ESCAPING SADDLE POINTS FASTER WITH STOCHASTIC MOMENTUM
Year: 2020

Abstract:
Stochastic gradient descent (SGD) with stochastic momentum is popular in nonconvex stochastic optimization and particularly for the training of deep neural networks. In standard SGD, parameters are updated by improving along the path of the gradient at the current iterate on a batch of examples, where the addition of a “momentum” term biases the update in the direction of the previous change in parameters. In non-stochastic convex optimization one can show that a momentum adjustment provably reduces convergence time in many settings, yet such results have been elusive in the stochastic and non-convex settings. At the same time, a widely-observed empirical phenomenon is that in training deep networks stochastic momentum appears to significantly improve convergence time, variants of it have flourished in the development of other popular update methods, e.g. ADAM (Kingma & Ba (2015)), AMSGrad (Reddi et al. (2018b)), etc. Yet theoretical justification for the use of stochastic momentum has remained a significant open question. In this paper we propose an answer: stochastic momentum improves deep network training because it modifies SGD to escape saddle points faster and, consequently, to more quickly find a second order stationary point. Our theoretical results also shed light on the related question of how to choose the ideal momentum parameter–our analysis suggests that β ∈ [0, 1) should be large (close to 1), which comports with empirical findings. We also provide experimental findings that further validate these conclusions.

heading: 1 INTRODUCTION
text: SGD with stochastic momentum has been a de facto algorithm in nonconvex optimization and deep learning. It has been widely adopted for training machine learning models in various applications. Modern techniques in computer vision (e.g.Krizhevsky et al. (2012); He et al. (2016); Cubuk et al. (2018); Gastaldi (2017)), speech recognition (e.g. Amodei et al. (2016)), natural language processing (e.g. Vaswani et al. (2017)), and reinforcement learning (e.g. Silver et al. (2017)) use SGD with stochastic momentum to train models. The advantage of SGD with stochastic momentum has been widely observed (Hoffer et al. (2017); Loshchilov & Hutter (2019); Wilson et al. (2017)). Sutskever et al. (2013) demonstrate that training deep neural nets by SGD with stochastic momentum helps achieving in faster convergence compared with the standard SGD (i.e. without momentum). The success of momentum makes it a necessary tool for designing new optimization algorithms in optimization and deep learning. For example, all the popular variants of adaptive stochastic gradient methods like Adam (Kingma & Ba (2015)) or AMSGrad (Reddi et al. (2018b)) include the use of momentum.
Despite the wide use of stochastic momentum (Algorithm 1) in practice, 1 justification for the clear empirical improvements has remained elusive, as has any mathematical guidelines for actually setting the momentum parameter—it has been observed that large values (e.g. β = 0.9) work well in practice. It should be noted that Algorithm 1 is the default momentum-method in popular software
1Heavy ball momentum is the default choice of momentum method in PyTorch and Tensorflow, instead of Nesterov’s momentum. See the manual pages https://pytorch.org/docs/stable/_modules/ torch/optim/sgd.html and https://www.tensorflow.org/api_docs/python/tf/ keras/optimizers/SGD.
Algorithm 1: SGD with stochastic heavy ball momentum 1: Required: Step size parameter η and momentum parameter β. 2: Init: w0 ∈ Rd and m−1 = 0 ∈ Rd. 3: for t = 0 to T do 4: Given current iterate wt, obtain stochastic gradient gt := ∇f(wt; ξt). 5: Update stochastic momentum mt := βmt−1 + gt. 6: Update iterate wt+1 := wt − ηmt. 7: end for
packages such as PyTorch and Tensorflow. In this paper we provide a theoretical analysis for SGD with momentum. We identify some mild conditions that guarantees SGD with stochastic momentum will provably escape saddle points faster than the standard SGD, which provides clear evidence for the benefit of using stochastic momentum. For stochastic heavy ball momentum, a weighted average of stochastic gradients at the visited points is maintained. The new update is computed as the current update minus a step in the direction of the momentum. Our analysis shows that these updates can amplify a component in an escape direction of the saddle points.
In this paper, we focus on finding a second-order stationary point for smooth non-convex optimization by SGD with stochastic heavy ball momentum. Specifically, we consider the stochastic nonconvex optimization problem, minw∈Rd f(w) := Eξ∼D[f(w; ξ)], where we overload the notation so that f(w; ξ) represents a stochastic function induced by the randomness ξ while f(w) is the expectation of the stochastic functions. An ( , )-second-order stationary point w satisfies
‖∇f(w)‖ ≤ and ∇2f(w) − I. (1)
Obtaining a second order guarantee has emerged as a desired goal in the nonconvex optimization community. Since finding a global minimum or even a local minimum in general nonconvex optimization can be NP hard (Anandkumar & Ge (2016); Nie (2015); Murty & Kabadi (1987); Nesterov (2000)), most of the papers in nonconvex optimization target at reaching an approximate second-order stationary point with additional assumptions like Lipschitzness in the gradients and the Hessian (e.g. Allen-Zhu & Li (2018); Carmon & Duchi (2018); Curtis et al. (2017); Daneshmand et al. (2018); Du et al. (2017); Fang et al. (2018; 2019); Ge et al. (2015); Jin et al. (2017; 2019); Kohler & Lucchi (2017); Lei et al. (2017); Lee et al. (2019); Levy (2016); Mokhtari et al. (2018); Nesterov & Polyak (2006); Reddi et al. (2018a); Staib et al. (2019); Tripuraneni et al. (2018); Xu et al. (2018)). 2 We follow these related works for the goal and aim at showing the benefit of the use of the momentum in reaching an ( , )-second-order stationary point.
We introduce a required condition, akin to a model assumption made in (Daneshmand et al. (2018)), that ensures the dynamic procedure in Algorithm 2 produces updates with suitable correlation with the negative curvature directions of the function f .
Definition 1. Assume, at some time t, that the Hessian Ht = ∇2f(wt) has some eigenvalue smaller than − and ‖∇f(wt)‖ ≤ . Let vt be the eigenvector corresponding to the smallest eigenvalue of ∇2f(wt). The stochastic momentum mt satisfies Correlated Negative Curvature (CNC) at t with parameter γ > 0 if
Et[〈mt, vt〉2] ≥ γ. (2)
As we will show, the recursive dynamics of SGD with heavy ball momentum helps in amplifying the escape signal γ, which allows it to escape saddle points faster.
Contribution: We show that, under CNC assumption and some minor constraints that upper-bound parameter β, if SGD with momentum has properties called Almost Positively Aligned with Gradient (APAG), Almost Positively Correlated with Gradient (APCG), and Gradient Alignment or Curvature Exploitation (GrACE), defined in the later section, then it takes T = O((1−β) log(1/(1−β) ) −10) iterations to return an ( , ) second order stationary point. Alternatively, one can obtain an ( , √ ) second order stationary point in T = O((1 − β) log(1/(1 − β) ) −5) iterations. Our theoretical result demonstrates that a larger momentum parameter β can help in escaping saddle points faster. As saddle points are pervasive in the loss landscape of optimization and deep learning (Dauphin et al.
2We apologize that the list is far from exhaustive.
(2014); Choromanska et al. (2015)), the result sheds light on explaining why SGD with momentum enables training faster in optimization and deep learning.
Notation: In this paper we use Et[·] to represent conditional expectation E[·|w1, w2, . . . , wt], which is about fixing the randomness upto but not including t and notice that wt was determined at t− 1.

