Universality Laws for High-Dimensional Learning with Random Features
Hong Hu, Yue M. Lu
I Introduction
The supervised learning process described above has two main performance metrics: the training error
which is simply a scaled version of the optimal value of (1), and the generalization error
where is some post-processing function (e.g., the sign function) and the expectation in (4) is taken over a fresh pair of samples that are independent of the training data. To carry out theoretical analysis of the training and generalization errors, it is necessary to make some further assumptions on how the training samples are generated. A classical model, which is also the one adopted in this work, is the so-called teacher-student framework. Specifically, we assume that and
In this paper, we study a particular case of the above setting, known in the literature as the random feature model . It corresponds to specializing the general regressors in (2) to
I-B The Gaussian Equivalence Conjecture
Fortunately, it has been observed by many authors (see, e.g., , and also in the context of random kernel matrices) that the random feature model considered above should be asymptotically equivalent to a Gaussian model, where we set the regressors in (2) to
In what follows, we shall refer to the setting where the regressors are in (6) as the nonlinear feature model, and refer to the one using in (7) as the linear Gaussian model. Let
The optimal weight vectors, the training and the generalization errors of these two formulations can then be written as , , and , respectively.
Roughly speaking, the Gaussian equivalence conjecture states that, under certain conditions on the feature matrix , we have
That the nonlinear feature model and the linear Gaussian model can be asymptotically equivalent has a simple intuitive explanation. Under certain conditions on the random feature matrix , one can show that the random vectors in (6) and in (7) have asymptotically matching first and second moments. (See Appendix -D for details.) Thus, the asymptotic equivalence in (10) points to the emergence of a universality phenomenon that is inherent in many large random systems: The macroscopic behaviors of such systems only depend on a few key parameters (the first two moments of and in our case), whereas the microscopic structures of the systems (i.e., the exact probability distributions of and ) are irrelevant.
Notice that the surrogate Gaussian formulation is much more amenable to theoretical analysis, as it only involves Gaussian vectors . Indeed, based on the Gaussian equivalence conjecture, the authors of provided a precise asymptotic characterization of maximum-margin linear classifiers in the overparameterized regime using Gaussian min-max theorems . The performance of the linear Gaussian model under more general settings, where one uses generic convex loss functions and ridge regularization in (1), was studied in by using the non-rigorous replica method from statistical physics. More recently, these replica predictions have been rigorously proved in .
I-C Main Contributions
The main contribution of this paper is to prove the aforementioned Gaussian equivalence conjecture. Our results are based on the following technical assumptions.
The latent input vectors in (6) and (7).
The dimension of the latent input vectors (denoted by ), the dimension of the regression vectors (denoted by ), and the number of training samples (denoted by ) tend to infinity at fixed ratios. Specifically, and as .
The unknown teacher vector in (5) is deterministic, with .
where is the function in (5).
The activation function is an odd function, with bounded first, second, and third derivatives.
The columns of the feature matrix are independent Gaussian random vectors: for . Moreover, is independent of the latent input variables .
We can verify that the conditions in Assumption (A.4) are satisfied by the quadratic loss function, the logistic loss function, and by any that grows no faster than some polynomial of as . Possible ways to generalize our results to non-differentiable loss functions (e.g. the hinge loss) will be discussed in Section IV. To simplify our analysis, we require in Assumption (A.6) that the activation function be odd, which then implies that in (8). This is merely a limitation of our current results, and the asymptotic equivalence in (10) is expected to hold for more general activation functions [such as the ReLU function as shown in Figure 1(a)]. Yet another limitation of our work is the Gaussian assumption on the feature vectors in Assumption (A.8). With some extra effort (mostly on generalizing the concentration inequalities in Appendix -E2), our proof can be easily extended to cases where the columns of the feature matrix are independent sub-Gaussian random vectors. However, we expect that the majority of our proof technique should work for deterministic feature matrices that satisfy the conditions in (64) and (65). We will elaborate on this point in Section IV and pinpoint the one technical difficulty that prevents us from working with deterministic matrices.
To state the results of our main theorem, we first introduce a perturbed version of the optimization problem in (1):
where are two parameters, is the teacher vector in (5), and
Note that and [with the regressor matrix specialized to and in (9)] are exactly the training errors associated with the feature and Gaussian formulations, respectively. The two extra terms and in (11) will be needed in our analysis of the generalization error. In particular, we shall consider different values of such that
The bound requires some explanation. At first glance, the possibility that can take negative values is worrisome, as will then be a concave function of . This concave term, however, will (most likely) not change the convexity of the overall objective function in (11). To see this, we recall from Assumption (A.8) that has a Wishart distribution and thus its spectral norm is bounded with high probability. Specifically, it is easy to show (see Appendix -E3) that
where and is some positive constant. By Assumption (A.5), the regularizer is strongly convex with parameter . It follows that, with , the overall objective function of (11) is -strongly convex with probability at least ,
Suppose Assumptions (A.1)–(A.8) hold. Fix and . For every and every finite constant , we have
for , where denotes some function that grows no faster than a polynomial of . Consequently,
where denotes convergence in probability as .
We prove this theorem in Section II-D. A special case, with , implies that the training errors of the nonlinear feature model and its Gaussian surrogate must necessarily have the same asymptotic limit.
The next result, whose proof can be found in Section II-E, establishes the universality for the generalization error, under one additional assumption:
There exists a limit function such that for all and . In addition, the partial derivatives of exist at . Let them be denoted by and , respectively. We further assume that .
and are two independent standard Gaussian random variables.
