Linear Probability Models for Purchase Outcomes

Why Model a Purchase Outcome?

A quote-level data set may record whether a customer purchased a policy, renewed, or declined. These outcomes are binary, but their probability can still be modeled as a function of quoted price and other recorded characteristics. A linear probability model is a direct, relatively simple way to estimate how those characteristics relate to the chance of purchase.

Its simplicity comes with important limits. The fitted relationship is linear, so a price coefficient has a direct probability-point interpretation, but the fitted values are not automatically valid probabilities. In addition, the outcome's variance depends on its probability, which matters when estimating uncertainty. These features make careful interpretation and inference essential.

Definition: Let $Y_i$ equal 1 if quote $i$ results in a purchase and 0 otherwise. A linear probability model (LPM) specifies the conditional purchase probability as a linear function of predictors: $$ P(Y_i=1\mid X_i)=E[Y_i\mid X_i]=\beta_0+\beta_1 x_{i1}+\cdots+\beta_k x_{ik}. $$ Equivalently, it can be written $Y_i=X_i^\mathsf{T}\beta+\varepsilon_i$, where $E[\varepsilon_i\mid X_i]=0$. The coefficients are commonly estimated by ordinary least squares (OLS).

Here, $X_i$ contains the predictors for quote $i$, such as price and customer characteristics. For a purchase outcome, $Y_i=1$ represents a purchase and $Y_i=0$ a nonpurchase. The model estimates a conditional mean, not an individual customer's certainty of purchase. A fitted value such as 0.60 describes an estimated purchase probability of 60% for quotes with those modeled characteristics.

Price Coefficients and Their Units

In an LPM, a coefficient is an additive change in fitted probability for a one-unit change in its predictor, holding the other included predictors fixed. If price is measured in dollars and its coefficient is $-0.001$, then a one-dollar price increase corresponds to a decrease of 0.001 in fitted probability, or 0.1 percentage point. The effect of a price change is the coefficient multiplied by the change in price.

Because the price unit can make a coefficient awkward to read, an analyst may define a scaled price variable, such as price in hundreds of dollars. The coefficient then describes the change in probability for a $100 price change. State the variable's definition and units before interpreting the estimate. A probability-point change is not the same as a percent change: a decrease from 60% to 55% is a 5-percentage-point decrease, or an 8.33% relative decrease.

Worked Example: Interpreting a Scaled Price Coefficient

Suppose a hypothetical insurer fits the model $\hat p=0.64-0.09z+0.04R$, where $z=(P-500)/100$, $P$ is the quoted annual premium in dollars, and $R=1$ for a specified customer category and 0 otherwise. Calculate the fitted purchase probability at a $650 premium for a customer with $R=1$. Then find the modeled change for a $50 premium increase, holding $R$ fixed.

Method and units. The price variable $z$ counts $100 increments relative to $500. Thus, its coefficient of $-0.09$ means a $100 increase in premium corresponds to a 0.09 decrease in fitted probability, or 9 percentage points, holding $R$ fixed.

Prediction at $650. First calculate $z=(650-500)/100=1.5$. Substitution gives $\hat p=0.64-0.09(1.5)+0.04(1)=0.64-0.135+0.04=0.545$, or 54.5%.

Change for $50. A $50 increase changes $z$ by $50/100=0.5$. The fitted probability change is $(-0.09)(0.5)=-0.045$, a decrease of 4.5 percentage points. At $700, $z=2$, so the fitted probability is $0.64-0.18+0.04=0.50$; the difference from 0.545 is indeed $-0.045$.

Common mistake avoided. Treating $-0.09$ as a 9% relative change is incorrect. It is a 9-percentage-point change for a $100 price increase, because the response is modeled on the probability scale.

Why Binary Outcomes Are Heteroskedastic

Conditional on predictors, a binary outcome is a Bernoulli random variable: it is 1 with probability $p_i=P(Y_i=1\mid X_i)$ and 0 with probability $1-p_i$. Its conditional mean is $p_i$. Its conditional variance follows directly from the two possible outcomes:

\operatorname{Var}(Y_i\mid X_i) =E[Y_i^2\mid X_i]-E[Y_i\mid X_i]^2 =p_i-p_i^2 =p_i(1-p_i).

