The finite-horizon five-expert prediction problem

We prove that no single rank ordered adversary strategy is globally optimal for the prediction with expert advice problem with six or more experts, in both the geometric stopping and finite time horizon settings. The proof is based on establishing a leading order correction when one expert moves far ahead of the others. This allows us to connect optimal strategies between $n$ and $j<n$ experts and utilize recent results on the exact optimality set for the five expert problem.

In this paper, we derive an explicit solution of the stationary prediction with expert advice PDE for five experts. The formula is given in three regions. In the first two regions, it is the four-expert solution plus a single integral with an elementary positive density. In the third region, it is a finite sum of hyperbolic products whose coefficients are determined by one scalar quadrature. Our formula establishes that the direction $(1,0,1,0,0)$ is optimal throughout the ordered sector, and that the COMB strategy $(1,0,1,0,1)$ is optimal only on a lower dimensional subset of the sector (where $x_1=x_2$ and $x_3=x_4$). This disproves the COMB optimality conjecture of Gravin, Peres and Sivan. The verification of the Hamiltonian inequalities is a tedious task, part of which is completed with a computer assisted proof. The verification reduces to 21 scalar inequalities, which we prove using 147 exact rational Bernstein polynomial certificates. The exact certificates and their independent arithmetic checks are included in a supplement to this paper, and a Lean 4 formalization machine-checks the verification and both main theorems, apart from the viscosity characterization.

Prediction with expert advice is a fundamental problem in online learning. When the time horizon $T$ is known in advance, the minimax cumulative regret over $n$ experts is asymptotically $\sqrt{\frac{T \ln n}{2}}$. This is achieved by the Multiplicative Weights Update algorithm with a learning rate tuned to $T$, and is known to be tight. If instead the regret bound is required to hold simultaneously at every time $t$, the best known guarantee has been $\sqrt{t \ln n}$---a factor of $\sqrt{2}$ worse---and it has remained unknown whether this factor of $\sqrt{2}$ is necessary. We show that it is not. We give an algorithm, requiring no knowledge of the horizon, whose cumulative regret satisfies $R_t \le \bigl(1 + O(\sqrt{\ln \ln n / \ln n})\bigr)\sqrt{t \ln n / 2}$ simultaneously for every $t \ge 1$.