Linear Probability Model (LPM)

Part of: Econometrics Lecture 02 — Applied Econometrics, Dr. Aluma Dembo Previous: Lec_01-Introduction & Treatment Effects | Next: Lec_03-Logit & Probit Models Key concepts: Binary Outcomes, LPM, Heteroskedasticity, Robust Standard Errors, Ultimatum Game

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Motivation — Binary Outcomes

Often the outcome of interest is binary: accept/reject, buy/don't buy, employed/unemployed, vote yes/no. We code it as $y_i \in \{0, 1\}$ and ask: what's the probability $y=1$, and how does it depend on $x$?

Today's agenda

1. Motivating example — the ultimatum game (Andersen et al. 2011)
2. Define the Linear Probability Model
3. Show $P(y=1 \mid x) = \beta_0 + \beta_1 x$ under A2
4. Interpret coefficients
5. Identify LPM's two big drawbacks — and how to patch one of them

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Worked Example: The Ultimatum Game

The Setup

Two players, one-shot, anonymous:

• Proposer receives a pot $S$ and offers share $p \in [0, 1]$ to the Responder.
• Responder sees the offer and either accepts (proposer keeps $(1-p)S$, responder gets $pS$) or rejects (both get 0).

Subgame-perfect Nash prediction: proposer offers the smallest positive $p$; responder accepts anything $> 0$. Real humans reject "unfair" offers — a classic violation of pure self-interest.

Andersen, Ertaç, Gneezy, Hoffman & List (2011)

Ran the ultimatum game in rural India with stakes varied from 20 Rs up to 20,000 Rs (roughly a year's income) to test whether "irrational" rejection persists when money actually matters.

Research question: Does the probability of acceptance rise with the offer, and does it change when the stakes are large?

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The Linear Probability Model

For a binary $y_i \in \{0,1\}$, run OLS directly:

The magic: under assumption A2 ($\mathbb{E}[u_i \mid x_i] = 0$), the fitted value is a conditional probability.

Proof — $\hat{y}$ is an estimate of $P(y=1 \mid x)$

Take conditional expectations of both sides:

Since $y_i$ is binary:

Therefore:

So $\hat{\beta}_0 + \hat{\beta}_1 x$ estimates the conditional probability of acceptance.

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Interpreting LPM Coefficients

Continuous regressor

A one-unit increase in $x$ changes the probability of $y=1$ by $\beta_1$ — in percentage points, not percent.

Dummy regressor

If $d \in \{0, 1\}$, then $\beta_d$ = difference in $P(y=1)$ between the treatment and reference group.

Andersen et al. results

Interpretation:

• $\hat{\beta}_1 = 0.583$: a 10 percentage-point larger offer raises acceptance probability by 5.83 pp.
• $\hat{\beta}_4 = 0.392$: at the 20,000 Rs stakes, acceptance probability is 39.2 pp higher than in the 20 Rs reference group — people are far more accepting when rejecting is very costly.

Key finding

High stakes do change behaviour — responders are much more willing to accept low offers when the amount at stake is meaningful. The "fairness" penalty shrinks as money grows.

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LPM Drawback #1 — Unbounded Predictions

Probabilities must lie in $[0, 1]$. OLS has no such constraint: $\hat{y}$ can be negative or greater than 1.

In Andersen et al., the max fitted value is 1.1943 — a "probability" of 119%, which is nonsense.

type: lpm-problems
mode: unbounded

Why this happens

LPM fits a straight line through a cloud of 0s and 1s. Extend the line far enough and it leaves the unit interval. The more extreme $x$ gets, the more often this happens. In the shaded regions the model predicts probabilities below 0 or above 1 — exactly the 119% fitted value Andersen et al. hit.

This is the motivation for Lec_03-Logit & Probit Models (next lecture), which squash $\hat{y}$ into $[0,1]$ via the logistic / normal CDF.

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LPM Drawback #2 — Built-in Heteroskedasticity

Proof — $\text{Var}(u \mid x) = p(x)(1 - p(x))$

Since $y \in \{0,1\}$, $u = y - p(x)$ is also binary, taking:

• $u = 1 - p(x)$ with probability $p(x)$
• $u = -p(x)$ with probability $1 - p(x)$

Then $\mathbb{E}[u \mid x] = 0$ (by A2), and:

What this violates

Since $\text{Var}(u \mid x)$ depends on $x$ through $p(x)$, assumption A4 (homoskedasticity) is always violated in LPM. The classical OLS standard errors are wrong — usually too small.

The shape of the variance

Variance is largest when $p(x) = 0.5$ (maximum uncertainty) and shrinks toward the extremes.

type: lpm-problems
mode: variance

Reading the frown

The error variance traces an inverted parabola in $p(x)$ — it has to, because a coin near 50/50 is the least predictable while one near 0 or 1 is nearly certain. Because the height of this curve changes with $x$, the homoskedasticity assumption (A4) is broken by construction in every LPM, which is why robust standard errors are non-negotiable.

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Fix: Robust (Heteroskedasticity-Consistent) Standard Errors

$\hat{\beta}$ from OLS is still unbiased — it's just the standard errors that are wrong. Compute HC/"sandwich" standard errors instead:

• In R base: lmtest::coeftest(m, vcov = sandwich::vcovHC(m, type = "HC1"))
• Cleaner: fixest::feols(accept ~ percent_offer + factor(stakes), data = df, se = "hetero")

Rule of thumb

Always report robust SEs for LPM. It costs you nothing and repairs the one assumption the model definitionally breaks.

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Summary — What Lecture 2 Teaches

1. Binary outcomes $\Rightarrow$ the OLS fitted value is a conditional probability (under A2).
2. LPM coefficients are marginal effects on $P(y=1)$, measured in percentage points.
3. Empirical lesson: Andersen et al. show fairness preferences survive small stakes but weaken sharply at life-changing amounts.
4. Drawback 1 (fatal-ish): $\hat{y}$ can leave $[0,1]$ — motivates logit/probit.
5. Drawback 2 (fixable): errors are always heteroskedastic with $\text{Var}(u \mid x) = p(x)(1-p(x))$ — use robust SEs.

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Related Notes

• Previous: Lec_01-Introduction & Treatment Effects
• Next: Lec_03-Logit & Probit Models — fixes the $\hat{y} \notin [0,1]$ problem
• Foundational: OLS Estimation, Classical Assumptions A1-A5, Hypothesis Testing
• Related: Heteroskedasticity, Robust Standard Errors, Game Theory