We look at a simple quadratic minimax problem on the Euclidean sphere, and consider what happens when it flips from convex to concave.

# Introduction

This is a sneak preview of a nice result with Malek, Bartlett and Abbasi-Yadkori to be presented in Koolen et al. (2015). Fix $$\alpha \in {\mathbb R}$$ and $${\boldsymbol{b}}\in {\mathbb R}^d$$ such that $${\|{\boldsymbol{b}}\|} \le 1$$. Let us consider the following min-max problem $V_\alpha ~:=~ \min_{{\boldsymbol{a}}\in {\mathbb R}^d} \max_{\substack{{\boldsymbol{x}}\in {\mathbb R}^d \\ {\|{\boldsymbol{x}}\|} \le 1}} ~ {\|{\boldsymbol{a}}-{\boldsymbol{x}}\|}^2 + (\alpha-1) {\|{\boldsymbol{x}}\|}^2 + 2 {\boldsymbol{b}}^{\intercal}{\boldsymbol{x}}$ as a function of $$\alpha$$. (Here $${\|{\boldsymbol{x}}\|}^2 = {\boldsymbol{x}}^{\intercal}{\boldsymbol{x}}$$ denotes the Ecludean norm squared.) In this post we establish that the value and minimiser are $V_\alpha ~=~ \begin{cases} \frac{{\|{\boldsymbol{b}}\|}^2}{1-\alpha} & \text{if \alpha \le 0,} \\[.2em] {\|{\boldsymbol{b}}\|}^2 + \alpha & \text{if \alpha \ge 0,} \end{cases} \qquad \text{and} \qquad {\boldsymbol{a}}~=~ \begin{cases} \frac{{\boldsymbol{b}}}{1-\alpha} & \text{if \alpha \le 0,} \\[.2em] {\boldsymbol{b}}& \text{if \alpha \ge 0.} \end{cases}$ The two pieces of $$V_\alpha$$ meet at $$\alpha = 0$$, where the value is $${\|{\boldsymbol{b}}\|}^2$$. The left and right derivatives at $$\alpha = 0$$ are ${\|{\boldsymbol{b}}\|}^2 \qquad \text{and} \qquad 1$ and so the function $$V_\alpha$$ is differentiable iff $${\|{\boldsymbol{b}}\|}=1$$.

Why is this min-max problem cool? Our problem is an easy example of a min-max function that flips (at $$\alpha=0$$) from convex-concave to convex-convex. The former type of functions admit nice minimax analysis. The latter is generally much more complicated to deal with. Yet in our example it is still analytically tractable.

# Analysis

Let us rewrite $$\label{eq:value} V_\alpha ~=~ \min_{{\boldsymbol{a}}\in {\mathbb R}^d} \max_{\substack{{\boldsymbol{x}}\in {\mathbb R}^d \\ {\|{\boldsymbol{x}}\|} \le 1}} ~ {\|{\boldsymbol{a}}\|}^2 + \alpha {\|{\boldsymbol{x}}\|}^2 + 2 ({\boldsymbol{b}}- {\boldsymbol{a}})^{\intercal}{\boldsymbol{x}}.$$

## The convex-concave case $$\alpha \le 0$$

The objective is concave in $${\boldsymbol{x}}$$. The maximum is hence found at zero derivative, i.e.${\boldsymbol{x}}~=~ \frac{{\boldsymbol{a}}- {\boldsymbol{b}}}{\alpha} ,$ (it remains to check this is feasible) where the value equals $V_\alpha ~=~ \min_{{\boldsymbol{a}}\in {\mathbb R}^d} ~ {\|{\boldsymbol{a}}\|}^2 - \frac{{\|{\boldsymbol{a}}- {\boldsymbol{b}}\|}^2}{\alpha} .$ As $$\alpha \le 0$$ this is convex in $${\boldsymbol{a}}$$, and hence minimised at vanishing derivative. That is, ${\boldsymbol{a}}~=~ \frac{{\boldsymbol{b}}}{1-\alpha} ,$ (back substitution reveals $${\boldsymbol{x}}= \frac{{\boldsymbol{b}}}{1-\alpha}$$, which is indeed feasible as $${\|{\boldsymbol{b}}\|} \le 1$$ by assumption and we are considering the case $$\alpha \le 0$$) where the value equals $V_\alpha ~=~ \frac{{\|{\boldsymbol{b}}\|}^2}{1-\alpha} .$ Note that $$\max_{{\boldsymbol{x}}} \min_{{\boldsymbol{a}}}$$ results in inner minimiser $${\boldsymbol{a}}={\boldsymbol{x}}$$ and then again in outer maximiser $${\boldsymbol{x}}= \frac{{\boldsymbol{b}}}{1-\alpha}$$. Therefore the pair $${\boldsymbol{a}}={\boldsymbol{x}}=\frac{{\boldsymbol{b}}}{1-\alpha}$$ form a saddle point or Nash equilibrium (are mutually best responses).

