GPT 6 Astra Pro attempt on Econ-1

With $r\in\mathbb N$, set $v_{i,r}=v_i^{1/r}$ and $x_r=x^{1/r}$. Multiplying (eq:bellmandisc) by $r$ yields \begin{equation}\label{wsref-0-65-71-3a-62-65-6c-6c-6d-61-6e-72} r v_{i,r}(s)\ge g_i(s,b_i,x_{-i,r}(s)) +(r-1)Q_{b_i,x_{-i,r}(s)}v_{i,r}. \end{equation} Let $\pi^N$ prescribe $x_r$ when $r$ stages remain in a block of length $N$. Define \[ \Delta_r=\max_{i,s}|v_{i,r-1}(s)-v_{i,r}(s)|,\qquad E_N=\sum_{r=2}^N(r-1)\Delta_r. \] Induction on $r$, using (eq:bellmanr), gives for all starting states and every unilateral deviation during the block \begin{equation}\label{wsref-0-65-71-3a-62-6c-6f-63-6b-73} {\mathbb{E}}_{\tau_i,\pi^N_{-i},s}\sum_{t=1}^N g_i \le N v_{i,N}(s)+E_N, \end{equation} and, using equality under the prescribed mixed actions, \begin{equation}\label{wsref-0-65-71-3a-62-6c-6f-63-6b-62-61-73-65-6c-69-6e-65} \left|{\mathbb{E}}_{\pi^N,s}\sum_{t=1}^N g_i-Nv_{i,N}(s)\right| \le E_N. \end{equation} The induction works for arbitrary within-block history-dependent deviations: a best-response continuation can be optimized at each successor history, while $\Delta_r$ bounds the error from replacing $v_{i,r-1}$ by $v_{i,r}$.

A bounded one-variable semialgebraic function with a finite limit has a Puiseux expansion near zero. In particular, for some $\alpha>0$ and $C<\infty$, \[ \left\|\frac{{\mathrm d} v_i^\lambda}{{\mathrm d}\lambda}\right\|_\infty \le C\lambda^{\alpha-1} \] for sufficiently small $\lambda>0$. Therefore $\Delta_r=O(r^{-1-\alpha})$ and \begin{equation}\label{wsref-0-65-71-3a-65-72-72-6f-72} E_N=O\left(\sum_{r=2}^Nr^{-\alpha}\right)+O(1)=o(N). \end{equation} Under (eq:flat), the errors \[ d_N=\max_{i,s}|v_{i,N}(s)-c_i|+E_N/N \] converge to zero. Consequently every $N$-block, from every initial state, has equilibrium expected average within $d_N$ of $c_i$, and a deviator's expected average is at most $c_i+d_N$.

Repeat $\pi^N$ periodically, restarting only the countdown after each block; the physical state is not reset. Call this profile ${\sigma}^N$. Conditioning at the start of each block and summing the bounds gives, for all $n$, \begin{align} {\mathbb{E}}_{s,{\sigma}^N}\bar g_{i,n}&\ge c_i-d_N-2MN/n,\label{wsref-0-65-71-3a-75-6e-69-66-6f-72-6d-6c-6f-77-65-72}\\ {\mathbb{E}}_{s,\tau_i,{\sigma}^N_{-i}}\bar g_{i,n} &\le c_i+d_N+2MN/n.\label{wsref-0-65-71-3a-75-6e-69-66-6f-72-6d-75-70-70-65-72} \end{align} The same argument gives an upper bound on the prescribed payoff. Since $N$ is fixed while $n\to\infty$, this proves the long finite-horizon part of uniform equilibrium.

