In linear regression (squared loss), using matrix we have a very concise notation for the objective

$$\text{minimize}~~ \|Ax-b\|^2$$

Where $A$ is the data matrix, $x$ is the coefficients, and $b$ is the response.

Is there similar a matrix notation for logistic regression objective? All the notations I have seen cannot get rid of the sum over all data points (something like $\sum_{\text data} \text{L}_\text{logistic}(y,\beta^Tx)$).

EDIT: thanks for joceratops and AdamO's great answer. Their answer helped me to realize that another reason linear regression have a more concise notation is because the definition of the norm, which encapsulate the square and the sum or $e^\top e$. But in logistic loss, there is not such definition, which makes notation a little bit more complicated.


In linear regression the Maximize Likelihood Estimation (MLE) solution for estimating $x$ has the following closed form solution (assuming that A is a matrix with full column rank):

$$\hat{x}_\text{lin}=\underset{x}{\text{argmin}} \|Ax-b\|_2^2 = (A^TA)^{-1}A^Tb$$

This is read as "find the $x$ that minimizes the objective function, $\|Ax-b\|_2^2$". The nice thing about representing the linear regression objective function in this way is that we can keep everything in matrix notation and solve for $\hat{x}_\text{lin}$ by hand. As Alex R. mentions, in practice we often don't consider $(A^TA)^{-1}$ directly because it is computationally inefficient and $A$ often does not meet the full rank criteria. Instead, we turn to the Moore-Penrose pseudoinverse. The details of computationally solving for the pseudo-inverse can involve the Cholesky decomposition or the Singular Value Decomposition.

Alternatively, the MLE solution for estimating the coefficients in logistic regression is:

$$\hat{x}_\text{log} = \underset{x}{\text{argmin}} \sum_{i=1}^{N} y^{(i)}\log(1+e^{-x^Ta^{(i)}}) + (1-y^{(i)})\log(1+e^{x^T a^{(i)}})$$

where (assuming each sample of data is stored row-wise):

$x$ is a vector represents regression coefficients

$a^{(i)}$ is a vector represents the $i^{th}$ sample/ row in data matrix $A$

$y^{(i)}$ is a scalar in $\{0, 1\}$, and the $i^{th}$ label corresponding to the $i^{th}$ sample

$N$ is the number of data samples / number of rows in data matrix $A$.

Again, this is read as "find the $x$ that minimizes the objective function".

If you wanted to, you could take it a step further and represent $\hat{x}_\text{log}$ in matrix notation as follows:

$$ \hat{x}_\text{log} = \underset{x}{\text{argmin}} \begin{bmatrix} 1 & (1-y^{(1)}) \\ \vdots & \vdots \\ 1 & (1-y^{(N)})\\\end{bmatrix} \begin{bmatrix} \log(1+e^{-x^Ta^{(1)}}) & ... & \log(1+e^{-x^Ta^{(N)}}) \\\log(1+e^{x^Ta^{(1)}}) & ... & \log(1+e^{x^Ta^{(N)}}) \end{bmatrix} $$

but you don't gain anything from doing this. Logistic regression does not have a closed form solution and does not gain the same benefits as linear regression does by representing it in matrix notation. To solve for $\hat{x}_\text{log}$ estimation techniques such as gradient descent and the Newton-Raphson method are used. Through using some of these techniques (i.e. Newton-Raphson), $\hat{x}_\text{log}$ is approximated and is represented in matrix notation (see link provided by Alex R.).