heading: 2 BACKGROUND
text: 2.1 A THOUGHT EXPERIMENT.
Let us provide some high-level intuition about the benefit of stochastic momentum with respect to escaping saddle points. In an iterative update scheme, at some time t0 the parameters wt0 can enter a saddle point region, that is a place where Hessian∇2f(wt0) has a non-trivial negative eigenvalue, say λmin(∇2f(wt0)) ≤ − , and the gradient ∇f(wt0) is small in norm, say ‖∇f(wt0)‖ ≤ . The challenge here is that gradient updates may drift only very slowly away from the saddle point, and may not escape this region; see (Du et al. (2017); Lee et al. (2019)) for additional details. On the other hand, if the iterates were to move in one particular direction, namely along vt0 the direction of the smallest eigenvector of ∇2f(wt0), then a fast escape is guaranteed under certain constraints on the
step size η; see e.g. (Carmon et al. (2018)). While the negative eigenvector could be computed directly, this 2nd-order method is prohibitively expensive and hence we typically aim to rely on gradient methods. With this in mind, Daneshmand et al. (2018), who study non-momentum SGD, make an assumption akin to our CNC property described above that each stochastic gradient gt0 is strongly non-orthogonal to vt0 the direction of large negative curvature. This suffices to drive the updates out of the saddle point region.
In the present paper we study stochastic momentum, and our CNC property requires that the update direction mt0 is strongly non-orthogonal to vt0 ; more precisely, Et0 [〈mt0 , vt0〉2] ≥ γ > 0. We are able to take advantage of the analysis of (Daneshmand et al. (2018)) to establish that updates begin to escape a saddle point region for similar reasons. Further, this effect is amplified in successive iterations through the momentum update when β is close to 1. Assume that at some wt0 we have mt0 which possesses significant correlation with the negative curvature direction vt0 , then on successive rounds mt0+1 is quite close to βmt0 , mt0+2 is quite close to β
2mt0 , and so forth; see Figure 1 for an example. This provides an intuitive perspective on how momentum might help accelerate the escape process. Yet one might ask does this procedure provably contribute to the escape process and, if so, what is the aggregate performance improvement of the momentum? We answer the first question in the affirmative, and we answer the second question essentially by showing that momentum can help speed up saddle-point escape by a multiplicative factor of 1− β. On the negative side, we also show that β is constrained and may not be chosen arbitrarily close to 1.

heading: 2.2 MOMENTUM HELPS ESCAPE SADDLE POINTS: AN EMPIRICAL VIEW
text: Let us now establish, empirically, the clear benefit of stochastic momentum on the problem of saddle-point escape. We construct two stochastic optimization tasks, and each exhibits at least one significant saddle point. The two objectives are as follows.
min w f(w) :=
1
n n∑ i=1 (1 2 w>Hw + b>i w + ‖w‖1010 ) , (3)
min w f(w) :=
1
n n∑ i=1 ( (a>i w) 2 − y )2 . (4)
Problem (3) of these was considered by (Staib et al. (2019); Reddi et al. (2018a)) and represents a very straightforward non-convex optimization challenge, with an embedded saddle given by the matrix
H := diag([1,−0.1]), and stochastic gaussian perturbations given by bi ∼ N (0, diag([0.1, 0.001])); the small variance in the second component provides lower noise in the escape direction. Here we have set n = 10. Observe that the origin is in the neighborhood of saddle points and has objective value zero. SGD and SGD with momentum are initialized at the origin in the experiment so that they have to escape saddle points before the convergence. The second objective (4) appears in the phase retrieval problem, that has real applications in physical sciences (Candés et al. (2013); Shechtman et al. (2015)). In phase retrieval3, one wants to find an unknown w∗ ∈ Rd with access to but a few samples yi = (a>i w
∗)2; the design vector ai is known a priori. Here we have sampled w∗ ∼ N (0, Id/d) and ai ∼ N (0, Id) with d = 10 and n = 200. The empirical findings, displayed in Figure 2, are quite stark: for both objectives, convergence is significantly accelerated by larger choices of β. In the first objective (Figure 4a), we see each optimization trajectory entering a saddle point region, apparent from the “flat” progress, yet we observe that large-momentum trajectories escape the saddle much more quickly than those with smaller momentum. A similar affect appears in Figure 4b. To the best of our knowledge, this is the first reported empirical finding that establishes the dramatic speed up of stochastic momentum for finding an optimal solution in phase retrieval.

heading: 2.3 RELATED WORKS.
text: Heavy ball method: The heavy ball method was originally proposed by Polyak (1964). It has been observed that this algorithm, even in the deterministic setting, provides no convergence speedup over standard gradient descent, except in some highly structure cases such as convex quadratic objectives where an “accelerated” rate is possible (Lessard et al. (2016); Goh (2017); Ghadimi et al. (2015); Sun et al. (2019); Loizou & Richtárik (2017; 2018); Gadat et al. (2016); Yang et al. (2018); Kidambi et al. (2018); Can et al. (2019)). We provide a comprehensive survey of the related works about heavy ball method in Appendix A.
Reaching a second order stationary point: As we mentioned earlier, there are many works aim at reaching a second order stationary point. We classify them into two categories: specialized algorithms and simple GD/SGD variants. Specialized algorithms are those designed to exploit the negative curvature explicitly and escape saddle points faster than the ones without the explicit exploitation (e.g. Carmon et al. (2018); Agarwal et al. (2017); Allen-Zhu & Li (2018); Xu et al. (2018)). Simple GD/SGD variants are those with minimal tweaks of standard GD/SGD or their variants (e.g. Ge et al. (2015); Levy (2016); Fang et al. (2019); Jin et al. (2017; 2018; 2019); Daneshmand et al. (2018);
3It is known that phase retrieval is nonconvex and has the so-called strict saddle property: (1) every local minimizer {w∗,−w∗} is global up to phase, (2) each saddle exhibits negative curvature (see e.g. (Sun et al. (2015; 2016); Chen et al. (2018)))
Staib et al. (2019)). Our work belongs to this category. In this category, perhaps the pioneer works are (Ge et al. (2015)) and (Jin et al. (2017)). Jin et al. (2017) show that explicitly adding isotropic noise in each iteration guarantees that GD escapes saddle points and finds a second order stationary point with high probability. Following (Jin et al. (2017)), Daneshmand et al. (2018) assume that stochastic gradient inherently has a component to escape. Specifically, they make assumption of the Correlated Negative Curvature (CNC) for stochastic gradient gt so that Et[〈gt, vt〉2] ≥ γ > 0. The assumption allows the algorithm to avoid the procedure of perturbing the updates by adding isotropic noise. Our work is motivated by (Daneshmand et al. (2018)) but assumes CNC for the stochastic momentum mt instead. In Appendix A, we compare the results of our work with the related works.

