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parse/dev/EcGGFkNTxdJ/EcGGFkNTxdJ.md
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@@ -621,7 +621,7 @@ where the last inequality follows from Equation (6). This proves that Algorithm
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Theorem 3. Supposing in Algorithm 1 any permutation of agents has a fixed non-zero probability to begin the update, a sequence $\left( \pi _ { k } \right) _ { k = 0 } ^ { \infty }$ of joint policies generated by the algorithm, in a cooperative Markov game, has a non-empty set of limit points, each of which is a Nash equilibrium.
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Proof. Step 1 (convergence). Firstly, it is clear that the sequence $( J ( \pi _ { k } ) ) _ { k = 0 } ^ { \infty }$ converges as, by Theorem 2, it is non-decreasing and bounded above by $\frac { R _ { \mathrm { m a x } } } { 1 - \gamma }$ . Let us denote the limit by $\bar { J }$ . For every $k$ , y enote the tuple of a, and we note that ording to whose order the agents perform the sequential updates,is a random process. Furthermore, we know that the sequence $i _ { 1 : n } ^ { k }$ $\breve { \left( i _ { 1 : n } ^ { k } \right) } _ { k \in \mathbb { N } }$
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of policies $\left( \pi _ { k } \right)$ is bounded, so by Bolzano-Weierstrass Theorem, it has at least one convergent subsequence. Let $\bar { \pi }$ be any limit point of the sequence (note that the set of limit points is a random set), and |