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md/dev/EcGGFkNTxdJ/EcGGFkNTxdJ.md

@@ -621,7 +621,7 @@ where the last inequality follows from Equation (6). This proves that Algorithm
 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.
 
 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 $i _ { 1 : n } ^ { k }$ $\breve { \left( i _ { 1 : n } ^ { k } \right) } _ { k \in \mathbb { N } }$   
-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 πkj 
 j=0 be a subsequence converging to $\bar { \pi }$ (which is a random subsequence as well). By continuity of $J$ in $\pi$ (Corollary 1), we have
+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 πkj 
 j=0 be a subsequence converging to $\bar { \pi }$ (which is a random subsequence as well). By continuity of $J$ in $\pi$ (Corollary 1), we have
 
 $$
 J ( \bar { \pmb { \pi } } ) = J ( \operatorname* { l i m } _ { j  \infty } \pmb { \pi } _ { k _ { j } } ) = \operatorname* { l i m } _ { j  \infty } J ( \pmb { \pi } _ { k _ { j } } ) = \bar { J } .

md/dev/taQ64d2KBX/taQ64d2KBX.md

@@ -159,7 +159,7 @@ $$
 118 Symplectic methods: The flow map of a Hamiltonian system is symplectic, meaning that its Jacobian   
 119 $\begin{array} { r } { \dot { \Upsilon _ { \varphi } } : = \frac { \partial } { \partial y } \varphi _ { h , f } ( y ) } \end{array}$ satisfies $\Upsilon _ { \varphi } ^ { T } J \Upsilon _ { \varphi } = J$ , where $J$ is the same matrix as in $\textcircled { 2 }$ . As explained in $\mathbb { B } ,$ Ch.   
 120 VI.2], this is equivalent to the preservation of a projected area in the phase space of $[ q , p ] ^ { T }$ . Similarly,   
-121 a numerical integrator is symplectic if its Jacobian ⌥ := @@yn $\begin{array} { r } { \Upsilon _ { \Phi } : = \frac { \partial } { \partial y _ { n } } \Phi _ { h , f } ( y _ { n } ) } \end{array}$ satisfies $\Upsilon _ { \Phi } ^ { T } J \Upsilon _ { \Phi } = J$ . It is   
+121 a numerical integrator is symplectic if its Jacobian ⌥ := @@yn $\begin{array} { r } { \Upsilon _ { \Phi } : = \frac { \partial } { \partial y _ { n } } \Phi _ { h , f } ( y _ { n } ) } \end{array}$ satisfies $\Upsilon _ { \Phi } ^ { T } J \Upsilon _ { \Phi } = J$ . It is   
 122 possible to prove $\mathbb { B } ,$ Ch. VI.4] that a Runge–Kutta method is symplectic if and only if the coeffients   
 123 satisfy
 
@@ -420,9 +420,9 @@ Figure 4: Average of $\overline { { \rho } }$ over 10 trajectories. Shaded area
 Methods and test problems: We train HNNs using different integrators and methods in the inverse problem $\textcircled{6}$ . We use MIRK4 together with the MII method and compare to the implicit midpoint method, RK4 and MIRK4 applied as one-step methods, as well as ISO followed by Störmer–Verlet and RK4 integrated over multiple time-steps. The latter strategy, illustrated in Figure $\bigtriangledown ,$ was suggested in [10], where Störmer–Verlet is used. Separable networks $H _ { \theta } ( q , p ) = H _ { 1 , \theta } ( q ) + H _ { 2 , \theta } ( p )$ are trained on data from the Fermi–Pasta–Ulam–Tsingou (FPUT) problem and the Hénon–Heiles system. For the double pendulum, which is non-separable, a fully connected Flow roll-out H´enon-Hnetwork is used for all methods except Störmer– Flow roll-out H´enon-Heiles h = 0.1, FVerlet, which requires separability in order to be explicit. The Hamiltonians are described in Appendix 0.2 0.0A and all systems have solutions $y ( t ) \not \in \mathbb { R } ^ { 4 }$ .
 
