Why is simulation used for probit choice probabilities?

As explained in this answer, the probability of a certain outcome in a Logit model can be written as $$P=\int_{\varepsilon=-\beta'x}^{\infty} f(\varepsilon)d\varepsilon\\ = 1- F(-\beta'x) = 1-\dfrac{1}{1+\exp(\beta'x)} = \dfrac{\exp(\beta'x)}{1+\exp(\beta'x)}$$ I think I understand the reasoning behind this; the logit model assumes that the error term is distributed logistically, so the CDF of the logistic distribution is used. The probit model assumes that the error term follows a normal distribution giving

$$\Pr(Y=1 \mid X) = \Phi(X'\beta) =\int_{\varepsilon=-\beta'x}^{\infty} f(\varepsilon)d\varepsilon = \Phi(x_{i}'\beta) = \int_{-\infty}^{x_{i}'\beta} \dfrac{1}{\sqrt{2\pi}}e^{-\dfrac{1}{2}t^{2}}dt$$ where $\Phi(x)$ is the CDF of the standard normal distribution. Now the referred answer, along with other sources, say that because these probabilities aren't closed-form, unlike those of the logit model, probit models require simulation.

But why use something complicated as simulation when there are a variety of methods that very accurately approximate $\Phi(x)$? It would seem to me that this would both be less complicated and more accurate.

• This is ancillary to your question, but neither the logit nor the probit can be properly said to have an error term. Mar 26, 2015 at 1:48
• @gung I agree, but there are people (economists) who think of logit/probit as a linear model for a latent propensity for success that includes an Gaussian/logisticly distributed error term, and we observe a success when the latent propensity is larger than 0. I suspect that that is what Max meant. This way of thinking is not to my taste: solid empirical research is about stuff we can see not about stuff we imagine. Nevertheless, it is in some disciplines a common way of thinking about these models. Mar 26, 2015 at 10:57
• @Maarten You're completely right in that suspicion - I was approaching it from a latent variable perspective.
– JH-
Mar 26, 2015 at 15:40

2 Answers

For a simple probit, software I am familiar with (Stata, R, SPSS) does not use simulation to compute that integral.

Simulation is sometimes used in more complicated variations, like a multinomial probit (this is what the answer you linked to refered to) or when random effects are included in the model. There the likelihood function contains an additional integral over the additional (group level) error term(s). For a list of examples see this special issue of the Stata Journal.

• Thank you, this answers my question. I got very confused because I found an exercise where one was asked to simulate the likelihood for a basic probit model with one regressor and I just couldn't see why one would be interested in doing that given the plentiful ways to solve the integral in question. I guess the exercise was purely meant as an introduction to simulating such choice probabilities.
– JH-
Mar 26, 2015 at 15:47

In both cases the model is $$Y\sim Be(p)$$ in which Y is the dependent variable. Here we model the effect of independent variables on Y through a link function. Logit and probit functions are used as the link function for logisic and probit regression models respectively $$logit(p)=X'\beta~~~~or~~~~~\Phi(p)=X'\beta$$ So it's easy to see if you want to simulate Y based on an observation of X, you have to compute p based on the $$X'\beta$$ In logistic regression it is straight forward as you have mentioned. But in probit regression model you have to compute $$p=\Phi^{-1}(X'\beta)$$ to be able to simulate Y. So it's up to you which method you like to use to compute this quantity. You can use numerical methods or interpolating from a table or any other method you know.