I-D Related Work
The Gaussian equivalence phenomenon studied in this paper was stated in , and explicitly exploited in to derive the asymptotic limits of several learning problems. Related phenomena also appear in the context of random kernel matrices , where it is shown that the impact of the nonlinear activation function [on the limiting singular value spectrum of the matrix in (9)] can be captured by the three parameters in (8). However, these results on the asymptotic spectrum are not sufficient for our purpose. Except for the special case of ridge regression, the training and generalization errors of the learning problem in (1) are not simple functions of the singular values/vectors of .
where , with defined in (12), and . This result is an important step towards a theoretical justification of the Gaussian equivalence, and indeed a quantitive version of (17) serves as a crucial ingredient of our proof. However, by itself the characterization in (17) does not imply the asymptotic equivalence stated in (10), as the training and generalization errors are all complicated functionals defined implicitly through the optimization problem (1). When calculating the generalization errors using (4), for example, one will be dealing with two different weight vectors and , respectively, as opposed to a single shared vector as in (17). Showing that and , which are the second-order statistics of the Gaussian distributions, is exactly among the technical challenges addressed in this work.
Our method for proving universality for the random feature model is based on the classical Lindeberg’s principle and a leave-one-out analysis of the optimization problem in (1). Similar approaches have been used before to establish universality for various estimation problems . As a technical challenge in our problem, the entries of the regression vectors have a particular correlation structure, due to the presence of the random feature matrix in (6) and (7). Thus, new techniques have to be developed to handle this correlation. Beyond the random feature model considered here, the Gaussian equivalence is a very general universality phenomenon that has been observed in many other models (see, e.g., ).
I-E Paper Outline
The rest of the paper is organized as follows. We prove Theorem 1 and Proposition 1 in Section II. To emphasize readability, we only highlight the central ideas and key intermediate results there. In Section III, we use Stein’s method to provide an alternative proof of the central limit theorem for the nonlinear feature model. Heavier technical details are left to the appendix, where we compile all the auxiliary results. We conclude the paper in Section IV with some additional remarks on how some of the technical assumptions in this work can be further relaxed.
II Proof of the Main Results
Notation: In our proof of Theorem 1, the parameters in (11) are always kept fixed. Thus, to streamline the notation, we will write and simply as and , when no confusion can arise. We will use and to denote generic constants that do not depend on the problem dimension . To reduce the burden of bookkeeping, the exact values of and can change from one line to the next. In addition, stands for any function that grows no faster than some polynomial of , i.e.,
We start by noting that, to prove the inequalities in (14) and (15), it suffices to show that
for every bounded test function that also has bounded first and second derivatives. The precise connection between (14), (15) and (18) will be made clear in Section II-D, when we prove Theorem 1. For now, we focus on showing (18).
In our analysis, we first show a conditional version of (18). Specifically, we will define a subset of all feature matrices, and show that
II-B The Admissible Set of Feature Matrices
Recall that , where are the feature vectors. For notational simplicity, we add one more vector by letting . The admissible set is constructed as
where is the constant in Assumption (A.2). Before defining , which requires some additional notation, we first note that are all high-probability events under Assumption (A.8). Specifically, standard concentration inequalities for sub-Gaussian random vectors give us
for some . (See Lemma 7 in Appendix -E1 for a proof.) Similarly, applying matrix concentration inequalities [(172) in Appendix -E3], we can conclude that
The definition of the last set in (21) is a bit technical. Consider a family of optimization problems
for , where and are the regressors in (6) and (7), respectively, and
The reason for considering this sequence of problems will become clear in Section II-C. For now, just note that our quantities of interest, namely and , are just the starting and end point of this sequence, i.e., and . We then have
where is the constant in Assumption (A.4).
Under Assumptions (A.1)–(A.8), there exists some such that
This result, whose proof can be found in Appendix -F5, shows that is still a high-probability event. In light of (24), (25) and (30), there exists such that
II-C The Lindeberg Method
In what follows, we prove (19) by using Lindeberg’s method . The idea is simple: The sequence shown in (26) serves as an interpolation path that allows us to go from to . To prove (19), it suffices to show that the difference between any two neighboring points on the interpolation path is small. Indeed, as there are only such pairwise comparisons, we just need to show that
uniformly over and .
By construction, the optimization problems associated with and differ only in their choice of the th regressor. The former uses , whereas the latter uses . Consequently, both and can be seen as a perturbation of a common “leave-one-out” problem:
As , it is natural to apply Taylor’s expansion around , which gives us
with denoting some value that lies between and . Writing an analogous expansion for around , and then subtracting it from (II-C), we can get
To make further progress, we need to introduce a surrogate optimization problem:
where is the leave-one-out optimal solution of (II-C), and
is the Hessian matrix of the objective function in (II-C) evaluated at . We note that has a simple interpretation: By setting , we can see that the optimization problem associated with is simply a quadratic approximation of the one associated with in (26). Similarly, is a quadratic approximation of . The following lemma, whose proof can be found in Appendix -F7, quantifies the accuracy of such approximation.
both of which hold uniformly over and .
Using this lemma, we can now bound the terms on the right-hand side of (33) as follows:
where to reach the last step we have used Hölder’s inequality and (37). Meanwhile, combining (37) and (36) gives us
In light of (II-C), (39), and (40), we just need to show that
to get a useful bound for the left-hand side of (33).
where is some fixed parameter. It is straightforward to show (see Lemma 15 in Appendix -F) that
By construction, both and are independent of the leave-one-out solution and the Hessian matrix . It is this independent structure that significantly simplifies our analysis.