heading: 3 MAIN RESULTS
text: We assume that the gradient∇f is L-Lipschitz; that is, f is L-smooth. Further, we assume that the Hessian∇2f is ρ-Lipschitz. These two properties ensure that ‖∇f(w)−∇f(w′)‖ ≤ L‖w−w′‖ and that ‖∇2f(w)−∇2f(w′)‖ ≤ ρ‖w−w′‖, ∀w,w′. The L-Lipschitz gradient assumption implies that |f(w′)−f(w)−〈∇f(w), w′−w〉| ≤ L2 ‖w−w
′‖2,∀w,w′, while the ρ-Lipschitz Hessian assumption implies that |f(w′)−f(w)−〈∇f(w), w′−w〉−(w′−w)>∇2f(w)(w′−w)| ≤ ρ6‖w−w
′‖3, ∀w,w′. Furthermore, we assume that the stochastic gradient has bounded noise ‖∇f(w)−∇f(w; ξ)‖2 ≤ σ2 and that the norm of stochastic momentum is bounded so that ‖mt‖ ≤ cm. We denote ΠiMi as the matrix product of matrices {Mi} and we use σmax(M) = ‖M‖2 := supx6=0 〈x,Mx〉 〈x,x〉 to denote the spectral norm of the matrix M .

heading: 3.1 REQUIRED PROPERTIES WITH EMPIRICAL VALIDATION
text: Our analysis of stochastic momentum relies on three properties of the stochastic momentum dynamic. These properties are somewhat unusual, but we argue they should hold in natural settings, and later we aim to demonstrate that they hold empirically in a couple of standard problems of interest. Definition 2. We say that SGD with stochastic momentum satisfies Almost Positively Aligned with Gradient (APAG) 4 if we have
Et[〈∇f(wt),mt − gt〉] ≥ − 1
2 ‖∇f(wt)‖2. (5)
We say that SGD with stochastic momentum satisfies Almost Positively Correlated with Gradient (APCG) with parameter τ if ∃c′ > 0 such that,
Et[〈∇f(wt),Mtmt〉] ≥ −c′ησmax(Mt)‖∇f(wt)‖2, (6)
where the PSD matrix Mt is defined as
Mt = (Π τ−1 s=1Gs,t)(Π τ−1 s=kGs,t) with Gs,t := I−η ∑s j=1 β s−j∇2f(wt) = I−η(1−β s) 1−β ∇ 2f(wt)
for any integer 1 ≤ k ≤ τ − 1, and η is any step size chosen that guarantees each Gs,t is PSD. Definition 3. We say that the SGD with momentum exhibits Gradient Alignment or Curvature Exploitation (GrACE) if ∃ch ≥ 0 such that
Et[η〈∇f(wt), gt −mt〉+ η 2 2 m > t ∇2f(wt)mt] ≤ η2ch. (7)
APAG requires that the momentum term mt must, in expectation, not be significantly misaligned with the gradient ∇f(wt). This is a very natural condition when one sees that the momentum term is acting as a biased estimate of the gradient of the deterministic f . APAG demands that the bias can not be too large relative to the size of ∇f(wt). Indeed this property is only needed in our analysis when the gradient is large (i.e. ‖∇f(wt)‖ ≥ ) as it guarantees that the algorithm makes progress; our analysis does not require APAG holds when gradient is small.
APCG is a related property, but requires that the current momentum term mt is almost positively correlated with the the gradient∇f(wt), but measured in the Mahalanobis norm induced by Mt. It
4Note that our analysis still go through if one replaces 1 2 on r.h.s. of (5) with any larger number c < 1; the resulted iteration complexity would be only a constant multiple worse.
s=1 Gs,t)(Π
3×105
s=1 Gs,t) and we only
report the values when the update is in the region of saddle points. For (f), we let Mt = (Π500s=1Gs,t)(Π500s=1Gs,t) and we observe that the value is almost always nonnegative. The figures implies that SGD with momentum has APAG and APCG properties in the experiments. Furthermore, an interesting observation is that, for the phase retrieval problem, the expected values might actually be nonnegative.
may appear to be an unusual object, but one can view the PSD matrix Mt as measuring something about the local curvature of the function with respect to the trajectory of the SGD with momentum dynamic. We will show that this property holds empirically on two natural problems for a reasonable constant c′. APCG is only needed in our analysis when the update is in a saddle region with significant negative curvature, ‖∇f(w)‖ ≤ and λmin(∇2f(w)) ≤ − . Our analysis does not require APCG holds when the gradient is large or the update is at an ( , )-second order stationary point.
For GrACE, the first term on l.h.s of (7) measures the alignment between stochastic momentum mt and the gradient ∇f(wt), while the second term on l.h.s measures the curvature exploitation. The first term is small (or even negative) when the stochastic momentum mt is aligned with the gradient∇f(wt), while the second term is small (or even negative) when the stochastic momentum mt can exploit a negative curvature (i.e. the subspace of eigenvectors that corresponds to the negative eigenvalues of the Hessian∇2f(wt) if exists). Overall, a small sum of the two terms (and, consequently, a small ch) allows one to bound the function value of the next iterate (see Lemma 8).
On Figure 3, we report some quantities related to APAG and APCG as well as the gradient norm when solving the previously discussed problems (3) and (4) using SGD with momentum. We also report a quantity regarding GrACE on Figure 4 in the appendix.

