# Why and what induces degeneracy in the solution space in logistic regression models with respect to the energy landscape of the logistic loss?

I was trying to understand the sources of having many (equivalent) solutions $w$ with respect to the training set in the context of logistic regression (with a predictor of the form $h_{w}(x) = \text{sigmoid}(y w^\top x)$ ). The conditions for degeneracy in the energy landscape is what interests me most and what conditions does the problem need to have (data, model complexity, the way true distribution looks like, etc) for there to be more than 1 unique minimizer (or more than 1 unique limit point minimizer). Thus, what matters is the set of parameters that achieves minimum loss (not zero classification error). There might be many and the conditions for this is what I am interested in.

Recall that the loss function (for the 2 classes cases, otherwise use softmax with cross-entropy loss) is:

$$J(w; X,Y) = \frac{1}{N} \sum^N_{n=1} \ln( 1 + e^{y^{(n)} w^\top x^{(n)}} )$$

in this problem I am wondering what induces having many solutions (or not).

My hypothesis are as follow:

1. The number of data points matter. Only when we have $N_{train} = \infty$ can we have a unique solution come out from logistic regression. This is obvious because then we essentially know the true hyperplane. If we lack data my intuition tells me we need some other sort regularizer to choose a unique solution. The funny thing is that because of the log the error keeps going down forever. So perhaps a better way to say "unique" solution is to say unique in the limit.
2. Even Bishop's book mentions that even if $N > D$ over constrained, its possible that there are infinite different solution (but the problem remains convex!?)
3. When the model is "too complex" then there are an infinite set of solutions. For example, if the data lies in $\mathbb{R}^{D^*}$ but the model we have has more features say lies in $\mathbb{R^D}$ where $D > D^*$, then things are problematic. This is because the new features can be changed to sum to zero without actually changing anything.
4. Since logistic regression, we only care about the sign. Thus we can always scale our solution $\hat w$ arbitrarily without changing the decision boundary. This seems to always be true no matter what...so maybe there are always infinite number of solutions in logistic regression?
5. Is there very a unique minimizer in logistic regression? The fact that the loss is always decreasing makes me think that no...(unless a limit exists?)
6. Assume separability. Even in this case, there could be an infinite number o

f solution unless we satisfy my first point $N_{train} = \infty$.

So my question is: when (and the whys) of when logistic regression has an infinite set of solution (and what conditions do we need)?

The reason the questions linked are not helpful (or wikipedia, which as a matter of fact does have a list where a model may not reach convergence which probably is related to my question) is because I don't just want to know what induces degeneracy, but also why.

For example, if a solution is possible to be completely separable, then how does that induce a degenerate sol? For example, in linear regression its clear that its induced due to a null space, so any solution can be expressed as

$$\hat x = x_{particular} + \alpha x_{nullspace}$$

which makes it easy to understand where the flatness of the landscape comes from. But for logistic regression, this isn't clear. How does say, rank directly affect the solution of the logistic regression model?

As I was reading Bishop section 4.3.2 I found the following statement:

Note that the problem will arise even if the number of data points is large compared with the number of parameters in the model, so long as the training data is linearly separable.

This means that the number of data points doesn't play an important role. A good answer should explain this too.

• @Sycorax why do you think those questions are helpful? My question is not about separation or separability. My question is about the energy landscape and how it looks like and the mathematical conditions for there to be a unique minimizer or not. Its obvious that the error can't be zero for the logistic loss because of $log(1+e^{-yw^\top x})$, so non-convergence is obvious too (which I point out in my question itself). In short, my question is more interested about mathematical conditions of how the landscape for logistic regression looks like when its degenerate. – Charlie Parker Jan 29 '18 at 21:00
• @Sycorax maybe this will help. I am not interested in the classification error. I am interested in the loss, so the actual landscape we are optimizing over and the number of minimizers that arise from that. Thats why separability is not necessarily interesting to me. If we need to impose it to make the analysis easier so be it, but separability is not the goal. I also remove the tag separation because its not relevant, this is a optimization problem technically. – Charlie Parker Jan 29 '18 at 21:32
• @Sycorax ah, that looks more promising. I will check that out later and see if it answers my Q. Thanks for taking the time to check my question btw :) – Charlie Parker Jan 29 '18 at 22:42
• @Sycorax I'm not sure how that question u linked is helpful. It doesn't explain anything, nor does it address any of the points I brought up well. It lacks a lot of quality. – Charlie Parker Jan 31 '18 at 16:48
• @Sycorax plus the question only talks about the data while I care to characterize the complexity of the model was well to specify the degeneracy of the solution space. – Charlie Parker Jan 31 '18 at 16:49

I think that your question hinges on two concepts.

