Non-strongly-convex smooth stochastic approximation with convergence rate O(1/n)

Introduction

Large-scale machine learning problems are becoming ubiquitous in many areas of science and engineering. Faced with large amounts of data, practitioners typically prefer algorithms that process each observation only once, or a few times. Stochastic approximation algorithms such as stochastic gradient descent (SGD) and its variants, although introduced more than 60 years ago , still remain the most widely used and studied method in this context (see, e.g., ).

We partially understand the properties of $f$ that affect the problem difficulty. Strong convexity (i.e., when $f$ is twice differentiable, a uniform strictly positive lower-bound $\mu$ on Hessians of $f$) is a key property. Indeed, after $n$ observations and with the proper step-sizes, averaged SGD achieves the rate of $O(1/\mu n)$ in the strongly-convex case , while it achieves only $O(1/\sqrt{n})$ in the non-strongly-convex case , with matching lower-bounds .

The main issue with strong convexity is that typical machine learning problems are high dimensional and have correlated variables so that the strong convexity constant $\mu$ is zero or very close to zero, and in any case smaller than $O(1/\sqrt{n})$. This then makes the non-strongly convex methods better. In this paper, we aim at obtaining algorithms that may deal with arbitrarily small strong-convexity constants, but still achieve a rate of $O(1/n)$.

Smoothness plays a central role in the context of deterministic optimization. The known convergence rates for smooth optimization are better than for non-smooth optimization (e.g., see ). However, for stochastic optimization the use of smoothness only leads to improvements on constants (e.g., see ) but not on the rate itself, which remains $O(1/\sqrt{n})$ for non-strongly-convex problems.

We show that for the square loss and for the logistic loss, we may use the smoothness of the loss and obtain algorithms that have a convergence rate of $O(1/n)$ without any strong convexity assumptions. More precisely, for least-squares regression, we show in Section 2 that averaged stochastic gradient descent with constant step-size achieves the desired rate. For logistic regression this is achieved by a novel stochastic gradient algorithm that (a) constructs successive local quadratic approximations of the loss functions, while (b) preserving the same running time complexity as stochastic gradient descent (see Section 3). For these algorithms, we provide a non-asymptotic analysis of their generalization error (in expectation, and also in high probability for least-squares), and run extensive experiments on standard machine learning benchmarks showing in Section 4 that they often outperform existing approaches.

Constant-step-size least-mean-square algorithm

In this section, we consider stochastic approximation for least-squares regression, where SGD is often referred to as the least-mean-square (LMS) algorithm. The novelty of our convergence result is the use of the constant step-size with averaging, leading to $O(1/n)$ rate without strong convexity.

${\mathcal{H}}$ is a $d$-dimensional Euclidean space, with $d\geqslant 1$.

The observations $(x_{n},z_{n})\in{\mathcal{H}}\times{\mathcal{H}}$ are independent and identically distributed.

We study the stochastic gradient (a.k.a. least mean square) recursion defined as

started from $\theta_{0}\in{\mathcal{H}}$. We also consider the averaged iterates $\bar{\theta}_{n}=(n+1)^{-1}\sum_{k=0}^{n}\theta_{k}$.

Assume (A1-6). For any constant step-size $\gamma<\frac{1}{R^{2}}$, we have

Proof technique.

We adapt and extend a proof technique from which is based on non-asymptotic expansions in powers of $\gamma$. We also use a result from which studied the recursion in Eq. (1), with $x_{n}\otimes x_{n}$ replaced by its expectation $H$. See the appendix for details.

Optimality of bounds.

Our bound in Eq. (2) leads to a rate of $O(1/n)$, which is known to be optimal for least-squares regression (i.e., under reasonable assumptions, no algorithm, even more complex than averaged SGD can have a better dependence in $n$) . The term $\sigma^{2}d/n$ is also unimprovable.

Initial conditions.

If $\gamma$ is small, then the initial condition is forgotten more slowly. Note that with additional strong convexity assumptions, the initial condition would be forgotten faster (exponentially fast without averaging), which is one of the traditional uses of constant-step-size LMS .

Specificity of constant step-sizes.

2 Convergence in higher orders

We are now going to consider an extra assumption in order to bound the $p$-th moment of the excess risk and then get a high-probability bound. Let $p$ be a real number greater than $1$.

There exists $R>0$, $\kappa>0$ and $\tau\geqslant\sigma>0$ such that, for all $n\geqslant 1$, $\|x_{n}\|^{2}\leqslant R^{2}$ a.s., and

The last condition in Eq. (4) says that the kurtosis of the projection of the covariates $x_{n}$ on any direction $z\in{\mathcal{H}}$ is bounded. Note that computing the constant $\kappa$ happens to be equivalent to the optimization problem solved by the FastICA algorithm , which thus provides an estimate of $\kappa$. In Table 1, we provide such an estimate for the non-sparse datasets which we have used in experiments, while we consider only directions $z$ along the axes for high-dimensional sparse datasets. For these datasets where a given variable is equal to zero except for a few observations, $\kappa$ is typically quite large. Adapting and analyzing normalized LMS techniques to this set-up is likely to improve the theoretical robustness of the algorithm (but note that results in expectation from Theorem 1 do not use $\kappa$). The next theorem provides a bound for the $p$-th moment of the excess risk.

Assume (A1-7). For any real $p\geqslant 1$, and for a step-size $\gamma\leqslant 1/(12p\kappa R^{2})$, we have:

Note that to control the $p$-th order moment, a smaller step-size is needed, which scales as $1/p$. We can now provide a high-probability bound; the tails decay polynomially as $1/(n\delta^{12\gamma\kappa R^{2}})$ and the smaller the step-size $\gamma$, the lighter the tails.

For any step-size such that $\gamma\leqslant 1/(12\kappa R^{2})$, any $\delta\in(0,1)$,

Beyond least-squares: M-estimation

Although this algorithm has shown to be the most efficient in practice (see Section 4) we currently have no proof of convergence. Given that the behavior of the algorithms does not change much when the support point is updated less frequently than each iteration, there may be some connections to two-time-scale algorithms (see, e.g., ). In Section 4, we also consider several other strategies based on doubling tricks.

Interestingly, for non-quadratic functions, our algorithm imposes a new bias (by replacing the true gradient by an approximation which is only valid close to $\bar{\theta}_{n-1}$) in order to reach faster convergence (due to the linearity of the underlying gradients).

