Can you provide me with an intuition about what it means to derive a probit or logit model from an underlying latent variable model?


1 Answer 1


Logit and probit models can be derived from an underlying latent variable model. To see this, let $y^{\ast}$ be an unobserved (i.e. latent) variable, and suppose that \begin{equation} y^{\ast} = \beta_0 + x\beta + \epsilon, ~y=1[y^{\ast}>\gamma] \label{latent} \end{equation} where $1[\cdot]$ is an indicator function taking value $1$ if $y^{\ast}>\gamma$ and zero otherwise. Here, $\gamma$ is a chosen threshold. If the latent unobserved utility $y^\ast$ is above the threshold, then $y$ takes the value one; otherwise, it takes the value zero.

The idea is to see the binary outcome $y=\{0,1 \}$ as a dichotomization of a latent continuous variable $y^{\ast}$ modeled with a canonical multiple linear regression model. For example, a bank can issue a loan ($y=1$) or choose not to ($y=0$). Think of it as having an underlying continuous latent outcome, $y^{\ast}$ which can be regarded as a measure of the choice utility.

In the model above, we assume that $\epsilon$ is independent of $x$ and that $\epsilon$ either has the standard logistic distribution or the standard normal distribution.

Recall, the standard logistic and standard normal distribution are symmetric about zero. Thus, $1-G(-z)=G(z)$.

From the model above, we can derive the response probability for $y$. First, we assumed that $y_i=1$ if and only if $y^{\ast}_i > \gamma$, with $\gamma$ set arbitrarily to zero. Thus, \begin{align*} \mathbb{P}(y=1|x) =& \mathbb{P}(y^{\ast} >0|x) = \mathbb{P}(\beta_0 + x\beta + \epsilon >0|x) = \\ =&\mathbb{P}[ \epsilon > -(\beta_0 + x\beta ) |x] =1 - \mathbb{P}[ \epsilon \le -(\beta_0 + x\beta ) |x] \\ =& 1 - G[ - (\beta_0 + x\beta)] = G(\beta_0 + x \beta) \end{align*} where $\mathbb{P}[ \epsilon \le -(\beta_0 + x\beta ) |x]$ is the cdf of the error term of the latent outcome.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.