Logistic Regression for Purchase and Renewal Demand

Why Use Logistic Regression for Demand?

Purchase and renewal outcomes are binary: a customer either buys a quoted policy or does not, and a current policyholder either renews or does not. A linear probability model can describe these outcomes, but its fitted values are not inherently restricted to the interval from zero to one. Logistic regression instead models the probability through a transformation that always produces a value strictly between zero and one.

For ratemaking demand analysis, the model can relate a purchase or renewal probability to quoted price and customer or quote characteristics. The output is a probability for a specified profile, not a guarantee about an individual customer. It can also be aggregated across customers to estimate expected demand. A price coefficient has a precise interpretation, but it is not itself a percentage-point change in probability or a price elasticity.

Definition: Let $Y_i=1$ if observation $i$ results in the outcome of interest, such as a purchase or renewal, and $Y_i=0$ otherwise. Write $p_i=P(Y_i=1\mid \mathbf{x}_i)$ for the probability of that outcome conditional on predictor values $\mathbf{x}_i$. A logistic regression specifies $$ \log\left(\frac{p_i}{1-p_i}\right)=\eta_i=\beta_0+\beta_1x_{i1}+\cdots+\beta_kx_{ik}, $$ where $p_i/(1-p_i)$ is the odds of the outcome and $\eta_i$ is the linear predictor. Equivalently, $$ p_i=\frac{1}{1+\exp(-\eta_i)}. $$

Specification, Coding, and Assumptions

The response must be defined before fitting a model. For a quote-level purchase model, $Y_i=1$ might mean that the quoted customer purchased within a stated observation window; for a renewal model, it might mean that the policy renewed at expiration. The population, outcome window, and treatment of incomplete or withdrawn quotes should be established consistently. A different outcome definition can produce a different demand model.

A basic model uses a linear predictor in price and other available predictors. For example, if premium is measured in units of $100, the coefficient on premium describes a one-unit change, or $100, holding the other included predictors fixed. A binary predictor can be coded as 1 for a specified condition and 0 for its reference condition. The intercept is the log odds when all numeric predictors equal zero and all binary predictors are at their reference values. That reference profile may or may not be practically meaningful.

The standard model treats each response, conditional on its predictors, as Bernoulli with probability $p_i$. In the usual independent-observation formulation, responses are also independent conditional on the predictors. The log odds are assumed to follow the stated linear predictor: this is an assumption about the model's form, not a claim that the probability is linear in price. A finite maximum-likelihood estimate also requires that the data do not exhibit complete or quasi-complete separation. When these conditions or the chosen predictor structure are inappropriate, fitted probabilities and coefficient interpretations may be unreliable.

Given observed outcomes $y_i$ and fitted probabilities $p_i$, the Bernoulli likelihood is

L(\boldsymbol{\beta})=\prod_i p_i^{y_i}(1-p_i)^{1-y_i}.

The coefficients are commonly estimated by maximizing this likelihood. Logistic regression does not require a normally distributed binary response. The model's probability specification and the suitability of its predictors are the important considerations here.

Why the Coefficients Are on the Log-Odds Scale

The inverse-logit converts any finite linear predictor into a valid probability. The coefficient interpretation follows directly from the transformation.

Start with the probability.
For a finite linear predictor $\eta$, the model gives $p=1/(1+\exp(-\eta))$. Therefore $1-p=\exp(-\eta)/(1+\exp(-\eta))$.

Form the odds.
Dividing $p$ by $1-p$ cancels the common denominator, so $p/(1-p)=\exp(\eta)$.

Take logarithms and change a predictor.
The log odds equal $\eta$. If $x_j$ increases by $\Delta$, with other predictors fixed, $\eta$ changes by $\beta_j\Delta$. The odds are multiplied by $\exp(\beta_j\Delta)$.

Thus, for a one-unit increase in $x_j$, the odds ratio is $\exp(\beta_j)$, conditional on the other modeled predictors. A negative price coefficient means that higher price is associated with lower odds, holding the other predictors fixed. It does not say by itself how many percentage points the probability changes. That probability change depends on the starting profile and its fitted probability.

Worked Example: A Purchase Probability at a Quoted Premium

A hypothetical insurer models whether a prospect purchases a quoted policy. Let $P$ be annual premium in hundreds of dollars, and let $L=1$ indicate that the customer has a specified loyalty discount; otherwise $L=0$. The fitted model is $$ \operatorname{logit}(p)=-1.10-0.35(P-10)+0.50L. $$ For a customer quoted $1,200 with $L=1$, calculate the predicted purchase probability. Also interpret the price coefficient.

