Universality of empirical risk minimization
Andrea Montanari, Basil Saeed
Introduction
Fit the parameters via (regularized) empirical risk minimization (ERM):
As a motivating example, consider a -layer network with two hidden layers of width and k:
Consider a learning procedure in which the first and last layers are not learnt from data, and we learn by minimizing the logistic loss for binary labels :
(Here we use instead of to denote composition.) This example fits in the general framework above, with featurization map , function , and loss .
From the point of view of theoretical analysis, we can replace by in Eq. (2), and redefine the empirical risk in terms of the feature vectors :
A significant line of recent work studies the asymptotic properties of the ERM (5) under the proportional asymptotics with . A number of phenomena have been elucidated by these studies [BM12, TOH15, TAH18], including the design of optimal loss functions and regularizers [DM16, EK18, CM22, AKLZ20], the analysis of inferential procedures [SCC19, CMW20], and the double descent behavior of the generalization error [HMRT19, DKT19, MRSY19, GLK+20]. However, these works often assume Gaussian feature vectors or feature vectors with independent coordinates, and the generalization to dependent non-Gaussian features is an open challenge.
Needless to say, both the Gaussian assumption and the assumption of independent covariates are highly restrictive. Neither corresponds to an actual nonlinear featurization map .
On the other hand, recent work has unveiled a remarkable phenomenon in the context of random features models, i.e. for . Under simple distributions on the covariates (for instances with i.i.d. coordinates) and for certain weight matrices , the asymptotic behavior of the ERM problem (5) appears to be identical to the one of an equivalent Gaussian model. In the equivalent Gaussian model, the feature vectors are replaced by Gaussian features:
(We refer to the next section for formal definitions.)
We stress that —in the proportional asymptotics — the test error is typically bounded away from zero as , and so is possibly the train error. Further, train and test error typically concentrate around different values. Existing proof techniques (for Gaussian) allow to compute the limiting values of these quantities. Insight into the ERM behavior is obtained by studying their dependence on various problem parameters, such as the overparameterization ratio or the noise level. When we say that the the non-Gaussian and Gaussian models have the same asymptotic behavior, we mean that the limits of the test and train errors coincide. This allows transferring rigorous results proven in the Gaussian model to similar statements for more realistic featurization maps.
In each case we fit the data by minimizing the empirical risk:
The agreement between the Gaussian and non-Gaussian models is excellent.
We follow the random matrix theory literature [Tao12] and refer to this as a universality phenomenon. When universality holds, the ERM behavior is roughly independent of the features distribution, provided their covariances are matched.
Universality is a more delicate phenomenon than concentration of the empirical risk around its expectation. Indeed, as emphasized above, it holds in the high-dimensional regime in which test error and train error do not match. Establishing universality requires understanding the dependence of the empirical risk minimizer on the data , as opposed to just bounding its distance from a population value via concentration.
Universality results for ERM were proven in the past for feature vectors with independent entries [KM11, MN17, PH17, HS22]. Related results for randomized dimension reduction were obtained in [OT18]. The case of general vectors is significantly more challenging. To the best of our knowledge, the first and only proof of universality beyond independent entries was given in the recent paper of Hu and Lu [HL20].
The result of [HL20] is limited to strongly convex ERM problems. Their proof uses a Lindeberg swapping argument, whereby the rows of are replaced one-by-one by Gaussian rows with the same mean and covariance. This requires bounding at each step the resulting change in train error , which the authors achieve by bounding the change in the minimizer. Strong convexity is crucial in this type of proof to control the change of minimizer under a perturbation of the cost.
Modern machine learning algorithms often use formulations that are either convex but not strongly convex, or non-convex, as in the example (4). Further, from a mathematical standpoint, there is no reason to believe that strong convexity should be the ‘right’ condition for universality.
In this paper, we present the following contributions:
Universality of test error. We prove that, under additional regularity conditions, the test error is also universal. We emphasized that these regularity conditions concern the asymptotics of the equivalent Gaussian model. Hence, they can be checked using existing techniques.
Applications. We prove that our results can be applied to feature vectors that are obtained by two interesting classes of featurization maps: random feature models (random one-layer neural networks) and neural tangent models (obtained by the first-order Taylor expansion of two-layer neural networks).
In the next section we state our main results. Then, in Section 3, we discuss our assumptions on the data distribution and prove that they are satisfied for random features and neural tangent models. In Section 4, we demonstrate via a counter-example that universality can fail to hold without this distributional assumption. Finally, in Section 5, we outline the proof of the main result. Most of the technical work is presented in the appendices.