# 1. Convexity

It can be simpler to consider logistic regression model as the composition of two steps. Suppose that $y\in\{0,1\}$ are the outcomes we'd like to model as a multilinear function of some features $X$.

\begin{align} z &= X\beta \\ p &= \text{logit}^{-1}(z) \\ y &\sim \text{Bernoulli}(p) \end{align}

From the definition of nullspace, if the nullspace of $X$ is nonempty, then one can choose some pair $\beta^\prime \neq \beta$ for which $X\beta = X\beta^\prime$.

Since the product is the same, so is $z=X\beta=X\beta^\prime$ also the same for either choice of coefficients.

Since $z$ is the same, either choice of coefficients has the same success probability $p$.

Since the success probability is the same, the likelihood (or cross-entropy loss, if you prefer) is the same for either choice of coefficients.

This implies that the optimization problem is not uniquely optimized when the nullspace of $X$ is nonempty.

# 2. Separability

Consider the binary cross-entropy loss. \begin{align} \mathcal{L}(\theta) &= -\frac{1}{n}\sum_{i=1}^n \left[y_i \log(p_i) + (1-y_i) \log(1-p_i)\right] \end{align}

For a single observation, for $y_j = 1$, the loss will get smaller as $p$ gets closer to $1$, so that means $z_j$ will have to get larger, implying that $\beta$ is moving in some direction to improve model fit. For $y_k = 0$, the loss will get smaller as $p$ gets closer to $0$, so $z_k$ will have to get smaller. (The logistic function has horizontal asymptotes at 0 and 1, so $\log(0)$ is never attained for $||\beta|| < \infty$.)

However, when separability is present, there is a set of features for which you can do both of these things at the same time.

These are some examples.

1. The simplest form is that a feature takes values $\pm 1$, and all positive values correspond to the positive class and all negative values to the negative class. Making its corresponding $\beta_i$ larger will always move $z$ in the "right" direction for all examples.

2. We can generalize the simple binary example to the case of a real-valued feature, if $x > 0$ corresponds to $y = 1$ and $x < 0$ to $y = 0$. It should be clear now that the same pathological behavior is present, since making the coefficient larger will always move $z$ in the "right" direction for both classes.

3. Finally, we can generalize to linear combinations of features. Suppose you're fitting a model with an intercept and a real-valued feature. If $x > c$ has $y=1$ and $x < c$ has $y=0$, choose the coefficient for the intercept as $-c$. In this case, we're back in the same situation as (2).

When separability is not present, this is not possible. Suppose that instead of $y$ always "matching" the feature, the label and feature "match" in all but 1 instance. In that case, making $\beta$ larger will not always improve the loss. In this case, $\beta$ is finite.

# 3. Convexity & Separability

It's possible that one, both or neither of these effects is present in any particular problem. Moreover, strongly convex, non-separable logistic regression has a unique, finite minimizer.

• are you saying that everything has to do with the invertibility of $X$? Like what if its perfectly invertible but the points are selected in a way such that they are separable (hope this is possible). Wouldn't this mean that there are an infinite number of solution still despite there being an empty nullspace? (I understand now based on ur answer that a non-empty nullspace has us have multiple identical products). – Charlie Parker Feb 1 '18 at 1:42
• (1) If $N \neq D,$ the matrix is not square and therefore is not invertible. (2) If the data are separable, then the loss function is unbounded, so there's a direction in which the loss can be improved arbitrarily. – Sycorax Feb 1 '18 at 2:00
• (2) is not obvious to me. I only know it cuz of the cartoon I sketch on a piece of paper. Do you know how to rigorously show thats true and how it affects (any high dimensional) energy landscape related to logistic regression (or even cross entropy loss + softmax)? – Charlie Parker Feb 1 '18 at 2:02
• I appreciate ur patience, bare with me. Maybe I have a misconception but the $- y_i log(p_i)$ is only infinity when $y_i=1$ and $p_i = 0$. Thats when the model is predicting probability zero for the label $y=1$ when $1$ is the correct label. If things are separable I'd assume that is easier not harder....Not sure how separability affects the energy landscape, but I'd guess that if its separable it should assign $p_i=1$ when $y=1$ and $1-p_i = 1$ when when $y=0$, so I am not sure why it would evaluate the negative log of zero. Do you mind clarifying that? – Charlie Parker Feb 1 '18 at 3:13
• Thanks for the updates. They were helpful. Your right since we can't achieve 0 or 1 the composition of -log and the sigmoid means that the loss will always keep decreasing. My (hopefully last) question is, if we were to run GD on separable data, should we actually be seeing the weights of the model grow to infinity? – Charlie Parker Feb 1 '18 at 5:01