I'm reviewing the very basics of discrete-choice models (binary choice, multinomial, tobit, etc.), and I seem to have taken for granted that latent variable models are just one (very convenient and intuitive) way of formulating discrete choice models. The "General Binary Outcome Model" as given by Cameron and Trivedi (2005) is

\begin{align*} y_i &= \begin{cases} 1 & \text{with probability }p_i \\ 0 & \text{with probability }1-p_i\end{cases}\\ p_i &= \Pr[y_i=1\mid \mathbf x] = F(\mathbf x_i'\boldsymbol\beta) \end{align*} for "some specified function" $F$, a vector of parameters $\boldsymbol\beta$, and regressors $\mathbf x$ . The authors go onto say "is is natural to specify $F(\cdot)$ to be a cumulative distribution function" such that $p_i\in[0,1]$. In practice, this isn't a strict requirement, because the linear probability model (LPM) arises from the function $F(\mathbf x_i'\boldsymbol\beta)=\mathbf x_i'\boldsymbol\beta$, but in this case would the probability instead be given as

$$ p_i = \max\{0, \min\{1,F(\mathbf x_i'\boldsymbol\beta)\}\}$$ to ensure the probability is in fact well defined? This is very pedantic, but I want to ensure I have a firm grasp on the appropriate DGP here.

  • $\begingroup$ Hi! The linear probability model assumes $Y_i$ to be normal, so the link is the identity function. $\endgroup$
    – utobi
    Nov 4, 2022 at 18:04

1 Answer 1


The linear probability model is just

$$ \Pr(Y=1) = \mathbf{X}\boldsymbol{\beta} $$

It is a very simple model that does not give you any guarantees of the probabilities being proper. Technically, it implicitly assumes that nothing could go beyond the $[0, 1]$ bound. This model simply does not try to give proper predictions (unless within some bounds), so trying to be pedantic here is hopeless. Truncating the probabilities is not a part of this model, though may be a practical workaround for its limitations. In most cases, however, it would make more sense to use a model that does give guarantees of making predictions that make sense, e.g. logistic regression.

  • $\begingroup$ So for the sake of illustration, if I were to simulate observations from the DGP, how would I draw observations of Y after drawing values of $\mathbf x_i\boldsymbol\beta$ if some probabilities fall outside the unit interval? Or is this not even possible, and the LPM is inherently misspecified? $\endgroup$
    – ECON10105
    Nov 4, 2022 at 19:50
  • $\begingroup$ @ECON10105 you can't. You need to have such $X$’s and $\beta$’s that it's not possible. Only for such a case the model makes sense. $\endgroup$
    – Tim
    Nov 5, 2022 at 7:35

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