heading: 3.2 CONVERGENCE RESULTS
text: The high level idea of our analysis follows as a similar template to (Jin et al. (2017); Daneshmand et al. (2018); Staib et al. (2019)). Our proof is structured into three cases: either (a) ‖∇f(w)‖ ≥ , or (b) ‖∇f(w)‖ ≤ and λmin(∇2f(w)) ≤ − , or otherwise (c) ‖∇f(w)‖ ≤ and λmin(∇2f(w)) ≥ − ,
Algorithm 2: SGD with stochastic heavy ball momentum 1: Required: Step size parameters r and η, momentum parameter β, and period parameter Tthred. 2: Init: w0 ∈ Rd and m−1 = 0 ∈ Rd. 3: for t = 0 to T do 4: Get stochastic gradient gt at wt, and set stochastic momentum mt := βmt−1 + gt. 5: Set learning rate: η̂ := η unless (t mod Tthred) = 0 in which case η̂ := r 6: wt+1 = wt − η̂mt. 7: end for
meaning we have arrived in a second-order stationary region. The precise algorithm we analyze is Algorithm 2, which identical to Algorithm 1 except that we boost the step size to a larger value r on occasion. We will show that the algorithm makes progress in cases (a) and (b). In case (c), when the goal has already been met, further execution of the algorithm only weakly hurts progress. Ultimately, we prove that a second order stationary point is arrived at with high probability. While our proof borrows tools from (Daneshmand et al. (2018); Staib et al. (2019)), much of the momentum analysis is entirely novel to our knowledge. Theorem 1. Assume that the stochastic momentum satisfies CNC. Set 5 r = O( 2), η = O( 5), and Tthred = c(1−β)η log( Lcmσ 2ρc′ch (1−β)δγ ) = O((1 − β) log( Lcmσ 2ρc′ch (1−β)δγ )
−6) for some constant c > 0. If SGD with momentum (Algorithm 2) has APAG property when gradient is large (‖∇f(w)‖ ≥ ), APCGTthred property when it enters a region of saddle points that exhibits a negative curvature (‖∇f(w)‖ ≤ and λmin(∇2f(w)) ≤ − ), and GrACE property throughout the iterations, then it reaches an ( , ) second order stationary point in T = 2Tthred(f(w0)−minw f(w))/(δFthred) = O((1− β) log(Lcmσ
2ρc′ch (1−β)δγ ) −10) iterations with high probability 1− δ, where Fthred = O( 4).
The theorem implies the advantage of using stochastic momentum for SGD. Higher β leads to reaching a second order stationary point faster. As we will show in the following, this is due to that higher β enables escaping the saddle points faster. In Subsection 3.2.1, we provide some key details of the proof of Theorem 1. The interested reader can read a high-level sketch of the proof, as well as the detailed version, in Appendix G.
Remark 1: (constraints on β) We also need some minor constraints on β so that β cannot be too close to 1. They are 1) L(1 − β)3 > 1, 2) σ2(1 − β)3 > 1, 3) c′(1 − β)2 > 1, 4) η ≤ 1−βL , 5) η ≤ 1−β , and 6) Tthred ≥ 1 + 2β 1−β . Please see Appendix E.1 for the details and discussions.
Remark 2: (escaping saddle points) Note that Algorithm 2 reduces to CNC-SGD of Daneshmand et al. (2018) when β = 0 (i.e. without momentum). Therefore, let us compare the results. We show that the escape time of Algorithm 2 is Tthred := Õ ( (1−β) η ) (see Appendix E.3.3, especially (81-82)).
On the other hand, for CNC-SGD, based on Table 3 in their paper, is Tthred = Õ ( 1 η ) . One can clearly see that Tthred of our result has a dependency 1 − β, which makes it smaller than that of Daneshmand et al. (2018) for any same η and consequently demonstrates escaping saddle point faster with momentum.
Remark 3: (finding a second order stationary point) Denote ` a number such that ∀t, ‖gt‖ ≤ `. In Appendix G.3, we show that in the high momentum regime where (1− β) << ρ 2`10
c9mc 2 hc ′ , Algorithm 2
is strictly better than CNC-SGD of Daneshmand et al. (2018), which means that a higher momentum can help find a second order stationary point faster. Empirically, we find out that c′ ≈ 0 (Figure 3) and ch ≈ 0 (Figure 4) in the phase retrieval problem, so the condition is easily satisfied for a wide range of β.