 0.2 0.0 0.0After using the specified integrators in training, a   
-294 0.0  0.2proximated solutions are computed for each learned   
-295 vector field $f _ { \theta }$ 0.2 0.4using the Scikit-learn implementation   
-296 0.2   0.6of DOP853 [35], which is also used to generate   
+294 0.0  0.2proximated solutions are computed for each learned   
+295 vector field $f _ { \theta }$ 0.2 0.4using the Scikit-learn implementation   
+296 0.2   0.6of DOP853 [35], which is also used to generate   
 297 0.0 2.5 5.0 7.5 training data. The error is averaged over $M = { \mathfrak { M } } =$   
 298 0.0 2.5 5.0 7.5 10.0 12.5 15.0 17points and we find what we call the flow error by
 

md/train/B1IDRdeCW/B1IDRdeCW.md

@@ -197,7 +197,7 @@ $$
 
 First, we note $E ( \eta ) = \sqrt { n } E ( | \rho | )$ . Then $\begin{array} { r } { E ( | \rho | ) = \int _ { 0 } ^ { 1 } d \rho \rho \frac { 2 } { \sqrt { \pi } } \frac { \Gamma ( n / 2 ) } { \Gamma ( ( n - 1 ) / 2 ) } ( 1 - \rho ^ { 2 } ) ^ { \frac { n - 3 } { 2 } } = } \end{array}$ $\frac { 2 } { \sqrt { \pi } } \ast \frac { 1 } { n - 1 } \frac { \Gamma ( n / 2 ) } { \Gamma ( ( n - 1 ) / 2 ) }$ (substitute $u \ : = \ : \rho ^ { 2 }$ and use $\Gamma ( x + 1 ) = x \Gamma ( x )$ ). Lemma
 
-two gives the $n  \infty$ limit. $\begin{array} { r l r } { \frac { 2 } { \sqrt { \pi } } \frac { \sqrt { n } } { n - 1 } \frac { \Gamma ( n / 2 ) } { \Gamma ( ( n - 1 ) / 2 ) } \ \approx \ \frac { 2 } { \sqrt { \pi } } \frac { \sqrt { n } } { n - 1 } \sqrt { \frac { n } { 2 } } \left[ 1 + \frac { 0 . 5 * 0 . 5 } { 2 ( n / 2 ) } \right] \ = } \end{array}$ q 2π 1 + 54 ∗ 1n 
 + O(1/n2)
+two gives the $n  \infty$ limit. $\begin{array} { r l r } { \frac { 2 } { \sqrt { \pi } } \frac { \sqrt { n } } { n - 1 } \frac { \Gamma ( n / 2 ) } { \Gamma ( ( n - 1 ) / 2 ) } \ \approx \ \frac { 2 } { \sqrt { \pi } } \frac { \sqrt { n } } { n - 1 } \sqrt { \frac { n } { 2 } } \left[ 1 + \frac { 0 . 5 * 0 . 5 } { 2 ( n / 2 ) } \right] \ = } \end{array}$ q 2π 1 + 54 ∗ 1n 
 + O(1/n2)
 
 $$
 { V a r } ( \eta ) = \frac { 1 } { n } \left( 1 - \frac { 1 } { \pi } \right) + { \cal O } ( 1 / n ^ { 2 } )

md/train/Re_VXFOyyO/Re_VXFOyyO.md

@@ -221,7 +221,7 @@ $$
 
 for some constants $C _ { 1 } , . . , C _ { 4 } > 0 .$ . In case $j = 0$ , we default $\textstyle \sum _ { j ^ { \prime } = 1 } ^ { 0 } \cdot = 0$
 