Main results
Throughout, the vectors are i.i.d. and . As mentioned above, we consider the proportional asymptotics whereby, assuming without loss of generality , we have
In fact most of our statements hold under the slightly more general assumption of
2 Assumptions
We will establish a general universality result under certain assumptions depending on the set , and then characterize the set on a case-by-case basis. In Section 3 we carry out this program by explicitly determining the set for models arising from the analysis of two-layer neural networks in the neural tangent regime.
The labels are binary: with
The set appearing in the constraint in (8) is a compact subset of .
Recall that the random vectors are assumed to be i.i.d. and that . We assume
Universality of the training error amounts to saying that is asymptotically distributed as . Namely, the two risks are similarly distributed at their respective, random, minimizers and in .
It is intuitively clear that for this to happen, their expectations must be close at a fixed, non-random point namely
Obviously, universality of the minimum of a random function is much stronger than universality of the the function evaluated at a single point, and therefore our main results require substantial technical work.
We will further discuss this assumption in Section 3. In particular, we will provide a counterexample showing that this or a similar assumption is necessary for universality to hold.
Also note that Eq. (11) states that the projections of and in the direction of are K-subgaussian, which is implied if and are K-subgaussian.
We additionally provide an alternative for Assumption 1 which is sufficient for our results to hold, but not as straightforward to check.
for some dependent only on .
3 Universality of the training error
Theorem 1 is the key technical result of this paper. While the training error is not as interesting as the test error, which is treated next, universality of the training error is more robust and we will build on it to establish universality of the test error.
The mathematical reason for the greater robustness of the training error is easy to understand. A small data perturbation, changing to , changes the value of the minimum by at most , but can change the minimizer by a large amount. The situation is of course significantly simpler if the cost is strongly convex, since in that case the change of the minimizer is controlled as well.
By Assumption 2, the ERM problem is subject to the constraint . In order to apply this theorem to unconstrained ERM problems, or to an ERM problem in which the constraint set is not a subset of , one can proceed in three steps: Prove that the unconstrained minimizer belongs, with high probability, to such a set ; Deduce that the unconstrained ERM problem is equivalent to the constrained one; Apply Theorem 1.
Proof technique. We outline the proof of Theorem 1 in Section 5. The proof is based on an interpolation method. Namely we consider an ERM problem with feature matrix that continuously interpolates between the two cases as goes from to . We then bound the change in the training error (minimum empirical risk) along this path.
This approach is analogous to the Lindeberg method [Lin22, Cha06], which was used in the context of statistical learning in [KM11] and subsequently in [MN17, OT18, HL20]. A direct application of the Lindeberg procedure would require to swap an entire row of with the corresponding row of and bound the effect on the minimum empirical risk (we cannot replace one entry at a time since these are dependent). We find the use of a continuous path more effective.
In [HL20], the effect of a swapping step is controlled by first bounding the change in the minimizer . This is achieved by assuming strong convexity of the empirical risk. The bound in the change of the minimizer immediately implies a bound in the change of the minimum value.
In the non-convex setting, we face the challenge of bounding the change of the minimum without bounding the change of the minimizer. We achieve this by using a differentiable approximation of the minimum. Even after this sequence of approximations, unlike in other universality proofs, the expectation one needs to bound is not obviously small. The key technical innovation is a polynomial approximation method which we believe can be of more general applicability.
4 Universality of the test error
We will state two theorems that provide sufficient conditions for universality of the test error. The first of these theorems concerns a scenario in which near interpolators (models achieving very small training error) exist. We are interested in this scenario because of its relevance to deep learning [BMR21], and because it is very different from the strongly convex one.
It is useful to denote the set of near empirical risk minimizers:
In other words, the minimum test error over all near-interpolators is universal (provided it does not change discontinuously with the accuracy of ‘near interpolation’). The same theorem holds (with identical proof) for the maximum test error over near interpolators, and if the level is replaced with any deterministic constant.
(In the first line denotes limit in probability.)
The next theorem provides alternative sufficient conditions that guarantee the universality of the test error. We emphasize that these are conditions on the Gaussian features only and it is therefore possible to check them on concrete models using existing techniques.
there exists a function differentiable at such that for all in a neighborhood of ,
for any minimizers of , , respectively.
Proof technique. The proofs of Theorems 2 and 3 are given in Appendix B and C. The basic technique can be gleaned from condition (23). We perturb the train error by a term proportional to the test error (this is only a proof device, not an actual algorithm). The test error can be related to the derivative with respect to of the resulting minimum value. The minimum value is universal by our results in the previous section. The technical challenge is therefore to control its derivative.