heading: 3.2.1 ESCAPING SADDLE POINTS
text: In this subsection, we analyze the process of escaping saddle points by SGD with momentum. Denote t0 any time such that (t0 mod Tthred) = 0. Suppose that it enters the region exhibiting a small
5See Table 3 in Appendix E for the precise expressions of the parameters. Here, we hide the parameters’ dependencies on γ, L, cm, c′, σ2, ρ, ch, and δ. W.l.o.g, we also assume that cm, L, σ2, c′, ch, and ρ are not less than one and ≤ 1.
gradient but a large negative eigenvalue of the Hessian (i.e. ‖∇f(wt0)‖ ≤ and λmin(∇2f(wt0)) ≤ − ). We want to show that it takes at most Tthred iterations to escape the region and whenever it escapes, the function value decreases at least by Fthred = O( 4) on expectation, where the precise expression of Fthred will be determined later in Appendix E. The technique that we use is proving by contradiction. Assume that the function value on expectation does not decrease at least Fthred in Tthred iterations. Then, we get an upper bound of the expected distance Et0 [‖wt0+Tthred − wt0‖2] ≤ Cupper. Yet, by leveraging the negative curvature, we also show a lower bound of the form Et0 [‖wt0+Tthred − wt0‖2] ≥ Clower. The analysis will show that the lower bound is larger than the upper bound (namely, Clower > Cupper), which leads to the contradiction and concludes that the function value must decrease at least Fthred in Tthred iterations on expectation. Since Tthred = O((1− β) log( 1(1−β) )
6), the dependency on β suggests that larger β can leads to smaller Tthred, which implies that larger momentum helps in escaping saddle points faster. Lemma 1 below provides an upper bound of the expected distance. The proof is in Appendix C.
Lemma 1. Denote t0 any time such that (t0 mod Tthred) = 0. Suppose that Et0 [f(wt0) − f(wt0+t)] ≤ Fthred for any 0 ≤ t ≤ Tthred. Then, Et0 [‖wt0+t − wt0‖2] ≤ Cupper,t := 8ηt ( Fthred+2r2ch+ ρ3 r 3c3m ) (1−β)2 + 8η 2 tσ2 (1−β)2 + 4η 2 ( β 1−β )2 c2m + 2r 2c2m.
We see thatCupper,t in Lemma 1 is monotone increasing with t, so we can defineCupper := Cupper,Tthred . Now let us switch to obtaining the lower bound of Et0 [‖wt0+Tthred − wt0‖2]. The key to get the lower bound comes from the recursive dynamics of SGD with momentum.
Lemma 2. Denote t0 any time such that (t0 mod Tthred) = 0. Let us define a quadratic approximation at wt0 , Q(w) := f(wt0) + 〈w − wt0 ,∇f(wt0)〉 + 12 (w − wt0)
>H(w − wt0), where H := ∇2f(wt0). Also, define Gs := (I − η ∑s k=1 β
s−kH). Then we can write wt0+t −wt0 exactly using the following decomposition.
qv,t−1︷ ︸︸ ︷( Πt−1j=1Gj )( − rmt0 ) +η qm,t−1︷ ︸︸ ︷ (−1)
t−1∑ s=1 ( Πt−1j=s+1Gj ) βsmt0
+ η qq,t−1︷ ︸︸ ︷ (−1)
t−1∑ s=1 ( Πt−1j=s+1Gj ) s∑ k=1 βs−k ( ∇f(wt0+k)−∇Q(wt0+s) )
+ η qw,t−1︷ ︸︸ ︷ (−1)
t−1∑ s=1 ( Πt−1j=s+1Gj ) s∑ k=1 βs−k∇f(wt0) +η
qξ,t−1︷ ︸︸ ︷ (−1)
t−1∑ s=1 ( Πt−1j=s+1Gj ) s∑ k=1 βs−kξt0+k .
The proof of Lemma 2 is in Appendix D. Furthermore, we will use the quantities qv,t−1, qm,t−1, qq,t−1, qw,t−1, qξ,t−1 as defined above throughout the analysis.
Lemma 3. Following the notations of Lemma 2, we have that
Et0 [‖wt0+t−wt0‖2] ≥ Et0 [‖qv,t−1‖2]+2ηEt0 [〈qv,t−1, qm,t−1+qq,t−1+qw,t−1+qξ,t−1〉] =: Clower.
We are going to show that the dominant term in the lower bound of Et0 [‖wt0+t − wt0‖2] is Et0 [‖qv,t−1‖2], which is the critical component for ensuring that the lower bound is larger than the upper bound of the expected distance.
Lemma 4. Denote θj := ∑j k=1 β j−k = ∑j k=1 β
k−1 and λ := −λmin(H). Following the conditions and notations in Lemma 1 and Lemma 2, we have that
Et0 [‖qv,t−1‖2] ≥ ( Πt−1j=1(1 + ηθjλ) )2 r2γ. (8)
Proof. We know that λmin(H) ≤ − < 0. Let v be the eigenvector of the Hessian H with unit norm that corresponds to λmin(H) so that Hv = λmin(H)v. We have (I − ηH)v = v − ηλmin(H)v =
(1− ηλmin(H))v. Then,
Et0 [‖qv,t−1‖2] (a) = Et0 [‖qv,t−1‖2‖v‖2]
(b) ≥ Et0 [〈qv,t−1, v〉2] (c) = Et0 [〈 ( Πt−1j=1Gj ) rmt0 , v〉2]
(d) = Et0 [〈 ( Πt−1j=1(I − ηθjH) ) rmt0 , v〉2] = Et0〈 ( Πt−1j=1(1− ηθjλmin(H)) ) rmt0 , v〉2]
(e) ≥ ( Πt−1j=1(1 + ηθjλ) )2 r2γ,
(9)
where (a) is because v is with unit norm, (b) is by Cauchy–Schwarz inequality, (c), (d) are by the definitions, and (e) is by the CNC assumption so that Et0 [〈mt0 , v〉2] ≥ γ.
Observe that the lower bound in (8) is monotone increasing with t and the momentum parameter β. Moreover, it actually grows exponentially in t. To get the contradiction, we have to show that the lower bound is larger than the upper bound. By Lemma 1 and Lemma 3, it suffices to prove the following lemma. We provide its proof in Appendix E.
Lemma 5. Let Fthred = O( 4) and η2Tthred ≤ r2. By following the conditions and notations in Theorem 1, Lemma 1 and Lemma 2, we conclude that if SGD with momentum (Algorithm 2) has the APCG property, then we have that Clower := Et0 [‖qv,Tthred−1‖2]+2ηEt0 [〈qv,Tthred−1, qm,Tthred−1 + qq,Tthred−1 + qw,Tthred−1 + qξ,Tthred−1〉] > Cupper.

heading: 4 CONCLUSION
text: In this paper, we identify three properties that guarantee SGD with momentum in reaching a secondorder stationary point faster by a higher momentum, which justifies the practice of using a large value of momentum parameter β. We show that a greater momentum leads to escaping strict saddle points faster due to that SGD with momentum recursively enlarges the projection to an escape direction. However, how to make sure that SGD with momentum has the three properties is not very clear. It would be interesting to identify conditions that guarantee SGD with momentum to have the properties. Perhaps a good starting point is understanding why the properties hold in phase retrieval. We believe that our results shed light on understanding the recent success of SGD with momentum in non-convex optimization and deep learning.

heading: ACKNOWLEDGMENTS
text: We gratefully acknowledge financial support from NSF IIS awards 1910077 and 1453304.

