Given a data set with binary outcomes $y\in\{0,1\}^n$ and some predictor matrix $X\in\mathbb{R}^{n\times p}$, the standard logistic regression model estimates coefficients $\beta_{MLE}$ which maximize the binomial likelihood. When $X$ is full rank $\beta_{MLE}$ is unique; when perfect separation is not present, it is finite.

Does this maximum likelihood model also maximize the ROC AUC (aka $c$-statistic), or does there exist some coefficient estimate $\beta_{AUC} \neq \beta_{MLE}$ which will obtain a higher ROC AUC? If it is true that the MLE does not necessarily maximize ROC AUC, then another way to look at this question is "Is there an alternative to likelihood maximization which will always maximize ROC AUC of a logistic regression?"

I am assuming that models are otherwise the same: we're not adding or removing predictors in $X$, or otherwise changing the model specification, and I'm assuming that the likelihood-maximizing and AUC-maximizing models are using the same link function.

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    $\begingroup$ Surely $\beta_{\text{AUC}} \neq \beta_{\text{MLE}}$ if, e.g., some link function generates a better fit than a logit? Other than that, good question, if the data generating process can be assumed as logit. $\endgroup$
    – runr
    Commented Aug 19, 2019 at 13:15
  • $\begingroup$ Good question but consider this. ROC and AUC are used to compare two different models, so if a solution for the MLE estimation of any model is unique, this means that you can get a different AUC only if you change the specification of the current model and you estimate a new different model via MLE. So at this point another question would be: is there any other “better” estimation method (maximization algorithm ecc) other than the simple MLE applicable to the same model such that I get to different estimates of the coefficients leading to new “better” betas with higher AUC? $\endgroup$
    – Fr1
    Commented Aug 19, 2019 at 13:18
  • $\begingroup$ @Nutle exactly, that would be a different specification $\endgroup$
    – Fr1
    Commented Aug 19, 2019 at 13:20
  • $\begingroup$ @Fr1 Yes, that is what unique means. What I'm implying in my question is something like "what if there is some alternative to the MLE which achieves a higher AUC?" If it is true that there is a different linear model (a model other than the MLE) which achieves a higher AUC, then that would be interesting to know about. $\endgroup$
    – Sycorax
    Commented Aug 19, 2019 at 13:22
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    $\begingroup$ @Sycorax what else do we assume?:) Assumptions are important, since if we know the true DGP with link and variables used, the MLE is uniformly most powerful unbiased statistic. $\endgroup$
    – runr
    Commented Aug 19, 2019 at 13:34

1 Answer 1


It is not the case that $\beta_{MLE} \equiv \beta_{AUC}$.

To illustrate this, consider that AUC can written as

$P(\hat y_1 > \hat y_0 | y_1 = 1, y_0 = 0)$

In otherwords, the ordering of the predictions is the only thing that affects AUC. This is not the case with the likelihood function. So as a mental exercise, suppose we had a single predictors and in our dataset, we don't see perfect separation (i.e., $\beta_{MLE}$ is finite). Now, if we simply take the value of the largest predictor and increase it by some small amount, we will change the likelihood of this solution, but it will not change the AUC, as the ordering should remain the same. Thus, if the old MLE maximized AUC, it will still maximize AUC after changing the predictor, but will no longer maximize the likelihood.

Thus, at the very least, it is not the case that $\beta_{AUC}$ is not unique; any $\beta$ that preserves the ordering of the estimates achieves the exact same AUC. In general, since the AUC is sensitive to different aspects of the data, I would believe that we should be able to find a case where $\beta_{MLE}$ does not maximize $\beta_{AUC}$. In fact, I'd venture a guess that this happens with high probability.

EDIT (moving comment into answer)

The next step is to prove that the MLE doesn't necessarily maximize the AUC (which isn't proven yet). One can do this by taking something like predictors 1, 2, 3, 4, 5, 6, $x$ (with $x > 6$) with outcomes 0, 0, 0, 1, 1, 1, 0. Any positive value of $\beta$ will maximize the AUC (regardless of the value of $x$), but we can chose an $x$ large enough that the $\beta_{MLE} < 0$.

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    $\begingroup$ (+1) Ah! Of course -- since it's about ordering, we could arbitrarily change the intercept which obviously must change the likelihood value, but the ordering must be the same because none of the feature coefficients have changed, so the AUC will remain fixed. $\endgroup$
    – Sycorax
    Commented Aug 19, 2019 at 13:42
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    $\begingroup$ +1. Does the edit example work with $n \rightarrow \infty$, though? If we need to take large enough $x$ for this to work with large $n$, doesn't the probability of such values existing quickly converge to 0, for some fixed logit? $\endgroup$
    – runr
    Commented Aug 19, 2019 at 14:17
  • $\begingroup$ @Nutle: well, depends what you mean about $n \rightarrow \infty$. If we took $n$ copies (predictors + outcomes) of my toy dataset, then yes the result would hold. However, if we took $n$ copies of those set of predictors, and the data really came from a logistic regression model, that would almost never happen (as you point out). Note, however, that something akin to this could happen with high probability if the relation between the predictors didn't really follow a logistic regression model. $\endgroup$
    – Cliff AB
    Commented Aug 19, 2019 at 14:28
  • $\begingroup$ Yes, thanks, was talking about the size. So, assuming such heavy tailed distribution is known, would the example still hold if the MLE estimate was adjusted for the true distribution? What I'm going for, is if the probability of such $x$ existing for any sample $n$ is not approaching 0, shouldn't the MLE estimate react to it accordingly and not act as it would with an outlier? Sorry If I'm not entirely clear here with the wording $\endgroup$
    – runr
    Commented Aug 19, 2019 at 14:37
  • $\begingroup$ AUROC as an optimization criteria was always a terrible idea. $\endgroup$ Commented Sep 24, 2023 at 12:25

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