Talk:Linear probability model

Why using LS vs ML as the default estimation method?
The current page introduces the concept of LPM by following the approach in the monograph "Analysis of binary data" by D.R. Cox, i.e. by using OLS as the default estimation method, then proposing iterated weighted least squares (ITLS) to get efficient estimates and stating their equivalence with ML method. However, that book doesn't mention the expression "Linear Probability Model", that instead is present in Agresti's "Introduction to categorical data". The LPM Section (3.2.1) in such book takes an opposite approach, by first talking about ML estimates to be gotten by ITLS, and then introducing the possibility to use a LS approach to avoid the problem of non-convergence due to the algorithm stopping when, in an iteration, a probability outside the 0-1 range is found. Why not considering Agresti's approach, or at least giving the same importance to the ML and the LS approach? Borisba (talk) 23:43, 1 January 2023 (UTC)

Considering LS with transformation of estimated probabilities to manage out-of range values
The page currently quotes the monograph from D.R. Cox several times. However, such book also describes the possibility (despite defining it as "usually rather unlikely") that "the underlying relationship has an 'inclined plane' form, i.e. a range with zero probability of success, a linear portion of positive slope and a range with unit probability of success" (thus, he describes the case of the probability increasing with the predictor: the one with a negative slope is symmetrical). In the case of multiple predictors, this could be generalized to: performing LS estimation first (given that the ML one wouldn't be feasible) and then moving to 1 the values above 1, and moving to 0 negative values. Therefore, I think we could add this as the only way to perform LPM in the case of out-of-range values. This would also be in line with the underlying latent variable foundation, since the probability of a Uniform (0,1) distribution to be below k1, with k1<0, is 0, and the one to be below k2, with k2>1, is 1. Borisba (talk) 21:35, 6 January 2023 (UTC)