heading: A LITERATURE SURVEY
text: Heavy ball method: The heavy ball method was originally proposed by Polyak (1964). It has been observed that this algorithm, even in the deterministic setting, provides no convergence speedup over standard gradient descent, except in some highly structure cases such as convex quadratic objectives where an “accelerated” rate is possible (Lessard et al. (2016); Goh (2017)). In recent years, some works make some efforts in analyzing heavy ball method for other classes of optimization problems besides the quadratic functions. For example, Ghadimi et al. (2015) prove an O(1/T ) ergodic convergence rate when the problem is smooth convex, while Sun et al. (2019) provide a non-ergodic convergence rate for certain classes of convex problems. Ochs et al. (2014) combine the technique of forward-backward splitting with heavy ball method for a specific class of nonconvex optimization problem. For stochastic heavy ball method, Loizou & Richtárik (2017) analyze a class of linear regression problems and shows a linear convergence rate of stochastic momentum, in which the linear regression problems actually belongs to the case of strongly convex quadratic functions. Other works includes (Gadat et al. (2016)), which shows almost sure convergence to the critical points by stochastic heavy ball for general non-convex coercive functions. Yet, the result does not show any advantage of stochastic heavy ball over other optimization algorithms like SGD. Can et al. (2019) show an accelerated linear convergence to a stationary distribution under Wasserstein distance for strongly convex quadratic functions by SGD with stochastic heavy ball momentum. Yang et al. (2018) provide a unified analysis of stochastic heavy ball momentum and Nesterov’s momentum for smooth non-convex objective functions. They show that the expected gradient norm converges at rate O(1/ √ t). Yet, the rate is not better than that of the standard SGD. We are also aware of the works (Ghadimi & Lan (2016; 2013)), which propose some variants of stochastic accelerated algorithms with first order stationary point guarantees. Yet, the framework in (Ghadimi & Lan (2016; 2013)) does not capture the stochastic heavy ball momentum used in practice. There is also a negative result about the heavy ball momentum. Kidambi et al. (2018) show that for a specific strongly convex and
strongly smooth problem, SGD with heavy ball momentum fails to achieving the best convergence rate while some algorithms can.
Reaching a second order stationary point: As we mentioned earlier, there are many works aim at reaching a second order stationary point. We classify them into two categories: specialized algorithms and simple GD/SGD variants. Specialized algorithms are those designed to exploit the negative curvature explicitly and escape saddle points faster than the ones without the explicit exploitation (e.g. Carmon et al. (2018); Agarwal et al. (2017); Allen-Zhu & Li (2018); Xu et al. (2018)). Simple GD/SGD variants are those with minimal tweaks of standard GD/SGD or their variants (e.g. Ge et al. (2015); Levy (2016); Fang et al. (2019); Jin et al. (2017; 2018; 2019); Daneshmand et al. (2018); Staib et al. (2019)). Our work belongs to this category. In this category, perhaps the pioneer works are (Ge et al. (2015)) and (Jin et al. (2017)). Jin et al. (2017) show that explicitly adding isotropic noise in each iteration guarantees that GD escapes saddle points and finds a second order stationary point with high probability. Following (Jin et al. (2017)), Daneshmand et al. (2018) assume that stochastic gradient inherently has a component to escape. Specifically, they make assumption of the Correlated Negative Curvature (CNC) for stochastic gradient gt so that Et[〈gt, vt〉2] ≥ γ > 0. The assumption allows the algorithm to avoid the procedure of perturbing the updates by adding isotropic noise. Our work is motivated by (Daneshmand et al. (2018)) but assumes CNC for the stochastic momentum mt instead. Very recently, Jin et al. (2019) consider perturbing the update of SGD and provide a second order guarantee. Staib et al. (2019) consider a variant of RMSProp (Tieleman & Hinton (2012)), in which the gradient gt is multiplied by a preconditioning matrix Gt and the update is wt+1 = wt − G−1/2t gt. The work shows that the algorithm can help in escaping saddle points faster compared to the standard SGD under certain conditions. Fang et al. (2019) propose average-SGD, in which a suffix averaging scheme is conducted for the updates. They also assume an inherent property of stochastic gradients that allows SGD to escape saddle points.
We summarize the iteration complexity results of the related works for simple SGD variants on Table 1. 6 The readers can see that the iteration complexity of (Fang et al. (2019)) and (Jin et al. (2019)) are better than (Daneshmand et al. (2018); Staib et al. (2019)) and our result. So, we want to explain the results and clarify the differences. First, we focus on explaining why the popular algorithm, SGD with heavy ball momentum, works well in practice, which is without the suffix averaging scheme used in (Fang et al. (2019)) and is without the explicit perturbation used in (Jin et al. (2019)). Specifically, we focus on studying the effect of stochastic heavy ball momentum and showing the advantage of using it. Furthermore, our analysis framework is built on the work of (Daneshmand et al. (2018)). We believe that, based on the insight in our work, one can also show the advantage of stochastic momentum by modifying the assumptions and algorithms in (Fang et al. (2019)) or (Jin et al. (2019)) and consequently get a better dependency on .
6We follow the work (Daneshmand et al. (2018)) for reaching an ( , )-stationary point, while some works are for an ( , √ )-stationary point. We translate them into the complexity of getting an ( , )-stationary point.

heading: B LEMMA 6, 7, AND 8
text: In the following, Lemma 7 says that under the APAG property, when the gradient norm is large, on expectation SGD with momentum decreases the function value by a constant and consequently makes progress. On the other hand, Lemma 8 upper-bounds the increase of function value of the next iterate (if happens) by leveraging the GrACE property. Lemma 6. If SGD with momentum has the APAG property, then, considering the update step wt+1 = wt − ηmt, we have that Et[f(wt+1)] ≤ f(wt)− η2‖∇f(wt)‖ 2 + Lη2c2m 2 .
Proof. By the L-smoothness assumption,
f(wt+1) ≤ f(wt)− η〈∇f(wt),mt〉+ Lη2
2 ‖mt‖2
≤f(wt)− η〈∇f(wt), gt〉 − η〈∇f(wt),mt − gt〉+ Lη2c2m
2 . (10)
Taking the expectation on both sides. We have
Et[f(wt+1)] ≤ f(wt)− η‖∇f(wt)‖2 − ηEt[〈∇f(wt),mt − gt〉] + Lη2c2m
2
≤ f(wt)− η
2 ‖∇f(wt)‖2 + Lη2c2m 2 . (11)
where we use the APAG property in the last inequality.
Lemma 7. Assume that the step size η satisfies η ≤ 2
8Lc2m . If SGD with momentum has the
APAG property, then, considering the update step wt+1 = wt − ηmt, we have that Et[f(wt+1)] ≤ f(wt)− η4 2 when ‖∇f(wt)‖ ≥ .
Proof. Et[f(wt+1)−f(wt)] Lemma 6 ≤ −η2‖∇f(wt)‖ 2+ Lη2c2m 2 ‖∇f(wt)‖≥ ≤ −η2 2+ Lη2c2m 2 ≤ − η 4
2, where the last inequality is due to the constraint of η.
Lemma 8. If SGD with momentum has the GrACE property, then, considering the update step wt+1 = wt − ηmt, we have that Et[f(wt+1)] ≤ f(wt) + η2ch + ρη 3 6 c 3 m.
Proof. Consider the update rule wt+1 = wt − ηmt, where mt represents the stochastic momentum and η is the step size. By ρ-Lipschitzness of Hessian, we have f(wt+1) ≤ f(wt)− η〈∇f(wt), gt〉+ η〈∇f(wt), gt−mt〉+ η 2 2 m > t ∇2f(wt)mt+ ρη3 6 ‖mt‖ 3. Taking the conditional expectation, one has
Et[f(wt+1)] ≤ f(wt)− Et[η‖∇f(wt)‖2] + Et[η〈∇f(wt), gt −mt〉+ η2
2 m>t ∇2f(wt)mt] +
ρη3
6 c3m.
≤ f(wt) + 0 + η2ch + ρη3
6 c3m.
(12)

