Nonlinear Price Response With Polynomial Terms

Why Model Curved Price Response?

A single price coefficient imposes a particular pattern on demand. For example, a linear model makes the change in expected demand per dollar constant, while a log-log model has constant elasticity. Actual price response may instead flatten, steepen, or change direction across the prices being studied. Polynomial price terms provide one way to represent such curvature without requiring the response to have the same slope everywhere.

A polynomial model does not, by itself, establish that the estimated curve is credible. The price at which a marginal effect is calculated matters, and the curve may behave poorly outside the data range. The central tasks are therefore to specify the scale on which the polynomial enters, calculate price-dependent effects correctly, and inspect fitted values over the observed prices.

Definition: A polynomial price-response model includes powers of price, or of a centered and scaled version of price, as predictors. A quadratic predictor has terms through degree two, such as $z$ and $z^2$; a higher-order predictor may also include $z^3$ or further powers. The polynomial may enter the outcome mean directly or enter a link function, so the resulting response curve depends on the model scale.

Centering, Scaling, and the Price Derivative

Let $p$ denote price and define $z=(p-c)/s$, where $c$ is a chosen reference price and $s>0$ is a price scale. Then $z$ is dimensionless and a one-unit change in $z$ corresponds to a change of $s$ dollars in price. Centering makes the intercept refer to price $c$, while scaling can make the coefficients easier to work with numerically. It does not change the fitted curve when the model is expressed consistently.

For a quadratic predictor $\eta(z)=\beta_0+\beta_1z+\beta_2z^2$, the price derivative is not just the coefficient of $z$. Applying the chain rule gives:

\frac{d\eta}{dp} =\frac{d\eta}{dz}\frac{dz}{dp} =\frac{\beta_1+2\beta_2z}{s}.

The factor $1/s$ converts a change per unit of $z$ into a change per dollar. For a polynomial of degree $K$, $\eta(z)=\sum_{k=0}^{K}\beta_kz^k$, the corresponding derivative is:

\frac{d\eta}{dp} =\frac{1}{s}\sum_{k=1}^{K}k\beta_kz^{k-1}.

These derivatives are local: they describe the response to a small change in price at the price being evaluated. They need not have the same value, or even the same sign, at other prices.

Direct Polynomial Models for a Continuous Outcome

Suppose $Y$ is a continuous demand outcome and its conditional mean at price $p$ is modeled directly as $\mu(p)=E[Y\mid p]=\beta_0+\beta_1z+\beta_2z^2$. This is an identity-link specification: the polynomial is on the mean scale. Its marginal effect is $(\beta_1+2\beta_2z)/s$, in outcome units per dollar. When $\mu(p)>0$, its signed price elasticity is the proportional change in the mean for a proportional change in price:

\varepsilon_\mu(p) =\frac{p}{\mu(p)}\frac{d\mu(p)}{dp} =\frac{p}{\mu(p)}\frac{\beta_1+2\beta_2z}{s}.

This identity follows from the definition of elasticity: a small price change $dp$ produces an approximate mean change $d\mu=(d\mu/dp)dp$. Dividing by the original mean gives $d\mu/\mu$, and dividing that proportional change by $dp/p$ gives $(p/\mu)(d\mu/dp)$. The formula is not meaningful when the fitted mean is zero and may be difficult to interpret for a mean that is negative.

Worked Example: Marginal Effect and Elasticity on the Mean Scale

Hypothetical scenario. An insurer models the expected number of completed quote requests per 1,000 eligible visitors as $Y\mid p\sim N(\mu(p),100)$, using variance 100 (standard deviation 10) and $\mu(p)=120-8z-2z^2$, where $z=(p-110)/20$. The fitted model is assessed over observed prices from $80 to $140. Interest and mortality assumptions are not applicable: this is a demand model, not a discounted cash-flow or life-contingency calculation. At $p=$130$, calculate the mean, marginal effect, and elasticity, then check the fitted mean and slope at the observed-range endpoints.

Method. This is a direct quadratic model for the conditional mean. At $130, $z=(130-110)/20=1$, so $\mu(130)=120-8(1)-2(1)^2=110$ quote requests per 1,000 visitors.