Relationship with one-step-estimators.

One-step estimators (see, e.g., ) typically take any estimator with $O(1/n)$-convergence rate, and make a full Newton step to obtain an efficient estimator (i.e., one that achieves the Cramer-Rao lower bound). Although our novel algorithm is largely inspired by one-step estimators, our situation is slightly different since our first estimator has only convergence rate $O(n^{-1/2})$ and is estimated on different observations.

1 Self-concordance and logistic regression

${\mathcal{H}}$ is a $d$-dimensional Euclidean space, with $d\geqslant 1$.

The observations $(x_{n},y_{n})\in{\mathcal{H}}\times\{-1,1\}$ are independent and identically distributed.

We denote by $\theta_{\ast}$ a global minimizer of $f$, which we thus assume to exist, and we denote by $H=f^{\prime\prime}(\theta_{\ast})$ the Hessian operator at a global optimum $\theta_{\ast}$.

We assume that there exists $R>0$, $\kappa>0$ and $\rho>0$ such that $\|x_{n}\|^{2}\leqslant R^{2}\mbox{ almost surely}$, and

We get an $O(1/n)$ convergence rate without assuming strong convexity, even locally, thus improving on results from where the the rate is proportional to $1/(n\lambda_{\min}(H))$. The proof relies on self-concordance properties and the sharp analysis of the Newton step (see appendix).

Experiments

We consider normally distributed inputs, with covariance matrix $H$ that has random eigenvectors and eigenvalues $1/k$, $k=1,\dots,d$. The outputs are generated from a linear function with homoscedastic noise with unit signal to noise-ratio. We consider $d=20$ and the least-mean-square algorithm with several settings of the step size $\gamma_{n}$, constant or proportional to $1/\sqrt{n}$. Here $R^{2}$ denotes the average radius of the data, i.e., $R^{2}=\mathop{\rm tr}H$. In the left plot of Figure 1, we show the results, averaged over 10 replications.

Without averaging, the algorithm with constant step-size does not converge pointwise (it oscillates), and its average excess risk decays as a linear function of $\gamma$ (indeed, the gap between each values of the constant step-size is close to $\log_{10}(4)$, which corresponds to a linear function in $\gamma$).

With averaging, the algorithm with constant step-size does converge at rate $O(1/n)$, and for all values of the constant $\gamma$, the rate is actually the same. Moreover (although it is not shown in the plots), the standard deviation is much lower.

With decaying step-size $\gamma_{n}=1/(2R^{2}\sqrt{n})$ and without averaging, the convergence rate is $O(1/\sqrt{n})$, and improves to $O(1/n)$ with averaging.

Logistic regression.

We consider the same input data as for least-squares, but now generates outputs from the logistic probabilistic model. We compare several algorithms and display the results in Figure 1 (middle and right plots).

On the middle plot, we consider SGD. Without averaging, the algorithm with constant step-size does not converge and its average excess risk reaches a constant value which is a linear function of $\gamma$ (indeed, the gap between each values of the constant step-size is close to $\log_{10}(4)$). With averaging, the algorithm does converge, but as opposed to least-squares, to a point which is not the optimal solution, with an error proportional to $\gamma^{2}$ (the gap between curves is twice as large).

On the right plot, we consider various variations of our Newton-approximation scheme. The “2-step” algorithm is the one for which our convergence rate holds ($n$ being the total number of examples, we perform $n/2$ steps of averaged SGD, then $n/2$ steps of LMS). Not surprisingly, it is not the best in practice (in particular at $n/2$, when starting the constant-size LMS, the performance worsens temporarily). It is classical to use doubling tricks to remedy this problem while preserving convergence rates , this is done in “2-step-dbl.”, which avoids the previous erratic behavior.

We have also considered getting rid of the first stage where plain averaged stochastic gradient is used to obtain a support point for the quadratic approximation. We now consider only Newton-steps but change only these support points. We consider updating the support point at every iteration, i.e., the recursion from Eq. (8), while we also consider updating it every dyadic point (“dbl.-approx”). The last two algorithms perform very similarly and achieve the $O(1/n)$ early. In all experiments on real data, we have considered the simplest variant (which corresponds to Eq. (8)).

2 Standard benchmarks

We have considered 6 benchmark datasets which are often used in comparing large-scale optimization methods. The datasets are described in Table 1 and vary in values of $d$, $n$ and sparsity levels. These are all finite binary classification datasets with outputs in $\{-1,1\}$. For least-squares and logistic regression, we have followed the following experimental protocol: (1) remove all outliers (i.e., sample points $x_{n}$ whose norm is greater than 5 times the average norm), (2) divide the dataset in two equal parts, one for training, one for testing, (3) sample within the training dataset with replacement, for 100 times the number of observations in the training set (this corresponds to $100$ effective passes; in all plots, a black dashed line marks the first effective pass), (4) compute averaged cost on training and testing data (based on 10 replications). All the costs are shown in log-scale, normalized to that the first iteration leads to $f(\theta_{0})-f(\theta_{\ast})=1$.

All algorithms that we consider (ours and others) have a step-size, and typically a theoretical value that ensures convergence. We consider two settings: (1) one when this theoretical value is used, (2) one with the best testing error after one effective pass through the data (testing powers of $4$ times the theoretical step-size).

Here, we only consider covertype, alpha, sido and news, as well as test errors. For all training errors and the two other datasets (quantum, rcv1), see the appendix.

We compare three algorithms: averaged SGD with constant step-size, averaged SGD with step-size decaying as $C/R^{2}\sqrt{n}$, and the stochastic averaged gradient (SAG) method which is dedicated to finite training data sets , which has shown state-of-the-art performance in this set-upThe original algorithm from is considering only strongly convex problems, we have used the step-size of $1/16R^{2}$, which achieves fast convergence rates in all situations (see \urlhttp://research.microsoft.com/en-us/um/cambridge/events/mls2013/downloads/stochastic_gradient.pdf).. We show the results in the two left plots of Figure 2 and Figure 3.

Averaged SGD with decaying step-size equal to $C/R^{2}\sqrt{n}$ is slowest (except for sido). In particular, when the best constant $C$ is used (right columns), the performance typically starts to increase significantly. With that step size, even after 100 passes, there is no sign of overfitting, even for the high-dimensional sparse datasets.