These last two terms are easy to control, due to the concentrations of and around . As shown in Lemma 24 in Appendix -F8, we have
uniformly over and .
That is due to the following fact: When conditioned on and , we have
Making (49) precise is the focus of Theorem 2 in Section III. It is easy to verify that the test function defined in (48) indeed satisfies the assumptions of Theorem 2. (See Lemma 25 in Appendix -F8.) Consequently, for every , Theorem 2 gives us
Note that the upper bound is uniform over all and all . Now let us recall the construction of the interpolation sequence in (26). Since and , we obtain (19) from (51) via triangle inequality. Finally, given the decomposition in (20) and the probability bound in (31), we establish the inequality in (18).
Before proceeding to the proof of Theorem 1, we pause and point out a subtle issue regarding the central limit theorem stated informally in (49). It is important that the weight vector in (49) is the leave-one-out solution , which is independent of both and . The situation will be very different if we use the original optimal solution instead. In this case, the asymptotic distribution of is not Gaussian (i.e., the central limit theorem is no longer valid), due to the weak yet non-negligible correlation between and .We illustrate this fact in Fig. 2. The theoretical prediction of the limit distributions shown in the figure can be found by using Lemma 15 and Lemma 16 in Appendix -F.
II-D Proof of Theorem 1
Equipped with (18), we just need to construct a suitable test function in order to complete the proof. For any fixed and , let
where is a scaled mollifier defined in (118) in Appendix -A. By properties of , it is easy to check that and . Moreover,
Letting and taking expectation over the functions in (53), we have
Changing to yields
which leads to (14) for and . The proof of (15) is analogous, as the above procedure is completely symmetric with respect to and .
II-E Proof of Proposition 1
Let be a Gaussian vector independent of the existing training samples and the feature matrix. Substituting (5) into (4), we can then write the generalization errors as
where is the matrix in (12). It is also easy to check that , where
with .
The rest of the proof falls naturally into three parts: (a) We will first show that and , where is the limit function in Assumption (A.9); (b) By replacing in (54) with , we introduce the analogous quantities and . We will show that have the same limits as ; (c) Finally, we will show that with high probability, where is the function in (55).
We start with part (a). By the definition of the optimization problem in (11), we have
for any . It follows that, for any ,
Fix . By Assumption (A.9), the limit function is differentiable at the origin. Thus, there is some such that
The first inequality in (56), with substituted by , then gives us
Next, we move on to part (b) and establish the limits for and . This is easy, in light of the universality laws given by Theorem 1. Specifically, (16) gives us . Replicating the same steps in part (a), with replaced by , allows us to conclude that
By Assumption (A.7), is differentiable with respect to except at a finite number of points. Moreover, it is easy to check that
where is the constant in Assumption (A.4) and
Also recall the admissible set defined in Section II-B. We can verify that the assumptions of Proposition 3 (as stated and shown in Section III-D) hold for any and . Thus, conditioned on , we can apply Proposition 3 to get
III A Central Limit Theorem for the Feature Model
In this section, we prove a central limit theorem (CLT) related to the nonlinear feature model. Let
as . Here, we consider the setting where and the feature vectors are all deterministic, and the only sources of randomness come from and . Thus, the right-hand side of (63) are just two jointly Gaussian random variables. CLT in the form of (63) was first studied and proved in (see our discussions in Section I-D and Remark 4 below). It will be useful in bounding the term in (44), a critical step in our application of the Lindeberg method. It also plays an important role in our proof of Proposition 1, where we establish the universality of the generalization error.
To state the theorem, we first need to put some restrictions on the feature vectors and the teacher vector . Let , and let denote the Kronecker delta function. We assume that
for some and . Moreover,
Note that, for the random feature vectors considered in this paper [see Assumption (A.8) and the admissible condition in (22)], the upper bound can actually be as small as , and the spectral norm can be set to be of . However, since we believe that the central limit theorem could be of independent interest in other problems beyond this paper, we are going to prove it under the more relaxed assumption in (64).
Suppose that the feature vectors satisfy (64) and (65), and the activation function satisfies the conditions in Assumption (A.6). Let be a sequence of two-dimensional test functions that are differentiable with respect to . Moreover, for each ,
where and .
The settings of the CLT shown in are also somewhat different from ours. On the one hand, the one in is more general in that it does not require the nonlinear activation function to be an odd function. On the other hand, Theorem 2 is more relaxed in terms of the test function , which only needs to be differentiable with respect to the first variable . In addition, we further relax this restriction in Section III-D, where a characterization similar to (67) is given for piecewise differentiable test functions, at the cost of a slower decay rate than the right-hand side of (67). This extension will be needed when we study the universality of the generalization error in (4). Finally, the new proof technique here, based on Stein’s method , might be of interest in its own right.
Consider a sequence of activation functions and differentiable test functions such that, for every ,
;
is compactly supported. Specifically, there is some threshold such that for all ;
for some .
Here, and , where and are two independent Gaussian vectors, is a collection of feature vectors satisfying (64) and (65), and
Lemma 2 is essentially a reduced form of Theorem 2. The characterization in (68) guarantees that has an asymptotical Gaussian law, whereas (67) needs to consider the joint distribution of and . Moreover, Lemma 2 puts some further constraints on and , requiring the former to have compact supports and the latter to be bounded and to have bounded derivatives.
