<p>This is a very interesting question. Let me do a little algebra.</p>
<p>Consider the latent variable model <span class="math-container">$y = I(a+xb &gt; u)$</span>, where <span class="math-container">$I(\cdot)$</span> is the indicator function. Let <span class="math-container">$F(\cdot)$</span> be the CDF of <span class="math-container">$u$</span>. Then <span class="math-container">$P(y=1|x)=F(a+xb)$</span>. For the probit, we have <span class="math-container">$F = \Phi$</span>; for the logit, we have <span class="math-container">$F=\Lambda$</span>.</p>
<p>The LPM specifies <span class="math-container">$F(a+xb)$</span> as <span class="math-container">$a^*+xb^*$</span>, which can happen only if <span class="math-container">$F$</span> is a linear function. But as we know, <span class="math-container">$F$</span> can't be (globally) linear because <span class="math-container">$F$</span> 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.</p>
<p>But what if <span class="math-container">$x$</span> is bounded so <span class="math-container">$a+xb$</span> is bounded? Then we can find an example of LPM consistent with unility maximization, as follows.</p>
<p>Suppose that <span class="math-container">$x \in [0,1]$</span> is a proportion (bounded). Then, if <span class="math-container">$b&gt;0$</span>, we have <span class="math-container">$a \le a+xb \le a+b$</span>. Suppose that the true parameters are <span class="math-container">$a=0$</span> and <span class="math-container">$b=1$</span> for simplicity. If <span class="math-container">$u$</span> is uniformly distributed in <span class="math-container">$[-2,2]$</span>, then the CDF is <span class="math-container">$F(t) = \frac14 t + \frac12$</span> (locally) on the support of <span class="math-container">$u$</span>. Then for all possible values of <span class="math-container">$a+xb$</span>,
<span class="math-container">$$P(y=1|x)=F(a+xb) = \frac{a+xb}{4} + \frac12 = \left( \frac{a}{4} + \frac12 \right) +\left( \frac{b}{4} \right) x = \frac12 + \frac14 x,$$</span>
which is certainly a linear probability model.</p>
<p>Here, the point is that the support of the uniform random variable <span class="math-container">$u$</span> is sufficiently wide so it covers the support of <span class="math-container">$a+xb$</span> and <span class="math-container">$F(a+xb)$</span> is linear in <span class="math-container">$(1,x)$</span>. We can find such examples. We can also find examples that are not consistent with the latent variable specification.</p>
<p>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.</p>
<p>Please note that the above is about the true model or the true DGP. The meaning of a &quot;model consistent with utility maximization&quot; should be defined before rigorously attending the question. If it means such a model for all possible parameter values (i.e., of <span class="math-container">$a$</span> and <span class="math-container">$b$</span>), the answer is no.</p>

CC BY-SA 4.0
2022-05-30T06:03:40.023Z
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".
ExcitedSnail
CC BY-SA 4.0
2022-05-30T04:13:24.337Z
2022-05-30T04:19:05.537Z
2022-05-30T04:19:05.537Z
chan1142
<p>We know that Logit/probit models are consistent with utility maximization in the sense that the model implied probability of choosing option <span class="math-container">$1$</span> equals the probability that option <span class="math-container">$1$</span> has larger latent utility than option <span class="math-container">$0$</span> (the outside option). Is this also true for the linear probability model? My guess is it's not, but I'm not sure.</p>

CC BY-SA 4.0
2022-05-29T09:06:28.683Z
The linear probability model is not even consistent with probability theory. I'm 102% certain about that.
Michael Greinecker
CC BY-SA 4.0
2022-05-29T09:08:04.493Z
@MichaelGreinecker I agree.
ExcitedSnail
<econometrics><utility><rationality><logit>
Is the linear probability model consistent with utility maximization?
ExcitedSnail
# Is the linear probability model consistent with utility maximization?

(Asked by: username_0 on 2022-05-29)

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.

**Tags:** econometrics, utility, rationality, logit
> **Comments:**
> **username_1**: The linear probability model is not even consistent with probability theory. I'm 102% certain about that.  
> **username_0**: @MichaelGreinecker I agree.  

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## 1 answers

### Answer by username_2 on 2022-05-30. Score: 5

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.

> **Comments:**
> **username_0**: 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".  

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