SAG and constant-step-size averaged SGD exhibit the best behavior, for the theoretical step-sizes and the best constants, with a significant advantage for constant-step-size SGD. The non-sparse datasets do not lead to overfitting, even close to the global optimum of the (unregularized) training objectives, while the sparse datasets do exhibit some overfitting after more than 10 passes.

Logistic regression.

We also compare two additional algorithms: our Newton-based technique and “Adagrad” , which is a stochastic gradient method with a form a diagonal scalingSince a bound on $\|\theta_{\ast}\|$ is not available, we have used step-sizes proportional to $1/\sup_{n}\|x_{n}\|_{\infty}$. that allows to reduce the convergence rate (which is still in theory proportional to $O(1/\sqrt{n})$). We show results in the two right plots of Figure 2 and Figure 3.

Averaged SGD with decaying step-size proportional to $1/R^{2}\sqrt{n}$ has the same behavior than for least-squares (step-size harder to tune, always inferior performance except for sido).

SAG, constant-step-size SGD and the novel Newton technique tend to behave similarly (good with theoretical step-size, always among the best methods). They differ notably in some aspects: (1) SAG converges quicker for the training errors (shown in the appendix) while it is a bit slower for the testing error, (2) in some instances, constant-step-size averaged SGD does underfit (covertype, alpha, news), which is consistent with the lack of convergence to the global optimum mentioned earlier, (3) the novel Newton approximation is consistently better.

On the non-sparse datasets, Adagrad performs similarly to the Newton-type method (often better in early iterations and worse later), except for the alpha dataset where the step-size is harder to tune (the best step-size tends to have early iterations that make the cost go up significantly). On sparse datasets like rcv1, the performance is essentially the same as Newton. On the sido data set, Adagrad (with fixed steps size, left column) achieves a good testing loss quickly then levels off, for reasons we cannot explain. On the news dataset, it is inferior without parameter-tuning and a bit better with. Adagrad uses a diagonal rescaling; it could be combined with our technique, early experiments show that it improves results but that it is more sensitive to the choice of step-size.

Overall, even with $d$ and $\kappa$ very large (where our bounds are vacuous), the performance of our algorithm still achieves the state of the art, while being more robust to the selection of the step-size: finer quantities likes degrees of freedom should be able to quantify more accurately the quality of the new algorithms.

Conclusion

In this paper, we have presented two stochastic approximation algorithms that can achieve rates of $O(1/n)$ for logistic and least-squares regression, without strong-convexity assumptions. Our analysis reinforces the key role of averaging in obtaining fast rates, in particular with large step-sizes. Our work can naturally be extended in several ways: (a) an analysis of the algorithm that updates the support point of the quadratic approximation at every iteration, (b) proximal extensions (easy to implement, but potentially harder to analyze); (c) adaptive ways to find the constant-step-size; (d) step-sizes that depend on the iterates to increase robustness, like in normalized LMS , and (e) non-parametric analysis to improve our theoretical results for large values of $d$.

This work was partially supported by the European Research Council (SIERRA Project 239993). The authors would like to thank Simon Lacoste-Julien and Mark Schmidt for discussions related to this work.

In the appendix we provide proofs of all three theorems, as well as additional experimental results (all training objectives, and two additional datasets quantum and rcv1).

Appendix A Proof of Theorem 1

We first denote by $\eta_{n}=\theta_{n}-\theta_{\ast}\in{\mathcal{H}}$ the deviation to $\theta_{\ast}$. Since we consider quadratic functions, it satisfies a simplified recursion:

We also consider $\bar{\eta}_{n}=\frac{1}{n+1}\sum_{k=0}^{n}\eta_{k}=\bar{\theta}_{n}-\theta_{\ast}$ the averaged iterate. We have $f(\theta_{n})-f(\theta_{\ast})=\frac{1}{2}\langle\eta_{n},H\eta_{n}\rangle$ and $f(\bar{\theta}_{n})-f(\theta_{\ast})=\frac{1}{2}\langle\bar{\eta}_{n},H\eta_{n}\rangle$.

The crux of the proof is to consider the same recursion as Eq. (12), but replacing $x_{n}\otimes x_{n}$ by its expectation $H$ (which is related to fixed design analysis in linear regression). This is of course only an approximation, and thus one has to study the remainder term; it happens to satisfy a similar recursion, on which we can apply the same technique, and so on. This proof technique is taken from . Here we push it to arbitrary orders with explicit constants for averaged constant-step-size stochastic gradient descent.

We denote by ${\mathcal{F}}_{n}$ the $\sigma$-algebra generated by $(x_{1},z_{1},\dots,x_{n},z_{n})$. Both $\theta_{n}$ and $\bar{\theta}_{n}$ are ${\mathcal{F}}_{n}$-measurable.

A.1 Two main lemmas

The proof relies on two lemmas, one that provides a weak result essentially equivalent (but more specific and simpler because the step-size is constant) to non-strongly-convex results from , and one that replaces $x_{n}\otimes x_{n}$ by its expectation $H$ in Eq. (12), which may then be seen as a non-asymptotic counterpart to the similar set-tup in .

Proof. We follow the proof technique of (which relies only on smoothness) and get:

This leads to the desired result, because, by convexity, $\langle\bar{\alpha}_{n-1},H\bar{\alpha}_{n-1}\rangle\leqslant\frac{1}{n}\sum_{k=0}^{n-1}\langle{\alpha}_{k},H{\alpha}_{k}\rangle$.

Proof. The proof relies on the fact that cost functions are quadratic and our recursions are thus linear, allowing to obtain $\alpha_{n}$ in closed form. The sequence $(\alpha_{n})$ satisfies a linear recursion, from which we get, for all $n\geqslant 1$:

which leads to the first result using classical martingale second moment expansions (which amount to considering $\xi_{i}$, $i=1,\dots,n$ independent, so that the variance of the sum is the sum of variances). Moreover, using the identity \sum_{k=0}^{n-1}(I-\gamma H)^{k}=\big{(}I-(I-\gamma H)^{n}\big{)}\big{(}\gamma H\big{)}^{-1}, we get:

We then get, using standard martingale square moment inequalities (which here also amount to considering $\xi_{i}$, $i=1,\dots,n$ independent, so that the variance of the sum is the sum of variances):

because for all $u\in$, $\frac{(1-(1-u)^{n})^{2}}{nu}\leqslant 1$ (see Lemma 3 in Section A.6), and the second term is the sum of terms which are all less than $\mathop{\rm tr}H^{-1}C$.