heading: C PROOF OF LEMMA 1
text: Lemma 1 Denote t0 any time such that (t0 mod Tthred) = 0. Suppose that Et0 [f(wt0) − f(wt0+t)] ≤ Fthred for any 0 ≤ t ≤ Tthred. Then,
Et0 [‖wt0+t − wt0‖2] ≤ Cupper,t := 8ηt ( Fthred + 2r2ch + ρ3r 3c3m )
(1− β)2 + 8η2
tσ2
(1− β)2 + 4η2 ( β 1− β )2 c2m + 2r 2c2m.
(13)
Proof. Recall that the update is wt0+1 = wt0 − rmt0 , and wt0+t = wt0+t−1 − ηmt0+t−1, for t > 1. We have that
‖wt0+t − wt0‖2 ≤ 2(‖wt0+t − wt0+1‖2 + ‖wt0+1 − wt0‖2) ≤ 2‖wt0+t − wt0+1‖2 + 2r2c2m, (14) where the first inequality is by the triangle inequality and the second one is due to the assumption that ‖mt‖ ≤ cm for any t. Now let us denote
• αs := ∑t−1−s j=0 β j
• At−1 := ∑t−1 s=1 αs
and let us rewrite gt = ∇f(wt) + ξt, where ξt is the zero-mean noise. We have that
Et0 [‖wt0+t − wt0+1‖2] = Et0 [‖ t−1∑ s=1 −ηmt0+s‖2] = Et0 [η2‖ t−1∑ s=1 ( ( s∑ j=1 βs−jgt0+j) + β smt0 ) ‖2]
≤ Et0 [2η2‖ t−1∑ s=1 s∑ j=1 βs−jgt0+j‖2 + 2η2‖ t−1∑ s=1 βsmt0‖2] ≤ Et0 [2η2‖ t−1∑ s=1 s∑ j=1 βs−jgt0+j‖2] + 2η2 ( β 1− β )2 c2m = Et0 [2η2‖ t−1∑ s=1 αsgt0+s‖2] + 2η2 ( β 1− β )2 c2m
= Et0 [2η2‖ t−1∑ s=1 αs ( ∇f(wt0+s) + ξt0+s ) ‖2] + 2η2 ( β 1− β )2 c2m
≤ Et0 [4η2‖ t−1∑ s=1 αs∇f(wt0+s)‖2] + Et0 [4η2‖ t−1∑ s=1 αsξt0+s‖2] + 2η2 ( β 1− β )2 c2m.
(15)
To proceed, we need to upper bound Et0 [4η2‖ ∑t−1 s=1 αs∇f(wt0+s)‖2]. We have that Et0 [4η2‖ t−1∑ s=1 αs∇f(wt0+s)‖2] (a) ≤ Et0 [4η2A2t−1 t−1∑ s=1 αs At−1 ‖∇f(wt0+s)‖2]
(b) ≤ Et0 [4η2 At−1 1− β t−1∑ s=1 ‖∇f(wt0+s)‖2] (c) ≤ Et0 [4η2 t (1− β)2 t−1∑ s=1 ‖∇f(wt0+s)‖2].
(16)
where (a) is by Jensen’s inequality, (b) is by maxs αs ≤ 11−β , and (c) is by At−1 ≤ t 1−β . Now let us switch to bound the other term.
Et0 [4η2‖ t−1∑ s=1 αsξt0+s‖2] = 4η2 ( Et0 [ t−1∑ i6=j αiαjξ > t0+iξt0+j ] + Et0 [ t−1∑ s=1 α2sξ > t0+sξt0+s] ) (a) = 4η2 ( 0 + Et0 [
t−1∑ s=1 α2sξ > t0+sξt0+s] ) ,
(b) ≤ 4η2 tσ 2
(1− β)2 . (17)
where (a) is because Et0 [ξ>t0+iξt0+j ] = 0 for i 6= j, (b) is by that ‖ξt‖ 2 ≤ σ2 and maxt αt ≤ 11−β . Combining (14), (15), (16), (17),
Et0 [‖wt0+t − wt0‖2] ≤ Et0 [8η2 t (1− β)2 t−1∑ s=1 ‖∇f(wt0+s)‖2] + 8η2 tσ2 (1− β)2 + 4η2 ( β 1− β )2 c2m + 2r 2c2m.
(18) Now we need to bound Et0 [ ∑t−1 s=1 ‖∇f(wt0+s)‖2]. By using ρ-Lipschitzness of Hessian, we have that
f(wt0+s) ≤ f(wt0+s−1)− η〈∇f(wt0+s−1),mt0+s−1〉+ 1
2 η2m>t0+s−1∇
2f(wt0+s−1)mt0+s−1 + ρ
6 η3‖mt0+s−1‖3.
(19)
By adding η〈∇f(wt0+s−1), gt0+s−1〉 on both sides, we have
η〈∇f(wt0+s−1), gt0+s−1〉 ≤ f(wt0+s−1)− f(wt0+s) + η〈∇f(wt0+s−1), gt0+s−1 −mt0+s−1〉
+ 1
2 η2m>t0+s−1∇
2f(wt0+s−1)mt0+s−1 + ρ
6 η3‖mt0+s−1‖3.
(20)
Taking conditional expectation on both sides leads to
Et0+s−1[η‖∇f(wt0+s−1)‖2] ≤ Et0+s−1[f(wt0+s−1)− f(wt0+s)] + η2ch + ρ
6 η3c3m, (21)
where Et0+s−1[η〈∇f(wt0+s−1), gt0+s−1 −mt0+s−1〉 + 12η 2m>t0+s−1∇ 2f(wt0+s−1)mt0+s−1] ≤ η2ch by the GrACE property. We have that for t0 ≤ t0 + s− 1
Et0 [η‖∇f(wt0+s−1)‖2] = Et0 [Et0+s−1[η‖∇f(wt0+s−1)‖2]] (21)
≤ Et0 [Et0+s−1[f(wt0+s−1)− f(wt0+s)]] + η2ch + ρ
6 η3c3m
= Et0 [f(wt0+s−1)− f(wt0+s)] + η2ch + ρ
6 η3c3m. (22)
Summing the above inequality from s = 2, 3, . . . , t leads to Et0 [ t−1∑ s=1 η‖∇f(wt0+s)‖2] ≤ Et0 [f(wt0+1)− f(wt0+t)] + η2(t− 1)ch + ρ 6 η3(t− 1)c3m
= Et0 [f(wt0+1)− f(wt0) + f(wt0)− f(wt0+t)] + η2(t− 1)ch + ρ
6 η3(t− 1)c3m
(a) ≤ Et0 [f(wt0+1)− f(wt0)] + Fthred + η2(t− 1)ch + ρ
6 η3(t− 1)c3m,
(23)
where (a) is by the assumption (made for proving by contradiction) that Et0 [f(wt0)− f(wt0+s)] ≤ Fthred for any 0 ≤ s ≤ Tthred. By (21) with s = 1 and η = r, we have
Et0 [r‖∇f(wt0)‖2] ≤ Et0 [f(wt0)− f(wt0+1)] + r2ch + ρ
6 r3c3m. (24)
By (23) and (24), we know that
Et0 [ t−1∑ s=1 η‖∇f(wt0+s)‖2] ≤ Et0 [r‖f(wt0)‖2] + Et0 [ t−1∑ s=1 η‖∇f(wt0+s)‖2]
≤ Fthred + r2ch + ρ
6 r3c3m + η
2tch + ρ
6 η3tc3m
(a) ≤ Fthred + 2r2ch + ρ
6 r3c3m +
ρ 6 r2ηc3m.
(b) ≤ Fthred + 2r2ch + ρ
3 r3c3m, (25)
where (a) is by the constraint that η2t ≤ r2 for 0 ≤ t ≤ Tthred and (b) is by the constraint that r ≥ η. By combining (25) and (18)
Et0 [‖wt0+t − wt0‖2] ≤ Et0 [8η2 t (1− β)2 t−1∑ s=1 ‖∇f(wt0+s)‖2] + 8η2 tσ2 (1− β)2 + 4η2 ( β 1− β )2 c2m + 2r 2c2m
≤ 8ηt ( Fthred + 2r2ch + ρ3r 3c3m )
(1− β)2 + 8η2
tσ2
(1− β)2 + 4η2 ( β 1− β )2 c2m + 2r 2c2m.
(26)