For completeness, the discounted criterion follows from the same block bounds. Let $\theta\downarrow0$ be an evaluation discount factor, distinct from the within-block parameters $1/r$. The normalized discount weights of successive complete blocks satisfy \[ \frac{\theta\sum_{t=kN+1}^{(k+1)N}(1-\theta)^{t-1}g_t} {\theta\sum_{t=kN+1}^{(k+1)N}(1-\theta)^{t-1}} =\frac1N\sum_{t=kN+1}^{(k+1)N}g_t+O(MN\theta), \] uniformly in the block index and realization; the geometric weights within one block have relative variation $O(N\theta)$. The block inequalities then imply that discounted prescribed payoffs are within $d_N+O(MN\theta)$ of $c_i$ and deviations yield at most $c_i+d_N+O(MN\theta)$. First take $N$ large, then $\theta$ small.

For the pathwise liminf/limsup criteria, let $Y_{i,k}$ be the total reward in block $k$. Under either prescribed play or one fixed arbitrary unilateral deviation, the bounded martingale differences \[ Y_{i,k}-{\mathbb{E}}[Y_{i,k}\mid{\mathcal H}_{kN}] \] have averages converging almost surely to zero. Under prescribed play, all conditional block expectations are within $Nd_N$ of $Nc_i$; under a unilateral deviation, they are at most $N(c_i+d_N)$. Thus almost surely under ${\sigma}^N$, \[ c_i-d_N\le\liminf_n\bar g_{i,n} \le\limsup_n\bar g_{i,n}\le c_i+d_N, \] and under any $\tau_i$, $\limsup_n\bar g_{i,n}\le c_i+d_N$ almost surely. Bounded convergence inequalities then show that ${\sigma}^N$ is a $2d_N$-equilibrium for both ${\mathbb{E}}\liminf$ and ${\mathbb{E}}\limsup$, with payoffs within $d_N$ of $c$. For formulations using $\liminf_n{\mathbb{E}}\bar g_{i,n}$ or $\limsup_n{\mathbb{E}}\bar g_{i,n}$, the same conditional-block estimates apply directly. For the prescribed periodic finite-memory profile, the ordinary limit $\lim_n{\mathbb{E}}\bar g_{i,n}$ exists by finite-state Markov-chain ergodic theory. An arbitrary deviating strategy need not have a well-defined ordinary limit of expected average payoffs; no unqualified assertion is made here for a criterion that is undefined under such deviations. This completes the claimed approximate-payoff conclusions. ∎

Step 2: a compact full-payoff-realization set. Under ${\sigma}$, the bounded martingale $V_i^{\sigma}(H_t)={\mathbb{E}}_{\sigma}[f_i^-\mid{\mathcal H}_t]$ converges almost surely to $f_i^-$. Accordingly, if \[ G=\{\omega: V_i^{\sigma}(H_t(\omega))\to f_i^-(\omega) \text{ for every }i\}, \] then ${\mathbb{P}}_{\sigma}(G)=1$. The infinite-play space is compact metrizable; regularity of its probability measures gives a compact $K\subset G$ with ${\mathbb{P}}_{\sigma}(K)>1-\delta$. Let $T$ be the first finite public history whose cylinder has empty intersection with $K$. Closedness of $K$ implies \begin{equation}\label{wsref-0-65-71-3a-65-78-69-74} \{T=\infty\}=K. \end{equation}

Step 3: identify an offending player at a finite exit prefix. Apply the closed-target finite-prefix blame theorem [alon2024, Section 4] to the prescribed law and target $K$. The standard reduction of stochastic transitions to an additional non-strategic Nature agent is performed by splitting each stage into (i) simultaneous player actions and (ii) Nature's publicly observed state draw according to $q$. Nature has a prescribed behavioral strategy and does not deviate. There are $m+1$ agents in the identification problem; map any attribution to Nature to an arbitrary genuine player. We obtain a finite-exit-prefix-measurable blame assignment $j(H_T)\in I$ such that, for each genuine sole deviator $i$, \begin{equation}\label{wsref-0-65-71-3a-62-6c-61-6d-65} {\mathbb{P}}_{\tau_i,{\sigma}_{-i}}\{T<\infty,\ j(H_T)\ne i\} \le 2\sqrt{m\delta}. \end{equation} At $T$, modify the continuation: the players other than $j(H_T)$ switch permanently to a $\rho$-minmax punishment of that player from the current state. Call the resulting profile $\widehat{\sigma}$. The event $T<\infty$ has prescribed probability at most $\delta$; therefore boundedness of $f_i^-$ gives (eq:convpay).