Note that we may replace the term $\frac{1}{n\gamma}\|\alpha_{0}\|^{2}$ by $\displaystyle\frac{1}{n^{2}\gamma^{2}}\langle\alpha_{0},H^{-1}\alpha_{0}\rangle$, which is only interesting when $\langle\alpha_{0},H^{-1}\alpha_{0}\rangle$ is small.

A.2 Proof principle

The proof relies on an expansion of $\eta_{n}$ and $\bar{\eta}_{n-1}$ as polynomials in $\gamma$ due to . This expansion is done separately for the noise process (i.e., when assuming $\eta_{0}=0$) and for the noise-free process that depends only on the initial conditions (i.e., when assuming that $\sigma=0$). The bounds may then be added.

Indeed, we have $\eta_{n}=M_{1}^{n}\eta_{0}+\gamma\sum_{k=1}^{n}M_{k+1}^{n}\xi_{k}$, with $M_{i}^{j}=(I-\gamma x_{j}\otimes x_{j})\cdots(I-\gamma x_{i}\otimes x_{i})$ and $M_{i}^{i-1}=I$, and thus \bar{\eta}_{n}=\frac{1}{n+1}\sum_{i=0}^{n}\bigg{[}M_{1}^{i}\eta_{0}+\gamma\sum_{k=1}^{i}M_{k+1}^{i}\xi_{k}\bigg{]}=\frac{1}{n+1}\sum_{i=0}^{n}M_{1}^{i}\eta_{0}+\gamma\sum_{k=1}^{n}\bigg{(}\sum_{i=k}^{n}M_{k+1}^{i}\bigg{)}\xi_{k}, leading to

for any $p\geqslant 2$ for which it is defined: the left term depends only on initial conditions and the right term depends only on the noise process (note the similarity with bias-variance decompositions).

A.3 Initial conditions

In this section, we assume that $\xi_{n}$ is uniformly equal to zero, and that $\gamma R^{2}\leqslant 1$.

We thus have $\eta_{n}=(I-\gamma x_{n}\otimes x_{n})\eta_{n-1}$ and thus

By taking expectations (first given ${\mathcal{F}}_{n-1}$, then unconditionally), we get:

from which we obtain, by summing from $1$ to $n$ and using convexity (note that Lemma 1 could be used directly as well):

Here, it would be interesting to explore conditions under which the initial conditions may be forgotten at a rate $O(1/n^{2})$, as obtained by in the strongly convex case.

A.4 Noise process

In this section, we assume that $\eta_{0}=\theta_{0}-\theta_{\ast}=0$ and $\gamma R^{2}\leqslant 1$ (which implies $\gamma H\preccurlyeq I$). Following , we recursively define the sequences $(\eta_{n}^{r})_{n\geqslant 0}$ for $r\geqslant 0$ (and their averaged counterparts $\bar{\eta}_{n}^{r}$):

The sequence $(\eta_{n}^{0})$ is defined as $\eta^{0}_{0}=\eta_{0}=0$ and for $n\geqslant 1$, $\eta^{0}_{n}=(I-\gamma H)\eta^{0}_{n-1}+\gamma\xi_{n}$.

The sequence $(\eta_{n}^{r})$ is defined from $(\eta_{n}^{r-1})$ as $\eta_{0}^{r}=0$ and, for all $n\geqslant 1$:

We now show that the sequence $\eta_{n}-\sum_{i=0}^{r}\eta_{n}^{i}$ then satisfies the following recursion, for any $r\geqslant 0$ (which is of the same type than $(\eta_{n})$):

In order to prove Eq. (16) by recursion, we have, for $r=0$,

Bound on covariance operators.

We now show that we also have a bound on the covariance operator of $\eta_{n-1}^{r}$, for any $r\geqslant 0$ and $n\geqslant 2$:

In order to prove Eq. (17) by recursion, we get for $r=0$:

In order to go from $r$ to $r+1$, we have, using Lemma 2 and the fact that $\eta_{k-1}^{r}$ and $x_{k}$ are independent:

Putting things together.

We may apply Lemma 1 to the sequence \big{(}\eta_{n}-\sum_{i=0}^{r}\eta_{n}^{i}\big{)}, to get

We may now apply Lemma 2 to Eq. (15), to get, with a noise process $\xi_{n}^{r}=(H-x_{n}\otimes x_{n})\eta_{n-1}^{r-1}$ which is such that

We thus get, using Minkowski’s inequality (i.e., triangle inequality for the norms $\|\cdot\|_{p}$):

This implies that for any $\gamma R^{2}<1$, we obtain, by letting $r$ tend to $+\infty$:

A.5 Final bound

We can now take results from Appendices A.3 and A.4, to get

A.6 Proof of Lemma 3

In this section, we state and prove a simple lemma.

For any $u\in$ and $n>0$, ${(1-(1-u)^{n})^{2}}{}\leqslant nu$.

Proof. Since $u\in$, we have, $1-(1-u)^{n}\leqslant 1$. Moreover, $n(1-u)^{n-1}\leqslant n$, and by integrating between and $u$, we get $1-(1-u)^{n}\leqslant nu$. By multiplying the two previous inequalities, we get the desired result.

Appendix B Proof of Theorem 2

We use the same notations than the proof of Theorem 1, and the same proof principle: (a) splitting the contributions of the initial conditions and the noise, (b) providing a direct argument for the initial condition, and (c) performing an expansion for the noise contribution.

B.1 Contribution of initial conditions

When the noise is assumed to be zero, we have $\eta_{n}=(I-\gamma x_{n}\otimes x_{n})\eta_{n-1}$ almost surely, and thus, since $0\preccurlyeq\gamma x_{n}\otimes x_{n}\preccurlyeq I$, $\|\eta_{n}\|\leqslant\|\eta_{0}\|$ almost surely, and

and $|M_{n}|\leqslant\gamma\|\eta_{0}\|^{2}R^{2}$. We may now apply the Burkholder-Rosenthal-Pinelis inequality in Eq. (B), to get:

We have used above that (a) $\sum_{k=1}^{n}\langle\eta_{k-1},H\eta_{k-1}\rangle\leqslant\frac{A_{n}}{\gamma}$ and that (b) \big{\|}A_{n}\big{\|}_{p/2}\leqslant\big{\|}A_{n}\big{\|}_{p}. This leads to

Finally, we obtain, for any $p\geqslant 2$

i.e., by a change of variable $p\rightarrow\frac{p}{2}$, for any $p\geqslant 4$, we get

By using monotonicity of norms, we get, for any $p\in$:

Note that the constants in the bound above could be improved by using a proof by recursion.