Our proof is based on Stein’s method . We start by observing that is a Gaussian random variable with zero mean and variance
It follows that we can rewrite the left-hand side of (68) as
for . Next, we introduce the following “Stein transform”:
Key to Stein’s method is the following identity
which can be directly verified from the definition of . Moreover, since , we have from [45, Lemma 2.4] that
By Stein’s identity, when follows the standard Gaussian distribution, the left-hand side of (76) exactly equals to zero. Intuitively, this quantity should be approximately equal to zero when is approximately standard Gaussian. This is what we are going to prove next. In what follows, we derive bounds for the two parts on the right-hand side of (76), separately.
We start with part (a). To simplify the notation, we let . Applying the bound on in (73) gives us
with the second inequality due to (65). Recall the definition of in (70). We then have
where in the last step we use a simple inequality () to bring the final bound to a convenient form.
Next, we consider the variance term in (78). Introducing the shorthand notation , we rewrite in (77) as
where to reach the second equality we have used Taylor’s expansion, with denoting a point between and . Substituting (80) into the expression for leads to
The term involving on the right-hand side of (84) is easy to bound, even deterministically. Using our assumptions about the function stated in the lemma, namely it has a compact support and bounded third derivatives, we have and . In addition, since the feature vectors satisfy (64), we can verify from the definition (74) that for some constant . It follows that
where the second inequality is due to the simple bound that .
Now we tackle the more challenging task of bounding in (84). We first note that, since and , we can view as a differentiable function of , denoted by , with . The Gaussian Poincaré inequality (see, e.g., [46, Theorem 3.20]) then gives us
where the gradient can be computed, with some diligence, as
where , , , , and . In light of (86), we just need to show that is properly bounded. We do so by controlling the norm of each term on the right-hand side of (87).
Note that our assumptions about the function implies that and for . Moreover, by assumption. Thus, the first term on the right-hand side of (87) can be bounded as
For the second term, we first rewrite it in the form of a matrix-vector multiplication as
where , , and . Clearly, and . We can also verify that
Similarly, the fourth term on the right-hand side of (87) can be rewritten as , where , , and , with denoting the Hadamard product of two matrices. The spectral norm of can be bounded as
for some constant , where the last inequality is due to (64). This then allows us to bound the norm of the fourth term of the gradient expression as
The situations for the third and fifth term on the right-hand side of (87) are completely analogous, and thus we avoid the repetitions. With the bounds in (88), (90) and (92), we can now apply (86) to get
Combining this bound with those in (85), (84), (79), we can retrace our steps back to (78) and conclude
where the last inequality also uses the fact that and thus
Now the remaining task is to bound the part (b) in (76) before we can complete the proof. Using Taylor’s expansion, we have
where is some point between and . By assumption, the function considered in this lemma is supported on for some . We can then write . This step of introducing an indicator function is not strictly necessary, but it helps to simplify some of our later arguments. We now have
where to reach the last inequality we have also used (73) and the boundedness of . Using a similar Taylor’s expansion as in (80) but only to the second order, we have
where are the matrices considered in (89) and (91), respectively, and . Using the spectral bounds given in (89) and (91), and the inequality (94), we get
Substituting this inequality and (93) into (76), and using the fact that , we are done. ∎
III-B Joint Distributions
Lemma 2 shows that has an asymptotically Gaussian distribution. Using this result, we can easily show that the asymptotic distribution of and is jointly Gaussian, via a conditioning technique.
Consider a sequence of activation functions and two-dimensional test functions such that, for every ,
;
is compactly supported. Specifically, there is some threshold such that for all ;
is differentiable with respect to . Moreover, there is a function such that
Here, , are defined the same way as in Lemma 2, and is a collection of feature vectors satisfying (64) and (65).
where and are two independent sets of Gaussian random variables. Let
We can then redefine the entries of and as
without changing their probability distributions. The reason we do such decomposition is that is independent of . This convenient independence structure allows us to calculate the expectations in (3) by first conditioning on .
Applying Taylor’s expansion to the expression for in (99), we get
where is some point between and . This expansion then leads to
Using the bounded derivative assumption in (97), we have
Next, we show that the terms involving and in (III-B) are small.
with the last step being the Gaussian Poincaré inequality. Recall the definition of and in (98). One can verify that
where to reach (102) we have used the bound due to (64). Substituting (102) into (101) then gives us
In light of (103) and (104), the left-hand side of (III-B) is well under control.
Using the equivalent representation for in (99), we have
where is simply a shifted version of . Combining this with (III-B), (103) and (104), we can now bound the left-hand side (LHS) of (3) as
Note that, for any fixed , we can use Lemma 2 to control the conditional expectation in the first term on the right-hand side of (105). Indeed, with fixed, can be viewed as a one-dimensional test function and it satisfies all the assumptions stated in Lemma 2. The only thing that is different here is that we are now using as the feature vectors. Thus, to apply Lemma 2, we need to check that this modified set of feature vectors still satisfy the condition in (64). But this is easy to do. Recall that , with satisfying (64) for some . Thus, for all ,
for some positive constant . Finally, by substituting the bounds (68) [with there replaced by ] and (106) into (105), we reach the target inequality in (3). ∎
III-C Proof of Theorem 2
To go from Lemma 3 to Theorem 2, we just need to remove the following two restrictions in the assumptions of Lemma 3: (1) is compactly supported on for some ; and (2) and its derivatives are bounded [see (97)]. The main ingredient of our proof is to show, via a standard truncation technique, that the central limit theorem characterization still holds even if we relax these two assumptions.