-The expression of constants $C _ { i }$ ’s are complicated, we provide their detailed formula in the appendix. If we set H = O log(1/) 
 and $\begin{array} { r } { \delta \le \frac { 1 } { 2 H \ell _ { \psi } } } \end{array}$ , then $C _ { i }$ only depends polynomially on the Lipschitz constants, $\log ( \epsilon ^ { - 1 } )$ , and $( 1 - \gamma ) ^ { - 1 }$ . Combining Lemma 5.5, 5.8, and 5.4 gives Theorem 5.9.
+The expression of constants $C _ { i }$ ’s are complicated, we provide their detailed formula in the appendix. If we set H = O log(1/) 
 and $\begin{array} { r } { \delta \le \frac { 1 } { 2 H \ell _ { \psi } } } \end{array}$ , then $C _ { i }$ only depends polynomially on the Lipschitz constants, $\log ( \epsilon ^ { - 1 } )$ , and $( 1 - \gamma ) ^ { - 1 }$ . Combining Lemma 5.5, 5.8, and 5.4 gives Theorem 5.9.
 
 Theorem 5.9. For Algorithm $^ { l }$ , we choose $\begin{array} { r } { H = \frac { 2 \log ( 1 / \epsilon ) } { 1 - \gamma } } \end{array}$ , $\begin{array} { r } { \delta = \frac { 1 } { 2 H \ell _ { \psi } } } \end{array}$ , $B = m = \epsilon ^ { - 1 }$ , $N = \epsilon ^ { - 2 }$ η = 11+(C3+C4)/L2θ · 12Lθ . After running the algorithm for T = −1 epochs and output θout from $\{ \theta _ { j } ^ { i } \} _ { j = 0 , \cdots , m - 1 } ^ { i = 1 , \cdots , T }$ uniformly at random, we have $\mathbb { E } [ \| \mathcal { G } _ { \eta } ( \theta _ { o u t } ) \| ] \le \mathcal { O } ( \epsilon )$ . The total number of samples is $\dot { T } \times ( ( m - 1 ) B + N ) \times H = \tilde { \mathcal { O } } ( \epsilon ^ { - 3 } )$ . By Lemma 5.4, we also have $\mathbb { E } [ \| \nabla _ { \theta } F ( \lambda ( \theta _ { o u t } ) ) \| ] \le { \mathcal { O } } ( \epsilon )$ .
 

md/train/jrA5GAccy_/jrA5GAccy_.md

@@ -105,7 +105,7 @@ Proof Sketch. The standard analysis in learning theory on ERM or regularized/con
 
 The above result holds for both square and cross-entropy loss. For ease of exposition, we use the standard setting of finite hypothesis class and extend all the results to infinite hypothesis classes in the supplement (summary of insights from the extension are in Section 3.3.2). Next, we state a standard result on ERM’s sample complexity. Define a $\Phi ^ { + }$ such that $\begin{array} { r } { \Phi ^ { + } \in \arg \operatorname* { m i n } _ { \Phi \in \mathcal { H } _ { \Phi } } R ( \Phi ) } \end{array}$
 
-Proposition 3. (Shalev-Shwartz & Ben-David, 2014) For every $\nu > 0$ and $\delta \in ( 0 , 1 )$ , if ${ \mathcal { H } } _ { \Phi }$ is a finite hypothesis class, Assumption 3 holds, and if the number of samples $| D |$ is greater than $\begin{array} { r } { \frac { 8 L ^ { 2 } } { \nu ^ { 2 } } \log \left( \frac { 2 | \mathcal { H } _ { \Phi } | } { \delta } \right) } \end{array}$  2|HΦ|δ 
, then with a probability at least 1 − δ, every solution Φ† to ERM is an ν approximation of expected risk minimization, i.e., $R ( \Phi ^ { + } ) \leq R ( \Phi ^ { \dagger } ) \leq R ( \Phi ^ { + } ) + \nu .$ .
+Proposition 3. (Shalev-Shwartz & Ben-David, 2014) For every $\nu > 0$ and $\delta \in ( 0 , 1 )$ , if ${ \mathcal { H } } _ { \Phi }$ is a finite hypothesis class, Assumption 3 holds, and if the number of samples $| D |$ is greater than $\begin{array} { r } { \frac { 8 L ^ { 2 } } { \nu ^ { 2 } } \log \left( \frac { 2 | \mathcal { H } _ { \Phi } | } { \delta } \right) } \end{array}$  2|HΦ|δ 
, then with a probability at least 1 − δ, every solution Φ† to ERM is an ν approximation of expected risk minimization, i.e., $R ( \Phi ^ { + } ) \leq R ( \Phi ^ { \dagger } ) \leq R ( \Phi ^ { + } ) + \nu .$ .
 
