Markov persuasion regret--violation tradeoffs: resolved gap
LLM status Open unverified
D82 — Asymmetric and Private Information - Mechanism Design Economics catalogue openaigpt 6 astra ultra 50% cataloguednot recordedOpenStatement
GT-051 | Learning and information design | H: the cited paper closes the advertised exponent gap.
Definitions and mathematical statement
An episodic Markov persuasion process has finite layer sets $X_0,\ldots,X_L$, finite outcomes $\Omega$ and actions $A$, unknown outcome kernels $\mu(\omega\mid x)$, unknown transition kernel $P(x'\mid x,\omega,a)$, and unknown mean rewards $r_S,r_R\in[0,1]$. Bounded reward vectors are drawn independently each episode with these means. The initial state and finite sets are known. In episode $t$ a learner chooses a recommendation rule $\phi_t(a\mid x,\omega)$; recommendations are followed during learning. The learner observes reached states, outcomes, and the two realized rewards for the chosen action, without seeing rewards for other actions. Write $q_t(x,\omega,a)$ for the resulting occupancy probability. For a reached recommendation, define \[ b_t(a,x)\in\operatorname*{arg\,max}_{b\in A} \sum_\omega\mu(\omega\mid x)\phi_t(a\mid x,\omega)r_R(x,\omega,b). \] For zero-probability recommendations its choice is immaterial. Let \[ U_S(\phi_t)=\sum_{x,\omega,a}q_t(x,\omega,a)r_S(x,\omega,a), \] \[ V_T=\sum_{t=1}^T\sum_{x,\omega,a}q_t(x,\omega,a) [r_R(x,\omega,b_t(a,x))-r_R(x,\omega,a)]. \] The sum over outcomes for each recommendation is nonnegative by best-response optimality. A persuasive rule has zero such violation at every reached recommendation. If $U_S^*$ is optimal value over persuasive rules under the true process, regret is $R_T=T U_S^*-\sum_tU_S(\phi_t)$. The original catalogue asks for matching bounds outside $\alpha\in[1/2,2/3]$. In the source's partial-feedback model, the advertised tradeoff is already established throughout \[ \alpha\in[1/2,1],\qquad R_T=\widetilde O(T^\alpha),\qquad V_T=\widetilde O(T^{1-\alpha/2}), \] with the corresponding simultaneous little-$o$ improvement ruled out by Theorem 5. Here $\widetilde O$ suppresses logarithmic factors and factors depending on the fixed finite process dimensions.Short English statement
The alleged lower-bound exponent gap is not open in the cited Markov persuasion model.Variants and scope boundaries
The source assumes recommendation obedience during learning and records violations relative to myopic best responses; it does not assume every exploratory policy is persuasive. Different feedback or strategically learning receivers define separate problems.Source relationship and status
Section 5.3 compares its full-interval result with a narrower guarantee in older work. The catalogue mistakenly assigns that older gap to the cited paper. This entry is preserved as a historical correction, not replaced by an invented new open question.References
[S1] Francesco Bacchiocchi, Francesco Emanuele Stradi, Matteo Castiglioni, Alberto Marchesi, and Nicola Gatti (2024). Markov Persuasion Processes: Learning to Persuade from Scratch. arXiv:2402.03077v2. Verified locator: Sections 2--3; Section 5.2, Theorems 3--4; Section 5.3, Theorem 5 and comparison paragraph. https://arxiv.org/html/2402.03077v2
Common conventions: Source mathematical conventions
Finite games. For a finite set $A$, $\Delta(A)$ is its probability simplex. Players choose independent mixed strategies in Nash equilibrium; correlation is allowed only when explicitly part of the solution concept. In a game with payoff functions $u_i$ and strategy profile $x$, an additive $\varepsilon$ Nash equilibrium satisfies \[ u_i(x)\geq u_i(x_i',x_{-i})-\varepsilon \quad\text{for every player $i$ and every admissible deviation $x_i'$}. \] For numerical approximation, payoffs are normalized as locally specified. An $\varepsilon$ payoff residual is distinct from distance at most $\varepsilon$ to the set of exact equilibria. Point convergence, convergence of empirical play, and convergence in distance to an equilibrium set are also distinct.
Computational representations. Unless stated otherwise, a complexity question takes rational data with binary encoding; polynomial time is measured in total bit length. A fixed-accuracy polynomial algorithm is not necessarily polynomial in $1/\varepsilon$, and neither is necessarily polynomial in $\log(1/\varepsilon)$. Succinct games and value, demand, or sampling oracles must declare their interfaces. Exact real solutions require an explicit representation; an unspecified list of arbitrary real numbers is not a finite computational output. A claimed algorithmic barrier is meaningful only for its specified model and reductions.
