Efficient and Optimal Fixed-Time Regret with Two Experts
Laura Greenstreet, Nick Harvey, Victor Sanches Portella
The Two-Experts' Problem
Prediction with Expert Advice
Player
Adversary
\(n\) Experts
\displaystyle
\mathrm{Regret}(T) = \sum_{t = 1}^T \langle \ell_t, x_t \rangle - \min_{i = 1, \dotsc, n} \sum_{t = 1}^T
\ell_t(i)
0.5
0.1
0.3
0.1
Probabilities
x_t
1
0
0.5
0.3
Costs
\ell_t
Player's loss:
\langle \ell_t, x_t \rangle
Loss of Best Expert
Player's Loss
Knows \(T\) (fixed-time)
Known and New Results
Multiplicative Weights Update method:
\displaystyle
\mathrm{Regret}(T) \leq \sqrt{\frac{T}{2} \ln n}
Optimal for \(n,T \to \infty\) !
If \(n\) is fixed, we can do better
Worst-case regret for 2 experts:
Cover's Algorithm
\(O(T)\) time per round
Dynamic Programming
\(\{0,1\}\) costs
\(O(1)\) time per round
Stochastic Calculus
\([0,1]\) costs
\displaystyle \sqrt{\frac{T}{2\pi}} + O(1)
[Cover '67]
Our Algorithm
Technique:
Discretize a solution to a stochastic calculus problem
[HLPR - FOCS '20]
How to exploit the knowledge of \(T\)?
We need to analyze the discretization error!
!
!
Online Learning
🤝
Stochastic Calculus
Our Results
Result:
An Efficient and Optimal Algorithm in Fixed-Time with Two Experts
\(O(1)\) time per round
was \(O(T)\) before
\displaystyle
\mathrm{Regret}(T) \leq \sqrt{\frac{T}{2 \pi}} + 1.3
Holds for general costs!
Technique:
Discretize a solution to a stochastic calculus problem
[HLPR '20]
How to exploit the knowledge of \(T\)?
Non-zero discretization error!
Insight:
Cover's algorithm has connections to stochastic calculus!
This connection seems to extend to more experts and other problems in online learning in general!
Gaps and Cover's Algorithm
Simplifying Assumptions
We will look only at \(\{0,1\}\) costs
1
0
0
1
0
0
1
1
Equal costs do not affect the regret
Cover's algorithm relies on these assumptions by construction
Our alg. and analysis extends to fractional costs
Gap between experts
Thought experiment: how much probability mass to put on each expert?
Cumulative Loss on round \(t\)
\(\frac{1}{2}\) is both cases seems reasonable!
Takeaway: player's decision may depend only on the gap between experts's losses
Gap = |42 - 20| = 22
Worst Expert
Best Expert
42
20
2
2
42
42
(and maybe on \(t\))
Cover's Dynamic Program
Player strategy based on gaps:
Choice doesn't depend on the specific past costs
p(t, g)
on the Worst expert
1 - p(t, g)
on the Best expert
We can compute \(V^*\) backwards in time via DP!
\displaystyle
V^*[t, g] =
Max regret to be suffered at time \(t\) with gap \(g\)
\(O(T^2)\) time to compute \(V^*\)
At round \(t\) with gap \(g\)
\displaystyle
V^*[0, 0] =
Max. regret for a game with \(T\) rounds
Computing the optimal strategy \(p^*\) from \(V^*\) is easy!
Cover's DP Table
(w/ player playing optimally)
Cover's Dynamic Program
Player strategy based on gaps:
Choice doesn't depend on the specific past costs
p(t, g)
on the Lagging expert
1 - p(t, g)
on the Leading expert
We can compute \(V^*\) backwards in time via DP!
Getting an optimal player \(p^*\) from \(V^*\) is easy!
