Marginal Effects and Elasticities in Logistic Models

From a Logistic Coefficient to a Price Response

A logistic demand model predicts the probability that a customer purchases or renews. Its price coefficient describes how price changes the model’s log odds, not directly how much purchase probability changes and not, by itself, the percentage response called an elasticity. To interpret price response, first convert the linear predictor into a probability, then differentiate that probability with respect to price.

The calculations below concern a binary outcome $Y$, with $Y=1$ denoting purchase or renewal. They give model-based responses for specified customer covariates. Interpreting a response as the effect of changing price, rather than as an association conditional on the model, additionally requires a suitable design and assumptions that support a causal interpretation.

Definition: In a logistic model, $P(Y=1\mid \mathbf{x})=p=1/(1+\exp(-\eta))$, where $\eta$ is the linear predictor. The marginal effect of price is the derivative $\partial p/\partial P$, holding the other modeled covariates fixed. The own-price elasticity of predicted probability is $(P/p)(\partial p/\partial P)$: the local percentage change in predicted purchase probability for a one-percent price change.

Deriving the Logistic Marginal Effect

Suppose price in dollars, $P$, enters the linear predictor linearly, with no price interactions:

\eta=\alpha+\beta_P P+\boldsymbol{\gamma}^{\mathsf T}\mathbf{z}, \qquad p=\frac{1}{1+\exp(-\eta)}.

Here $\mathbf{z}$ denotes the other customer characteristics, which are held fixed when taking the price derivative. The coefficient $\beta_P$ is the change in log odds per dollar. Applying the chain rule gives the marginal effect:

\frac{\partial p}{\partial P} = \frac{\partial p}{\partial \eta}\frac{\partial \eta}{\partial P} = p(1-p)\beta_P.

To see why the first derivative is $p(1-p)$, differentiate $p=(1+\exp(-\eta))^{-1}$. This gives $\partial p/\partial\eta=\exp(-\eta)/(1+\exp(-\eta))^2$. Since $p=1/(1+\exp(-\eta))$ and $1-p=\exp(-\eta)/(1+\exp(-\eta))$, their product is exactly that derivative. The result then follows because $\partial\eta/\partial P=\beta_P$.

Multiplying the marginal effect by $P/p$ converts the change per dollar into a local percentage response:

\varepsilon(P) = \frac{P}{p}\frac{\partial p}{\partial P} = \beta_P P(1-p).

When a price increase reduces demand, $\beta_P<0$, so both the marginal effect and elasticity are negative. The marginal effect is measured in probability points per dollar; elasticity is dimensionless. In particular, multiplying a marginal effect by 100 gives percentage points per dollar, not an elasticity.

Worked Examples

Worked Example: One Customer, Price in Dollars

A hypothetical insurer fits a logistic renewal model. For one customer at an annual premium of $P=\$1{,}000$, the fitted renewal probability is $p=0.40$, and the raw-price coefficient is $\beta_P=-0.004$ per dollar. Find the local marginal effect and own-price elasticity. Then calculate the model’s exact predicted probability after a $\$50$ increase, assuming all other covariates are unchanged.

Method. Use $p(1-p)\beta_P$ for the derivative and $\beta_P P(1-p)$ for the elasticity. First, $p(1-p)=0.40(0.60)=0.24$.

\frac{\partial p}{\partial P} =(-0.004)(0.24)=-0.00096 \quad\text{per dollar}.

Thus the local probability change is $-0.00096$ per dollar, or $-0.096$ percentage points per dollar. The elasticity at the current premium is

\varepsilon=(-0.004)(1{,}000)(0.60)=-2.4.

For the exact $\$50$ change, the linear predictor changes by $\Delta\eta=\beta_P\Delta P=(-0.004)(50)=-0.20$. The starting odds are $p/(1-p)=0.40/0.60=2/3$. The new odds are $(2/3)\exp(-0.20)$, approximately $0.5458$, so the new probability is $0.5458/(1+0.5458)$, approximately $0.3531$.