B.2 Contribution of the noise

We follow the same proof technique than for Theorem 1 and consider the expansion based on the sequences $(\eta_{n}^{r})_{n}$, for $r\geqslant 0$. We need (a) bounds on $\eta_{n}^{0}$, (b) a recursion on the magnitude (in $\|\cdot\|_{p}$ norm) of $\eta_{n}^{r}$ and (c) a control of the error made in the expansions.

We start by a lemma similar to Lemma 2 but for all moments. This will show a bound for the sequence $\bar{\eta}_{n}^{0}$.

Proof. We have, from the proof of Lemma 2:

with \beta_{j}=\big{(}{I-(I-\gamma H)^{n-j}}\big{)}\big{(}{\gamma H}\big{)}^{-1}H^{1/2}\xi_{j}. We have

using Lemma 3 in Section A.6, and assumption (A7).

Using Burkholder-Rosenthal-Pinelis inequality in Eq. (B), we then obtain

Following the same proof technique as above, we have

from which we get, for any positive semidefinite operator $M$ such that $\mathop{\rm tr}M=1$, using BRP’s inequality:

We introduce the following quantity to control the deviations of $\eta_{n}^{r}$:

We have from Eq. (20), $A_{0}\leqslant\frac{1}{R}\sqrt{p\gamma R^{2}}(\sigma+\tau\sqrt{p\gamma R^{2}})$.

Since $\eta_{n}^{r}=(I-\gamma H)\eta_{n-1}^{r}+\gamma(H-x_{n}\otimes x_{n})\eta_{n-1}^{r-1}$, for all $n\geqslant 1$, we have the closed form expression

and we may use BRP’s inequality in Eq. (B) to get, for any $M$ such that $\mathop{\rm tr}M=1$:

This implies that A_{r}\leqslant A_{0}\big{(}\sqrt{p\gamma R^{2}\kappa}+p\gamma R^{2}+p(\gamma R^{2})^{1-1/p}\big{)}^{r}, which in turn implies, from Eq. (20),

The condition on $\gamma$ will come from the requirement that $\sqrt{p\gamma R^{2}\kappa}+p\gamma R^{2}+p(\gamma R^{2})^{1-1/p}<1$.

leading to, using BRP’s inequality in Eq. (B), similar arguments than in the previous bounds, \big{(}{I-(I-\gamma H)^{n-j}}\big{)}^{2}\big{(}{\gamma H}\big{)}^{-2}H\preccurlyeq H^{-1}\gamma and \big{(}{I-(I-\gamma H)^{n-j}}\big{)}^{2}\big{(}{\gamma H}\big{)}^{-2}H\preccurlyeq\frac{n}{\gamma}I:

We may then impose a restriction on $\gamma R^{2}$, i.e., $\gamma R^{2}\leqslant\frac{1}{\alpha\kappa p}$ with $\alpha>1$. We then have

With $\alpha=12$, we obtain a bound of $0.781\leqslant\frac{8}{10}$ above.

From Eq. (16) and the fact that $0\preccurlyeq I-\gamma x_{n}\otimes x_{n}\preccurlyeq I$ almost surely, we get:

Putting things together.

We get by combining Lemma 4 with Eq. (22) and Eq. (23) and then letting $r$ tends to infinity,

B.3 Final bound

For $\gamma\leqslant\frac{1}{12\kappa pR^{2}}$, we obtain, from the last equations of Section B.1 and Section B.2,

Moreover, when $\gamma=\frac{1}{12\kappa pR^{2}}$, we have:

B.4 Proof of Corollary 1

We have from the previous proposition, for $\gamma\leqslant\frac{1}{12\kappa pR^{2}}$:

with $\eta=12\gamma\kappa R^{2}\leqslant 1/p$, and $\square=7\tau\sqrt{d}+R\|\theta_{0}-\theta_{\ast}\|\sqrt{3}$ and $\triangle=R\|\theta_{0}-\theta_{\ast}\|\sqrt{24\kappa}$.

This leads to, using Markov’s inequality:

Thus the large deviations decay as power of $t$, with a power that decays as $1/(12\gamma\kappa R^{2})$. If $\gamma$ is small, the deviations are lighter.

In order to get the desired result, we simply take $t=\frac{1}{12\gamma\kappa R^{2}}\delta^{-12\kappa\gamma R^{2}}$.

Appendix C Proof of Theorem 3

In terms of convergence rates, $\theta_{1}$ will be $(1/\sqrt{n})$-optimal, while $\theta_{3}$ will be $(1/n)$-optimal for the quadratic problem because of previous results on averaged LMS. A classical property is that a single Newton step squares the error. Therefore, the full Newton step should have an error which is the square of the one of $\theta_{1}$, i.e., $O(1/n)$. Overall, since $\theta_{3}$ approaches the full Newton step with rate $O(1/n)$, this makes a bound of $O(1/n)$.

In Section C.1, we provide a general deterministic result on the Newton step, while in Section C.2, we combine with two stochastic approximation results, making the informal reasoning above more precise.

In this section, we study the effect of an approximate Newton step. We consider $\theta_{1}\in{\mathcal{H}}$, the Newton iterate $\theta_{2}=\theta_{1}-f^{\prime\prime}(\theta_{1})^{-1}f^{\prime}(\theta_{1})$, and an approximation $\theta_{3}$ of $\theta_{2}$. In the next proposition, we provide a bound on $f(\theta_{3})-f(\theta_{\ast})$, under different conditions, whether $\theta_{1}$ is close to optimal for $f$, and/or $\theta_{3}$ is close to optimal for the quadratic approximation around $\theta_{1}$ (i.e., close to $\theta_{2}$). Eq. (25) corresponds to the least-favorable situations where both errors are small, while Eq. (26) and Eq. (27) consider cases where $\theta_{1}$ is sufficiently good. See proof in Section E.3. These three cases are necessary for the probabilistic control.