Let be a test function satisfying (66). We can construct a smoothly truncated version of this function via
where is the smooth window function defined in (119) in Appendix -A and
for some positive constant . The threshold in (107) is chosen strategically. With this choice, we can show
The detailed proof of (108) and (109) are provided in Appendix -B Together, (108) and (109) show that replacing the original test function with its smoothly truncated approximation only incurs a small price of .
Next, we consider the activation function . Using the smooth window function in (119) again, we can build a truncated approximation
for some positive constant . It is easy to verify that satisfies all the assumptions stated in Lemma 3 concerning the activation functions. With this truncated activation function, define
as the counterparts of and in (62). Here, are the constants defined in (69). Our goal is to show that and . Specifically, we can get (details are relegated to Appendix -B)
Given the inequalities in (108), (109), (112) and (113), we have
We can use Lemma 3 to bound the second term on the right-hand side, since its test function and the activation function satisfy the assumptions stated in that lemma. Using (3) and the property that , we reach the main result (67) of the theorem.
III-D Extension to Piecewise Smooth Test Functions
In what follows, we generalize Theorem 2 to test functions that are only piecewise differentiable. This auxiliary result will be needed in our proof of Proposition 1 for the case where the “output function” in the generalization error (4) lacks smoothness [e.g., .
Consider the same assumptions of Theorem 2 with “ is differentiable with respect to ” replaced by “ differentiable with respect to except at a finite number of points ”. Additionally, we also assume that
The upper bound in (64).
.
Let , where is the covariance matrix in (12). Then for some constant .
It is possible to improve the convergence rate on the right-hand side of (114) from to , by requiring higher moments of to be bounded. We do not pursue this optimization as the current form is sufficient for our proof of Proposition 1.
IV Conclusion and Final Remarks
In this paper, we have proved the asymptotic equivalence of a nonlinear random feature model and a surrogate linear Gaussian models in terms of their training and generalization errors. As a consequence of this universality theorem, the learning performance of high-dimensional random feature models can be precisely characterized by studying their linear Gaussian counterparts, which are much more amenable to theoretical analysis. Our proof, which builds on the classical Lindeberg approach, makes several technical assumptions on the loss function, the nonlinear activation function, and the feature matrix. We close the paper by discussing how some of these assumptions can be further relaxed.
In our proofs, we often need to apply smoothing and truncation to certain functions. This appendix collects the background and auxiliary results associated with such operations. First, we recall the construction of a standard mollifier
for some numerical constant . For each , we can rescale the mollifier as
so that the resulting function is supported on . For any piecewise-smooth function , we can obtain a smooth approximation by convolving it with a mollifier, i.e.,
A special case, frequently used in our proofs, is when is the indicator function defined on certain intervals. In particular, for , we define
as a smooth “window function”. It is easy to check that for , for , and for in the smooth “transition bands”. Moreover, it follows from (117) that .
Let be a function that is differentiable everywhere except at a finite number of points . If there is a function such that
where and is a smoothed window function as defined in (119). Moreover,
Let . For any , the function is differentiable on the interval . For such , we have
The desired inequality in (121) then follows from the simple observation that , which can be easily verified from the definition in (119).
The first inequality in (122) is obvious. To get the second inequality, we have
-B Auxiliary Results for the Proof of Theorem 2
The standard trick in a truncation method is to introduce two indicator functions defined on and , respectively. Since \big{|}\varphi\big{(}\tfrac{1}{\sqrt{p}}\boldsymbol{a}^{\mkern-1.5mu\mathsf{T}}\boldsymbol{\beta};\boldsymbol{g}^{\mkern-1.5mu\mathsf{T}}\boldsymbol{\xi}\big{)}-\widehat{\varphi}_{p}\big{(}\tfrac{1}{\sqrt{p}}\boldsymbol{a}^{\mkern-1.5mu\mathsf{T}}\boldsymbol{\beta};\boldsymbol{g}^{\mkern-1.5mu\mathsf{T}}\boldsymbol{\xi}\big{)}\big{|}\mathds{1}_{\mathcal{B}}\equiv 0,
where to reach (126) we have used Hölder’s inequality and the fact that . To bound the first term on the right-hand side of (126), we can use (66) and get
where the last inequality is obtained by using the moment estimate (157) in Lemma 8. Substituting (125) and (127) into (126), we can get (108). The steps leading to (109) are completely analogous to what we did to reach (108), so we omit the details here.
for all sufficiently large . [Without loss of generality, we should also assume that , as this is needed in the proof of an auxiliary result in Appendix -D.] On the other hand, by the construction of and the assumption in (66), we can easily verify that
where is the constant in (66). Then using the boundedness of given in (130) and defining as the indicator function supported on , we have
Next we prove (113). It follows from the definition in (111) that
where the last inequality uses the estimate given in Lemma 6 in Appendix -D. We now have
-C Proof of Proposition 3
be a smoothed version of the test function, where is the mollifier introduced in Appendix -A. The main idea of the proof is choosing a diminishing sequence of so that the left-hand side of (114) is well-approximated by a similar term involving the smooth function . To shorten notation, in what follows, we abbreviate \varphi_{p}\big{(}\tfrac{1}{\sqrt{p}}\boldsymbol{a}^{\mkern-1.5mu\mathsf{T}}\boldsymbol{\beta};\boldsymbol{g}^{\mkern-1.5mu\mathsf{T}}\boldsymbol{\xi}\big{)} and \varphi_{\delta_{p}}\big{(}\tfrac{1}{\sqrt{p}}\boldsymbol{a}^{\mkern-1.5mu\mathsf{T}}\boldsymbol{\beta};\boldsymbol{g}^{\mkern-1.5mu\mathsf{T}}\boldsymbol{\xi}\big{)} to and , respectively. The meaning of the notation and should also be clear. Since
we just need to bound the three terms on the right-hand side.