Checking pointwise normality
In this section we study some concrete examples for the distribution of the feature vectors . In each case, we characterize the set of parameter vectors for which the pointwise normality condition of Eq. (12) holds. For simplicity of exposition, we use throughout this section.
We first consider examples of featurization maps from the deep learning literature. Section 3.1 analyzes the featurization map that is obtained by linearizing a two-layer neural network around a random initialization. This is also known as the ‘neural tangent model.’ We establish asymptotic equivalence (in distributional sense) of ERM under the neural tangent model, to ERM under the Gaussian model with matching covariance structure. Comparable universality results were not known in this model, even in the case of convex losses. Indeed, checking the pointwise normality condition of Eq. (12) is challenging in this case.
Next, in Section 3.2, we consider the featurization map that is obtained by applying a one-layer network with random weights. This is equivalent to the ‘random features’ model of [RR07]. Pointwise normality (along the lines of Eq. (12)) and universality of the expected risk at a fixed for this model was first shown in [GRM+20]. Universality of test and train error for ridge regression was established in [MM19], while [HL20] proved universality of the ERM for strongly convex losses. Finally, [LGC+21] presented empirical evidence and conjectured that universality holds for a wide class of such featurization maps and loss functions.
In the setting of Section 3.2, our main contribution is the generalization of the results of [HL20] to non-convex losses.
Finally, in Section 3.3, we consider the case in which has i.i.d. entries: this is a standard model in random matrix theory. This data distribution was studied in the past mostly for convex or strongly convex losses [MN17, PH17, OT18]. The only exceptionAfter a first posting of the present manuscript, [HS22] also analyzed non-convex losses with having i.i.d. entries. is provided by [KM11] which studies certain non-convex losses when .
As we will see, the set typically excludes parameters that are too aligned with an element of the canonical basis. In other words, the parameters needs to be ‘incoherent’ with respect to the canonical basis.
In specific applications, if the constraint set is not a subset of , in order to apply our general theorems, it will be necessary to prove that a minimizer actually belongs to . In general, this will require a case-by-case analysis. However, Section 3.4 shows that a minimizer satisfies this condition for a broad class of overparametrized models. In these cases, no further analysis is required.
where are the first layer weights at initializations , and . As in the rest of the paper, we assume to be given training samples and to compute feature vectors .
Here we are not concerned with the connection between the original neural network and its neural tangent model, for which we refer to the literature [JGH18, DLL+19, LXS+19, BMR21, MZ20]. We will instead focus on the neural tangent model, and show that it can be approximated by an equivalent Gaussian model. Let us emphasize once more that –despite the neural tangent approximation– the loss function which we assume for the neural tangent model is not necessarily convex.
We have the following universality result for the neural tangent model (24).
Under the additional conditions of Theorem 2, Corollary 1 or Theorem 3, the universality results for the test error stated there hold.
Theorem 4 does not hold if we relax the set to . Indeed, for , the random variable is not asymptotically Gaussian. Clearly, this choice of is not in the set defined in (25).
Proof technique. We prove Theorem 4 in Appendix E by using Theorem 1. The key technical challenge is to establish that Assumption (5) for the distribution of the feature vectors , cf. Eq. (24). We Stein’s method as done in [HL20] for the random features model. However, treating the neural tangent features of Eq. (24) requires extra care due to the more complex covariance structure.
2 Random features
Let be the matrix whose columns are the weights . We have the following corollary of Theorem 1.
In Appendix F, we derive this corollary as a consequence of Theorem 1. To do so, we use a result established by [HL20] implying that the feature vectors satisfy Assumption 5, for every in a high probability set.
3 Linear functions of vectors with independent entries
We have therefore the following corollary of Theorem 1.
4 Controlling a minimizer in the overparametrized setting
The general universality results of Theorem 1 to Theorem 3 are stated for the ERM problem of Eq. (8), where we constrain , with satisfying Assumption 5. As discussed in Remark 2.4, these theorems can be applied to unconstrained ERM problems, or to ERM problems in which the constraint set is not a subset of , by separately proving that the minimizer belongs, with high probability, to a suitable compact set .
Assume for some , have i.i.d., mean , unit variance and subgaussian entries. Further assume that there exist constants such that
Then for any , there exists depending only on such that
That is, condition (12) of Assumption 5 holds in this case. In particular, under Assumptions 1 to 4 and the subgaussian condition of Eq. (11), we have
where is the optimum of the unconstrained ERM problem.
The proof of this result is deferred to Appendix D.
Necessity of pointwise normality
Let us now give a counterexample demonstrating that universality does not hold for general ERM problems, unless we restrict the optimization to subsets of where the latter satisfies the pointwise normality condition (12).