 Proposition 2 vs. 3 Since $\kappa \leq \epsilon$ , the sample complexity of EIRM grows at least as $\mathcal { O } ( \operatorname* { m a x } \{ \frac { 1 } { \epsilon ^ { 2 } } , \frac { 1 } { \nu ^ { 2 } } \} )$ . Let us look at the two terms inside max- i) $\scriptstyle { \frac { 1 } { \nu ^ { 2 } } }$ growth term is similar to ERM, it ensures $\nu$ approximate optimality in the overall risk $R$ , ii) $\frac { 1 } { \epsilon ^ { 2 } }$ growth ensures the IRM penalty $R ^ { ' }$ is less than . A direct comparison of sample complexities in Propositions 2 and 3 suggests that the sample complexity of EIRM is higher than ERM, which is not the complete picture. The two approaches may not converge to the same solutions and IRM may converge to a solution with better OOD behavior than one achieved by ERM. Therefore, a fair comparison is only possible when we also study the OOD properties of the solutions achieved by the two approaches, which is the subject of the next section.
 
@@ -159,7 +159,7 @@ Assumption 7. Inductive bias. ${ \mathcal { H } } _ { \Phi }$ is a finite set of
 
 Informally stated, the above assumption requires the OOD optimal predictor $\tilde { S } ^ { \top } \gamma$ to lie in the interior of the search space and not on the boundary. If Assumptions 5, 7 hold, then Assumption 3 holds. Hence, we can use the bounds in our next result. Define the minimum eigen $L$ and lue a $L ^ { \prime }$ on ss a $\ell$ d a $\frac { \partial \ell ( w \cdot \Phi ( \cdot ) , \cdot ) } { \partial w } | _ { w = 1 . 0 }$ , $\Sigma ^ { e }$ $\begin{array} { r } { \lambda _ { \sf m i n } = \operatorname* { m i n } _ { e \in \mathcal { E } _ { t r } } \lambda _ { \sf m i n } ( \Sigma _ { e } ) } \end{array}$ $\begin{array} { r } { \epsilon _ { \mathsf { t h } } = \frac { ( 2 4 - 1 6 \sqrt { 2 } ) } { 3 } \frac { \pi ^ { \mathsf { m i n } } } { | \mathcal { E } _ { t r } | } ( \omega \lambda _ { \mathsf { m i n } } ) ^ { 2 } } \end{array}$ and $\begin{array} { r } { \tau = \frac { 1 } { 2 \omega \lambda _ { \mathrm { m i n } } } \sqrt { \frac { 3 | \mathcal { E } _ { t r } | } { 2 \pi ^ { \mathrm { m i n } } } } } \end{array}$ . Next, we analyze how EIRM learns $\tilde { S } ^ { \top } \gamma$ .
 