The first term can be controlled by Theorem 2, as is differentiable. By assumption, for some . Using the simple bound (122) in Lemma 4 (see Appendix -A), we can check that, for any ,
where is some numerical constant and . Theorem 2 then gives us
where we have simplified the term in (67) by using the additional assumption that and .
To control the second term on the right-hand side of (133), we apply Lemma 4 again. Using a shorthand notation \widehat{B}_{p}(\boldsymbol{a})=C^{\prime}B_{p}(s)(1+\big{\lvert}\tfrac{1}{\sqrt{p}}\boldsymbol{a}^{\mkern-1.5mu\mathsf{T}}\boldsymbol{\beta}\big{\rvert}^{K}), we have, from (121),
where in reaching the last step we have used the moment bound obtained in (127). The same reasoning also yields
Note that is a Gaussian random variable with zero mean and variance . (Recall the definition of in the statement of the proposition.) As the function with a compact support of width , we have
Substituting (138), (137), (135), (136), (134) into (133), and after some simplifications, we get
The convergence rate of the right-hand side can be optimized by setting . This then leads to the claim in (114).
-D Asymptotic Equivalence of the Covariance Matrices
Consider a sequence of activation functions such that, for every , is an odd function and
Suppose that the feature vectors satisfy (64) with some . We have
For the case of , we define .
Using (141), we can verify the following decomposition of :
Since , we must have
Recall the assumptions about the feature vectors in (64). It then follows from (139) that . Similarly, we also have . Controlling requires a few more steps. Let and .
It follows that . Substituting our bounds for and into (143), we then reach the bound (140) in the statement of the lemma. ∎
Next, we prove an auxiliary result that will be used in the proof of Theorem 2. Here, we consider a particular sequence of activation functions as defined in (110). They form a family of smoothly truncated versions of a fixed activation function .
Let and be the constants associated with and , respectively. If the threshold in (110) is chosen with a constant , then
By construction, and for . Let . We then have
Combining this bound with (146) and recall the definitions of and , we have . Finally, the second bound in (145) can be obtained from the following inequality: for any two nonnegative numbers and . ∎
-E Some Concentration Results
Let be the event defined in (22). There exists a constant such that
We start by stating the following simple result: for \boldsymbol{f}_{1},\boldsymbol{f}_{2}\overset{i.i.d.}{\sim}\mathcal{N}\Big{(}0,\tfrac{1}{d}\boldsymbol{I}_{d}\Big{)}, there exists positive constants and such that for any
Indeed, for any , and are both sub-Gaussian random variables with sub-Gaussian norm bounded by , for some [47, Example 2.5.8], so is a sub-exponential random variable with sub-exponential norm [47, Lemma 2.7.7]. Then we can apply Bernstein’s inequality [47, Corollary 2.8.3] to get (147). Also, (148) can be proved in the same way. Then we can let in (147) and (148) and use union bound to get for any ,
For any , we have . Thus, for any , the standard Gaussian tail bound gives us
By setting and applying union bound, we can obtain (151). Recall the definition of in (22). Combining (149), (150) and (151), we complete the proof. ∎
-E2 Concentration of Lipschitz Functions of Gaussian Vectors
The results presented in this section are all consequences of the following well-known theorem about the concentration of Lipschitz functions of independent Gaussian random variables. See e.g., [48, Theorem 1.3.4] for a proof.
It follows that the function is -Lipschitz continuous. Therefore, using (152) we have
To establish (156), we observe that can be represented as , where . It follows that can also be seen as a Lipschitz function of a standard normal vector, with a Lipschitz constant equal to . Therefore (156) is again a consequence of (152). Finally, the moment bounds in (157) and (158) can be obtained by applying (154). ∎
There exists such that for any and ,
Similarly, for any , we have
Correspondingly, there exists such that
Recall that and is a -Lipschitz continuous mapping. It follows that is a -Lipschitz continuous function. From (152), there exists such that for any ,
For any , we can use (165) and (164) to deduce that
The proof of (161) is analogous. We write , where . Therefore, similar to what we did to reach (164), we can show there exists such that for any ,
where the last step follows from the fact that is a -Lipschitz function of . Meanwhile,
It follows that, for any ,
Let be the admissible set of feature matrices defined in (21), and the leave-one-out Hessian matrix defined in (35). There exists such that, for every , and ,
Note that, conditioned on , is independent of for . For any ,
where step (a) follows from the fact that for (see Remark 2) and hence and step (b) follows from the concentration inequalities in (155) and (160).
To optimize the bound on the right-hand side of (170), we choose different values of according to . For , we let and get
For , we let , which gives us
Combining these two inequalities and using Assumption (A.6) that and the fact that for , we get (166). The proofs of (167)-(169) follow exactly the same procedure, and we omit them. ∎
-E3 The Spectral Norm of Random Matrices
We first recall a well-known result on the spectral norm of Gaussian random matrices, the proof of which can be found in [47, Corollary 7.3.3].