The pointwise normality condition (12) is not satisfied for this distribution of the feature vectors and this choice of . Indeed . However , while under the Gaussian model with the same covariance —namely, for — we have , for all . In other words, Assumption 5 does not hold in this case.
We next construct an ERM problem whose minimum value under this features distribution is different from the value under the Gaussian model. Consider the non-negative, Lipschitz continuous loss function
We then have the following minima of the two empirical risk problems:
In the non-Gaussian case, we clearly have for all , since by construction, while follows by evaluating the cost at . Hence, will be a minimizer which achieves a training loss of for all . However, in the Gaussian model (defined by ), there exist , such that if ,
This can shown by a uniform convergence argument as we detail in Appendix G.1.
Proof outline for Theorem 1
We redefine the vector from our assumptions to include and : . We will use …etc, to denote constants that depend only on , often without explicit definition. If a constant depends additionally on some variable, say , we write .
Under Assumption 1’ along with Assumptions 2-5, for any fixed and any bounded differentiable function with bounded Lipschitz derivative we have
Here, we outline the proof of this lemma deferring several technical details to Appendix A.3 where we present the complete proof. A standard estimate bounds the difference between the free energy and the minimum empirical risk (see Appendix): For .
Hence, Theorem 1 follows from Lemma 1 via an approximation argument detailed in Appendix A.
We assume, without loss of generality, that and are defined on the same probability space and are independent, and define the interpolating paths
for and . We use to denote the matrix whose th row is ; note that these rows are i.i.d. since the rows of and are so. Noting that for all , and are subgaussian with subgaussian norms bounded by RK uniformly over , it is easy to see that
It is convenient to define the probability mass function over :
for . With this notation, we can write
Via a leave-one-out argument detailed in Appendix A.3, we show that this form allows us to control
where is the expectation respect seen as independent samples from . The next lemma then states that the right-hand side in (40) can be controlled via its Gaussian equivalent.
from which the statement of Lemma 1 can be deduced.
Acknowledgements
This work was supported by the NSF through award DMS-2031883, the Simons Foundation through Award 814639 for the Collaboration on the Theoretical Foundations of Deep Learning, the NSF grant CCF-2006489, the ONR grant N00014-18-1-2729, and an NSF GRFP award.
References
Appendix A Proof of Theorem 1
In this section, we complete the proof of Theorem 1 by deducing it from Lemma 1 and give a complete proof of this lemma.
Recall that for , in Section 5 we let be a minimal net of , so that for some depending only on and . Let us define the discretized minimization over
We have the following consequence of Lemma 1.
Under Assumption 1’ along with Assumptions 2-5, we have for any bounded differntiable function with bounded Lipschitz derivative
The proof of this result is deferred to Section A.2. Here, we show that Theorem 1, under the alternative Assumption 1’ is a direct consequence of this lemma. First, we need a few technical lemmas. Let us define the restricted operator norm
For as in Assumption 5, we have for some depending only on ,
Under Assumptions 1’, 3, 4 and 5, we have for all
for some depending only on . A similar bound also holds for the Gaussian model.
The proofs are deferred to Sections G.2 and G.3 respectively. Here, we derive Theorem 1.
where in we used Lemma 5 and in the subgaussianity conditions in Assumptions 1 and 5 along with the condition on . An analogous argument then shows that
where the last equality is by Lemma 4. Now using that and sending concludes the proof of Eq. (16) for bounded differentiable with bounded Lipschitz derivative. To extend it to bounded Lipschitz, it is sufficient to find a sequence of bounded differentiable functions with bounded Lipschitz derivative approximating uniformly (see for example the following section for a similar argument).
A.1.2 Proof of Eq. (16) of Theorem 1 under Assumption 1
The proof under Assumption and is via an approximation argument. The proof under is a straightforward modification of that under , hence, we omit the former and only prove the latter.
is infinitely differentiable (see [Eva10], Appendix C.4.). Additionally, we have the following properties of .
for some . Then for , we have
for some . Furthermore, if for some positive integer , satisfies
Optimizing over gives the claim. Meanwhile, the bound in (46) is obtained as
Finally, the last property is obtained via a similar argument, namely,
Now for the labels , note that we can write where for and
Once again, is locally Lipschitz, differentiable and has
where in we used that is continuous and supported on a bounded interval, along with Lemma 7 applied to . This implies that satisfies the conditions on the labeling function in Assumption 1’. Furthermore, are i.i.d. subgaussian, and finally, we have for all , and random variables as in Eq. (14),
For any and bounded Lipschitz test functions , there exists a constant depending only on such that
where
Let and denote the minimizers of and respectively. Since is Lipschitz, it is sufficient to bound
for depending only on . First, let us obtain an upper bound on
For the term in (53), letting be the columns of ,
By symmetry, we can obtain a similar lower bound on the left-hand side of (54) (by replacing throughout with ), which allows us to write
for large enough and depending only on . Here, in we used Lemma 5.