-Proposition 5. Let $\ell$ be the square loss. For every $\epsilon \in ( 0 , \epsilon _ { \mathtt { t h } } )$ and $\delta \in ( 0 , 1 )$ , if Assumptions 5, $6$ (with $r = 1 \AA$ ), 7 hold and if the number of data points $| D |$ is greater tha n 16L042 l $\begin{array} { r } { \frac { 1 6 L ^ { \prime 4 } } { \epsilon ^ { 2 } } \log \left( \frac { 2 | \mathcal { H } _ { \Phi } | } { \delta } \right) } \end{array}$  2|HΦ| 
, then with a probability at least $1 - \delta$ , every solution $\hat { \Phi }$ to EIRM (equation $6$ ) satisfies $\hat { \Phi } = ( \tilde { S } ^ { \top } \gamma ) \alpha$ , where $\begin{array} { r } { \alpha \in [ \frac { 1 } { 1 + \tau \sqrt { \epsilon } } , \frac { 1 } { 1 - \tau \sqrt { \epsilon } } ] } \end{array}$ .
+Proposition 5. Let $\ell$ be the square loss. For every $\epsilon \in ( 0 , \epsilon _ { \mathtt { t h } } )$ and $\delta \in ( 0 , 1 )$ , if Assumptions 5, $6$ (with $r = 1 \AA$ ), 7 hold and if the number of data points $| D |$ is greater tha n 16L042 l $\begin{array} { r } { \frac { 1 6 L ^ { \prime 4 } } { \epsilon ^ { 2 } } \log \left( \frac { 2 | \mathcal { H } _ { \Phi } | } { \delta } \right) } \end{array}$  2|HΦ| 
, then with a probability at least $1 - \delta$ , every solution $\hat { \Phi }$ to EIRM (equation $6$ ) satisfies $\hat { \Phi } = ( \tilde { S } ^ { \top } \gamma ) \alpha$ , where $\begin{array} { r } { \alpha \in [ \frac { 1 } { 1 + \tau \sqrt { \epsilon } } , \frac { 1 } { 1 - \tau \sqrt { \epsilon } } ] } \end{array}$ .
 
 Proof Sketch. In learning theory it is common to analyze the concentration of empirical risks around the expected risks. In our case, we have a target ideal solution to equation 1 $( \tilde { S } ^ { \top } \gamma )$ and we want our empirical solutions to concentrate around that. A direct finite sample approximation of equation 4 is hard to analyze. Therefore, we introduce an intermediate problem in equation 5 and then develop a finite sample approximation of it in equation 6. We first show that solving equation 5 leads to solutions in the neighborhood of the target. To show this we use the linear general position assumption. Next, we connect equation 6 and equation 5 using our new estimator for $R ^ { ' }$ and Hoeffding’s inequality.
 
@@ -1001,7 +1001,7 @@ Define th = 24−16 2 $\begin{array} { r } { \epsilon _ { \mathsf { t h } } = \
 
 Next, we prove Proposition 5 from the main body of the manuscript.
 
-Proposition 12. Let $\ell$ be the square loss. For every $\epsilon \in ( 0 , \epsilon _ { \mathtt { t h } } )$ and $\delta \in ( 0 , 1 )$ , if Assumptions 5, 6 (with $r = 1 \AA$ ), 7 hold and if the number of data points $| D |$ is greater than $\begin{array} { r } { \frac { 1 6 L ^ { \prime 4 } } { \epsilon ^ { 2 } } \log \left( \frac { 2 | \mathcal { H } _ { \Phi } | } { \delta } \right) } \end{array}$  2|HΦ| 
, then with a probability at least $1 - \delta$ , every solution $\hat { \Phi }$ to EIRM (equation $6$ ) satisfies $\hat { \Phi } = ( \tilde { S } ^ { \mathsf { T } } \gamma ) \alpha$ , where $\begin{array} { r } { \alpha \in [ \frac { 1 } { 1 + \tau \sqrt { \epsilon } } , \frac { 1 } { 1 - \tau \sqrt { \epsilon } } ] } \end{array}$ .
+Proposition 12. Let $\ell$ be the square loss. For every $\epsilon \in ( 0 , \epsilon _ { \mathtt { t h } } )$ and $\delta \in ( 0 , 1 )$ , if Assumptions 5, 6 (with $r = 1 \AA$ ), 7 hold and if the number of data points $| D |$ is greater than $\begin{array} { r } { \frac { 1 6 L ^ { \prime 4 } } { \epsilon ^ { 2 } } \log \left( \frac { 2 | \mathcal { H } _ { \Phi } | } { \delta } \right) } \end{array}$  2|HΦ| 
, then with a probability at least $1 - \delta$ , every solution $\hat { \Phi }$ to EIRM (equation $6$ ) satisfies $\hat { \Phi } = ( \tilde { S } ^ { \mathsf { T } } \gamma ) \alpha$ , where $\begin{array} { r } { \alpha \in [ \frac { 1 } { 1 + \tau \sqrt { \epsilon } } , \frac { 1 } { 1 - \tau \sqrt { \epsilon } } ] } \end{array}$ .
 