The derivative with respect to price is $(-8-4z)/20$. At $z=1$, the marginal effect is $(-8-4)/20=-0.6$ expected requests per 1,000 visitors per dollar. Therefore, $\varepsilon_\mu(130)=(130/110)(-0.6)=-0.7090909$, or $-0.7091$ to four decimal places.

For the range check, $p=$80$ gives $z=-1.5$, $\mu=120+12-4.5=127.5$, and slope $(-8+6)/20=-0.1$. At $p=$140$, $z=1.5$, $\mu=120-12-4.5=103.5$, and slope $(-8-6)/20=-0.7$. The mean is positive at both endpoints; since this quadratic opens downward, its minimum on the closed interval occurs at an endpoint, so the mean is positive throughout. The slope is negative throughout the interval because $-8-4z\leq -2$ for $-1.5\leq z\leq1.5$. These checks agree with a decreasing fitted response over the observed range.

Common mistake avoided. The derivative $-8-4z$ is per unit of $z$, not per dollar. Dividing by 20 is essential for both the marginal effect and the elasticity.

Polynomials on Logistic and Log-Mean Scales

For a binary purchase outcome, a logistic model can place the polynomial on the log-odds scale: $\log(q/(1-q))=\eta(p)$, where $q(p)$ is the purchase probability. Differentiation gives $dq/dp=q(1-q)(d\eta/dp)$. Thus the probability’s marginal price effect depends on both the polynomial slope and the probability at that price. Its signed elasticity is $p(1-q)(d\eta/dp)$, provided $q>0$.

For a positive conditional mean, a log-mean model uses $\log\mu(p)=\eta(p)$, so $\mu(p)=\exp(\eta(p))$. Differentiating gives $d\mu/dp=\mu(d\eta/dp)$, and hence $\varepsilon_\mu(p)=p(d\eta/dp)$. The log link ensures a positive fitted mean, but it does not ensure that the mean increases or decreases monotonically with price.

Worked Example: Price-Dependent Logistic Elasticity

Hypothetical scenario. A carrier models whether a visitor purchases a specified insurance quote as a Bernoulli outcome, with $P(\text{purchase}\mid p)=q(p)$ and logit $q(p)=-1-0.4z-0.1z^2$, where $z=(p-90)/15$. The observed price range is $75 to $120. At $p=$105$, calculate the purchase probability and elasticity. The Bernoulli distribution has parameter $q(p)$; interest and mortality assumptions are not applicable.

Method. At $105, $z=(105-90)/15=1$. The logit is $\eta=-1-0.4-0.1=-1.5$, so $q=1/(1+\exp(1.5))=0.1824255$ before rounding, or 0.1824 to four decimal places.

The logit derivative is $d\eta/dp=(-0.4-0.2z)/15$. At $z=1$, it equals $-0.6/15=-0.04$ per dollar. By the logistic derivative, $dq/dp=q(1-q)(-0.04)$; the elasticity is therefore $105(1-q)(-0.04)$. Using the unrounded probability gives $105(0.8175744762)(-0.04)=-3.4338128$, or $-3.4338$ to four decimal places.

The fitted probability remains between zero and one by the logistic transformation. To check direction over the observed range, the endpoints correspond to $z=-1$ and $z=2$. The logit derivative has numerator $-0.4-0.2z$, which is negative throughout that interval: it ranges from $-0.2$ to $-0.8$. Thus the fitted purchase probability decreases throughout the observed range.

Common mistake avoided. Rounding the probability to 0.1824 before calculating elasticity can change the last reported digit. Keep full precision through the calculation, then round the final result only.

Worked Example: Elasticity from a Quadratic Log-Mean Model

Hypothetical scenario. An insurer models the number of policies purchased in a market cell as $N\mid p\sim\operatorname{Poisson}(\mu(p))$, with $\log\mu(p)=\log(50)-0.2z-0.1z^2$, where $z=(p-60)/10$. The observed range is $50 to $90. Calculate the fitted mean, marginal effect, and elasticity at $p=$80$. The Poisson parameter is the stated mean; interest and mortality assumptions are not applicable.