Assume (B3-4), and $\theta_{1},\theta_{2},\theta_{3}\in{\mathcal{H}}$ such that $f(\theta_{1})-f(\theta_{\ast})=\varepsilon_{1}$, $\theta_{2}=\theta_{1}-f^{\prime\prime}(\theta_{1})^{-1}f^{\prime}(\theta_{1})$ and $\frac{1}{2}\langle\theta_{3}-\theta_{2},f^{\prime\prime}(\theta_{1})(\theta_{3}-\theta_{2})\rangle=\varepsilon_{2}$. Then, if $t^{2}=\varepsilon_{1}\kappa\rho$,

If $t=\sqrt{\varepsilon_{1}\kappa\rho}\leqslant 1/16$, then

Moreover, if $t=\sqrt{\varepsilon_{1}\kappa\rho}\leqslant 1/16$ and $\varepsilon_{2}\kappa\rho\leqslant 1/16$, then

Note that in the favorable situation in Eq. (26), we get error of the form $O(\varepsilon_{1}^{2}+\varepsilon_{2})$. It essentially suffices now to show that in our set-up, in a probabilistic sense to be determined, $\varepsilon_{1}=O(1/\sqrt{n})$ and $\varepsilon_{2}=O(1/{n})$, while controlling the unfavorable situations.

C.2 Stochastic analysis

We consider the following two-step algorithm:

Starting from any initialization $\theta_{0}$, run $n$ iterations of averaged stochastic gradient descent to get $\theta_{1}$,

Run from $\theta_{1}$ $n$ steps of LMS on the quadratic approximation around $\theta_{1}$, to get $\theta_{3}$, which is an approximation of the Newton step $\theta_{2}$.

We denote by $\mathcal{G}_{1}$ the $\sigma$-field generated by the first $n$ observations (the ones used to define $\theta_{1}$). We have, by separating all events, i.e., using $1=1_{A_{1}}1_{A_{2}}+1_{A_{1}}1_{A_{2}^{\sf c}}+1_{A_{1}^{\sf c}}$:

We now need to control $\varepsilon_{2}$, i.e., the error made by the LMS algorithm started from $\theta_{1}$.

We consider the quadratic approximation around $\theta_{1}\in{\mathcal{H}}$, and write is as an expectation, i.e.,

$\|x_{n}\|^{2}\leqslant R^{2}/4$ almost surely.

We may thus apply the previous results, i.e., Theorem 1, to obtain with the LMS algorithm a $\theta_{3}\in{\mathcal{H}}$ such that, with $\gamma=\frac{1}{R^{2}}$:

which are such that $\triangle_{2}(t)\leqslant 0.6$ and $\triangle_{3}(t)\leqslant 1.2$ if $t\leqslant 1/16$.

The last inequalities may be checked graphically.

We now need to use bounds on the behavior of the first $n$ steps of regular averaged stochastic gradient descent.

Fine results on averaged stochastic gradient descent.

In order to get error bounds on $\theta_{1}$, we run $n$ steps of averaged stochastic gradient descent with constant-step size $\gamma=1/(2R^{2}\sqrt{n})$. We need the following bounds from [21, Appendix E and Prop. 1]:

However, we need a finer control of the deviations in order to bound quantities of the form $e^{\alpha\varepsilon_{1}}$. In Section D, extending results from , we show in Prop. 3 that if $\frac{\alpha(10+2R^{2}\|\theta_{0}-\theta_{\ast}\|^{2})}{\sqrt{n}}\leqslant\frac{1}{2e}$, then $Ee^{\alpha(f(\theta_{1})-f(\theta_{\ast}))}\leqslant 1$.

Putting things together.

From Eq. (30), we then get, if $\frac{6(10+2R^{2}\|\theta_{0}-\theta_{\ast}\|^{2})}{\sqrt{n}}\leqslant\frac{1}{2e}$:

Using the notation $D=(R^{2}\|\theta_{0}-\theta_{\ast}\|^{2}+\frac{3}{2})$, we obtain:

The condition $\frac{6(10+2R^{2}\|\theta_{0}-\theta_{\ast}\|^{2})}{\sqrt{n}}\leqslant\frac{1}{2e}$ is implied by $n\geqslant(19+9R\|\theta_{0}-\theta_{\ast}\|)^{4}$.

Appendix D Higher-order bounds for stochastic gradient descent

In this section, we provide high-order bounds for averaged stochastic gradient for logistic regression. The first proposition gives a finer result than , with a simpler proof, while the second proposition is new.

Proof. Following , we have the recursion:

Thus if B=\bigg{\|}\sup_{k\in\{0,\dots,n\}}A_{k}\bigg{\|}_{p}, we have

By solving this quadratic inequality, we get:

The previous statement leads to the desired result if $p\geqslant 2$. For $p\in$, we may bound it by the value at $p=2$, and a direct calculation shows that the bound is still correct.

Proof. Using that almost surely, $\|\bar{\theta}_{n-1}-\theta_{\ast}\|\leqslant\|\theta_{0}-\theta_{\ast}\|+n\gamma R$ we obtain that almost surely $f(\bar{\theta}_{n-1})-f(\theta_{\ast})\leqslant R\|\theta_{0}-\theta_{\ast}\|+n\gamma R^{2}$.

Moreover, from the previous proposition, we have for $p\leqslant\frac{n}{4}$,

For $\gamma=\frac{1}{2R^{2}\sqrt{n}}$, we get:

This leads to the bound valid for all $p$:

Appendix E Properties of self-concordance functions

In this section, we review various properties of self-concordant functions, that will prove useful in proving Theorem 3. All these properties rely on bounding the third-order derivatives by second-order derivatives. More precisely, from assumptions (B3-4), we have for any $\theta,\delta,\eta\in{\mathcal{H}}$, where $f^{(r)}[\delta_{1},\dots,\delta_{k}]$ denotes the $k$-th order differential of $f$:

In this section, we derive global non-asymptotic Taylor expansions for self-concordant functions, which show that they behave similarly to like quadratic functions.

The following proposition shows that having a small excess risk $f(\theta)-f(\theta_{\ast})$ implies that the weighted distance to optimum $\langle\theta-\theta_{\ast},H(\theta-\theta_{\ast})\rangle$ is small. Note that for quadratic functions, these two quantities are equal and that throughout this section, we always consider norms weighted by the matrix $H$ (Hessian at optimum).