\displaystyle
V^*[t, g] =
Max regret-to-be-suffered at round \(t\) with gap \(g\)
\(O(T^2)\) time to compute the table — \(O(T)\) amortized time per round
p^*(t,g) = \frac{1}{2} \big( V^*[t, g-1] - V^*[t, g+1]\big)
At round \(t\) with gap \(g\)
\displaystyle
V^*[0, 0] =
Optimal regret for 2 experts
Connection to Random Walks
Optimal player \(p^*\) is related to Random Walks
\displaystyle
p^*(t,g)
For \(g_t\) following a Random Walk
\displaystyle
\approx\mathbb{P}\Big(\mathcal{N}(0,T - t) > g\Big)
Central Limit Theorem
Not clear if the approximation error affects the regret
The DP is defined only for integer costs!
Lagging expert finishes leading
= \mathbb{P} \Big(
\Big)
Let's design an algorithm that is efficient and works for all costs
Bonus: Connections of Cover's algorithm with stochastic calculus
Connection to Random Walks
Theorem
\displaystyle
\Big]
\displaystyle
= \sqrt{\frac{T}{2\pi}} + O(1)
Player \(p^*\) is also connected to RWs
\displaystyle
p^*(t,g)
For \(g_t\) following a Random Walk
\displaystyle
\approx\mathbb{P}\Big(\mathcal{N}(0,T - t) > g\Big)
Central Limit Theorem
Not clear if the approximation error affects the regret
The DP is defined only for integer costs!
\displaystyle
V^*[0,0] =
\displaystyle
\frac{1}{2}
\displaystyle
\mathbb{E}\Big[
Lagging expert finishes leading
= \mathbb{P} \Big(
\Big)
[Cover '67]
# of 0s of a Random Walk of len \(T\)
Let's design an algorithm that is efficient and works for all costs
Bonus: Connections of Cover's algorithm with stochastic calculus
Continuous Regret
A Probabilistic View of Regret Bounds
Formula for the regret based on the gaps
\displaystyle
\mathrm{Regret(T)}
= \sum_{t = 1}^{T}
p(t, g_{t-1})(g_t - g_{t-1})
Discrete stochastic integral
Moving to continuous time:
Random walk \(\longrightarrow\) Brownian Motion
\(g_0, \dotsc, g_t\) are a realization of a random walk
\displaystyle
\Bigg\{
\displaystyle
\Delta g_t = \pm 1
Useful Perspective:
Deterministic bound = Bound with probability 1
A Probabilistic View of Regret Bounds
Formula for the regret based on the gaps
\displaystyle
\mathrm{Regret(T)}
= \sum_{t = 1}^{T}
p(t, g_{t-1})(g_t - g_{t-1})
Random walk \(\longrightarrow\) Brownian Motion
\displaystyle
\mathrm{ContRegret(p, T)}
= \int_{0}^{T}
p(t, |B_t|)\mathrm{d}|B_t|
Reflected Brownian motion (gaps)
Conditions on the continuous player \(p\)
Continuous on \([0,T) \times \mathbb{R}\)
p(t,0) = \frac{1}{2}
for all \(t \geq 0\)
Stochastic Integrals and Itô's Formula
How to work with stochastic integrals?
\displaystyle
R(T, |B_T|) - R(0, 0) =
Itô's Formula:
\(\overset{*}{\Delta} R(t, g) = 0\) everywhere
ContRegret \( = R(T, |B_T|) - R(0,0)\)
\displaystyle
\implies
Goal:
Find a "potential function" \(R\) such that
(1) \(\partial_g R\) is a valid continuous player
(2) \(R\) satisfies the Backwards Heat Equation
\displaystyle
+ \int_{0}^T \overset{*}{\Delta} R(t, |B_t|) \mathrm{d}t
Different from classic FTC!
