We know that Logit/probit models are consistent with utility maximization in the sense that the model implied probability of choosing option $1$ equals the probability that option $1$ has larger latent utility than option $0$ (the outside option). Is this also true for the linear probability model? My guess is it's not, but I'm not sure.

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    $\begingroup$ The linear probability model is not even consistent with probability theory. I'm 102% certain about that. $\endgroup$ May 29, 2022 at 9:06
  • $\begingroup$ @MichaelGreinecker I agree. $\endgroup$ May 29, 2022 at 9:08

1 Answer 1


This is a very interesting question. Let me do a little algebra.

Consider the latent variable model $y = I(a+xb > u)$, where $I(\cdot)$ is the indicator function. Let $F(\cdot)$ be the CDF of $u$. Then $P(y=1|x)=F(a+xb)$. For the probit, we have $F = \Phi$; for the logit, we have $F=\Lambda$.

The LPM specifies $F(a+xb)$ as $a^*+xb^*$, which can happen only if $F$ is a linear function. But as we know, $F$ can't be (globally) linear because $F$ is a CDF (can't exceed 1, can't be smaller than 0). That's what @MichaelGreinecker is 102% certain about (hats off to you with such a sense of humor), I guess.

But what if $x$ is bounded so $a+xb$ is bounded? Then we can find an example of LPM consistent with unility maximization, as follows.

Suppose that $x \in [0,1]$ is a proportion (bounded). Then, if $b>0$, we have $a \le a+xb \le a+b$. Suppose that the true parameters are $a=0$ and $b=1$ for simplicity. If $u$ is uniformly distributed in $[-2,2]$, then the CDF is $F(t) = \frac14 t + \frac12$ (locally) on the support of $u$. Then for all possible values of $a+xb$, $$P(y=1|x)=F(a+xb) = \frac{a+xb}{4} + \frac12 = \left( \frac{a}{4} + \frac12 \right) +\left( \frac{b}{4} \right) x = \frac12 + \frac14 x,$$ which is certainly a linear probability model.

Here, the point is that the support of the uniform random variable $u$ is sufficiently wide so it covers the support of $a+xb$ and $F(a+xb)$ is linear in $(1,x)$. We can find such examples. We can also find examples that are not consistent with the latent variable specification.

So in general, no, LPM is not always consistent with the latent variable model, but it does not mean LPM cannot be consistent with the latent variable model.

Please note that the above is about the true model or the true DGP. The meaning of a "model consistent with utility maximization" should be defined before rigorously attending the question. If it means such a model for all possible parameter values (i.e., of $a$ and $b$), the answer is no.

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    $\begingroup$ Thank you very much! Your answer is extremely clear and helpful. In particular, I like the fact that you pointed out the need to further clarify the question as "there exists a latent variable model of the form y=1(a+bx>u)" vs "for all such models". $\endgroup$ May 30, 2022 at 6:03

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