Worked Example of a Nonlinear Relativity Curve

From a Fitted Spline to a Rating Relativity

A continuous rating variable rarely has to affect expected claim frequency in a straight line. A log-link model with spline terms can represent a changing curve while keeping the fitted frequency positive. The practical task is to move carefully from the model specification to the spline basis values, then from the linear predictor to a fitted frequency and a relativity.

The examples use a hypothetical insurance portfolio and a Poisson claim-frequency model. The fitted coefficient values are hypothetical output from fitting the stated model; they are not presented as results for a real insurer or as coefficient estimates calculated from an unstated data set.

Definition: A fitted relativity at rating value $x$, relative to reference value $x_0$, is the ratio of the model’s fitted mean response at $x$ to its fitted mean response at $x_0$, with all other predictors held fixed. For a log-link model with linear predictors $\eta(x)$ and $\eta(x_0)$, this ratio is $\exp\{\eta(x)-\eta(x_0)\}$.

Specify the Model Before Evaluating the Curve

Let $N_j$ be the claim count for policy record $j$, and let $e_j$ be its exposure measured in policy-years. Suppose counts are conditionally independent and $N_j$ has a Poisson distribution with mean $\mu_j$, so that $E[N_j]=\operatorname{Var}(N_j)=\mu_j$. The model uses the exposure as a fixed offset: doubling exposure doubles expected count, all else equal.

For a continuous rating variable $x$, use the scaled value $z=(x-40)/10$. In this example, the spline has a knot at $x=50$, or $z=1$. Define the positive-part function by $a_+=\max(a,0)$. The model is

N_j\mid x_j,e_j \sim \operatorname{Poisson}(\mu_j), \qquad \log \mu_j=\log e_j+\eta(x_j)

\eta(x)=\beta_0+\beta_1z+\beta_2z^2+\beta_3z^3+\gamma(z-1)_+^3

The first four terms form a cubic polynomial. The final, truncated cubic term adds flexibility above the knot but is zero at and below it. Because the added term and its first two derivatives are zero at $z=1$, the fitted curve is continuous there, with continuous first and second derivatives. The third derivative can change at the knot. This is a one-knot cubic spline basis, not a natural cubic spline; its behavior outside the range of the data should therefore be treated cautiously.

The coefficients are estimated by fitting a Poisson GLM to the observed counts, with $\log e_j$ as the offset. Under the stated independence and Poisson assumptions, the likelihood is the product of Poisson probabilities with means $e_j\exp\{\eta(x_j)\}$. Fitting estimates the coefficients jointly; the calculations that follow use the resulting coefficient estimates rather than trying to recover them from raw observations.

Why a Relativity Is an Exponential Difference

The offset is the same for two otherwise identical records when the comparison is made at the same exposure. Their fitted mean counts are $e\exp\{\eta(x)\}$ and $e\exp\{\eta(x_0)\}$. Dividing cancels $e$, and the exponential rule turns a difference in linear predictors into a ratio of fitted means:

\frac{e\exp\{\eta(x)\}}{e\exp\{\eta(x_0)\}} =\exp\{\eta(x)-\eta(x_0)\}.

This is why a coefficient in a log-link model is not itself a percentage change in frequency. The entire basis-weighted linear predictor must be evaluated, and its difference from the reference predictor must be exponentiated. In particular, the cubic hinge term contributes only when $z>1$.

Worked Example: Evaluate a Fitted Relativity Above the Knot

A hypothetical insurer fits the model above to claim counts and policy-year exposures. The fitted coefficients are $\beta_0=\log(0.08)$, $\beta_1=0.12$, $\beta_2=-0.04$, $\beta_3=0.02$, and $\gamma=0.03$. Calculate the fitted frequency and the relativity at $x=60$, using $x=40$ as the reference. Also find the expected count for 2.5 policy-years at $x=60$.