To conclude the proof, we show that the expectation on line (55) is bounded by a positive constant times . This follows via the following computation:
for some depending only on . Here, in we used for all and for all , in we used Lemma 7 with , and in we used subgaussianity of .
A.1.3 Proof of the bounds in Eq. (17) of Theorem 1
and that for some constant depending only on . Hence, we can apply (16) with to conclude
which establishes the first bound in (17). The second bound follows via a similar argument. ∎
A.2 Universality of the minimum over the discretized space: Proof of Lemma 4
Recall the minimization problem over the set defined in (42). We show in this section that Lemma 4 is a direct consequence of Lemma 1.
Fix . Let us first bound the derivative of the free energy. Define the probability mass function for ,
and define similalry for the Gaussian model. Recall that the Shannon entropy of a distribution satisfies
where depends only on and . Therefore, the derivative of the free energy with respect to can be bounded as
Clearly, a similar bound holds with replacing . Hence, we have
where follows from Lemma 1 along with the assumption that . Sending completes the proof. ∎
A.3 Complete proof of universality of the free energy: Proof of Lemma 1
Let us recall the interpolating paths and defined in (35) for and , and the associated matrix whose th row is . Further, recall the gradient notation introduced in Section 5:
Finally, recall the probability mass function and its associated expectation defined in (37)
where all sums are implicitly over ; the minimal net of introduced in Section 5.
where is the free energy defined in (34).
Using the interpolator , we can write
where follows via an application of the dominated convergence theorem along with Lemma 9. So it is sufficient to show that for all ,
With the notation previously defined, we can compute the deriative of the free energy as
Since our goal is to establish (59), let us fix some and suppress it in the notation. We use the previous display to bound the expectation of the derivative as
The term in (60) can be controlled via a simple leave-one-out argument. Indeed, since the samples are i.i.d, it is sufficient to control the term in the sum:
Meanwhile, to control the term (61), it is sufficient to establish that
To see that this is sufficient, note that with (62), we can control (61) as
where follows by the i.i.d assumption on the samples and follows by reverse Fatou’s and Lemma 9.
In order to prove (59), fix and let be the polynomial from Lemma 2, where and depend only on and . Then the this lemma yields the bound
where is the statement of Lemma 2 and in we defined the expectation with respect to independent samples from . Now recall the definitions of and from Lemma 3. Note that and are jointly Gaussian with means
for all , and hence they are independent. And since is independent of by definition, the assertion of Lemma 3 implies that the summands in (63) converge to . Indeed, for :
where in we applied Lemma 3, in we used the independence of and and in we used that the mean of is . Combining this with the bound in (63) yields, for all ,
Taking then establishes (62) for any and concludes the proof. ∎
Appendix B Proof of Theorem 2
for all , where we set the value of the minimum to whenever the constraints are not feasible. First we give the following lemma.
For all and any , we have
Fix . On , let
be any minimizers of the respective functions so that and . Then note that we can upper bound
where in we used that on . An analogous argument with the roles of and exchanged shows that we also have
where follows from the subgaussianity in Assumption 5 and the assumption on the labels and noise in Assumption 1, and hence a similar bound holds for and . Now define
on . Letting denote a minimizer of this problem we write
where in we used that is nonincreasing in and the definition of , and in that by (67). Meanwhile we can obtain an upper bound for
Hence, for and , we have
Using a similar argument with the roles of and exchanged gives the second statement.
Appendix C Proof of Theorem 3
We prove the statement under each condition separately in the subsections that follow. We will use for simplicity, and without losing generality, since the arguments that follow can be directly extended to the setting where as long as it is a fixed constant.
(note the asymmetry), and use to denote their unique minimizers respectively. Furthermore, we write and for the minima.
First, we show that the convexity assumptions imply the following lemma.
for some depending only on . A similar inequality also holds for .
where follows from the KKT conditions for ; namely, for some , we have
where follows by noting that minimizes , and follows since is Lipschitz with bounded Lipschitz modulus under Assumption 1: Indeed we have
Combining the upper and lower bounds and rearranging gives
This proves the lemma for . A similar argument clearly holds for the Gaussian model. ∎
Now let us define, for , the differences
where in we used and , in we used that that is Lipschitz with bounded Lipschitz modulus and in we used Lemma 11. A similar argument then shows the same property for .