Assume (B3-4). Then, for any $\theta\in{\mathcal{H}}$:

Proof. Let \varphi:t\mapsto f\big{[}\theta_{\ast}+t(\theta-\theta_{\ast})\big{]}. Denoting $d=\sqrt{\langle\theta-\theta_{\ast},f^{\prime\prime}(\theta_{\ast})(\theta-\theta_{\ast})\rangle}$, we have:

from which we obtain $\varphi^{\prime\prime}(t)\geqslant\varphi^{\prime\prime}(0)e^{-\sqrt{\kappa\rho}dt}$. Following , by integrating twice (and noting that $\varphi^{\prime}(0)=0$ and $\varphi^{\prime\prime}(0)=d^{2}$), we get

The identity $\kappa^{-1}(v)\leqslant\sqrt{3v+v^{2}}$ is equivalent to $e^{-u}+u-1\geqslant\sqrt{u^{2}+\alpha^{2}}-\alpha$, for $\alpha=\frac{3}{2}$. It then suffices to show that $1-e^{-u}\geqslant\frac{u}{\sqrt{u^{2}+\alpha^{2}}}$. This can be shown by proving the monotonicity of $u\mapsto e^{-u}+u-1-\sqrt{u^{2}+\alpha^{2}}+\alpha$, and we leave this exercise to the reader.

The next proposition shows that Hessians between two points which are close in weighted distance are close to each other, for the order between positive semi-definite matrices.

Assume (B3-4). Then, for any $\theta_{1},\theta_{2}\in{\mathcal{H}}$:

Proof. Let $z\in{\mathcal{H}}$ and $\psi(t)=z^{\top}f^{\prime\prime}(\theta_{1}+t(\theta_{2}-\theta_{1}))z$. We have:

with $d_{12}=\sqrt{\langle\theta_{2}-\theta_{1},H(\theta_{2}-\theta_{1})\rangle}$. Thus $\psi(0)e^{\sqrt{\kappa\rho}d_{12}t}\geqslant\psi(t)\geqslant\psi(0)e^{-\sqrt{\kappa\rho}d_{12}t}$. This implies, for $t=1$, that

which implies the desired results. $\|\cdot\|_{\rm op}$ denotes the operator norm (largest singular value).

The following proposition gives an approximation result bounding the first order expansion of gradients by the first order expansion of function values.

Assume (B3-4). Then, for any $\theta_{1},\theta_{2}\in{\mathcal{H}}$ and $\Delta\in{\mathcal{H}}$:

Proof. Let $\varphi(t)=\langle\Delta,f^{\prime}(\theta_{1}+t(\theta_{2}-\theta_{1}))-f^{\prime}(\theta_{1})-tf^{\prime\prime}(\theta_{1})(\theta_{2}-\theta_{1})\rangle$. We have $\varphi^{\prime}(t)=\langle\Delta,f^{\prime\prime}(\theta_{1}+t(\theta_{2}-\theta_{1}))(\theta_{1}-\theta_{2}\rangle-\langle\Delta,f^{\prime\prime}(\theta_{1})(\theta_{1}-\theta_{2})\rangle$ and $|\varphi^{\prime\prime}(t)|=|f^{\prime\prime\prime}(\theta_{1}+t(\theta_{2}-\theta_{1}))[\theta_{2}-\theta_{1},\theta_{2}-\theta_{1},\Delta]|\leqslant\sqrt{\kappa\rho}\langle\Delta,H\Delta\rangle^{1/2}\langle\theta_{1}-\theta_{2},f^{\prime\prime}(\theta_{1}+t(\theta_{2}-\theta_{1}))(\theta_{1}-\theta_{2})\rangle$. This leads to

The following proposition considers a global Taylor expansion of function values. Note that when ${\kappa\rho}{\langle\theta_{2}-\theta_{1},H(\theta_{2}-\theta_{1})\rangle}$ tends to zero, we obtain exactly the second-order Taylor expansion. For more details, see . This is followed by a corrolary that upper bounds excess risk by distance to optimum (this is thus the other direction than Prop. 4).

Assume (B3-4). Then, for any $\theta_{1},\theta_{2}\in{\mathcal{H}}$ and $\Delta\in{\mathcal{H}}$:

Proof. Let \varphi(t)=f\big{[}\theta_{1}+t(\theta_{2}-\theta_{1})\big{]}-f(\theta_{1})-t\langle f^{\prime}(\theta_{1}),\theta_{2}-\theta_{1}\rangle. We have \varphi^{\prime}(t)=\langle f^{\prime}\big{[}\theta_{1}+t(\theta_{2}-\theta_{1})\big{]},\theta_{2}-\theta_{1}\rangle-\langle f^{\prime}(\theta_{1}),\theta_{2}-\theta_{1}\rangle and \varphi^{\prime\prime}(t)=\langle\theta_{2}-\theta_{1},f^{\prime\prime}\big{[}\theta_{1}+t(\theta_{2}-\theta_{1})\big{]}(\theta_{2}-\theta_{1})\rangle. Moreover, $\varphi^{\prime\prime\prime}(t)\leqslant\sqrt{\kappa\rho}\varphi^{\prime\prime}(t)\sqrt{\langle\theta_{2}-\theta_{1},H(\theta_{2}-\theta_{1})\rangle}$, leading to $\varphi^{\prime\prime}(t)\leqslant e^{\sqrt{\kappa\rho}t\sqrt{\langle\theta_{2}-\theta_{1},H(\theta_{2}-\theta_{1})\rangle}}\varphi^{\prime\prime}(0)$. Integrating twice between 0 and 1 leads to the desired result.

Assume (B3-4), and $\theta_{1}\in{\mathcal{H}}$ and $\theta_{2}=\theta_{1}-f^{\prime\prime}(\theta_{1})^{-1}f^{\prime}(\theta_{1})$. Then

where $d=\sqrt{\langle\theta-\theta_{\ast},H(\theta-\theta_{\ast})\rangle}$.

Proof. Applying Prop. 7 to $\theta_{2}=\theta$ and $\theta_{1}=\theta_{\ast}$, we get the desired result.

The following proposition looks at a similar type of bounds than Prop. 7; it is weaker when $\theta_{2}$ and $\theta_{1}$ are close (it does not converge to the second-order Taylor expansion), but stronger for large values (it does not grow exponentially fast).

Assume (B3-4), and $\theta 1,\theta_{2}\in{\mathcal{H}}$. Then

E.2 Analysis of Newton step

Self-concordance has been traditionally used in the analysis of Newton’s method (see ). In this section, we adapt classical results to our specific notion of self-concordance (see also ). A key quantity is the so-called “Newton decrement” at a certain point $\theta_{1}$, equal to $\langle f^{\prime}(\theta_{1}),f^{\prime\prime}(\theta_{1})^{-1}f^{\prime}(\theta_{1})\rangle$, which governs the convergence behavior of Newton methods (this is the quantity which is originally shown to be quadratically convergent). In this paper, we consider a slightly different version where the Hessian is chosen to be the one at $\theta_{\ast}$, i.e., $\langle f^{\prime}(\theta_{1}),H^{-1}f^{\prime}(\theta_{1})\rangle$.

