2
$\begingroup$

Can someone please provide a definition of the ordered probit model so that I can calculate the value of $\Pr(y \le k | \mathbf{x})$, i.e. probability of observation falling in the $k$-th class or worse given the observations $x$.

I am trying to follow the definition of the ordered logit in this wikipedia article, which I have reproduced here.

However, in the probit case, I cannot find out what the function $\sigma$ should be even after following the links in the article.

$\endgroup$
3
$\begingroup$

In this article $\sigma$ is any function which maps an individual's score, which can typically be any number along $(- \infty, \infty)$, to the interval $(0, 1)$. The most commonly-used function is the sigmoid (the inverse logit) which looks like

$$ \frac{e^{\beta^T x}}{e^{\beta^T x} + 1} $$

and this is mentioned in the article. The function $\sigma$ will usually be a distribution function (that is, it will express the probability that some random variable is less than or equal to the argument of the function) because these have the properties of mapping to the unit interval while being monotone increasing, the two main characteristics we want inverse link functions to have. In the probit case $\sigma(x) = \Phi(x)$ where $\Phi$ is the standard normal distribution function. That is, if $Z \sim$ normal$(0, 1)$ then $\Phi(x) = P(Z \leq x) = \int_{-\infty}^{x} e^{-t^2 / 2} / \sqrt{2 \pi} \, dt$.

$\endgroup$
  • $\begingroup$ thanks for your answer. I am specifically looking for the function for the probit case. $\endgroup$ – Alex Jan 13 '16 at 2:44
  • $\begingroup$ @Alex Ok, I added a bit about the probit link function. $\endgroup$ – dsaxton Jan 13 '16 at 2:48
  • 1
    $\begingroup$ @Alex Yes, the cumulative distribution function of the standard normal, and any distribution function would work. There would just be subtle differences in the relationship between $x$ and the fitted probabilities. However, if you're only interested in putting observations in different groups and not the class probabilities themselves, then it's entirely irrelevant which one you use. $\endgroup$ – dsaxton Jan 13 '16 at 2:59
  • 1
    $\begingroup$ Just to point out that $\sigma()$ is not the standard deviation here. $\endgroup$ – Nick Cox Jan 13 '16 at 10:31
  • 1
    $\begingroup$ @NickCox Right. Probably not the best choice of notation by the authors of that article. $\endgroup$ – dsaxton Jan 13 '16 at 14:59

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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