In particular, choosing gives us
Recall the definitions of and in (6) and (7), respectively. Next, we show that the spectral norms of \big{\|}\frac{1}{p}\sum_{t=1}^{n}\boldsymbol{a}_{t}\boldsymbol{a}_{t}^{\mkern-1.5mu\mathsf{T}}\big{\|} and \big{\|}\frac{1}{p}\sum_{t=1}^{n}\boldsymbol{b}_{t}\boldsymbol{b}_{t}^{\mkern-1.5mu\mathsf{T}}\big{\|} are bounded with high probability.
There exists some positive constant such that, for any fixed , the following holds.
for any , and
for any .
Let and be two fixed vectors with unit norms. For any , we have from (155) that
and thus is a sub-Gaussian random variable. Then by the independence of ,
where the last step follows from Hoeffding’s inequality for sub-Gaussian random variables [47, Theorem 2.6.3].
Next, we construct two -nets: on and on , with . It can be shown [47, Corollary 4.2.13] that the cardinality of and satisfies: and . Let be the matrix defined in (9). Its operator norm can be bounded as follows [47, Lemma 4.4.1]:
where to reach the second inequality we have used (175). Since , the desired inequality in (173) immediately follows if we choose . We omit the proof of (174) as it is completely analogous. ∎
-E4 Concentration of Quadratic Forms
for every , and . Correspondingly, there exists such that
Similarly, there exists and , such that
We first recall the definition of in (35). Since is -strongly convex, and for , , we must have and thus . (See Remark 2 for additional details.)
The concentration inequality (177) then directly follows from [8, Lemma 1] and the fact that for . To show (179), we note that . Thus, can be represented as , where . It follows that . Since , we have for . Applying the Hanson-Wright inequality (see, e.g., [47, Theorem 6.2.1]) then gives us the concentration inequality in (179).
By applying the inequalities in (153) and (154), we can obtain the moment bounds (178) and (180) from (177) and (179), respectively. ∎
Let or . We first show there exists such that
As is shown in (155), for any , is a sub-Gaussian variable, with a sub-Gaussian norm proportional to . It follows from (154) that
for some , where the last step is due to (23). Substituting this inequality into (184) and (183), we have verified (182) for .
Applying (178), (180) and (182), we reach the desired bounds in (181). ∎
-F Characterizations of the Optimization Problems
In this appendix, we collect some useful properties of the optimization problems that we encounter when constructing and analyzing the interpolation path based on Lindeberg’s method.
where is the function defined in (28), for , and for . Let
As explained in Section II-C, the optimization problems formulated in (188)-(190) can be referred to as the “original problem”, the “leave-one-out problem” and the “quadratic approximation problem”, respectively.
We first show that the quadratic approximation problem (190) allows for convenient closed-form solutions.
By the definition of Moreau envelopes, we immediately get (191). Besides, the optimal solution of (194) is
Since , we then get (193). Finally, by using the first order optimality condition , we can directly get (192). ∎
The next result is a deterministic bound for , i.e., the distance between the true optimal solution and the solution to the quadratic approximation problem.
For any and , there exists such that
We follow the proof technique of [29, Proposition 3.4]. For notational simplicity, we write and in the proof. We start by noting that, since is -strongly convex for , we have
where the first inequality is a property of strongly-convex functions (see, e.g., [49, pp. 112–113]), and the last equality is due to the optimality condition . Therefore, to prove (196), it suffices to control .
where in reaching the last step we have used the intermediate value theorem, with being some number that lies between and . From (192) and the definition of in (35), we have
Substituting this inequality into (199) then gives us
where is some number lying between and , and the last step follows from Assumption (A.4). From (201) and (202), there exists ,
where in the last step, we have used (192) and the assumption that . Substituting this inequality into (198) and using the fact that , we conclude the proof. ∎
We will introduce a function to wrap up all the terms in (186), except the loss function, i.e.,
where is defined in (28).
Let denote either or , and be the leave-one-out solution in (189). There exists such that for every ,
Recall the definition of the set in (23). We start by noting that
where the last step is due to the fact that is the optimal solution. On the other hand, for , is -strongly convex. This then gives us
Combining the above upper and lower bounds for , we have
By its definition in (203), , and thus
where and we have used (23). It then follows from (207) that
From Assumption (A.4), we know there exists some such that for any
for some . Combining (208) and (211) and choosing , we get
As this holds uniformly over all , we get (204) from (206). The proof of (205) follows exactly the same steps, and we omit it. ∎
where is the constant defined in Assumption (A.4).
Using the simple inequality for , we can deduce from (208) that
According to Assumption (A.4), there exists such that for any ,
Combining (214) and (216) gives us (212). The proof of (213) follows the same steps, and we omit it. ∎
Let be the optimal solution to the quadratic optimization problem as defined in (190). There exists such that for any and
We first show there exists such that for any ,
where is the leave-one-out solution defined in (189).
Note that is independent of . From (155), there exists such that for any , when conditioned on and ,
where is the same constant as the one in (205). Then it holds that
It then follows from (171), (205), (219) and Assumption (A.6) that there exists such that
for every and . The case of for (218) can be proved in the same way and we omit its proof.
Next, we show (217) by using the characterization in (193). Since
On the other hand, from Lemma 13 and Lemma 14, there exists such that for any
Then it follows from (222), (223), (224) and (218) that there exists such that for any ,
where and are two constants that depend on . Using (222), we have for ,
where is a constant that depend on and the last step follows from (181), (227) and Assumption (A.4).
