The latent variable motivation for losgistic regression goes thus. There exist $Y^*=\beta^tX+\epsilon$ which is continuous. We can only observe $Y$ at specific thresholds of $Y^*$, say at $Y^*\leq \alpha_1$, $\alpha_1<Y^*\leq \alpha_2$, $\alpha_2<Y^*\leq \alpha_3$ and $\alpha_3<Y^*\leq \alpha_4$, with $Y=1, 2, 3,4$ respectively

Therefore $P(Y\leq j)=P(Y^*\leq \alpha_j)=P(\beta^tX+\epsilon\leq \alpha_j)=P(\epsilon\leq \alpha_j-\beta^tX)$

Assuming that $\epsilon$ is logistic then we have: logit$(P(Y\leq j))=\alpha_j-\beta^tX$

My question goes like this: With only one threshold say say $\alpha$: We get a binary logistic model of the form


and with $n$ thresholds for large $n$ we observe almost all of $Y^*$ so we can use ordinary least squares regression for modeling. Can someone illustrate to me in a nice way what we gain and what we loose for large $n$ and for small $n$ where $n$ refers to the number of thresholds?.

  • $\begingroup$ You have an issue of scale. You are essentially cutting up the real line into $n$ intervals, possibly of unequal length. There's no reason to believe that the first interval corresponds to 1 on the latent scale, even if the intervals are thin. $\endgroup$
    – dimitriy
    Commented May 14, 2014 at 15:26
  • $\begingroup$ Well I do not see a problem with that, the model transform what ever real values you give to clusters to the latent scale. The intercepts you get are in the latent scale. $\endgroup$ Commented May 14, 2014 at 16:03
  • $\begingroup$ So if you increase $n$ your model will give you $n-1$ values of the latent response scale, that is the intercepts. $\endgroup$ Commented May 14, 2014 at 16:05
  • $\begingroup$ If your variable is a "coarsened" latent variable, then I would agree. But take bond ratings. There's an underlying latent variable called creditworthiness that some agency has divided into many bins, which range from AAA, AA, A, BBB, and so on to D. You can imagine coding these as 12, 11, 10,.... But you could just as well use another coding scheme. Which one should you use? $\endgroup$
    – dimitriy
    Commented May 14, 2014 at 17:45
  • 1
    $\begingroup$ I see your point and agree with you. So the problem of scale only arises when you use OLS. And using the proportional odds model with large $n$, there is no scale problem but then the number of parameters increase drastically. Do you agree?. $\endgroup$ Commented May 14, 2014 at 19:13