Check. The fitted probability falls from $0.4000$ to $0.3531$, consistent with the negative price coefficient. The derivative’s linear approximation for a $\$50$ change is $0.40+(-0.00096)(50)=0.3520$, close to, but not exactly, the logistic prediction. The derivative is local; the odds update gives the exact finite change. The common mistake is to treat a local derivative as an exact probability change over a large price movement.

Worked Example: Average Marginal Effect and Two Average Elasticities

For three hypothetical quoted risks, a model has raw-price coefficient $\beta_P=-0.003$ per dollar. At their current prices, the predicted probabilities and prices are respectively $(p_i,P_i)=(0.20,\$800),(0.50,\$1{,}000),(0.80,\$1{,}200)$. Calculate the average marginal effect, the average of the three individual elasticities, and the elasticity of mean predicted probability under a common proportional increase in all three prices.

Average marginal effect. For each customer, the derivative is $\beta_Pp_i(1-p_i)$. The probability factors are $0.20(0.80)=0.16$, $0.50(0.50)=0.25$, and $0.80(0.20)=0.16$. Their mean is $(0.16+0.25+0.16)/3=0.19$. Therefore,

\operatorname{AME} =(-0.003)(0.19)=-0.00057 \quad\text{per dollar}.

Mean individual elasticity. For each customer, use $\varepsilon_i=\beta_P P_i(1-p_i)$:

\begin{aligned} \varepsilon_1&=(-0.003)(800)(0.80)=-1.92,\\ \varepsilon_2&=(-0.003)(1{,}000)(0.50)=-1.50,\\ \varepsilon_3&=(-0.003)(1{,}200)(0.20)=-0.72. \end{aligned}

The arithmetic mean is $(-1.92-1.50-0.72)/3=-4.14/3=-1.38$.

Elasticity of mean probability. Mean predicted probability is $\bar p=(0.20+0.50+0.80)/3=0.50$. If every price is multiplied by the same factor, differentiating the mean probability with respect to that factor’s logarithm gives the mean of the customer-level terms $\beta_P P_i p_i(1-p_i)$. Those terms are $-0.384$, $-0.750$, and $-0.576$; their mean is $-1.710/3=-0.570$. Divide by $\bar p$ to obtain the elasticity:

\frac{-0.570}{0.50}=-1.14.

Check. The three quantities answer different questions: the AME is $-0.00057$ probability per dollar, the mean individual elasticity is $-1.38$, and the elasticity of mean predicted probability under common proportional price scaling is $-1.14$. They need not agree. Averaging individual elasticities and taking the elasticity of an average are not interchangeable operations. The common mistake is to report one of these averages without stating which response it summarizes.

Worked Example: Price Entered as Log Price

A hypothetical purchase model uses $x=\ln(P)$ as its price predictor, with coefficient $\beta_{\ln P}=-1.5$. For a customer with current premium $P=\$1{,}200$, the fitted probability is $p=0.60$. Find the local marginal effect per dollar and the own-price elasticity. Then estimate the effect of a one-percent premium increase using the local elasticity.

Method. The linear predictor contains $\beta_{\ln P}\ln(P)$, so $\partial\eta/\partial P=\beta_{\ln P}/P$. Applying the logistic derivative gives

\frac{\partial p}{\partial P} = p(1-p)\frac{\beta_{\ln P}}{P} = \frac{(0.60)(0.40)(-1.5)}{1{,}200} = -0.00030 \quad\text{per dollar}.

The elasticity is

\varepsilon = \frac{P}{p}\frac{\partial p}{\partial P} = \beta_{\ln P}(1-p) = (-1.5)(0.40) = -0.60.

A one-percent price increase therefore corresponds locally to a $0.60$-percent decrease in predicted probability. At $p=0.60$, that is an approximate absolute decrease of $0.006(0.60)=0.0036$, or $0.36$ percentage points.

Check. The marginal effect is negative, and the elasticity is dimensionless. The coefficient $-1.5$ is not itself the elasticity: the elasticity at this customer’s probability is $-0.60$. The common mistake is to use the raw-price formula when the modeled predictor is log price.