The equality $Y_i^2=Y_i$ holds because both 0 and 1 are unchanged when squared. Unless the conditional probability is constant across observations, $p_i(1-p_i)$ generally varies with the predictors. Therefore, the LPM error variance is generally not constant. This is heteroskedasticity.

Heteroskedasticity does not by itself make OLS coefficients biased when the conditional mean is correctly specified and $E[\varepsilon_i\mid X_i]=0$. However, the conventional OLS standard errors rely on a constant-variance assumption. When that assumption fails, those standard errors need not be reliable. A heteroskedasticity-robust covariance estimator, such as the HC0 sandwich estimator, allows the error variance to differ across observations. It does not change the OLS coefficient estimates.

HC0 robust covariance: For the OLS design matrix $X$, fitted residuals $\hat\varepsilon_i$, and diagonal matrix $\widehat{\Omega}$ with entries $\hat\varepsilon_i^2$, the HC0 covariance estimate is $$ \widehat{\operatorname{Var}}_{\mathrm{HC0}}(\hat\beta) =(X^\mathsf{T}X)^{-1}X^\mathsf{T}\widehat{\Omega}X(X^\mathsf{T}X)^{-1}. $$ The square root of a diagonal element gives the HC0 standard error for the corresponding coefficient. Robust standard errors rely on suitable large-sample conditions and independent observations; they do not repair a misspecified conditional mean or address dependence between observations.

Worked Example: Calculating an HC0 Price Standard Error

Consider four hypothetical, independent quotes with centered price predictor $x=-1,-1,1,1$, where a one-unit increase represents a $100 premium increase. Their purchase outcomes, in the same order, are $0,1,0,1$. Fit an intercept and price slope, then calculate the HC0 standard error for the slope and compare it with the conventional OLS standard error.

Method. At each price level, the mean outcome is $(0+1)/2=0.5$. Both fitted group means are therefore 0.5. The OLS intercept is $\hat\beta_0=0.5$, and the slope is $\hat\beta_1=0$. The four residuals are $-0.5,0.5,-0.5,0.5$.

HC0 calculation. With an intercept and centered, symmetric $x$, the slope variance estimate simplifies to $\sum_i x_i^2\hat\varepsilon_i^2/(\sum_i x_i^2)^2$. Here, $\sum_i x_i^2=4$, and $\sum_i x_i^2\hat\varepsilon_i^2=4(1)(0.25)=1$. Thus the HC0 slope variance estimate is $1/4^2=1/16=0.0625$, and its standard error is $\sqrt{0.0625}=0.25$.

Conventional calculation. The residual sum of squares is $4(0.25)=1$. With four observations and two fitted coefficients, the conventional residual variance estimate is $1/(4-2)=0.5$. Since $\sum_i(x_i-\bar x)^2=4$, the conventional slope variance estimate is $0.5/4=0.125$, giving a standard error of $\sqrt{0.125}\approx0.3536$.

The two standard errors differ because they use different variance assumptions and estimators. This four-observation calculation illustrates the arithmetic, not a recommendation to rely on large-sample robust inference from such a small data set. Common mistake avoided: using the conventional OLS standard error automatically, without considering the outcome's probability-dependent variance.

Out-of-Range Predictions and Model Limits

A probability must lie between 0 and 1, inclusive. An LPM does not impose that restriction on its fitted values. A fitted value below zero or above one is therefore an invalid probability, even though it is a valid numerical result from the linear equation. Such values can arise especially when predictions are made outside the range of predictors represented in the data.

Do not silently truncate an invalid fitted value to 0 or 1 and then treat the result as though it came from the fitted LPM. Truncation changes the prediction rule and its implications. Instead, report the limitation, check whether the prediction is an extrapolation, and reconsider the model or the range over which predictions are used. A binary-response model with a bounded mean is one possible alternative, but choosing and fitting one is a separate modeling task.

