Why Use Polynomial Rating Effects?
A continuous rating variable does not always have a constant effect on expected claim count. For example, the change in expected claims associated with one more unit of a hypothetical vehicle-use measure might differ between low and high values of that measure. A polynomial effect lets the fitted relationship curve while retaining a compact model.
Under a log link, polynomial terms act on the logarithm of the conditional mean. Their coefficients therefore combine to determine the fitted mean, the relativity between two predictor values, and the local rate of change. A coefficient on a squared or cubed term is not, by itself, a multiplicative rating factor.
For illustration, suppose the response \(Y\) is a claim count over a fixed exposure period and \(Y\mid z\sim\operatorname{Poisson}(\mu(z))\). The log-link mean model is \(\log\mu(z)=\beta_0+g(z)\), so \(\mu(z)=\exp(\beta_0+g(z))\). The polynomial interpretation below depends on the mean model and log link, not on Poisson variance specifically. If a different response distribution is used, the same mean and relativity calculations apply when it has the same log-link mean specification.
Centering and scaling, as introduced in the earlier material on centering and scaling continuous rating predictors, can make the terms easier to work with and the coefficients easier to interpret numerically. The fitted curve does not change merely because the same polynomial is expressed in a different coordinate system; the coefficient values do change. In this tutorial, \(c\) and \(s\) are stated whenever they are needed to interpret a coefficient per unit of the original variable.
Relativities and Changing Slopes
For two values \(z_a\) and \(z_b\), hold all other predictors and the offset fixed. The fitted relativity at \(z_b\) relative to \(z_a\) is the ratio of the two fitted conditional means. Since the link is logarithmic, this ratio is the exponential of the difference in polynomial contributions.
For a quadratic effect, the log-mean slope with respect to \(z\) is \(\beta_1+2\beta_2z\). For a cubic effect it is \(\beta_1+2\beta_2z+3\beta_3z^2\). If \(z=(x-c)/s\), the slope with respect to the original \(x\) is the corresponding derivative with respect to \(z\), divided by \(s\). These derivatives describe the local change in log mean; the derivative of the mean itself is the mean multiplied by that log-mean slope.
A positive log-mean slope means the fitted mean is locally increasing; a negative one means it is locally decreasing. A zero slope is a stationary point, not automatically a maximum. For a quadratic, the sign of \(\beta_2\) identifies the curvature: if \(\beta_2>0\), the stationary point is a minimum; if \(\beta_2<0\), it is a maximum. For a cubic, stationary points require solving a quadratic equation in \(z\), and the curve can change direction more than once.
State the log-mean polynomial and identify the transformed predictor \(z=(x-c)/s\).
Calculate \(g(z_b)-g(z_a)\); exponentiate this difference to obtain the fitted relativity.
Differentiate every included power of \(z\), then divide by \(s\) to express the slope per unit of \(x\).
Confirm the comparison holds other rating variables fixed and is within a range supported by the modeled data.
Worked Quadratic Effects
Worked Example: Comparing Adjacent Values and a Curved Slope
A hypothetical insurer models claim count \(Y\) over a fixed exposure period using \(Y\mid z\sim\operatorname{Poisson}(\mu(z))\). The fitted mean is \(\log\mu(z)=\log(120)+0.08z-0.02z^2\), where \(z=(x-40)/10\). All other predictors and exposure are held fixed. Find the fitted means at \(x=40,50,60\), the relativity at \(x=50\) versus \(x=40\), and the local log-mean slope per unit of \(x\) at each point. No interest or mortality assumption enters this claim-count calculation.
Method. Evaluate the polynomial at each transformed value. For the relativity, subtract the polynomial values and exponentiate. For the slopes, differentiate with respect to \(z\) and divide by the scale \(s=10\).
At \(x=40\), \(z=(40-40)/10=0\), so \(g(0)=0.08(0)-0.02(0)^2=0\) and \(\mu(0)=120e^0=120.00\).
At \(x=50\), \(z=(50-40)/10=1\), so \(g(1)=0.08(1)-0.02(1)^2=0.08-0.02=0.06\). Thus \(\mu(1)=120e^{0.06}=120(1.0618365)=127.42\). Relative to \(x=40\), the fitted relativity is \(e^{g(1)-g(0)}=e^{0.06}=1.0618\).
At \(x=60\), \(z=(60-40)/10=2\), so \(g(2)=0.08(2)-0.02(2)^2=0.16-0.08=0.08\). Thus \(\mu(2)=120e^{0.08}=120(1.0832871)=130.00\).
