Take $x \in \{0,1\}^d$ and $y \in \{0,1\}$ and suppose we model the task of predicting y given x using logistic regression. When can logistic regression coefficients be written in closed form?

One example is when we use a saturated model.

That is, define $P(y|x) \propto \exp(\sum_i w_i f_i(x_i))$, where $i$ indexes sets in the power-set of $\{x_1,\ldots,x_d\}$, and $f_i$ returns 1 if all variables in the $i$'th set are 1, and 0 otherwise. Then you can express each $w_i$ in this logistic regression model as a logarithm of a rational function of statistics of the data.

Are there other interesting examples when closed form exists?

  • 6
    $\begingroup$ I assume you mean "when are the MLEs of the parameters in closed form?" $\endgroup$
    – Glen_b
    Commented Aug 23, 2010 at 0:43
  • $\begingroup$ Can you give more detail what you did? Your question reads as if you tried to derive the ordinary least squares estimator for a logistic regression problem? $\endgroup$
    – Momo
    Commented Sep 19, 2012 at 23:00
  • 1
    $\begingroup$ Thanks for the interesting post/question, Yaroslav. Do you have a reference for the example that you show? $\endgroup$
    – Bitwise
    Commented Apr 27, 2014 at 19:49
  • 1
    $\begingroup$ It's been a while, but possibly it was in Lauritzen's "Graphical Models" book. The broader foundations of the answer for this question are there -- you get closed form solution when the (hyper)graph formed by sufficient statistics is chordal $\endgroup$ Commented May 13, 2014 at 19:52
  • 1
    $\begingroup$ This might be interesting tandfonline.com/doi/abs/10.1080/… I believe this is a special case of an analytical solution when you only have a 2x2 table $\endgroup$
    – Austin
    Commented Dec 20, 2017 at 2:28

2 Answers 2


As kjetil b halvorsen pointed out, it is, in its own way, a miracle that linear regression admits an analytical solution. And this is so only by virtue of the linearity of the problem (with respect to the parameters). In OLS, you have $$ \sum_i (y_i - x_i \beta)^2 \to \min_\beta, $$ which has the first order conditions $$ -2 \sum_i (y_i - x_i\beta) x_i = 0 $$ For a problem with $p$ variables (including constant, if needed— there are some regression through the origin problems, too), this is a system with $p$ equations and $p$ unknowns. Most importantly, it is a linear system, so you can find a solution using the standard linear algebra theory and practice. This system will have a solution with probability 1 unless you have perfectly collinear variables.

Now, with logistic regression, things aren't that easy anymore. Writing down the log-likelihood function, $$ l(y;x,\beta) = \sum_i y_i \ln p_i + (1-y_i) \ln(1-p_i), \quad p_i = (1+\exp(-\theta_i))^{-1}, \quad \theta_i = x_i \beta, $$ and taking its derivative to find the MLE, we get $$ \frac{\partial l}{\partial \beta'} = \sum_i \frac{{\rm d}p_i}{{\rm d}\theta}\Bigl( \frac{y_i}{p_i} - \frac{1-y_i}{1-p_i} \Bigr)x_i = \sum_i \Bigl[y_i-\frac1{1+\exp(x_i\beta)}\Bigr]x_i $$ The parameters $\beta$ enter this in a very nonlinear way: for each $i$, there's a nonlinear function, and they are added together. There is no analytical solution (except probably in a trivial situation with two observations, or something like that), and you have to use nonlinear optimization methods to find the estimates $\hat\beta$.

A somewhat deeper look into the problem (taking the second derivative) reveals that this is a convex optimization problem of finding a maximum of a concave function (a glorified multivariate parabola), so either one exists, and any reasonable algorithm should be finding it rather quickly, or things blow off to infinity. The latter does happen to logistic regression when ${\rm Prob}[Y_i=1|x_i\beta > c] = 1$ for some $c$, i.e., you have a perfect prediction. This is a rather unpleasant artifact: you would think that when you have a perfect prediction, the model works perfectly, but curiously enough, it is the other way round.

  • 1
    $\begingroup$ the question is why your last equation is not solvable. is it due to the logistic function's inverse diverging at 0 and 1, or is this due to the nonlinearity in general? $\endgroup$ Commented Sep 20, 2012 at 12:17
  • 8
    $\begingroup$ (+1) Regarding your last paragraph: From a mathematical perspective it does work "perfectly" in the sense that an MLE will yield a perfect separating hyperplane. Whether your numerical algorithm behaves sensibly in that circumstance is a separate matter. Laplace smoothing is often used in such situations. $\endgroup$
    – cardinal
    Commented Sep 20, 2012 at 13:27
  • $\begingroup$ @eyaler, I would say this is due to nonlinearity in general. My understanding is that there is a limited set of circumstances when this can be solved, although I don't know what these circumstances are. $\endgroup$
    – StasK
    Commented Sep 21, 2012 at 3:54
  • 1
    $\begingroup$ I don't understand, what mathematical condition is present that makes the system not have a closed form solution? Is there a general condition where things in general don't have closed form solutions? $\endgroup$ Commented Jan 31, 2018 at 21:59
  • 1
    $\begingroup$ @CharlieParker, in general, nonlinear equations do not have closed form solutions. (See however a comment to the original question that refers to a paper deriving the score equations for categorical predictors, and solving for parameters explicitly.) $\endgroup$
    – StasK
    Commented Jun 16, 2020 at 17:37

This post was originally intended as a long comment rather than a complete answer to the question at hand.

From the question, it's a little unclear if the interest lies only in the binary case or, perhaps, in more general cases where they may be continuous or take on other discrete values.

One example that doesn't quite answer the question, but is related, and which I like, deals with item-preference rankings obtained via paired comparisons. The Bradley–Terry model can be expressed as a logistic regression where $$ \mathrm{logit}( \Pr(Y_{ij} = 1) ) = \alpha_i - \alpha_j , $$ and $\alpha_i$ is an "affinity", "popularity", or "strength" parameter of item $i$ with $Y_{ij} = 1$ indicating item $i$ was preferred over item $j$ in a paired comparison.

If a full round-robin of comparisons is performed (i.e., a pairwise preference is recorded for each unordered $(i,j)$ pair), then it turns out that the rank order of the MLEs $\hat{\alpha}_i$ correspond to the rank order of $S_i = \sum_{j \neq i} Y_{ij}$, the sum total of times each object was preferred over another.

To interpret this, imagine a full round-robin tournament in your favorite competitive sport. Then, this result says that the Bradley–Terry model ranks the players/teams according to their winning percentage. Whether this is an encouraging or disappointing result depends on your point of view, I suppose.

NB This rank-ordering result does not hold, in general, when a full round-robin is not played.

  • 4
    $\begingroup$ I was interested in binary because it was easiest to analyze. I have found a very broad sufficient condition in works of Lauritzen -- you get closed form if a corresponding log-linear model is decomposable $\endgroup$ Commented Jan 29, 2011 at 22:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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