=(0+1+3)/3=4/3\). The centered times are \(-4/3,-1/3,5/3\), so the denominator is \[ \frac{16}{9}+\frac{1}{9}+\frac{25}{9}=\frac{14}{3}. \] The geometric mean of the three frequencies is 6, so the centered log frequencies are \(\ln(4/6)=-L,\ 0,\ \ln(9/6)=L\), where \(L=\ln(3/2)=0.405465\). The slope numerator is \[ \left(-\frac43\right)(-L) +\left(-\frac13\right)(0) +\left(\frac53\right)L =3L=1.216395. \] Thus \[ \widehat{\beta} =\frac{3L}{14/3} =\frac{9\ln(3/2)}{14} =0.260656. \]
The annual factor is \(\exp(0.260656)=1.297781\), giving an estimated annual increase of \((1.297781-1)\times100\%=29.7781\%\), or approximately 29.78% per year. The calculation uses each actual time, including \(t=3\); treating the observations as equally spaced would change the estimate.
Common mistake avoided: dividing the endpoint log change by the elapsed three years is not the OLS slope when an intermediate observation is included. OLS uses the centered values of all three times and all three log frequencies.
Interpretation, Limitations, and Exam Traps
Log-linear regression is useful when a constant proportional trend is a reasonable summary of the observed frequencies. It also provides a consistent way to use several experience periods rather than relying only on the first and last observations. But the fitted slope answers a specific question: what is the linear OLS trend in the supplied log frequencies against the supplied time values? It does not, by itself, establish why the observed pattern occurred or whether it should be projected forward.
A few distinctions matter in exam calculations:
- Use the stated time scale. If time is measured in years, the exponentiated slope is an annual factor. If time is measured in months, it is a monthly factor; it must be converted appropriately before calling it an annual trend.
- Use actual time gaps. Unequally spaced periods must be entered at their actual time values. The last observation’s position is not determined by its row number.
- Exponentiate before converting to a percentage. The annual percentage change is \((\exp(\widehat{\beta})-1)\times100\%\), not \(\widehat{\beta}\times100\%\) except as a small-slope approximation.
- Keep the direction. A negative slope gives a factor below 1 and a percentage decrease. A positive slope gives a factor above 1 and a percentage increase.
- Do not confuse a log-scale fit with an arithmetic mean prediction. Exponentiating a fitted log value gives the fitted value on the original scale under the specified log-linear calculation. If a separate stochastic model is used to estimate an arithmetic conditional mean, any needed retransformation adjustment must be justified by that model.
- Check for a trend that is not constant. A straight line on the log scale assumes a constant multiplicative factor per year. A pronounced curve, abrupt operational change, or other instability can make one slope an inadequate summary.
When the input consists of observed counts divided by exposure, the resulting frequencies may have different levels of statistical precision. The unweighted OLS calculations in this tutorial give each listed period equal influence on the log scale; they do not automatically account for that difference. The regression should therefore be interpreted in light of how its observations were defined and the purpose of the estimate.
Check Your Understanding
For each problem, compute the OLS trend on the log-frequency scale and interpret the result in context.
- A hypothetical service line has frequencies of 4, 6, and 9 claims per 1,000 exposure units at \(t=0,1,2\). Find the OLS slope and annual percentage change. (Answer: \(\widehat{\beta}=\ln(1.5)=0.405465\); 50.00% increase per year.)
- For a hypothetical product, frequencies are 20, 18, and 16 claims per 1,000 exposure units at \(t=0,1,2\). Calculate the OLS slope, annual factor, and percentage change. (Answer: \(\widehat{\beta}=\tfrac12\ln(0.8)=-0.111572\); factor \(0.894427\); 10.56% decrease per year.)
- A hypothetical portfolio has frequencies 4, 6, and 9 at \(t=0,1,3\). Use the actual time spacing to estimate the slope and annual trend. (Answer: \(\widehat{\beta}=\tfrac{9}{14}\ln(3/2)=0.260656\); approximately 29.78% increase per year.)
- Frequencies are 10, 14.4, and 24.8832 claims per 1,000 exposure units at \(t=0,2,5\). The figures follow an exact constant annual multiplicative pattern. Estimate the annual factor and percentage change using log-linear regression. (Answer: \(\widehat{\beta}=\ln(1.2)=0.182322\); factor \(1.2\); 20.00% increase per year.)
- A hypothetical period has zero claims and therefore zero observed frequency. Can its log frequency be included directly in this OLS calculation, and is adding 1 a standard correction? (Answer: No; \(\ln(0)\) is undefined, and adding 1 changes the data and is not a standard correction to this method.)