# Logistic regression and latent data

Assume a simple logistic regression model: given binary data $y_1,\ldots,y_N$ where for each $1 \leq i \leq N$ the outcome of $y_i$ depends on one variable. The succes probability is $p_i = \mathbb{P}(y_i = 1|x_i)$ is then modeled as a function of $x_i$ by the following relation $$\ln\left(\frac{p_i}{1-p_i}\right) = \beta_0+\beta_1 x_i$$ In some cases, they use latent variables $Z$ by defining $Z_i \geq 0 \Leftrightarrow y_i = 1$ and $Z_i <0 \Leftrightarrow y_i = 0$ and then define the regression model $$Z_i = \beta_0+\beta_1 x_i + \epsilon_i$$ Is there any particular reason why the latent variable approach is more useful? Furthermore, when using the original logistic model above we can plot $p_i$ in function of $x_i$. How does that work for the latent variable approach? I don't fully understand the main idea behind this approach.

• There's a missing error term $e_i$ in your latent regression model. This is sometimes assumed to follow a normal distribution which would lead to $\text{probit}(p_i)=\beta_0 + \beta_1 x_i$. The logit link implicitly assumes that $e_i$ follows a logistic distribution. – Jarle Tufto Jun 13 '16 at 10:18
• Thanks for the comment. I forgot the error term, but does it not appear in the $\mbox{probit}(p_i) = \beta_0+\beta_1 x_i$ as well? So, the use of latent variables is useful since we can say in the probit model : $\mathbb{P}(Y_i = 1 | x_i) = \mathbb{P}(Z_i \geq 0 | x_i) = \Phi(\beta_0+\beta_1x_i)$, where $\Phi$ is the cdf of the standard normal. – Cavents Jun 13 '16 at 11:21
• It should be better called "underlying variable" model, not "latent variable" model. – ttnphns Jun 13 '16 at 11:55