Step 4: bound an arbitrary infinite deviation. Fix $i$ and an unrestricted $\tau_i$. Before $T$, the opponents use ${\sigma}_{-i}$; hence $W_i(H_{t\wedge T})$ is a bounded supermartingale, converging almost surely to some $Z_i$ with ${\mathbb{E}} Z_i\le W_i(s_0)$. On $\{T=\infty\}$, the path lies in $K$, so \begin{equation}\label{wsref-0-65-71-3a-6f-6e-4b} f_i^-=\lim_tV_i^{\sigma}(H_t) \le\lim_tW_i(H_t)=Z_i. \end{equation} On $\{T<\infty,j(H_T)=i\}$, the punishment yields conditional expected payoff at most \begin{equation}\label{wsref-0-65-71-3a-63-6f-72-72-65-63-74-62-6c-61-6d-65} v_i(s_{H_T})+\rho \le V_i^{\sigma}(H_T)+\alpha+\rho \le W_i(H_T)+\alpha+\rho. \end{equation} On incorrect blame use the payoff bound $1$ and (eq:blame). Because the law of the stopped history is unaffected by the punishment that follows the exit, \[ u_i^-(\tau_i,\widehat{\sigma}_{-i}) \le{\mathbb{E}} Z_i+\alpha+\rho+2\sqrt{m\delta} \le u_i^-({\sigma})+\beta+\alpha+\rho+2\sqrt{m\delta}. \] Together with (eq:convpay), this proves (eq:conveps). ∎

A deviation ending by date $N$ cannot improve player $i$'s expected tail payoff: conditional on the date-$N$ history, the remaining payoff is precisely $V_i^\pi(h_N)$. Thus $R_N({\sigma}^{(N)})=0$.

To propagate acceptability backward, use the dynamic-programming identity for the state-dependent minmax level with shift-invariant payoff: \begin{equation}\label{wsref-0-65-71-3a-6d-69-6e-6d-61-78-42-65-6c-6c-6d-61-6e} v_i(s)=\min_{x_{-i}\in\prod_{j\ne i}\Delta(A_j(s))} \max_{a_i\in A_i(s)} \sum_{a_{-i},s'}x_{-i}(a_{-i})q(s'\mid s,a_i,a_{-i})v_i(s'). \end{equation} For clarity, (eq:minmaxBellman) uses product mixed actions of the opponents. It follows by decomposing an arbitrary opponent strategy into its first-stage product mixed action and continuation profiles and using the shift-invariance of $f_i^-$. The upper bound concatenates near-minmax opponent continuation profiles after each observed action-state history; the lower bound responds optimally at the first stage and then nearly best-responds to the opponents' continuations. Only finitely many first-stage successors arise, so the approximation errors can be made arbitrarily small.

If all next-history continuation values $y_i(h')$ satisfy $y_i(h')\ge v_i(s_{h'})-\alpha$, the current one-shot Nash equilibrium with opponent product mixed action $x_{-i}$ gives \[ \begin{split} y_i(h) &\ge\max_{a_i}{\mathbb{E}}[y_i(H_{t+1})\mid h,a_i,x_{-i}]\\ &\ge\max_{a_i}{\mathbb{E}}[v_i(S_{t+1})\mid h,a_i,x_{-i}]-\alpha\\ &\ge v_i(s_h)-\alpha, \end{split} \] by (eq:minmaxBellman). This inducts backward from (eq:acceptablepi). After date $N$, acceptability again follows from (eq:acceptablepi). ∎