First, note that that for , is sandwhiched between and . Indeed, we have
For any , take where is the constant appearing in Eq. (73) of Lemma 12 and write
To conclude the proof, note that Lemma 30 implies that
C.2 Proof of Theorem 3 under the condition b
Let be the event in condition b, namely,
and take and to be any minimizers of and respectively.
C.3 Proof of Theorem 3 under the condition c
shown in (76) hold generally without the convexity assumption. Hence, using
we can write, for any and ,
for in some neighborhood of . Combining this with the previous display gives
where follows by differentiability of at . By exchanging the roles of and in this argument we additionally obtain
Appendix D Proof of Theorem 5
The claim of the theorem is a direct corollary of the following lemma.
Assume and that the feature vectors have i.i.d. mean , unit variance and subgaussian entries. Fix . Then the following holds with probability at least for some constants , : For any , there exists such that satisfying
for some depending only on .
Let and denote by the set of indices corresponding to the entries in with largest absolute value. Namely if , then we let (ties are broken arbitrarily). We also let denote the set of indices of ‘small’ entries.
Note that , whence
Postponing the proof of this claim, we define by
Further , and , thus proving the lemma.
We are left with the task of proving the existence of with the properties stated above. We construct by setting and
This vector satisfies the condition (80) by construction, and we are therefore left with the task of proving that it satisfies the norm constraints, with the claimed probability.
Recalling that , we define the
Here is a constant that will be specified below. On the intersection of these events, we have
where is the -th column of . We therefore have, on the event ,
In order to conclude the proof of the lemma, we need to prove that each of events , , , holds with probability at least , for a suitable choice of .
For event , by Theorem 1.1 of [RV09], for any set , , we have
for and any , where is the -th largest singular value. Hence, for a suitable choice of , with probability at least . The claim follows by taking a union bound over the choices of set .
For event , the bound follows in the same way (the only difference being that the union bound is over terms).
Next note that, defining , we have
Appendix E The neural tangent model: Proof of Corollary 4
Note that is symmetric, convex, and . Furthermore, for all we have for all .
The key to proving Corollary 4 is showing that the distribution of the feature vectors satisfy, on a high probability set, Assumption 5. Our proof here is analogous to that of [HL20] for the random features model. Let us begin our treatment by defining the event
For a given , let us define the set
Our goal in this subsection is to prove the following lemma.
Define, for the notation
For a fixed bounded Lipschitz function , let be the solution to Stein’s equation for , namely, the function satisfying
(see [CGS11] for more on Stein’s method and properties of the solution .). In order to prove Lemma 14, it is sufficient to show that
In Section E.1.3, we upper bound the quantity (84) as
So first, let us control the terms on the right hand side: We do this in Sections E.1.1 and E.1.2, respectively. Before doing this, we make the following definitions which will be used throughout. Define along with the matrix notation
For , we have for any fixed integers and
for some constants depending only on .
Using Lemma 21, the first five inequalities are direct. Indeed, recalling that , we have
where the last equality holds because is a diagonal matrix. Now recall that for the two square matrices and , we have (see for example [Joh90], (3.7.9))
where follows using the same bound we applied to (88). This establishes the seventh bound in the lemma.
For the ninth bound, we first note that by definition of and , so that
Fix throughout. Define for convenience
Let us compute the expectation of and control its variance.
Now, we control . First, note that we can write as
Taylor expanding to the third order gives
for some between and . Using this expansion and the notation defined earlier, can be re-written as
Using the expansion (92) in the expression for gives
Let us write for the terms on the right-hand on lines (93), (94), (95), (96) respectively. Observe that
where follows from the Gaussian Poincaré inequality. We control each summand directly. In doing so, we will make heavy use of the bounds in Lemma 15 and hence we will often do so without reference. First let us bound the expected norm of the gradients in the above display.
Now, the gradient of can be computed as
where we recall that denotes the vector whose th entry is . We have the following bounds on the expected norm squared of each term in (99): for the first of these terms,
for all , where depends only on , and depends only on and . Note that in we used is finite.
Moving on to bound the norm squared of the second term in (99), we have
Similarly, the expected norm squared of the third term in (99) is bounded as
and finally, for the fourth term in (99) we have
Now moving on to , we can write
Let us again bound the expected norm squared of each of the terms in the previous display.