The following proposition shows how a full Newton step improves the Newton decrement (by taking a square).

Assume (B3-4), and $\theta_{1}\in{\mathcal{H}}$ and $\theta_{2}=\theta_{1}-f^{\prime\prime}(\theta_{1})^{-1}f^{\prime}(\theta_{1})$. Then

where $d_{12}^{2}={\langle\theta_{2}-\theta_{1},H(\theta_{2}-\theta_{1})\rangle}\leqslant e^{\sqrt{\kappa\rho}d_{1}}\langle f^{\prime}(\theta_{1}),H^{-1}f^{\prime}(\theta_{1})\rangle$ and $d_{1}=\langle\theta_{1}-\theta_{\ast},H(\theta_{1}-\theta_{\ast})\rangle^{1/2}$.

Proof. When applying the two previous propositions to the Newton step $\theta_{2}=\theta_{1}-f^{\prime\prime}(\theta_{1})^{-1}f^{\prime}(\theta_{1})$, we get:

We then optimize with respect to $\Delta$ to obtain the desired result.

The following proposition shows how the Newton decrement is upper bounded by a function of the excess risk.

Assume (B3-4), and $\theta_{1}\in{\mathcal{H}}$, then,

with $d_{1}=\sqrt{\langle\theta_{1}-\theta_{\ast},H(\theta_{1}-\theta_{\ast})\rangle}$ and $\Delta_{1}=f(\theta_{1})-f(\theta_{\ast})$.

Proof. We may bound the Newton decrement as follows:

The following proposition provides a bound on a quantity which is not the Newton decrement. Indeed, this is (up to the difference in the Hessians), the norm of the Newton step. This will be key in the following proofs.

Assume (B3-4), and $\theta_{1}\in{\mathcal{H}}$, then,

with $d_{1}=\sqrt{\langle\theta_{1}-\theta_{\ast},H(\theta_{1}-\theta_{\ast})\rangle}$.

The next proposition shows that having a small Newton decrement implies that the weighted distance to optimum is small.

Assume (B3-4). If we have ${\sqrt{\kappa\rho}e^{\sqrt{\kappa\rho}d}\langle f^{\prime}(\theta),H^{-1}f^{\prime}(\theta)\rangle^{1/2}}\leqslant\frac{1}{2}$, with $d=\sqrt{\langle\theta-\theta_{\ast},H(\theta-\theta_{\ast})\rangle}$, then

Proof. For any $\Delta\in{\mathcal{H}}$ such that $\langle\Delta,H\Delta\rangle=1$, and $t\geqslant 0$, we have, following the same reasoning than for Prop. 7:

with $v=\sqrt{\kappa\rho}\sqrt{\langle\Delta,H\Delta\rangle}=\sqrt{\kappa\rho}$ and

This implies that if $s\leqslant 1/2$, for $t=\frac{2\sqrt{\kappa\rho}^{-1}s}{1-s}$, $f(\theta_{2}+t\Delta)\geqslant f(\theta_{2})$. Thus,

Note that the quantity $d$ appears twice in the result above.

E.3 Proof of Prop. 1

In this section, we prove Prop. 1 using tools from self-concordance analysis. These tools are described in the previous Sections E.1 and E.2. In order to understand the proof, it is preferable to read these sections first.

We use the notation $t^{2}=\kappa\rho\varepsilon_{1}$. We then get $d_{1}^{2}\stackrel{{\scriptstyle\rm def}}{{=}}\langle\theta_{1}-\theta_{\ast},H(\theta_{1}-\theta_{\ast})\rangle\leqslant(3+t^{2})\varepsilon_{1}$ from Prop. 4.

Moreover, we have, also from Prop. 8, $f(\theta_{2})-f(\theta_{\ast})\leqslant\sqrt{\rho}\|H^{1/2}f^{\prime\prime}(\theta_{1})^{-1}f^{\prime}(\theta_{1})\|$, and using Prop. 5, we get

We may now use Prop. 10 and use the bound:

Combining with Eq. (46) and Eq. (47), we get

Proof of Eq. (26) and Eq. (27).

For these two inequalities, the starting point is the same. Using Eq. (48) (i.e., the Newton decrement at $\theta_{1}$), we first show that the distances $d_{12}$ and $d_{2}$ are bounded. Using $f^{\prime\prime}(\theta_{1})\succcurlyeq e^{-\sqrt{\kappa\rho}d_{1}}H$ (Prop. 5):

Now, we can bound the Newton decrement at $\theta_{2}$, using Prop. 9:

Thus, using Prop. 12, if $\kappa\rho e^{2\sqrt{\kappa\rho}d_{2}}\langle f^{\prime}(\theta_{2}),H^{-1}f^{\prime}(\theta_{2})\rangle\leqslant t^{4}e^{2t[\sqrt{3+t^{2}}+\square_{2}(t)]}\square_{3}(t)\leqslant\frac{1}{4}$, then

where $\varepsilon_{2}\kappa\rho\leqslant u^{2}$.

We now have two separate paths to obtain Eq. (26) and Eq. (27).

If we assume that $\varepsilon_{2}$ is bounded, i.e., with $t=1/16$ and $u=1/4$, then, one can check computationally that we obtain $d_{3}\sqrt{\kappa\rho}\leqslant 0.41$ and thus $b=0.576$ below:

If we only assume $\varepsilon_{1}$ bounded, then we have (from the beginning of the proof):

because we may use the earlier reasoning with $\varepsilon_{3}=0$. This is Eq. (26).

Appendix F Additional experiments

In Table 2, we describe the datasets we have used in experiments and where they were downloaded from.

In Figure 4, we provide similar results than in Section 4, for two additional datasets, quantum and rcv1, while in Figure 5, Figure 6 and Figure 7, we provide training objectives for all methods. We can make the following observations:

For non-sparse datasets, SAG manages to get the smallest training error, confirming the results of .

For the high-dimensional sparse datasets, constant step-size SGD is performing best (note that as shown in Section 3, it is not converging to the optimal value in general, this happens notably for the alpha dataset).

References