Method and basis values. First calculate $z=(60-40)/10=2$. The linear, quadratic, and cubic basis values are $2$, $2^2=4$, and $2^3=8$. Since $z>1$, the hinge value is $(2-1)^3=1$.

Evaluate the non-intercept predictor. Its value at $x=60$ is $0.12(2)-0.04(4)+0.02(8)+0.03(1)=0.24-0.16+0.16+0.03=0.27$. At the reference $x=40$, $z=0$, so every non-intercept basis term is zero.

Calculate the relativity and frequency. The relativity to $x=40$ is $\exp(0.27)=1.3100$, rounded to four decimal places. The fitted frequency at $x=40$ is $\exp\{\beta_0\}=0.08$ claims per policy-year. Thus the frequency at $x=60$ is $0.08(1.3100)=0.1048$ claims per policy-year, rounded to four decimal places.

Apply exposure to obtain expected count. For 2.5 policy-years, the expected count is $2.5(0.104797\ldots)=0.261992\ldots$, or $0.2620$ claims to four decimal places. This is an expected count, not a prediction that the record will have a fraction of a claim.

Check and common mistake. The relative frequency is greater than one, as expected from a positive predictor difference of 0.27. A common error is to report $0.27$ as the relativity; it is the log-frequency difference. Exponentiating gives the frequency ratio, approximately $1.3100$.

Interpret Fitted Frequencies and Their Relativities

For this fitted model, the non-intercept predictor values at selected rating values are as follows. The displayed relativities use $x=40$ as the reference, where the non-intercept predictor is zero.

Rating value $x$	$z$	Non-intercept predictor	Relativity to $x=40$
30	-1	-0.18	0.8353
40	0	0.00	1.0000
50	1	0.10	1.1052
60	2	0.27	1.3100
70	3	0.78	2.1815

For example, at $x=30$, the hinge is zero and the non-intercept predictor is $-0.12-0.04-0.02=-0.18$, giving $\exp(-0.18)=0.8353$. At $x=70$, the hinge basis value is $(3-1)^3=8$; the predictor is $0.36-0.36+0.54+0.24=0.78$, giving $\exp(0.78)=2.1815$. The steep increase in this example is a consequence of the hypothetical fitted coefficients, not a general property of spline models.

A relativity compares fitted means under a specified reference. It does not directly give a premium unless the modeled response and any other rating components are accounted for. For instance, a frequency relativity is not automatically a pure-premium relativity if expected claim severity also varies with the rating variable.

Worked Example: Convert Frequency Relativities to Hypothetical Loss Costs

Suppose the same hypothetical model is used for a product whose expected claim severity is assumed constant at $6,000 for every value of $x$. Calculate the fitted pure premiums at $x=50$ and $x=70$, and compare them. Use the $x=40$ fitted frequency of 0.08 claims per policy-year.

Method. Under the stated constant-severity assumption, expected loss cost per policy-year is fitted frequency multiplied by $6,000. First find each frequency relativity to $x=40$.

At $x=50$, $z=1$, so the hinge is zero and the non-intercept predictor is $0.12-0.04+0.02=0.10$. The frequency relativity is $\exp(0.10)=1.105170\ldots$, or 1.1052 to four decimal places. The fitted frequency is $0.08(1.105170\ldots)=0.0884136$ claims per policy-year. The pure premium is $0.0884136(6{,}000)=\$530.48$, rounded to the nearest cent.

At $x=70$, $z=3$ and the hinge is $(3-1)^3=8$. The non-intercept predictor is $0.12(3)-0.04(9)+0.02(27)+0.03(8)=0.36-0.36+0.54+0.24=0.78$. The relativity is $\exp(0.78)=2.181472\ldots$, or 2.1815 to four decimal places. The fitted frequency is $0.08(2.181472\ldots)=0.1745178$ claims per policy-year. The pure premium is $0.1745178(6{,}000)=\$1{,}047.11$, rounded to the nearest cent.