For the terms on lines (101) and (102) we have
where in we used that . A similar calculation shows that
For the term on line (104), an analogous calculation shows that
and then similarly for (105), and (106) we have
What remains is the term . However, this can be bounded naively as
where follows from an application of Hölder’s and Lemma 15. Hence we have
Combining this with (98), (100), and (107) gives
E.1.2 Bounding the second term in Eq. (86)
Using that for , not necessarily independent, subgaussian with subgaussian norm
for some universal constant . Hence, it is sufficient to establish the desired bound on the set . Indeed, suppose
where follows by a naive bound on and follows by an application of Hölder’s. Hence, throughout we work on the event .
By Lemma 2.4 of [CGS11], is differentiable and since is assumed to be differntiable with bounded derivative. Hence,
where follows from (111), follows from boundedness of , and follows from and the definition of . Now recall the form of introduced in Eq. (91) and let us again Taylor expand to write
for some between and . We show that for each ,
For the contributions of , we have
uniformly over . Taking supremum over and sending proves (114) for .
uniformly over , where holds by the definition of . Sending shows (114) for .
uniformly over , establishing (114) for .
Finally, can be bounded almost surely on :
uniformly over , where follows from the definition of the event and follows because is a projection matrix for all and that . Therefore, we have
where follows from (112), follows from (113) and follows from (114) holding for . Hence, we have shown (110) and completed the proof. ∎
E.1.3 Proof of Lemma 14
Recall the definition of in (85) and note that for all ,
Since is Gaussian, is independent of any function of and hence is independent of . Therefore, we have
where follows by Eq. (83) and follows by Eq. (116). Taking the supremum over then and applying Lemmas 16 and 17 completes the proof. ∎
We give the following consequence of Lemma 14.
and take to be bounded differentiable with bounded derivative. Then for we have
where follows from Lemma 14 and follows from the definition of . Now sending proves the lemma for differentiable Lipschitz functions, which can then be extended to Lipschitz functions via a standard uniform approximation argument.
E.3 Truncation
Let us define and the random variable The following Lemma establishes the subgaussianity condition of Assumption 5 for .
Conditional on we have
for some constant depending only on .
Take arbitrary . Let
then consider the function . Note that is continuous and differentiable almost everywhere with gradient
almost everywhere. Noting that and we can bound
where follows by nothing that . This shows that is subgaussian with subgaussian norm constant in and . Since was arbitrary, this proves the claim.
Now, let us show that the condition of Eq. (12) holds for the truncated variables .
E.4 Proof of Corollary 4
Now, note that we have for some ,
Combining the displays (121), (123) and (E.4) gives
where follows by dominated convergence. ∎
E.5 Auxiliary lemmas
We include the following auxiliary lemmas for the sake of completeness.
Let be mean zero subgaussian random variables with . We have for all integer ,
There exist constants depending only on such that
where we used that and are independent for , and that is subgaussian with subgaussian norm . Hence, we have
This proves the existence of the constant in the statement of the lemma. Meanwhile the existence of is a consequence of Theorem 4.6.1 in [Ver18]. ∎
Appendix F The random features model: Proof of Corollary 2
The following lemma is a direct consequence of Theorem 2 and Lemma 8 from [HL20].
Furthermore, conditional on , is subgaussian with subgaussian norm constant in .
We remark that the setting of [HL20] differs slightly from the one considered above. Indeed, they take
the activation function to be odd and the weight vectors to be , and
the “asymptotically equivalent” Gaussian vectors to be for instead of , where and are defined so that
Theorem 2 of [HL20] prove a more general result than the one stated here for their setting. Additionally, they give bounds for the rate of convergence for a fixed in terms of and (and other parameters irrelevant to our setting.). However, here we are only interested in the consequence given above.
First note that via a standard argument uniformly approximating Lipschitz functions wtih differentiable Lipschitz functions, Lemma 23 can be extended to hold for that are bounded Lipschitz.
where follows by the dominated convergence theorem and follows from Eq. (127).
Appendix G Deferred proofs
Let be a minimal net of so that . It is easy to show that for centered isotropic Gaussian,
for some constants , where the last inequality holds on . (A similar argument was carried out in the proof of Lemma 6.)
Choose . By union bound over , for sufficiently large , the following holds with probability at least :
Let be the event that this inequality holds. Having chosen , choose to satisfy and so that we have
Finally notice that, for any two matrices , ,
In conjunction with Eq. (128), this proves the claim of Eq. (33).