Method and units. Since $P$ is in hundreds of dollars, $1,200 corresponds to $P=12$. The centered price term is $12-10=2$, so this is a $200 increase above the $1,000 reference premium.

Calculate the linear predictor. $$ \eta=-1.10-0.35(2)+0.50(1)=-1.10-0.70+0.50=-1.30. $$

Convert to a probability. $$ p=\frac{1}{1+\exp(-\eta)} =\frac{1}{1+\exp(1.30)} \approx\frac{1}{1+3.6693} \approx 0.2142. $$ The predicted purchase probability is 0.2142, or about 21.42%.

Interpret the price coefficient. A $100 increase in premium increases $P$ by one unit. Holding loyalty status fixed, it multiplies purchase odds by $$ \exp(-0.35)\approx 0.7047. $$ That is about a 29.53% reduction in odds for a $100 increase. It is not a 35-percentage-point reduction in purchase probability.

Check. The negative price coefficient gives a lower predicted probability at a higher premium when other predictors are fixed. The computed probability is between zero and one, as the logistic model requires. The calculation avoids treating dollars as the model's premium units.

Comparing Renewal Profiles

Renewal models use the same link and probability conversion, but their response and predictors are defined for the renewal decision. Consider a model with a numeric price-change predictor and two binary customer characteristics. A coefficient describes the change in log odds associated with a predictor change, conditional on the others included in the model. To compare profiles, calculate each profile's linear predictor first; do not compare coefficients directly as if they were probability changes.

Worked Example: Two Renewal Quotes With Different Price Changes

For a hypothetical renewal model, let $q$ be the increase in quoted premium measured in $10 units relative to the expiring premium. Let $A=1$ indicate autopay and $C=1$ indicate a prior claim. The model is $$ \operatorname{logit}(p)=0.30-0.08q+0.50A-0.40C. $$ Compare the predicted renewal probabilities for two customers who both have autopay and no prior claim. Customer 1 has a $20 increase; Customer 2 has a $50 increase.

Method and coding. A $20 increase is $q=2$; a $50 increase is $q=5$. Both customers have $A=1$ and $C=0$.

Customer 1. $$ \eta_1=0.30-0.08(2)+0.50(1)-0.40(0)=0.64, $$ so $$ p_1=\frac{1}{1+\exp(-0.64)} \approx \frac{1}{1+0.5273} \approx 0.6548. $$

Customer 2. $$ \eta_2=0.30-0.08(5)+0.50(1)-0.40(0)=0.40, $$ so $$ p_2=\frac{1}{1+\exp(-0.40)} \approx \frac{1}{1+0.6703} \approx 0.5987. $$

The predicted renewal probabilities are 0.6548 and 0.5987, respectively. The $30 difference in increases is three $q$-units, so the odds ratio for Customer 2 relative to Customer 1 is $$ \exp[-0.08(5-2)]=\exp(-0.24)\approx 0.7866. $$ The second customer's renewal odds are about 78.66% of the first customer's odds, conditional on the model. This is an odds comparison, not a ratio of probabilities.

Check. The linear predictors differ by $0.40-0.64=-0.24$, matching the stated log-odds change. The larger premium increase corresponds to the lower fitted probability, consistent with the negative price coefficient. The calculation avoids using a $30 change as three dollars or interpreting $-0.08$ as a direct probability decrement.

From Customer Probabilities to Expected Demand

For a set of customers with fitted purchase probabilities $p_i$, let $Y_i$ be each customer's binary purchase indicator. By linearity of expectation, the expected number of purchases is $\sum_i E[Y_i]=\sum_i p_i$. This result does not require the purchase indicators to be independent, although independence is part of the usual basic logistic model for estimating the probabilities. For a group of $n$ customers sharing a profile with probability $p$, the expected count is $np$.

Worked Example: Aggregating Predicted Purchases Across Quote Groups

A hypothetical quote model uses $P$, annual premium in hundreds of dollars, and $D=1$ for a digital quote, with $D=0$ otherwise: $$ \operatorname{logit}(p)=-0.60-0.20(P-10)+0.40D. $$ The portfolio has 20 non-digital quotes at $1,000, 30 digital quotes at $1,000, and 15 non-digital quotes at $1,200. Calculate the predicted probability for each group and the total expected number of purchases.