Method. At $80, $z=(80-60)/10=2$. The log-mean is $\log(50)-0.2(2)-0.1(2^2)=\log(50)-0.8$, so $\mu(80)=50\exp(-0.8)=22.4664$ purchases, to four decimal places.

The derivative of the log-mean is $(-0.2-0.2z)/10$. At $z=2$, this is $-0.6/10=-0.06$ per dollar. The marginal effect on the mean is $22.4664(-0.06)=-1.347984$ purchases per dollar. The elasticity is $p(d\log\mu/dp)=80(-0.06)=-4.800$.

Over the observed range, $z$ runs from $-1$ to $3$. The derivative of the log-mean is zero at $z=-1$ and negative for larger $z$, so the mean is nonincreasing across the range. The log link keeps the fitted mean positive. This check does not imply that the same pattern holds outside the observed range.

Common mistake avoided. Elasticity uses the price at which it is evaluated: here $80(-0.06)=-4.8$. The value $50\exp(-0.8)$ is the fitted mean, not a factor to multiply into the elasticity.

Checking Plausibility and Interpreting the Curve

Polynomial terms can fit curved responses efficiently, but polynomial behavior can become extreme outside the prices represented in the data. A good fit near the center does not guarantee plausible predictions at the edges. Review the full observed price range, not just a single reference price or the point at which an effect is most convenient to calculate.

Set the price scale.
Record $c$, $s$, and the observed price range, then translate that range into values of $z$.

Differentiate on the correct scale.
Differentiate the polynomial with respect to $z$, then multiply by $1/s$. Apply the link-function derivative if the polynomial is on a link scale.

Evaluate fitted outcomes across the range.
Check boundary prices and any interior turning points. For probabilities, inspect both the probability levels and the sign of their marginal effects.

Separate in-range interpretation from extrapolation.
Do not treat a polynomial’s behavior far beyond the observed prices as reliable evidence about demand.

A logistic transformation constrains probabilities to the interval from zero to one, while a log link constrains means to be positive. Neither transformation guarantees a substantively sensible curve: the response may turn upward, become nearly flat, or change sharply across the observed range. An identity-link polynomial has no such automatic bounds, so its fitted means should also be checked for impossible values.

Centering and scaling help prevent unit mistakes, but do not remove the need to interpret coefficients carefully. In a quadratic model, $\beta_1$ is the slope with respect to $z$ at $z=0$, not a universal effect per dollar. The derivative at a particular price includes the contribution $2\beta_2z$ and the conversion factor $1/s$. In higher-order models, every included power can contribute to the local slope.

Key takeaway: A polynomial price term allows the response to curve, so its marginal effect and elasticity generally depend on price. Differentiate with respect to the actual price, account for the model’s link, retain full precision in calculations, and check fitted values and slopes across the observed range.

Check Your Understanding

For each problem, calculate the requested quantities using the stated price scale and check any stated range condition.

A hypothetical continuous demand mean is $\mu=90-5z-z^2$, where $z=(p-100)/20$. At $p=$120$, find the mean and elasticity. (Answer: mean $=84$; elasticity $=-0.5000$.)
A hypothetical Bernoulli purchase model has logit $\eta=-0.8+0.3z-0.15z^2$, where $z=(p-50)/10$. At $p=$60$, calculate the probability and elasticity. (Answer: probability $=0.3430$; elasticity $=0.0000$.)
A hypothetical Poisson demand model has $\log\mu=\log(40)-0.1z-0.05z^2$, where $z=(p-70)/10$. At $p=$90$, calculate the mean and elasticity. (Answer: mean $=26.8128$; elasticity $=-2.700$.)
An identity-link mean is $\mu=30+2z-3z^2$ over $-2\leq z\leq2$. Is the mean positive throughout, and does it increase throughout? (Answer: Yes, it is positive throughout; no, it increases only until $z=1/3$ and decreases thereafter.)
A logistic model has logit $\eta=-2+0.5z-0.2z^2$ over $0\leq z\leq3$. At $z=3$, calculate the probability and determine the direction of the marginal effect. (Answer: probability $=0.0911$; the marginal effect is negative because $d\eta/dz=-0.7$.)

What You'll Learn