I want to predict a health problem. I have 3 outcome categories that are ordered: 'normal', 'mild', and 'severe'. I wish to predict this from two predictor variables, a test result (a continuous, interval covariate) and family history with this problem (yes or no). In my sample, the probabilities are 55% (normal), 35% (mild), and 10% (severe). In this sense, I could always just predict 'normal' and be right 55% of the time, although this would give me no information about individual patients. I fit the following model:

\begin{align} \text{the cut point for }\widehat{(y \ge 1)} &= -2.18 \\ \text{the cut point for }\widehat{(y \ge 2)} &= -4.27 \\ \hat\beta_{\rm test} &= 0.60 \\ \hat\beta_{\rm family\ history} &= 1.05 \end{align}

Assume there is no interaction and everything is fine with the model. The concordance, c, is 60.5%, which I understand to be the maximum predictive accuracy the model affords.

I come across two new patients with the following data: 1. test = 3.26, family = 0; 2. test = 2.85, family = 1. I want to predict their prognosis. Using the formula: $$ \frac{\exp(-X\beta - {\rm cutPoint})}{(1+\exp(-X\beta - {\rm cutPoint}))} $$ (and then taking the differences amongst the cumulative probabilities), I can calculate the probability distribution over the response categories conditional on the model. R code (n.b., due to rounding issues, the output doesn't match perfectly):

cut1 <- -2.18
cut2 <- -4.27
beta <- c(0.6, 1.05)
X    <- rbind(c(3.26, 0), c(2.85, 1))

pred_cat1      <- exp(-1*(X%*%beta)-cut1)/(1+exp(-1*(X%*%beta)-cut1))
pred_cat2.temp <- exp(-1*(X%*%beta)-cut2)/(1+exp(-1*(X%*%beta)-cut2))
pred_cat3      <- 1-pred_cat2.temp
pred_cat2      <- pred_cat2.temp-pred_cat1

predicted_distribution <- cbind(pred_cat1, pred_cat2, pred_cat3)

Namely: 1. 0 = 55.1%, 1 = 35.8%, 2 = 9.1%; and 2. 0 = 35.6%, 1 = 46.2%, 2 = 18.2%. My question is, how do I go from the probability distribution to a predicted response category?

I have tried several possibilities using the sample data, where the outcome is known. If I just pick max(probabilities), accuracy is 57%, a slight improvement over the null, but below the concordance. Moreover, in the sample, this approach never picks 'severe', which is what I really want to know. I tried a Bayesian approach by converting null and model probabilities into odds and then picking the max(odds ratio). This does pick 'severe' occasionally, but yields worse accuracy 49.5%. I also tried a sum of the categories weighted by the probabilities and rounding. This, again, never picks 'severe', and has low accuracy 51.5%.

What is the equation that takes the information above and yields optimal accuracy (60.5%)?


2 Answers 2


You are making a leap that you need to classify predicted values. The fact that your method never picks the "severe" category is a consequence of the discrete nature of the problem and that "severe" is infrequent. With ordinal response models you can just use exceedance probabilities on their own (for all but one category) or just quote the individual probabilities. If $Y$ is roughly interval scaled you can also use the predicted mean. These are all available in the R rms package lrm and associated function predict.lrm. Many people assume that classification is the goal when in fact risk prediction is the underlying goal.

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    $\begingroup$ Thanks for your help. I suspected the low freq of severe was part of the issue. I think my raw Y, 0 1 2, is insufficiently equal interval. I gather my goal is mistaken. Unfortunately, I think I want to know what category a new patient will fall into / don't fully understand what my goal should be. Is it possible to provide a little more insight? (Actually, I suspect CV isn't the forum for a complete lesson; alternatively, do you know where I could learn about this issue? I have read the sections from Agresti's Intro & Hosmer & Lemeshow's Logistic, but to no avail.) $\endgroup$ Commented Nov 9, 2011 at 15:16
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    $\begingroup$ The goal is driven by the needed decision or by the subject matter. If you state the ultimate goal I might be able to comment. $\endgroup$ Commented Nov 9, 2011 at 23:11
  • $\begingroup$ Sorry about my lack of clarity, it seems to have been a problem above also. At present, I want to be able to predict the outcome for new cases. In the long run, I want to understand ord log reg better, e.g. how do you get residuals if you don't have predicted categories? I gather slightly better accuracy is possible, but I don't know how to get it. I'm sure you don't have time to explain everything, but neither Agresti, nor H&L say anything about predictions or residuals, etc. and I couldn't find anything w/ google. Thus, I asked on CV. I do appreciate your continued assistance. $\endgroup$ Commented Nov 10, 2011 at 2:50
  • $\begingroup$ Please explain your need for residuals. These are not in intrinsic part of the model. In terms of how to state predictions, providing the predicted probability that $Y\geq j$ for various $j$ is a good way, and requires no arbitrary decisions. Looking at the histogram of predicted probabilities you can judge how useful the model is, i.e., how often it provides more definitive probabilities that are closer to 0 or 1 than to 0.5. $\endgroup$ Commented Nov 12, 2011 at 14:28

Concordance is defined by checking the mean score, not the maximum score.

So for your examples, mean scores for 1 is 0*55.1% + 1*35.8% + 2*9.1% = 0.54, and 2 is (through similar calculations) 0.826.

It is this value that you should compare to get the concordance or any other association statistics.

Ref - http://support.sas.com/documentation/cdl/en/statug/63347/HTML/default/viewer.htm#statug_logistic_sect042.htm

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    $\begingroup$ No, concordance is computed using the raw outcome variable $Y$ and the linear predictor $X\beta$ or any of the predicted probabilities (as they are all monotonically related to each other, i.e., just shifted by in intercept before computing the expit). Somers' $D_{xy}$ rank correlation coefficient uses this concordance measure. $\endgroup$ Commented Nov 9, 2011 at 14:02
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    $\begingroup$ P.S. Note that the SAS documentation, which is a re-write of my original documentation for the precursor SAS PROC LOGIST I wrote many years ago, is now incorrect, and its formula for the mean is incorrect unless $Y$ consists of consecutive integers. $\endgroup$ Commented Nov 9, 2011 at 14:13

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