According to section 4.4.5 of "The Elements of Statistical Learning: Data Mining, Inference, and Prediction. Second Edition" by Trevor Hastie & Robert Tibshirani & Jerome Friedman:

... if the data in a two-class logistic regression model can be perfectly separated by a hyperplane, the maximum likelihood estimates are undefined (i.e. infinite)

I can see how this is the case in $\mathbb{R}$ as:

$$l(\beta) = \sum_{x_i<x_0}\{-log(1+\exp{(\beta_0 + \beta_1x_i)}\} + \sum_{x_i>x_0}\{\beta_0 + \beta_1x_i -log(1+\exp{(\beta_0 + \beta_1x_i)}\}$$ where the two classes are separated by point $x_0$ and coding classes as $y_1=1$ and $y_2=0$.


$$l(\beta) = \sum_{x_i<x_0}\{-log(1+\exp{(\beta_0 + \beta_1x_0 + \beta_1(x_i - x_0))}\} + \sum_{x_i>x_0}\{\beta_0 + \beta_1x_0 + \beta_1(x_i - x_0) -log(1+\exp{(\beta_0 + \beta_1x_0 + \beta_1(x_i - x_0))}\}$$


$x_1 - x_0 > 0$ when $x_i>0$,

$x_1 - x_0 < 0$ when $x_i<0$,

and setting $\beta_0 = -\beta_1x_0$,

this reduces to:

$$l(\beta) = \sum_{x_i<x_0}\{-log(1+\exp{(\beta_1(x_i - x_0))}\} + \sum_{x_i>x_0}\{\beta_1(x_i - x_0) -log(1+\exp{(\beta_1(x_i - x_0))}\}$$

which is maximised as $\beta_1 \to \infty$

Giving the following (undefined) MLE's: $\beta_0 = -sign(x_0)\infty$ and $\beta_1 = \infty$

How can this be proven in $p$ dimensional space ($\mathbb{R}^p$) and thus a separating hyperplane?

  • $\begingroup$ There's no difference at all, because modulo the hyperplane, you're back down to one dimension. $\endgroup$
    – whuber
    Commented Feb 24, 2020 at 17:31
  • $\begingroup$ Thanks @whuber . Would you mind expanding on that? I have updated my question with how I approached it in $\mathbb{R}$ but I'm not entirely clear on how to extend to $\mathbb{R}^p$ $\endgroup$
    – Seraf Fej
    Commented Feb 24, 2020 at 18:04

1 Answer 1


The data are a sequence of observations of vectors of explanatory variables $x_i$ in $p$ dimensions, to which are associated binary responses $y_i\in\{0,1\}.$ "Separated by a hyperplane" means there exists a nonzero $p$-vector $v$ for which $$x_i^\prime v \gt 0 \text{ when } y_i=1 \text{ and otherwise }x_i^\prime v \lt 0.$$

By minimizing $x_i^\prime v$ over all observations with $y_i=1$ and minimizing $-x_i^\prime v$ over all other observations we obtain a positive number $a\gt 0$ for which

$$x_i^\prime v \ge a \text{ when } y_i=1 \text{ and otherwise }x_i^\prime v \le -a.\tag{*}$$

It turns out the issue is reduced to analyzing what happens in the direction of $v,$ which essentially reproduces the one-dimensional situation. The details follow. They reveal that the quotation is a little misleading: there is no such thing as "the" maximum likelihood estimate and they aren't all characterized by vectors of "undefined" coefficients.

Let's pause a moment to discuss Bernoulli variables, or "unfair coins." Logistic regression relies on a probability model of $y_i$ as an independent flip of an unfair coin, but it does so in a special way. It posits an increasing "link function" $h$ that continuously maps the open interval of possible probabilities $(0,1)$ one-to-one onto the real numbers. A formula for $h$ needn't concern us here; all that matters is the implication that probabilities approaching $0$ and $1$ correspond under $h$ to real numbers that get arbitrarily large in size: that is, "approach infinity." Specifically, $h$ has an inverse $h^{-1}$ and

$$\lim_{x\to\infty}h^{-1}(x) = 1;\quad \lim_{x\to-\infty}h^{-1}(x) = 0.$$

Back to the question. The likelihood for coefficients $\beta=(\beta_1, \ldots, \beta_p)$ is the product, over all observations, of the chance that a Bernoulli variable with parameter $x_i\beta$ actually equals $y_i:$

$$\mathcal{L}(\beta) = \prod_{i\mid y_i = 1} h^{-1}(x_i^\prime \beta)\, \prod_{i\mid y_i = 0} (1-h^{-1}(x_i^\prime \beta)).$$

Consider a positive real number $\lambda$ and use the inequalities $(*)$ and the fact $h^{-1}$ is increasing (since $h$ is) to bound the likelihood for $v\lambda$ by

$$\mathcal{L}(v\lambda) = \prod_{i\mid y_i = 1} h^{-1}(x_i^\prime v\lambda)\, \prod_{i\mid y_i = 0} (1-h^{-1}(x_i^\prime v\lambda)) \ge \prod_{i\mid y_i=1} h^{-1}(a\lambda)\, \prod_{i\mid y_i = 0} (1-h^{-1}(-a\lambda)).$$

Because $a\gt 0,$ $a\lambda\to\infty$ and $-a\lambda\to-\infty$ as $\lambda\to\infty,$ showing

$$\lim_{\lambda\to\infty} h^{-1}(a\lambda) = 1 = \lim_{\lambda\to\infty} 1 - h^{-1}(-a\lambda),$$


$$\lim_{\lambda\to\infty} \mathcal{L}(v\lambda) \ge \prod_{i\mid y_i=1} 1\, \prod_{i\mid y_i=0} 1 = 1.$$

This is a global maximum because the likelihood, being a product of probabilities, cannot exceed $1.$

We're not quite done, because it's logically possible for most of the components of $v$ to be zero. When multiplied by large $\lambda,$ they would remain zero and not "become undefined" as claimed in the quotation.

The last step is to note that we may always arrange for every component of $v$ to be nonzero. One proof of this notices that the values $x_i v$ are continuous functions of the components of $v$ and therefore the smallest of all the $|x_i v|$ (equal to $a$ above) is a continuous function of the components. Since $a\gt 0,$ we may change every nonzero component of $v$ by a sufficiently small amount to assure that the smallest of all the $|x_i v|$ is still greater than, say, $a/2.$ This new reduced value for $a$ will serve just as well in the foregoing analysis. Now, using this adjusted $v$ (which defines a different separating hyperplane), all the components of $\beta=v\lambda$ diverge as $\lambda\to\infty.$

A careful statement of the conclusion, then, is

When a separating hyperplane exists, the likelihood can be maximized in a manner that causes all the coefficients of $\beta$ to diverge.

It does not rule out solutions in which some components of $\beta$ are finite: but the argument clearly implies that

  1. any such components must be zero;

  2. at least one component will diverge; and

  3. there can be infinitely many directions $v$ for which the likelihood of $\beta=v\lambda$ is maximized as $\lambda\to\infty.$


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