 Proof. Define an event $A$ : $\{ \tilde { D } : \forall \Phi \in \mathcal { H } _ { \Phi } , | \hat { R } ^ { \prime } ( \Phi ) - R ^ { \prime } ( \Phi ) | \leq \frac { \epsilon } { 2 } \}$ . If event $A$ happens, then
 
@@ -1183,7 +1183,7 @@ Define the minimum eigenvalue over all the matrices $\bar { \Sigma } _ { e }$ as
 
 From the analysis in this section, we see that we have been able to construct a linear model identical to 5, where the role of $X ^ { e } , Z ^ { e } , S , \tilde { S } , \Sigma ^ { e } , \lambda _ { \sf m i n } , \epsilon _ { \sf t h }$ is taken by $\bar { X ^ { e } } , \bar { Z ^ { e } } , \bar { S } , \bar { \bar { S } } , \bar { \Sigma ^ { e } } , \bar { \lambda } _ { \sf m i n } , \bar { \epsilon } _ { \sf t h } .$ . Now we are ready to use the result already proven for linear model and state the next Proposition in terms of the parameters for the polynomial model.
 
-Proposition 13. Let \` be the square loss. For every $\epsilon \in ( 0 , \bar { \epsilon } ^ { \mathsf { t h } } )$ , $\delta \in ( 0 , 1 )$ , if Assumptions 8, 9, $I O$ (with $r = 1$ ), $1 l$ hold and if the number of data points $| D |$ is greater than $\begin{array} { r } { \frac { 1 6 L ^ { \prime 4 } } { \epsilon ^ { 2 } } \log \left( \frac { 2 | \mathcal { H } _ { \Phi } | } { \delta } \right) } \end{array}$  2|HΦ| 
, then with a probability at least $1 - \delta$ , every solution $\hat { \Phi }$ to EIRM (equation $6$ ) satisfies $\hat { \Phi } = ( \bar { \tilde { S } } ^ { \mathsf { T } } \gamma ) \alpha$ , where $\begin{array} { r } { \alpha \in [ \frac { 1 } { 1 + \tau \sqrt { \epsilon } } , \frac { 1 } { 1 - \tau \sqrt { \epsilon } } ] } \end{array}$ .
+Proposition 13. Let \` be the square loss. For every $\epsilon \in ( 0 , \bar { \epsilon } ^ { \mathsf { t h } } )$ , $\delta \in ( 0 , 1 )$ , if Assumptions 8, 9, $I O$ (with $r = 1$ ), $1 l$ hold and if the number of data points $| D |$ is greater than $\begin{array} { r } { \frac { 1 6 L ^ { \prime 4 } } { \epsilon ^ { 2 } } \log \left( \frac { 2 | \mathcal { H } _ { \Phi } | } { \delta } \right) } \end{array}$  2|HΦ| 
, then with a probability at least $1 - \delta$ , every solution $\hat { \Phi }$ to EIRM (equation $6$ ) satisfies $\hat { \Phi } = ( \bar { \tilde { S } } ^ { \mathsf { T } } \gamma ) \alpha$ , where $\begin{array} { r } { \alpha \in [ \frac { 1 } { 1 + \tau \sqrt { \epsilon } } , \frac { 1 } { 1 - \tau \sqrt { \epsilon } } ] } \end{array}$ .
 
 # 7.4.2 INFINITE HYPOTHESIS CLASSES