## The convex-convex case $$\alpha \ge 0$$

The objective is convex in $${\boldsymbol{x}}$$, and it is hence maximised on the unit ball at the point of the unit sphere that maximises the linear term, i.e. at ${\boldsymbol{x}}~=~ \frac{{\boldsymbol{b}}-{\boldsymbol{a}}}{{\|{\boldsymbol{b}}-{\boldsymbol{a}}\|}} ,$ Plugging this in, we find $V_\alpha ~=~ \min_{{\boldsymbol{a}}\in {\mathbb R}^d} ~ {\|{\boldsymbol{a}}\|}^2 + \alpha + 2 {\|{\boldsymbol{b}}- {\boldsymbol{a}}\|} .$ This is a concave function of $${\boldsymbol{a}}$$, although it is not differentiable due to the norm. Yet it is minimised where zero is a sub-gradient. In our case this happens at ${\boldsymbol{a}}~=~ {\boldsymbol{b}},$ where we used the assumption that $${\|{\boldsymbol{b}}\|} \le 1$$. We conclude $V_\alpha ~=~ {\|{\boldsymbol{b}}\|}^2 + \alpha .$ Notice that the choice $${\boldsymbol{a}}={\boldsymbol{b}}$$ kills the linear term in $${\boldsymbol{x}}$$ of the objective $$\eqref{eq:value}$$, and hence equalises the objective over the unit sphere.

# Using a single algorithm throughout

We found two algorithms, one for $$\alpha \le 0$$ and one for $$\alpha \ge 0$$. Now suppose we can only play one of them in all cases. How bad would that be?

First suppose we always play $${\boldsymbol{a}}={\boldsymbol{b}}$$. This kills the linear term in $${\boldsymbol{x}}$$. So the objective is maximised in $${\boldsymbol{x}}$$ either anywhere on the unit sphere if $$\alpha \ge 0$$ or at $${\boldsymbol{x}}=0$$ if $$\alpha \le 0$$. The worst-case value is hence $\begin{cases} {\|{\boldsymbol{b}}\|}^2 & \text{if \alpha \le 0,} \\ {\|{\boldsymbol{b}}\|}^2 + \alpha & \text{if \alpha \ge 0.} \end{cases}$

Now suppose we always play $${\boldsymbol{a}}= \frac{{\boldsymbol{b}}}{1-\alpha}$$. Then $\max_{\substack{{\boldsymbol{x}}\in {\mathbb R}^d \\ {\|{\boldsymbol{x}}\|} \le 1}} ~ \frac{{\|{\boldsymbol{b}}\|}^2}{(1-\alpha)^2} + \alpha {\|{\boldsymbol{x}}\|}^2 - \frac{2 \alpha}{1-\alpha} {\boldsymbol{b}}^{\intercal}{\boldsymbol{x}}$ is maximised at $${\boldsymbol{x}}= \frac{ - \frac{\alpha}{1-\alpha} {\boldsymbol{b}}}{{\| - \frac{\alpha}{1-\alpha} {\boldsymbol{b}}\|}}$$ if $$\alpha \ge 0$$ where it takes value $\frac{{\|{\boldsymbol{b}}\|}^2}{(1-\alpha)^2} + \alpha + 2 \frac{\alpha}{{|1-\alpha|}} {\|{\boldsymbol{b}}\|}$ or at $${\boldsymbol{x}}=\frac{{\boldsymbol{b}}}{1-\alpha}$$ if $$\alpha \le 0$$ where it takes value $\frac{{\|{\boldsymbol{b}}\|}^2}{1-\alpha}$ as before.

# Violated norm bound

We now look at what happens if we play an algorithm designed to operate against data satisfying the constraint $${\|{\boldsymbol{x}}\|} \le 1$$, and encounter data violating that constraint.

## Convex-concave

Say $$\alpha \le0$$ and we play $${\boldsymbol{a}}= \frac{{\boldsymbol{b}}}{1-\alpha}$$. Then the objective is $\frac{{\|{\boldsymbol{b}}\|}^2 }{1-\alpha} + \alpha {\Big\|\frac{{\boldsymbol{b}}}{1-\alpha} - {\boldsymbol{x}}\Big\|}^2 ~\le~ \frac{{\|{\boldsymbol{b}}\|}^2 }{1-\alpha}$ regardless of $${\boldsymbol{x}}$$.

## Convex-convex

Say $$\alpha \ge 0$$ and we play $${\boldsymbol{a}}={\boldsymbol{b}}$$. Then the objective is ${\|{\boldsymbol{b}}\|}^2 + \alpha {\|{\boldsymbol{x}}\|}^2$ which scales nicely with the actual data norm.

So in both cases we see that we do not need to assume the norm bound on the data. The convex-concave worst case is independent of any norm bound. The convex-convex case adapts to the norm bound without needing to know it.

Koolen, Wouter M., Alan Malek, Peter L. Bartlett, and Yasin Abbasi-Yadkori. 2015. “Minimax Time Series Prediction.” In Advances in Neural Information Processing Systems (NIPS) 28, to appear.