\displaystyle
\mathrm{ContRegret}(\;\;\;\;\;\;, T)
\displaystyle
\partial_g R
\;\;\; \vphantom{\overset{*}{\Delta}} R = \;\;\;R + \;\;\;\;\;\;\; R
\displaystyle
\partial_t
\displaystyle
\tfrac{1}{2}\partial_{gg}
\displaystyle
\overset{*}{\Delta}
Backwards Heat Equation
Stochastic Integrals and Itô's Formula
\displaystyle
R(T, |B_T|) - R(0, 0) =
Goal:
Find a "potential function" \(R\) such that
(1) \(\partial_g R\) is a valid continuous player
(2) \(R\) satisfies the Backwards Heat Equation
\displaystyle
\mathrm{ContRegret}(\;\;\;\;\;\;, T)
\displaystyle
\partial_g R
How to find a good \(R\)?
?
Suffices to find a player \(p\) satisfying the BHE
p(t,g) = \mathbb{P}(\mathcal{N}(0, T - t) > g)
\(\approx\) Cover's solution!
Also a solution to an ODE
\displaystyle R(T, |B_T|) - R(0,0) \leq \sqrt{\frac{T}{2\pi}}
R(t,g) \approx \int p(t,g)
Then setting
preserves BHE and
p = \partial_g R
Stochastic Integrals and Itô's Formula
How to work with stochastic integrals?
\displaystyle
R(T, |B_T|) - R(0, |B_0|) = \int_{0}^T \partial_g R(t, |B_t|) \mathrm{d}|B_t|
Itô's Formula:
\(\overset{*}{\Delta} R(t, g) = 0\) everywhere
ContRegret is given by \(R(T, |B_T|)\)
\displaystyle
\implies
Goal:
Find a "potential function" \(R\) such that
(1) \(\partial_g R\) is a valid continuous player
(2) \(R\) satisfies the Backwards Heat Equation
\displaystyle
+ \int_{0}^T \overset{*}{\Delta} R(t, |B_t|) \mathrm{d}t
Different from classic FTC!
\displaystyle
\mathrm{ContRegret}(\;\;\;\;\;\;, T)
\displaystyle
\partial_g R
\;\;\; \vphantom{\overset{*}{\Delta}} R = \;\;\;R + \;\;\;\;\;\;\; R
\displaystyle
\partial_t
\displaystyle
\tfrac{1}{2}\partial_{gg}
\displaystyle
\overset{*}{\Delta}
Backwards Heat Equation
[C-BL 06]
A Solution Inspired by Cover's Algorithm
From Cover's algorithm, we have
p^*(t,g) \approx
\mathbb{P}\Big(\mathcal{N}(0,T - t) > g\Big)
\displaystyle
\Big\} = \text{player}~Q(t, g)
We can find \(R(t,g)\) such that
\displaystyle
\Bigg\{
\(\overset{*}{\Delta} R = 0\)
\(\partial_g R = Q\)
Potential \(R\) satisfying BHE?
Player \(Q\) satisfies the BHE!
By Itô's Formula:
\displaystyle
\mathrm{ContRegret}(Q, T) = R(T, |B_T|) - R(0,0)
\displaystyle
\leq \sqrt{\frac{T}{2\pi}}
(BHE)
Discretization
Discrete Itô's Formula
\displaystyle
R(T, |B_T|) - R(0, 0) = \displaystyle
\mathrm{ContRegret}(\partial_g R, T) + \int_{0}^T \overset{*}{\Delta} R(t, |B_t|) \mathrm{d}t
\displaystyle
R(T, g_T) - R(0, 0) =\;\;\; \mathrm{Regret}(R_g, T)
How to analyze a discrete algorithm coming from stochastic calculus?
Discrete Itô's Formula!
\displaystyle
+ \sum_{t = 1}^T \Big(R_t(t, g_t) + \frac{1}{2} R_{gg}(t, g_t)\Big)
Discrete Derivatives
Surprisingly, we can analyze Cover's algorithm with discrete Itô's formula
Itô's Formula
Discrete Itô's Formula
Discrete Algorithms
p^*(t,g)
= V_g^*[t,g]
V_t^*[t, g] + \frac{1}{2} V_{gg}^*[t,g] = 0
\(V^*\) satisfies the "discrete" Backwards Heat Equation!