Now we are ready to obtain (226). By Assumption (A.4), there exists such that for any ,
where are two constants that depend on . Then (226) can be proved by using (228) and standard moment bounds for . ∎
There exists such that for every and ,
For notational simplicity, we write and in the proof. and denote constants whose values can change from one line to the other. From (196), we have that
where or and denotes the th column of . Therefore, to show (230), it suffices to control each term on the right-hand side of (231).
(I) \tfrac{1}{p}\big{[}\sum_{i=1}^{p}(\boldsymbol{h}^{\mkern-1.5mu\mathsf{T}}_{\backslash k,i}\boldsymbol{r})^{4}\big{]}^{\tfrac{1}{2}}. Conditioned on , and hence , for any . Applying (155) and (156) and taking into account the independence between and , we can find a constant such that for any , and ,
for some . Therefore, there exists such that for any sufficiently large ,
where to reach the last step we have used (217) and the standard tails bound for Gaussian random variables , together with union bound. Then, by choosing a large enough , we can make (235) hold for any .
(III) . Notice that
From Lemma 12, we can then find two constants and such that
Moreover, as when , we have from Lemma 9 that
for some . Combining (237), (238), and using Lemma 11, we have
(IV) By Lemmas 11, 10 and the union bound, we have, for every ,
Substituting the bounds (233), (235), (239) and (240) into (231), we have for any ,
where constant does not depend on and . This completes our proof. ∎
From (196), there exists a function such that
where or . It follows that
Let be the optimal solution to the optimization problem defined in (27). There exists some such that for every and ,
The general strategy of our proof is as follows. To bound , we just need to show that any given coordinate of , e.g., its last entry, is bounded with high probability. By symmetry, all the coordinates have the same marginal distribution. Consequently, each coordinate of can be analyzed in the same way and can then be controlled by using the union bound.
where (with , independent of ) and . From (26), the th coordinate can be expressed as
The rest of the proof consists of two steps. First, we will show
Second, we show that (249) holds with high probability and that each term on the right-hand side of (248) is also bounded with high probability.
We start by proving the bound in (248). Let denote the objective function of in (247), i.e., . We first derive a lower bound for . To that end, we note from the convexity of the loss function that
Moreover, recall from Remark 2 that is -strongly convex when satisfies . It follows that . Furthermore, being -strongly convex gives us . Substituting these inequalities into (247) and using the first-order optimality condition of , we have
(I) Since , there exists such that
where the second equality is due to symmetry of . Similarly, when , we can get
By Lemma 19, there exists such that
for or . Moreover, there exists such that
where in reaching the last step we have used (230), Lemma 11 and Lemma 9. Substituting these two bounds into (254) and (255), we get there exists such that for every ,
On the other hand, since , we have
Recall the definition of in (246). We have
where , is the constant in (258), is the matrix of the latent input vectors in Assumption (A.1), and is some sufficiently large constant. Notice that is a high probability event. Indeed, from (171) and (258), there exists such that for every large enough,
Conditioned on any and in , the two terms of the right-hand side of (259) can be easily bounded. Specifically, let
Since is a set of i.i.d. standard normal random variables independent of , we have
Since is the last coordinate of the optimal weight vector, and since all the coordinates have the same distribution by symmetry, we get from the union bound that
Note that there exists such that for any , . We can get (245) by choosing to be the smallest number satisfying and . ∎
-F6 Proof of Proposition 2
We write as , where
To show (30), it suffices to show that each has high probability. Consider the following set of :
where is the constant in (245). From (245), we have
Let be the set defined in (23). From Lemma 18, we know there exists such that, for every , and ,
Therefore, for every , it holds that for ,
for every and . Finally, (30) can be obtained by applying the union bound.
-F7 Proof of Lemma 1
Recall the definitions of and in (188) and (190) of Appendix -F. The corresponding optimal solutions and are also defined in (188) and (190), respectively. We first show (36). Let or . It follows from (191) that
where in step (a) is some finite degree polynomial. To reach (a), we have used (229) and Lemma 8 and (b) follows from (213).
We now move on to showing (37). By applying Taylor expansion, in (186) can be written as
where is the Hessian matrix defined in (35), denotes some point that lies between and , with , and denotes some point that lies between and . By recalling the definition of in (187) and that of in (197), we have
for some constants , where the first step is obtained similar as (202).
Let . It is easy to verify that
This then allows us to apply (268) to get
-F8 Two Auxiliary Lemmas for Proving Theorem 1
Let and be the quantities defined in (45) and (46), respectively. It holds that and , uniformly over and .
To bound the right-hand side of (272), first note from (222) that
Thus, under Assumptions (A.4), there exists such that
where and and the first inequality follows from (229). From (272), there exists such that for any between and ,
Then using (181), (158) and Assumption (A.4), we can get
where is a finite degree polynomial. Therefore, for some between and ,
where is some constant. Here, (a) follows from (275); in (b), we use (180); in (c), we use (212).
The term can be bounded similarly. Following the same steps as above, we can show there exists some polynomial such that
for any between and . It follows that
where in the last step, we use Lemma 5 and the fact that for . Plugging (278), (178) and (213) into (277), we conclude that . ∎
Combining (280), (282) with Assumption (A.4) allows us to show that \mathcal{M}_{k}\big{(}x;\gamma_{k}\big{)} satisfies (279). Indeed, similar as (229), we can get
for some . Then from (280) and (283), there exists such that
Similarly, there exist such that