Dynamic games. Histories, information, observation, and allowable randomization are part of the model. A stationary strategy depends only on the current state; a positional strategy in a deterministic graph game chooses one action at each controlled vertex. Nash equilibrium at an initial state and equilibrium after every history are different requirements. An expectation of a pathwise $\liminf$ payoff, a $\liminf$ of expected averages, a limiting expected average, and a uniform finite-horizon equilibrium are different criteria. None is silently substituted for another.
Allocation and mechanisms. A complete allocation assigns every item exactly once unless otherwise stated. Zero-valued goods, negative values, capacity constraints, feasibility, lotteries, and copies of goods affect fairness definitions and are specified locally. Dominant-strategy incentive compatibility, Bayesian incentive compatibility, universal truthfulness, and truthfulness in expectation impose different inequalities. Whenever an objective ratio has zero denominator, its convention is stated or those instances are excluded.
Infinite-dimensional models. Expectations, integrals, stochastic equations, and optimization objectives require their local regularity, integrability, filtrations, and admissible domains. A backward--forward mean-field system is a boundary-value problem; it is not an initial-value semiflow without an additional construction. Long-time, large-population, and vanishing-noise limits require an order of limits and a convergence topology. If a minimum need not be attained, the objective is its infimum or supremum and the approximation requirement is stated separately.
Statement licence: CC-BY-4.0 · Source file
Contributions 1
GPT 6 Astra Ultra attempt on Econ-654
Markov persuasion regret--violation tradeoffs: resolved gap: a research attempt
Status and the source's full exponent interval
The alleged tradeoff gap is already resolved in the cited partial-feedback model. Bacchiocchi, Stradi, Castiglioni, Marchesi, and Gatti establish upper bounds throughout $\alpha\in[1/2,1]$, with the corresponding simultaneous little-$o$ improvement excluded by their lower bound. Theorems 3--4 and Theorem 5 are the relevant inspected statements. This attempt derives the exponent arithmetic and gives a self-contained one-state statistical experiment explaining it. It does not claim the full unknown-transition theorem as a new result or independently prove all of that theorem's state-visitation bounds.Suppressing fixed process dimensions and logarithmic factors, the source's exploration parameter $N$ gives the structure \[ R_T\lesssim N+\sqrt T,\qquad V_T\lesssim\frac{T}{\sqrt N}+\sqrt N+\sqrt T. \tag{1} \] For $N=\lceil T^\alpha\rceil$, $1/2\leq\alpha\leq1$, the first term dominates regret and $T^{1-\alpha/2}$ dominates violation, yielding \[ R_T=\widetilde O(T^\alpha),\qquad V_T=\widetilde O(T^{1-\alpha/2}). \tag{2} \] The equal-exponent point is $\alpha=2/3$. At $\alpha=1$ the regret bound is linear; (2) should not be described as simultaneously sublinear at that endpoint.
A restricted experiment retaining the information tension
Consider one decision per episode, a single state, no informative outcome, and two actions. Sender rewards are known and deterministic: $1$ for action 1 and $0$ for action 2. Receiver reward from action 1 is deterministically $1/2$; reward from action 2 is an independent Bernoulli variable of unknown mean $1/2+\theta$, where $\theta\in[-1/4,1/4]$. Only the chosen action's reward is observed. This is a restricted submodel of the canonical feedback experiment. Recommendations are obeyed during learning, even before their persuasiveness has been established.For $\theta<0$, the unique persuasive action is 1, so the benchmark sender value is one. If $N_2$ is the cumulative probability of recommending action 2, then \[ R_T=N_2,\qquad V_T=(-\theta)N_2. \tag{3} \] For $\theta>0$, only action 2 is persuasive; the benchmark sender value is zero, and \[ R_T=-(T-N_2),\qquad V_T=\theta(T-N_2). \tag{4} \] At $\theta=0$, either action is persuasive and the optimal sender value is one. Regret can be negative in (4) because an unpersuasive policy can outperform the feasible persuasive benchmark. Replacing regret by its absolute value changes the problem.
A sequential procedure and finite-sample bounds
Choose an integer exploration cap $1\leq N\leq T$ and confidence level $\delta\in(0,1)$. After $s$ observations of action 2, write $\widehat\mu_s$ for their mean and set \[ L=\log(2N/\delta),\qquad r_s=\sqrt{L/(2s)}. \] Initially recommend action 2. After every such observation, if $\widehat\mu_s+r_s<1/2$, stop exploration and recommend action 1 thereafter. Otherwise continue until the cap. At the cap, recommend action 2 forever if $\widehat\mu_N-r_N>1/2$, and action 1 otherwise. There is no assumption of an unobserved reward for the other action.Regard the observations of action 2 as an initial segment of a pre-generated independent sequence. Hoeffding's inequality and a union bound give, with probability at least $1-\delta$, \[ |\widehat\mu_s-(1/2+\theta)|\leq r_s \quad\hbox{for every }s=1,\ldots,N. \tag{5} \] This simultaneous event remains valid at a data-dependent stopping time.