Comparison and check. The ratio of the two pure premiums is $1{,}047.11/530.48$, approximately 1.9739. Directly, their frequency ratio is $\exp(0.78-0.10)=\exp(0.68)=1.9739$; constant severity cancels, so the two calculations agree. If severity varied with $x$, the frequency relativity alone would not determine the pure-premium ratio.

Common mistake. Using the $x=70$ frequency relativity as a dollar premium would confuse a dimensionless ratio with a loss cost. The conversion requires the reference frequency and the assumed severity; the constant-severity assumption is essential to this example.

Changing the Reference Does Not Change the Fitted Curve

The reference value is a reporting choice. A different reference changes the numerical relativities but not the fitted means. For example, a curve originally expressed relative to $x=40$ can be restated relative to $x=50$ by dividing every original relativity by the relativity at $x=50$. Equivalently, subtract the predictor at $x=50$ before exponentiating.

Worked Example: Re-Reference the Curve and Check an Exposure Offset

Using the same fitted model, express the $x=60$ relativity relative to $x=50$, rather than $x=40$. Then calculate the expected number of claims for 3 policy-years at $x=60$. Identify the error made by someone who uses the $x=40$ relativity as though it were relative to $x=50$.

Calculate both predictor values. At $x=60$, the non-intercept predictor is 0.27. At $x=50$, it is 0.10. Their difference is $0.27-0.10=0.17$, so the relativity to $x=50$ is $\exp(0.17)=1.1853$, rounded to four decimal places.

Check by dividing original relativities. The $x=60$ and $x=50$ relativities to $x=40$ are $\exp(0.27)=1.3100$ and $\exp(0.10)=1.1052$. Their ratio is $1.3100/1.1052$, approximately 1.1853, consistent with the predictor-difference calculation.

Calculate expected count. The fitted frequency at $x=60$ remains $0.08\exp(0.27)=0.104797\ldots$ claims per policy-year. For 3 policy-years, the expected count is $3(0.104797\ldots)=0.314392\ldots$, or $0.3144$ to four decimal places. Re-referencing does not alter this count.

Common mistake and correction. Reporting 1.3100 as the relativity to $x=50$ incorrectly retains the $x=40$ reference. The correction is to divide by the $x=50$ relativity, or subtract its predictor: $\exp(0.27-0.10)=1.1853$. Exposure affects expected count through the offset; it is not part of a relativity between rating values.

Practical Checks When Using a Spline Model

Confirm the variable scaling and knot.
Convert the observed rating value to the exact scaled value used in the fitted model. A different scaling or knot changes every basis value.

Evaluate each basis term separately.
In particular, calculate the positive-part term as zero when $z\leq 1$, and as $(z-1)^3$ when $z>1$.

Form the predictor difference.
For a relativity, compare the target value with the specified reference while holding other predictors fixed. The intercept cancels in the difference.

Exponentiate, then interpret.
The exponential of the predictor difference is the fitted mean ratio. Apply exposure to obtain expected counts; do not confuse that count with the relativity.

A frequent exam trap is to treat spline coefficients as if each were a direct relativity. They multiply basis functions, and the effect at a value depends on all active terms. Another is to forget that the comparison depends on its reference. State the reference explicitly whenever reporting a fitted relativity. These calculations are aligned with the CAS Exam 8 (Advanced Ratemaking) syllabus.

Check Your Understanding

Use the fitted coefficients and basis specified in this tutorial. Round relativities and expected counts to four decimal places unless a question requests otherwise.

Calculate the relativity at $x=45$ relative to $x=40$. (Answer: 1.0539)
Calculate the relativity at $x=55$ relative to $x=40$. (Answer: 1.1750)
Calculate the relativity at $x=65$ relative to $x=40$. (Answer: 1.5900)
If the fitted frequency at $x=40$ is 0.06 claims per policy-year, find the expected claim count for 4 policy-years at $x=60$. (Answer: 0.3144 claims)
Calculate the relativity at $x=30$ relative to $x=60$. (Answer: 0.6376)

What You'll Learn