Worked Example: Detecting an Invalid Extrapolated Prediction

A hypothetical purchase model is $\hat p=0.68-0.0011(P-400)+0.06R$, where $P$ is annual premium in dollars and $R$ is a binary customer indicator. The data used for fitting cover premiums from $300 through $700. Calculate predictions for $R=0$ at $300, $700, and $1,100, and assess whether the last prediction can be used as a purchase probability.

At $P=300$, $\hat p=0.68-0.0011(-100)=0.68+0.11=0.79$. At $P=700$, $\hat p=0.68-0.0011(300)=0.68-0.33=0.35$. Both predictions lie between 0 and 1.

At $P=1{,}100$, $\hat p=0.68-0.0011(700)=0.68-0.77=-0.09$. This is not a valid purchase probability. The $1,100 quote is also outside the observed premium range, so the prediction is an extrapolation. The correct conclusion is that this LPM does not provide a usable probability for that case—not that the customer has a $-9\%$ chance of purchase or that the fitted probability is automatically 0%.

Common mistake avoided. Clipping $-0.09$ to zero hides both the invalid LPM prediction and the extrapolation. The model, its prediction range, or the intended decision use must instead be reconsidered.

Interpretation, Inference, and Practical Checks

An LPM price coefficient describes a conditional association in the modeled purchase probability, holding the other included predictors fixed. Interpreting it as the effect of changing a price requires more than fitting OLS: the model must support that interpretation, including appropriate assumptions about how price was set and which relevant factors are accounted for. Robust standard errors address a variance-estimation issue; they do not establish a causal interpretation.

Before using LPM results in a pricing analysis, check the response coding, price units, fitted values over the intended prediction range, and uncertainty method. Also examine whether the linear conditional mean is a reasonable approximation over that range. If observations are dependent, for example because multiple quotes belong to the same customer, a variance estimator that accounts for the dependence may be needed; ordinary HC0 errors alone do not do so.

Define the outcome and predictors.
Code purchase consistently as 1 and nonpurchase as 0. State the units and any scaling used for price.

Fit and interpret the linear mean.
Read the price coefficient as an additive probability change for one unit of its predictor, holding other included predictors fixed.

Check uncertainty and prediction range.
Use a suitable robust variance estimate when errors may be heteroskedastic, and check that fitted probabilities are valid where they will be used.

Key takeaway: The LPM gives a simple, direct probability-point interpretation of price coefficients. Its binary response generally creates heteroskedastic errors, and its unconstrained linear predictions can fall outside zero to one. Interpret coefficients with their units, use appropriate standard errors, and do not treat invalid fitted values as probabilities.

Check Your Understanding

Work through each problem, paying attention to price units, the probability scale, and the limits of robust inference.

A fitted LPM is $\hat p=0.70-0.04z$, where $z=(P-300)/100$ and $P$ is annual premium in dollars. Find the fitted probability at $450 and the modeled change for a $50 price increase. (Answer: $0.64$, or 64%; a decrease of 0.02, or 2 percentage points.)
For two quote groups, the fitted purchase probabilities are 0.30 and 0.70. Calculate the conditional Bernoulli variance in each group. What does the comparison imply? (Answer: both variances are $0.30(0.70)=0.2100$; the variances are equal in these two groups, though binary outcomes generally have probability-dependent variance.)
An LPM price coefficient is $-0.003$ per dollar. Holding other predictors fixed, what is the modeled probability-point change for a $40 increase? If the starting fitted probability is 0.58, what is the new fitted probability? (Answer: a decrease of 0.12, or 12 percentage points; the new fitted probability is 0.46.)
For a binary outcome, one observation has fitted probability 0.20 and another has fitted probability 0.50. Calculate each conditional variance. Which observation has the larger variance? (Answer: $0.20(0.80)=0.1600$ and $0.50(0.50)=0.2500$; the observation with fitted probability 0.50 has the larger variance.)
A model was fitted using premiums from $250 through $650. For $R=0$, it predicts $\hat p=0.62-0.0015(P-350)$. Calculate the prediction at $650 and $850. Explain which result should not be interpreted as a purchase probability and why. (Answer: at $650, $\hat p=0.17$; at $850, $\hat p=-0.13$, which is invalid and is also an extrapolation beyond the observed premium range.)

What You'll Learn