Comparing Customers and Interpreting Coefficients

For a raw-price predictor, $\beta_P$ is a change in log odds per dollar. For a log-price predictor, $\beta_{\ln P}$ is a change in log odds per unit change in log price. Neither coefficient directly gives the probability-point change. The logistic transformation makes that change depend on the predicted probability, and the price scale also matters when calculating elasticity.

With a fixed coefficient and a raw-price predictor, the marginal effect has magnitude $|\beta_P|p(1-p)$. The factor $p(1-p)$ is largest at $p=0.50$, where it equals $0.25$, and is smaller for probabilities near zero or one. Thus customers with the same price coefficient can have different probability responses because their baseline probabilities differ. For a log-price predictor, the elasticity is $\beta_{\ln P}(1-p)$, so its value also varies with baseline probability even though the price coefficient is constant.

An average marginal effect is usually calculated by finding each observation’s derivative at its own covariate values and then taking the arithmetic mean. It is a probability change per unit of price. An average of individual elasticities instead gives equal weight to each customer’s proportional response. The elasticity of mean predicted demand under a common proportional price change uses the derivative of total or mean predicted probability relative to that mean. If portfolio weights are used, apply those same weights consistently to the predicted probabilities and their derivatives.

Key result: If price enters the logit linearly in dollars, then $\partial p/\partial P=\beta_Pp(1-p)$ and $\varepsilon=\beta_PP(1-p)$. If log price enters linearly, then $\partial p/\partial P=\beta_{\ln P}p(1-p)/P$ and $\varepsilon=\beta_{\ln P}(1-p)$. These formulas assume the stated linear predictor and hold the other covariates fixed.

Practical Checks and Common Traps

Check the price scale. A coefficient per dollar differs numerically from one per hundred dollars. Convert the coefficient to the units used for the price before applying the formula.
Check the predictor. A coefficient on $P$ and a coefficient on $\ln(P)$ have different units and lead to different marginal-effect formulas.
Do not confuse a derivative with a finite change. A derivative is local. For a finite change under a linear logit predictor, update $\eta$ by the coefficient times the change in the predictor, then transform back to a probability.
Name the average. An average marginal effect, a mean individual elasticity, and an elasticity of average predicted demand are distinct summaries.
Separate prediction from causation. A fitted price response is conditional on the model. A causal interpretation requires assumptions that justify comparing demand at different prices; the logistic formula alone does not establish them.

These distinctions matter in pricing analysis because a coefficient can appear large or small only relative to its predictor’s units, while the business question may concern probability points or percentage response. Reporting the price specification, baseline probabilities, and the precise average being calculated makes a model-based demand response interpretable and comparable.

Check Your Understanding

For each problem, state the relevant price specification and use the local formulas unless a finite change is explicitly requested.

A raw-price logit has $\beta_P=-0.0025$. At $P=\$800$, a customer’s predicted purchase probability is $0.30$. Find the marginal effect per dollar and own-price elasticity. (Answer: $-0.000525$ probability per dollar; elasticity $-1.400$)
A log-price logit has $\beta_{\ln P}=-0.9$. At $P=\$1{,}500$, the predicted probability is $0.40$. Find the marginal effect per dollar and elasticity. (Answer: $-0.000144$ probability per dollar; elasticity $-0.540$)
A raw-price model has $\beta_P=-0.002$. Three customers have predicted probabilities $0.20,0.50,0.70$. Find their average marginal effect in probability per dollar. (Answer: $-0.0004133$ probability per dollar, approximately)
A raw-price model has $\beta_P=-0.0015$. Two customers have $(P,p)=(\$600,0.25)$ and $(\$900,0.75)$. Calculate each individual elasticity and their unweighted mean. (Answer: $-0.675$ and $-0.3375$; mean elasticity $-0.50625$)
A log-price model has $\beta_{\ln P}=-0.8$, and three customers have predicted probabilities $0.10,0.50,0.90$. Find the mean individual elasticity. Separately find the elasticity of mean predicted probability under a common proportional increase in all three prices. (Answer: mean individual elasticity $-0.4000$; elasticity of mean predicted probability $-0.2293$, approximately)

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