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@@ -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