The derivative with respect to \(z\) is \(g'(z)=0.08-0.04z\). Therefore the local log-mean slopes per unit of \(x\) are: at \(x=40\), \(0.08/10=0.008\); at \(x=50\), \((0.08-0.04)/10=0.004\); and at \(x=60\), \((0.08-0.08)/10=0\). The fitted curve is still increasing between \(x=50\) and \(x=60\), but its slope has fallen to zero at \(x=60\).
Check and trap avoided. The relativity at \(x=50\) is \(1.0618\), consistent with \(127.42/120.00\). A common mistake is to use \(e^{0.08}=1.0833\), treating \(\beta_1\) as the whole effect at \(z=1\). The correction is to include the squared-term contribution: \(g(1)=0.08-0.02=0.06\). Values are rounded to two decimals for means and four decimals for the relativity.
The example illustrates why polynomial coefficients must be interpreted jointly. For a quadratic, the linear coefficient is the log-mean slope at \(z=0\), while the squared coefficient controls how that slope changes as \(z\) moves away from zero. Neither coefficient alone gives the relativity between arbitrary values.
The derivative and relativity formulas follow directly from the model. If \(\log\mu(z)=\beta_0+g(z)\), then \(\mu(z)=e^{\beta_0}e^{g(z)}\). Taking the ratio at \(z_b\) and \(z_a\) cancels \(e^{\beta_0}\), leaving \(e^{g(z_b)-g(z_a)}\). Differentiating \(\log\mu=\beta_0+g(z)\) gives \(d\log\mu/dz=g'(z)\); applying \(dz/dx=1/s\) gives \(d\log\mu/dx=g'(z)/s\). Since \(d\mu/dx=\mu(d\log\mu/dx)\), the mean-scale derivative follows as well.
Cubic Effects and Turning Points
Worked Example: A Cubic Relativity Across the Rating Range
A hypothetical insurer models claim count \(Y\) over a fixed exposure period with \(Y\mid z\sim\operatorname{Poisson}(\mu(z))\) and \(\log\mu(z)=\log(80)+0.05z-0.01z^2+0.002z^3\), where \(z=(x-30)/5\). Find the fitted means at \(x=25\) and \(x=40\), the relativity at \(x=40\) versus \(x=25\), and the local log-mean slopes per unit of \(x\) at those values. No interest or mortality assumption is relevant.
Method. The cubic contribution includes all three powers. Evaluate it at each endpoint, then subtract the contributions before exponentiating the relativity.
At \(x=25\), \(z=(25-30)/5=-1\), and \(g(-1)=0.05(-1)-0.01(-1)^2+0.002(-1)^3=-0.05-0.01-0.002=-0.062\). The fitted mean is \(80e^{-0.062}=80(0.939882)=75.19\).
At \(x=40\), \(z=(40-30)/5=2\), and \(g(2)=0.05(2)-0.01(2)^2+0.002(2)^3=0.10-0.04+0.016=0.076\). The fitted mean is \(80e^{0.076}=80(1.078963)=86.32\).
The difference in polynomial contributions is \(0.076-(-0.062)=0.138\), so the relativity is \(e^{0.138}=1.1480\). Checking by division, \(86.32/75.19\) is approximately \(1.148\), subject to rounding the displayed means.
The derivative is \(g'(z)=0.05-0.02z+0.006z^2\). At \(z=-1\), it is \(0.05+0.02+0.006=0.076\) per unit of \(z\), or \(0.076/5=0.0152\) per unit of \(x\). At \(z=2\), it is \(0.05-0.04+0.024=0.034\) per unit of \(z\), or \(0.034/5=0.0068\) per unit of \(x\). The local slope is positive at both evaluated points and smaller at the higher value.
Check and trap avoided. The endpoint comparison uses \(g(2)-g(-1)=0.138\), so the fitted mean at \(x=40\) exceeds the mean at \(x=25\). Omitting the cubic term, or using only \(\beta_1(z_b-z_a)\), would give the wrong relativity. Mean values are rounded to two decimals and the relativity and slopes to four significant figures.
A cubic can provide a changing slope that a quadratic cannot capture, but this flexibility does not guarantee a sensible curve outside the observed rating range. Its derivative is quadratic, so it may have zero, one, or two stationary points. Those points should be assessed in context; an apparent turning point far beyond the range of observed \(x\) values is not strong evidence about real risk at that value.
Finding a Quadratic Stationary Point
Worked Example: Locating the Fitted Maximum
For a hypothetical fixed-exposure claim-count model, assume \(Y\mid z\sim\operatorname{Poisson}(\mu(z))\) and \(\log\mu(z)=\log(200)+0.12z-0.02z^2\), with \(z=(x-45)/5\). Find the value of \(x\) at the stationary point, determine whether it is a maximum or minimum, and calculate the fitted mean there and at \(x=55\). No interest or mortality assumption enters.
