In statistics, a probit model is a type of regression where the dependent variable can take only two values, for example married or not married. The word is a portmanteau, coming from probability + unit. The purpose of the model is to estimate the probability that an observation with particular characteristics will fall into a specific one of the categories; moreover, classifying observations based on their predicted probabilities is a type of binary classification model. A probit model is a popular specification for a binary response model. As such it treats the same set of problems as does logistic regression using similar techniques. When viewed in the generalized linear model framework, the probit model employs a probit link function. It is most often estimated using the maximum likelihood procedure, such an estimation being called a probit regression.
Conceptual framework Suppose a response variable Y is binary, that is it can have only two possible outcomes which we will denote as 1 and 0. For example, Y may represent presence/absence of a certain condition, success/failure of some device, answer yes/no on a survey, etc. We also have a vector of regressors X, which are assumed to influence the outcome Y. Specifically, we assume that the model takes the form
P ( Y = 1 ∣ X ) = Φ ( X T β ) , {\displaystyle P(Y=1\mid X)=\Phi (X^{\operatorname {T} }\beta ),}
where P is the probability and Φ {\displaystyle \Phi } is the cumulative distribution function (CDF) of the standard normal distribution. The parameters β are typically estimated by maximum likelihood. It is possible to motivate the probit model as a latent variable model. Suppose there exists an auxiliary random variable
Y ∗ = X T β + ε , {\displaystyle Y^{\ast }=X^{T}\beta +\varepsilon ,}
where ε ~ N(0, 1). Then Y can be viewed as an indicator for whether this latent variable is positive:
Y = { 1 Y ∗ > 0 0 otherwise } = { 1 X T β + ε > 0 0 otherwise } {\displaystyle Y=\left.{\begin{cases}1&Y^{*}>0\\0&{\text{otherwise}}\end{cases}}\right\}=\left.{\begin{cases}1&X^{\operatorname {T} }\beta +\varepsilon >0\\0&{\text{otherwise}}\end{cases}}\right\}}
The use of the standard normal distribution causes no loss of generality compared with the use of a normal distribution with an arbitrary mean and standard deviation, because adding a fixed amount to the mean can be compensated by subtracting the same amount from the intercept, and multiplying the standard deviation by a fixed amount can be compensated by multiplying the weights by the same amount. To see that the two models are equivalent, note that
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