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.

References

Francesco Bacchiocchi, Francesco Emanuele Stradi, Matteo Castiglioni, Alberto Marchesi, and Nicola Gatti, Markov Persuasion Processes: Learning to Persuade from Scratch, arXiv:2402.03077v2, 2024. Inspected locators: Sections 2--3, Theorems 3--4 in Section 5.2, and Theorem 5 in Section 5.3. https://arxiv.org/html/2402.03077v2.