Group 1: 20 non-digital quotes at $1,000. Here $P=10$, $D=0$, and $$ \eta_1=-0.60-0.20(10-10)+0.40(0)=-0.60, \qquad p_1=\frac{1}{1+\exp(0.60)}\approx 0.3543. $$ Using the unrounded probability, the expected purchases are $20(0.3543437)\approx 7.0869$.

Group 2: 30 digital quotes at $1,000. Here $P=10$, $D=1$, and $$ \eta_2=-0.60-0.20(10-10)+0.40(1)=-0.20, \qquad p_2=\frac{1}{1+\exp(0.20)}\approx 0.4502. $$ The expected purchases are $30(0.4501660)\approx 13.5050$.

Group 3: 15 non-digital quotes at $1,200. Here $P=12$, $D=0$, and $$ \eta_3=-0.60-0.20(12-10)+0.40(0)=-1.00, \qquad p_3=\frac{1}{1+\exp(1.00)}\approx 0.2689. $$ The expected purchases are $15(0.2689414)\approx 4.0341$.

Total. Adding the expectations calculated from unrounded probabilities gives $$ 7.0869+13.5050+4.0341=24.6260. $$ The predicted total is approximately 24.626 expected purchases across 65 quotes. This is an expectation, not a claim that a fractional number of customers will purchase. Rounding the final expected count to three decimal places gives 24.626.

Check. The digital indicator raises the linear predictor at the same premium, while the higher premium lowers it for a non-digital quote. The resulting group probabilities reflect those directions. Using unrounded probabilities for aggregation avoids small discrepancies caused by rounding each probability first.

Common Mistakes and Practical Interpretation

A coefficient must be interpreted with the predictor's coding and unit in view. If premium is measured in hundreds of dollars, a one-unit change is $100; if it is measured in $10 units, a one-unit change is $10. Centering a predictor changes the interpretation of the intercept, but does not change the slope coefficient or its one-unit odds ratio.

A frequent mistake is to read a price coefficient of $-0.35$ as a 35-percentage-point decrease in purchase probability per $100. That is wrong: the coefficient is a change in log odds. For the model in the first worked example, the correct one-unit odds ratio is $\exp(-0.35)=0.7047$. To find a probability for a particular customer, substitute that customer's predictor values into the linear predictor and apply the inverse-logit. The probability change will depend on the customer's starting profile.

Logistic regression also does not establish that price caused a change in demand. The coefficient describes the modeled conditional association. A causal interpretation requires a suitable design and assumptions about how prices and customer characteristics were determined. For pricing work, the definition of the quoted price, the outcome window, and the relevant customer population should be consistent with the intended use of the model. These distinctions support interpretation; the coefficient alone does not resolve them.

Key takeaway: Specify a binary outcome and its predictor units, calculate the linear predictor for the profile of interest, then convert it with $p=1/(1+\exp(-\eta))$. Interpret a coefficient as a change in log odds, or exponentiate it to obtain an odds ratio—not as a direct probability change.

Check Your Understanding

For each problem, state the key model calculation and check that your interpretation matches the predictor units.

A fitted model has price coefficient $-0.12$, where price is measured in $50 units. What is the odds ratio for a $50 increase, holding other predictors fixed? (Answer: $\exp(-0.12)\approx 0.8869$.)
A purchase model has linear predictor $\eta=-0.70+0.30D$, where $D=1$ for a digital quote. What is the predicted probability for a digital quote? (Answer: $\eta=-0.40$, so $p=1/(1+\exp(0.40))\approx 0.4013$.)
A renewal model is $\operatorname{logit}(p)=0.50-0.10q+0.40A$, where $q$ is the premium increase in $10 units and $A=1$ for autopay. Find the renewal probability for $q=3$, $A=1$. (Answer: $\eta=0.60$, so $p\approx 0.6457$.)
A group of 28 customers has the same fitted purchase probability, obtained from $\eta=-0.60$. Find the expected number of purchases, using the unrounded probability and rounding the count to three decimals. (Answer: $28/[1+\exp(0.60)]\approx 9.922$.)
A model's price coefficient is $-0.25$ per $100. What is the odds ratio for a $300 increase, with other predictors fixed? (Answer: $\exp[(-0.25)(3)]=\exp(-0.75)\approx 0.4724$.)

What You'll Learn