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I estimated a linear probability model (LPM) $P(y=1|x_1) = b_0 +b_1x_1 + u $ and a probit model $P(y=1|x_1) = \Phi(b_0 +b_1x_1 + u) $, where $\Phi()$ denotes the cumulative normal distribution. The regressor $x_1$ is the same binary variable.

I observed that the predicted probabilities for both models $\hat{y} =\hat{b_0} +\hat{b_1}x_1 $ and $\hat{y} =\Phi(\hat{b_0} + \hat{b_1}x_1)$ are identical across both models. Why is this the case from a theoretical point of view?

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    $\begingroup$ Assuming $y$ is binary, $y = b_0 +b_1x_1 + u$ is not a linear probability model (LPM), just a usual linear regression model. $P(y=1\mid x_1) = b_0 +b_1x_1 + u$ is the specification for the LPM. Also $y = \Phi(b_0 +b_1x_1 + u)$ is not a probit model - $P(y=1\mid x_1) = \Phi_u(b_0 +b_1x_1)$ is. Are these just typos of the question, or you actually estimated the specifications that you write? $\endgroup$ – Alecos Papadopoulos Dec 18 '13 at 19:52
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Because the model's saturated: with a binary response & a single binary predictor you get a perfect fit no matter what link function you use, with the estimated probability being the observed probability for each value of $x_1$. Think of fitting a line to two points: it doesn't matter whether it's straight or curvy; if it has two parameters it'll go through them.

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  • $\begingroup$ thanks a lot! if i add more explanatory variables (continuous and binary) the fitted probabilities are not identical anymore, but still highly correlated (~98% of correlation). Why is this the case? $\endgroup$ – Daniel Ryback Dec 18 '13 at 18:19
  • $\begingroup$ well, it's to be hoped that the fitted probabilities aren't wildly different for sensible link functions within the range of predictors in the sample. $\endgroup$ – Scortchi - Reinstate Monica Dec 18 '13 at 18:27

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