Not Efficient
Efficient
Discrete Itô \(\implies\)
Regret of \(p^* \leq V^*[0,0]\)
BHE = Optimal?
Hopefully, \(R\) satisfies the discrete BHE
Discretized player:
R_g(t,g)
We show the total is \(\leq 1\)
Cover's strategy
Bounding the Discretization Error
In the work of Harvey et al., they had
In this fixed-time solution, we are not as lucky.
Negative discretization error!
g
t
T = 1000
We show the total discretization error is always \(\leq 1\)
Our Results
An Efficient and Optimal Algorithm in Fixed-Time with Two Experts
Technique:
Solve an analogous continuous-time problem, and discretize it
[HLPR '20]
How to exploit the knowledge of \(T\)?
Discretization error needs to be analyzed carefully.
BHE seems to play a role in other problems in OL as well!
Solution based on Cover's alg
Or inverting time in an ODE!
We show \(\leq 1\)
\(V^*\) and \(p^*\) satisfy the discrete BHE!
Insight:
Cover's algorithm has connections to stochastic calculus!
Questions?
Known Results
Multiplicative Weights Update method:
\displaystyle
\mathrm{Regret}(T) \leq \sqrt{\frac{T}{2} \ln n}
Optimal for \(n,T \to \infty\) !
If \(n\) is fixed, we can do better
\(n = 2\)
\(n = 3\)
\(n = 4\)
\sqrt{\frac{T}{2\pi}} + O(1)
\sqrt{\frac{8T}{9\pi}} + O(\ln T)
\sim \sqrt{\frac{T \pi}{8}}
Player knows \(T\) !
Minmax regret in some cases:
What if \(T\) is not known?
\displaystyle
\frac{\gamma}{2} \sqrt{T}
Minmax regret
\(n = 2\)
[Harvey, Liaw, Perkins, Randhawa FOCS 2020]
They give an efficient algorithm!
\displaystyle
\gamma \approx 1.307
A Dynamic Programming View
Optimal regret (\(V^* = V_{p^*}\))
\displaystyle
V^*[t,g] = \frac{1}{2}(V^*[t+1, g-1] + V^*[t+1, g + 1])
\displaystyle
V^*[t,0] = \frac{1}{2} + V^*[t+1, 1]
For \(g > 0\)
For \(g = 0\)
g
t
4
3
2
1
0
0
0
0
0
0
0
0
\frac{1}{2}
0
0
0
0
0
0
0
0
0
0
0
0
0
1
2
3
Regret and Player in terms of the Gap
Path-independent player:
If
round \(t\) and gap \(g_{t-1}\) on round \(t-1\)
p(t, g_{t-1})
1 - p(t, g_{t-1})
on the Lagging expert
on the Leading expert
Choice doesn't depend on the specific past costs
p(t, 0) = 1/2
for all \(t\), then
\displaystyle
\mathrm{Regret(T)}
= \sum_{t = 1}^{T}
p(t, g_{t-1})(g_t - g_{t-1})
gap on round \(t\)
A discrete analogue of a Riemann-Stieltjes integral
A formula for the regret
A Dynamic Programming View
\displaystyle
V_p[t, g] =
Maximum regret-to-be-suffered on rounds \(t+1, \dotsc, T\) when gap on round \(t\) is \(g\)
Path-independent player \(\implies\) \(V_p[t,g]\) depends only on \(\ell_{t+1}, \dotsc, \ell_T\) and \(g_t, \dotsc, g_{T}\)
\displaystyle
V_p[t, 0] = \max\{p(t+1,0), 1 - p(t+1,0)\} + V_p[t+1, 1]
Regret suffered on round \(t+1\)
Regret suffered on round \(t + 1\)
\displaystyle
V_p[t, g] = \max \Bigg\{
\displaystyle
V_p[t+1, g+1] + p(t + 1,g)
\displaystyle
V_p[t+1, g-1] - p(t + 1,g)
A Dynamic Programming View
\displaystyle
V_p[t, g] =
Maximum regret-to-be-suffered on rounds \(t+1, \dotsc, T\) if gap at round \(t\) is \(g\)
We can compute \(V_p\) backwards in time!