  • $\begingroup$ Great. Thanks. I think the reason we do not have something like solving $A^\top A x=A^\top b$ is the reason we do not take that step more to make the matrix notation and avoid sum symbol. $\endgroup$
    – Haitao Du
    Aug 9 '16 at 17:28
  • $\begingroup$ We do have some advantage of taking one step further, making it into matrix multiplication would make the code simpler, and in many platforms such as matlab, for loop with sum over all data, is much slower than matrix operations. $\endgroup$
    – Haitao Du
    Aug 9 '16 at 17:33
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    $\begingroup$ @hxd1011: Just a small comment: reducing to matrix equations is not always wise. In the case of $A^TAx=A^Tb$, you shouldn't actually try looking for matrix inverse $A^TA$, but rather do something like a Cholesky decomposition which will be much faster and more numerically stable. For logistic regression, there are a bunch of different iteration schemes which do indeed use matrix computations. For a great review see here: research.microsoft.com/en-us/um/people/minka/papers/logreg/… $\endgroup$
    – Alex R.
    Aug 9 '16 at 17:43
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    $\begingroup$ @AlexR. thank you very much. I learned that using normal equation will make the matrix conditional number squared. And QR or Cholesky would be much better. Your link is great, such review with numerical methods is always what I wanted. $\endgroup$
    – Haitao Du
    Aug 9 '16 at 17:54

@joceratops answer focuses on the optimization problem of maximum likelihood for estimation. This is indeed a flexible approach that is amenable to many types of problems. For estimating most models, including linear and logistic regression models, there is another general approach that is based on the method of moments estimation.

The linear regression estimator can also be formulated as the root to the estimating equation:

$$0 = \mathbf{X}^T(Y - \mathbf{X}\beta)$$

In this regard $\beta$ is seen as the value which retrieves an average residual of 0. It needn't rely on any underlying probability model to have this interpretation. It is, however, interesting to go about deriving the score equations for a normal likelihood, you will see indeed that they take exactly the form displayed above. Maximizing the likelihood of regular exponential family for a linear model (e.g. linear or logistic regression) is equivalent to obtaining solutions to their score equations.

$$0 = \sum_{i=1}^n S_i(\alpha, \beta) = \frac{\partial}{\partial \beta} \log \mathcal{L}( \beta, \alpha, X, Y) = \mathbf{X}^T (Y - g(\mathbf{X}\beta))$$

Where $Y_i$ has expected value $g(\mathbf{X}_i \beta)$. In GLM estimation, $g$ is said to be the inverse of a link function. In normal likelihood equations, $g^{-1}$ is the identity function, and in logistic regression $g^{-1}$ is the logit function. A more general approach would be to require $0 = \sum_{i=1}^n Y - g(\mathbf{X}_i\beta)$ which allows for model misspecification.

Additionally, it is interesting to note that for regular exponential families, $\frac{\partial g(\mathbf{X}\beta)}{\partial \beta} = \mathbf{V}(g(\mathbf{X}\beta))$ which is called a mean-variance relationship. Indeed for logistic regression, the mean variance relationship is such that the mean $p = g(\mathbf{X}\beta)$ is related to the variance by $\mbox{var}(Y_i) = p_i(1-p_i)$. This suggests an interpretation of a model misspecified GLM as being one which gives a 0 average Pearson residual. This further suggests a generalization to allow non-proportional functional mean derivatives and mean-variance relationships.

A generalized estimating equation approach would specify linear models in the following way:

$$0 = \frac{\partial g(\mathbf{X}\beta)}{\partial \beta} \mathbf{V}^{-1}\left(Y - g(\mathbf{X}\beta)\right)$$

With $\mathbf{V}$ a matrix of variances based on the fitted value (mean) given by $g(\mathbf{X}\beta)$. This approach to estimation allows one to pick a link function and mean variance relationship as with GLMs.

In logistic regression $g$ would be the inverse logit, and $V_{ii}$ would be given by $g(\mathbf{X}_i \beta)(1-g(\mathbf{X}\beta))$. The solutions to this estimating equation, obtained by Newton-Raphson, will yield the $\beta$ obtained from logistic regression. However a somewhat broader class of models is estimable under a similar framework. For instance, the link function can be taken to be the log of the linear predictor so that the regression coefficients are relative risks and not odds ratios. Which--given the well documented pitfalls of interpreting ORs as RRs--behooves me to ask why anyone fits logistic regression models at all anymore.

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    $\begingroup$ +1 great answer. formulate it as a root finding on derivative is really new for me. and the second equation is really concise. $\endgroup$
    – Haitao Du
    Aug 9 '16 at 23:00

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