Method. Set the derivative of the log mean to zero. The stationary point of the log mean is also the stationary point of the mean because \(\mu\) is positive and \(d\mu/dx=\mu\,d\log\mu/dx\).
The derivative with respect to \(z\) is \(g'(z)=0.12-0.04z\). Setting it to zero gives \(0.12-0.04z=0\), so \(z=3\). Converting back, \(x=45+5(3)=60\). Since \(\beta_2=-0.02<0\), the quadratic opens downward on the log-mean scale, and this stationary point is a maximum.
At \(x=60\), \(g(3)=0.12(3)-0.02(3)^2=0.36-0.18=0.18\). Hence \(\mu(3)=200e^{0.18}=200(1.197217)=239.44\).
At \(x=55\), \(z=(55-45)/5=2\). Then \(g(2)=0.12(2)-0.02(2)^2=0.24-0.08=0.16\), so \(\mu(2)=200e^{0.16}=200(1.173511)=234.70\). The relativity at \(x=60\) versus \(x=55\) is \(e^{0.18-0.16}=e^{0.02}=1.0202\).
Check and trap avoided. The derivative changes from positive before \(z=3\) to negative after \(z=3\), confirming a maximum. The fitted mean at the stationary point is about 2.02% greater than at \(x=55\), consistent with the relativity \(1.0202\). A zero derivative alone does not identify a maximum; the sign of the quadratic coefficient supplies that check. Means are rounded to two decimals and the relativity to four decimals.
Modeling and Exam Considerations
Before interpreting fitted polynomial effects, confirm what the response mean represents. For example, a frequency model with a log-exposure offset has \(\log\mu=\log(e)+\) the rating linear predictor, where \(e\) is exposure. The comparison formulas above assume exposure and other predictors are the same at both points, so the offset cancels. If exposure differs, the ratio of fitted counts also reflects that exposure difference; it is not solely a rating relativity.
Keep the polynomial terms in a coherent model. If a cubic term is included, the linear and quadratic terms are ordinarily retained as well, even if an individual lower-order coefficient is not statistically significant. This preserves a complete polynomial specification and makes the fitted curve's behavior interpretable under changes of origin. The coefficients depend on the chosen centering and scale, so report those conventions when interpreting or implementing the effect.
- Do not call \(\beta_2\) a standalone factor. It multiplies \(z^2\) on the log scale; its contribution at a particular value is \(\beta_2z^2\).
- Distinguish a local slope from a finite relativity. A derivative describes an instantaneous change; a comparison between two distinct values requires the difference \(g(z_b)-g(z_a)\).
- Convert units correctly. When \(z=(x-c)/s\), a derivative per unit of \(z\) must be divided by \(s\) to obtain a derivative per unit of \(x\).
- Be cautious about extrapolation. A polynomial can rise or fall rapidly outside the range where the data support the fitted curve.
Check Your Understanding
For each problem, hold other predictors and exposure fixed unless the problem states otherwise.
- A hypothetical Poisson claim-count model has \(\log\mu=\log(50)+0.10z-0.02z^2\), where \(z=(x-20)/4\). Find the relativity at \(x=24\) versus \(x=20\), and the local log-mean slope per unit of \(x\) at \(x=24\). (Answer: \(e^{0.08}=1.0833\); slope \(0.015\) per unit of \(x\).)
- A quadratic effect is \(g(z)=0.06z-0.01z^2\), with \(z=(x-50)/10\). Find the stationary point and state whether it is a maximum or minimum. (Answer: \(z=3\), so \(x=80\); it is a maximum.)
- A hypothetical claim-count model has \(g(z)=0.03z-0.004z^2+0.001z^3\). Calculate the relativity at \(z=2\) versus \(z=0\). (Answer: \(e^{0.052}=1.0534\).)
- A quadratic effect has \(g(z)=-0.04z+0.01z^2\), where \(z=(x-30)/5\). Find the stationary point in \(x\) and identify its type. (Answer: \(z=2\), so \(x=40\); it is a minimum.)
- A model uses \(g(z)=0.05z-0.01z^2+0.002z^3\), with \(z=(x-10)/2\). Find the relativity at \(x=12\) versus \(x=10\), and the local log-mean slope per unit of \(x\) at \(x=12\). (Answer: \(z\) changes from 0 to 1, so the relativity is \(e^{0.042}=1.0429\); the slope is \((0.05-0.02+0.006)/2=0.018\) per unit of \(x\).)