Path-independent player \(\implies\)
\(V_p[t,g]\) depends only on \(\ell_{t+1}, \dotsc, \ell_T\) and \(g_t, \dotsc, g_{T}\)
We then choose \(p^*\) that minimizes \(V^*[0,0] = V_{p^*}[0,0]\)
\displaystyle
V_p[0, 0] =
Maximum regret of \(p\)
A Dynamic Programming View
For \(g > 0\)
\displaystyle
p^*(t,g) = \frac{1}{2}(V_{p^*}[t, g-1] - V_{p^*}[t, g + 1])
Optimal player
\displaystyle
p^*(t,0) = \frac{1}{2}
Optimal regret (\(V^* = V_{p^*}\))
\displaystyle
V^*[t,g] = \frac{1}{2}(V^*[t+1, g-1] + V^*[t+1, g + 1])
For \(g = 0\)
\displaystyle
V^*[t,0] = \frac{1}{2} + V^*[t+1, 1]
For \(g > 0\)
For \(g = 0\)
Discrete Derivatives
p^*(t,g) = \frac{1}{2} \big( V^*[t, g-1] - V^*[t, g+1]\big)
\approx \partial_g V^*[t,g]
\eqqcolon V_g^*[t,g]
V^*[t, g] - V^*[t-1, g]
= \frac{1}{2}( V^*[t, g-1] - V^*[t, g]) - \frac{1}{2}(V^*[t, g] - V^*[t,g+1])
\coloneqq V_t^*[t,g]
\coloneqq \frac{1}{2}V_{gg}^*[t,g]
\partial_t V^*[t,g] \approx
\approx \frac{1}{2}\partial_g V^*[t,g-1]
\approx \frac{1}{2} \partial_g V^*[t,g]
\approx \frac{1}{2} \partial_{gg} V^*[t,g]
Bounding the Discretization Error
Main idea
\(R\) satisfies the continuous BHE
\implies
R_t(t,g) + \frac{1}{2} R_{gg}(t,g) \approx
Approximation error of the derivatives
\implies
\leq 0.74
\displaystyle
\mathrm{Regret(T)} \leq \frac{1}{2} + \sqrt{\frac{T}{2\pi}} + \sum_{t = 1}^T O\Bigg( \frac{1}{(T - t)^{3/2}}\Bigg)
Lemma
\partial_{gg}R(t,g) - R_{gg}(t,g)
\partial_t R(t,g) - R_t(t,g)
\in O\Big( \frac{1}{(T - t)^{3/2}}\Big)
\Bigg\}
Known and New Results
Multiplicative Weights Update method:
\displaystyle
\mathrm{Regret}(T) \leq \sqrt{\frac{T}{2} \ln n}
Optimal for \(n,T \to \infty\) !
If \(n\) is fixed, we can do better
Worst-case regret for 2 experts
Player knows \(T\) (fixed-time)
Player doesn't know \(T\) (anytime)
Question:
Is there an efficient algorithm for the fixed-time case?
Ideally an algorithm that works for general costs!
\(O(T)\) time per round
Dynamic Programming
\(\{0,1\}\) costs
\(O(1)\) time per round
Stochastic Calculus
\([0,1]\) costs
[Harvey, Liaw, Perkins, Randhawa FOCS 2020]
\displaystyle
\approx
0.65 \sqrt{T}
\displaystyle \sqrt{\frac{T}{2\pi}} + O(1)
[Cover '67]
ALT 2022 - Two Experts
By Victor Sanches Portella