Suppose first $\theta=-\Delta<0$. Event (5) prevents a false positive certificate at the cap. Thus $R_T\leq N$. Let $K\leq N$ be the number of exploratory actions. If $K\geq2$, the procedure did not stop at $K-1$, so (5) implies \[ \Delta\leq2r_{K-1},\qquad V_T=\Delta K\leq2K\sqrt{L/(2(K-1))} \leq2\sqrt{NL}. \tag{6} \] For $K=1$ the violation is at most $1/4$. When $\theta=0$, violation is zero and regret is again at most $N$.
For $\theta>0$, event (5) prevents a negative certificate, so exploration lasts exactly $N$ rounds and causes no violation. If action 1 is chosen after the cap, its failure to trigger a positive certificate and (5) imply $\theta\leq2r_N$. Consequently \[ R_T\leq0,\qquad V_T\leq2Tr_N=T\sqrt{2L/N}. \tag{7} \] Together these cases give a uniform high-probability bound \[ R_T\leq N,\qquad V_T\leq\max\{1/4,\,2\sqrt{NL},\,T\sqrt{2L/N}\}. \tag{8} \] Substituting $N=\lceil T^\alpha\rceil$ explains the exponents in the restricted model. The confidence guarantee is conditional only on the specified independent reward law, with the sampling adaptivity already covered by (5).
A two-instance lower bound
The same experiment shows why faster learning cannot uniformly remove the tradeoff. Compare $\theta=-\varepsilon$ and $\theta=+\varepsilon$, for $0<\varepsilon\leq1/4$. Let $P_-,P_+$ be the complete history laws, including the learner's private randomization. For an arbitrary adaptive algorithm, write $p_t$ for its conditional probability of recommending action 2 and let $N_2=\sum_t p_t$. This quantity matches cumulative occupancy, rather than a realized sample count when the rule randomizes. Conditional likelihood factors for chosen action 1 are identical in the two instances, and the chain rule for relative entropy gives \[ D(P_-\Vert P_+) =\mathbb E_-N_2\, \operatorname{kl}(1/2-\varepsilon,1/2+\varepsilon) \leq16\varepsilon^2\mathbb E_-N_2. \tag{9} \] The scalar bound follows by expanding the binary divergence as $2\varepsilon\log((1+2\varepsilon)/(1-2\varepsilon))$ and using $\log((1+z)/(1-z))\leq2z/(1-z)$ for $0\leq z\leq1/2$.Assume $\mathbb E_-R_T=\mathbb E_-N_2\leq r$, where $1\leq r\leq T/8$, and choose $\varepsilon=1/(16\sqrt r)$. Pinsker's inequality gives \[ \|P_--P_+\|_{\mathrm{TV}} \leq\sqrt{8\varepsilon^2r}<1/4. \] Markov's inequality gives $P_-(N_2\leq4r)\geq3/4$, and hence $P_+(N_2\leq4r)\geq1/2$. On this event, (4) yields \[ V_T\geq\varepsilon(T-4r)\geq\frac{T}{32\sqrt r}. \tag{10} \] In particular $\mathbb E_+V_T\geq T/(64\sqrt r)$. A guarantee of small expected regret in the negative instance therefore forces violation of order $T/\sqrt r$ in a statistically nearby positive instance. This is an expectation-regret, constant-probability-violation statement for the restricted experiment. It is not a substitution for the precise uniform high-probability quantifiers of the source's full Theorem 5.
Scope and outcome
The calculation preserves the important distinction between obeyed recommendations during learning and retrospective violations of myopic incentives. Individual outcome terms in the canonical violation formula can be negative; the sum for each recommendation is nonnegative by best-response optimality. In the one-outcome experiment that distinction collapses to the nonnegative gaps in (3)--(4).Unknown transitions, state reachability, and estimating outcome kernels require additional machinery beyond (8)--(10). That machinery and the full-interval resolution are credited to the inspected source. The attempt's unresolved label records incomplete independent verification of that full result, not an asserted open exponent gap. Strategically learning receivers or other feedback conventions would be different problems and are not substituted for the canonical target here.
Completion Estimate
50%This is a post hoc, heuristic estimate for the full catalogue target. The attempt independently proves the regret-violation information tension through matching one-state statistical constructions and correctly explains the source's full exponent interval. The historical general gap is resolved by the cited work, while this attempt still lacks the unknown-transition, state-visitation and full-quantifier arguments needed to reconstruct that result.