G.2 Proof of Lemma 5
Assume satisfies Assumption 5. Then there exist constants depending only on such that for all ,
Letting be the rows of , note that by Lemma 2.6.8 of [Ver18] we have
Recall that , and hence there exists an -net of of size for some constant depending only on . Fix and note that
By Assumption 5, are squares of i.i.d subgaussian random variables with subgaussian norm uniformly in , and with means
where the last inequality holds uniformly over (see Proposition 2.5.2 of [Ver18] for the properties of subgaussian variables). Hence, via Bernstein’s inequality (2.8.3 of [Ver18]), we have for any ,
where for we used that and . Taking and , we have via a union bound over
where for we used that , for we used the definition of , and for that Now via a standard epsilon net argument (see for example the proof of Theorem 4.6.1 in [Ver18]), one can show that
for some depending only on R and . Combining this with (129) gives the desired result. ∎
There exist constants depending only on such that for all ,
Let be the high probability event of Lemma 24, i.e.
where for the constant appearing in the statement of the lemma. Next define the event
We have on ,
holding for , . Meanwhile, holds on and is from the definition of . Hence, by the definition of in Lemma 24 we have
Meanwhile, from the definition of , we directly have
By an application of Lemma 25 with , we have for all
Hence, we can bound the desired expectation as
for some sufficiently large depending only on since Using that are i.i.d. subgaussian for , we have
for some depending only on . ∎
G.3 Proof of Lemma 6
where in we used that the regulizer is assumed to be locally Lipschitz in Frobenius norm and that for . Now using to denote the partial derivative with respect to the th entry, we compute the gradient
for some depending only on . Combining equations (131) and (132) allows us to bound the first term in (130) as
where follows from Eq. (131), follows from the assumption that is symmetric and convex, and follows from (132). Finally, combining with (130) we obtain
for some constant depending only on . This concludes the proof.
G.4 Proof of Lemma 9
Let us expand this via the definition of in (36):
for depending only on . However, for any fixed and we have
for depending only on , since by Assumption 5. A similar bound clearly holds for . Hence, using that are assumed to be fixed, an application of Hölder’s gives
for some depending only on , where we also used that is assumed to be subgaussian by Assumption 1’. Therefore, we have
To establish the second inequality, recall the explicit form of the derivative from (38) and note that are i.i.d. for different so that
G.5 Proof of Lemma 2
For any , we have constants depending only on such that
From the definition of along with Assumption 5, we have , and similarly for Furthermore, Assumption 1 asserts that . So a union bound directly gives
for some universal constants ∎
Let us now consider the power series of centered at 1, and its associated remainder
We have the following properties of and , whose proofs are elementary and are included here for the sake of completeness.
for ;
For any and , there exists such that .
For ii, the convexity of can be shown by noting that i gives
Finally, iii can be shown by verifying that is indeed the power series of with a radius of convergence of . ∎
The following lemma bounds the error in the approximation, and is the key for proving Lemma 2.
For any and , there exists some finite integer , depending only on and such that
Recall the definition of in (G.5) for and and write for arbitrary integer ,
where follows from Jensen and point ii of Lemma 27 asserting the convexity of on . The expectation in the second term can be bounded uniformly over , namely
Therefore, for any ,
Then, by points iii of Lemma 27, we can choose a sufficiently large integer so that
which when combined with the bound on yields the claim of the lemma. ∎
Finally, let us complete the proof of Lemma 2.
Let be the constant in Lemma 9 guaranteeing that
Fix , and let so that Lemma 28 holds with replaced by Then, we directly have via an application of Cauchy-Schwarz
G.6 Proof of Lemma 3
This section is dedicated to proving Lemma 3. The first step is extending Eq. (12) as follows.
Fix be arbitrary. Let and define
Both terms on the right hand side on line (139) are similar and can be bounded in an analogous manner. Namely, we can write for the first of these
where . Note that in we used
where follows from the decomposition in (139) and the bounds in (141) and (142), and follows from the dominated convergence theorem along with the limit in (144) and domination of the integrand . Sending completes the proof. ∎
Now, via a truncation argument, we show that this can be extended to square integrable locally Lipschitz functions.
and define . Noting that and that is Lipschitz, we see that is bounded and Lipschitz. To see that it is indeed Lipschitz, take with ,
for depending only on since is locally Lipschitz. We can now write
for some depending only on . Here, follows from Lemma 29 and that , and follows from the tail bounds in equations (146) and (G.6) along with the square integrability assumption of . Sending completes the proof. ∎
Recall equations (35) and (36) defining and , respectively, in terms of and . Further, recall the definitions of and in the statement of the lemma. Define and the function
i.e., the expectation is with respect to . Since has the same distribution as , we have
Assumption 5 is satisfied for replacing . Hence we similarly have
for some . Therefore, satisfies the square integrability condition in (145) of Lemma 30. An application of this lemma then yields the claim of Lemma 3. ∎