heading: D PROOF OF LEMMA 2 AND LEMMA 3
text: Lemma 2 Denote t0 any time such that (t0 mod Tthred) = 0. Let us define a quadratic approximation at wt0 , Q(w) := f(wt0) + 〈w − wt0 ,∇f(wt0)〉 + 12 (w − wt0)
>H(w − wt0), where H := ∇2f(wt0). Also, define Gs := (I − η ∑s k=1 β s−kH) and
• qv,t−1 := ( Πt−1j=1Gj )( − rmt0 ) . • qm,t−1 := − ∑t−1 s=1 ( Πt−1j=s+1Gj ) βsmt0 . • qq,t−1 := − ∑t−1 s=1 ( Πt−1j=s+1Gj )∑s k=1 β
s−k(∇f(wt0+k)−∇Q(wt0+s)). • qw,t−1 := − ∑t−1 s=1 ( Πt−1j=s+1Gj )∑s k=1 β s−k∇f(wt0). • qξ,t−1 := − ∑t−1 s=1 ( Πt−1j=s+1Gj )∑s k=1 β s−kξt0+k.
Then, wt0+t − wt0 = qv,t−1 + ηqm,t−1 + ηqq,t−1 + ηqw,t−1 + ηqξ,t−1.
Notations: Denote t0 any time such that (t0 mod Tthred) = 0. Let us define a quadratic approximation at wt0 ,
Q(w) := f(wt0) + 〈w − wt0 ,∇f(wt0)〉+ 1
2 (w − wt0)>H(w − wt0), (27)
where H := ∇2f(wt0). Also, we denote
Gs := (I − η s∑
k=1
βs−kH)
vm,s := β smt0
vq,s := s∑ k=1 βs−k ( ∇f(wt0+k)−∇Q(wt0+s) ) vw,s :=
s∑ k=1 βs−k∇f(wt0)
vξ,s := s∑ k=1 βs−kξt0+k
θs := s∑ k=1 βs−k.
(28)
Proof. First, we rewrite mt0+j for any j ≥ 1 as follows.
mt0+j = β jmt0 + j∑ k=1 βj−kgt0+k
= βjmt0 + j∑ k=1 βj−k ( ∇f(wt0+k) + ξt0+k ) .
(29)
We have that
wt0+t − wt0 = wt0+t−1 − wt0 − ηmt0+t−1
(a) = wt0+t−1 − wt0 − η ( βt−1mt0 + t−1∑ k=1 βt−1−k ( ∇f(wt0+k) + ξt0+k )) (b) = wt0+t−1 − wt0 − η
t−1∑ k=1 βt−1−k∇Q(wt0+t−1)
− η ( βt−1mt0 + t−1∑ k=1 βt−1−k ( ∇f(wt0+k)−∇Q(wt0+t−1) + ξt0+k )) (c) = wt0+t−1 − wt0 − η
t−1∑ k=1 βt−1−k ( H(wt0+t−1 − wt0) +∇f(wt0) ) − η ( βt−1mt0 +
t−1∑ k=1 βt−1−k ( ∇f(wt0+k)−∇Q(wt0+t−1) + ξt0+k )) = (I − η
t−1∑ k=1 βt−1−kH) ( wt0+t−1 − wt0 ) − η ( βt−1mt0 +
t−1∑ k=1 βt−1−k ( ∇f(wt0+k)−∇Q(wt0+t−1) +∇f(wt0) + ξt0+k )) ,
(30) where (a) is by using (29) with j = t− 1, (b) is by subtracting and adding back the same term, and
(c) is by ∇Q(wt0+t−1) = ∇f(wt0) +H(wt0+t−1 − wt0). To continue, by using the nations in (28), we can rewrite (30) as
wt0+t − wt0 = Gt−1 ( wt0+t−1 − wt0 ) − η ( vm,t−1 + vq,t−1 + vw,t−1 + vξ,t−1 ) . (31)
Recursively expanding (31) leads to wt0+t − wt0 = Gt−1 ( wt0+t−1 − wt0 ) − η ( vm,t−1 + vq,t−1 + vw,t−1 + vξ,t−1 ) = Gt−1 ( Gt−2 ( wt0+t−2 − wt0 ) − η ( vm,t−2 + vq,t−2 + vw,t−2 + vξ,t−2
)) − η ( vm,t−1 + vq,t−1 + vw,t−1 + vξ,t−1
) (a) = ( Πt−1j=1Gj )( wt0+1 − wt0)− η
t−1∑ s=1 ( Πt−1j=s+1Gj )( vm,s + vq,s + vw,s + vξ,s ) ,
(b) = ( Πt−1j=1Gj )( − rmt0 ) − η t−1∑ s=1 ( Πt−1j=s+1Gj )( vm,s + vq,s + vw,s + vξ,s ) ,
(32)
where (a) we use the notation that Πt−1j=sGj := Gs × Gs+1 × . . . . . . Gt−1 and the notation that Πt−1j=tGj = 1 and (b) is by the update rule. By using the definitions of {q?,t−1} in the lemma statement, we complete the proof.
Lemma 3 Following the notations of Lemma 2, we have that
Et0 [‖wt0+t − wt0‖2] ≥ Et0 [‖qv,t−1‖2] + 2ηEt0 [〈qv,t−1, qm,t−1 + qq,t−1 + qw,t−1 + qξ,t−1〉] := Clower (33)
Proof. Following the proof of Lemma 2, we have wt0+t − wt0 = qv,t−1 + η ( qm,t−1 + qq,t−1 + qw,t−1 + qξ,t−1 ) . (34)
Therefore, by using ‖a+ b‖2 ≥ ‖a‖2 + 2〈a, b〉,
Et0 [‖wt0+t − wt0‖2] ≥ Et0 [‖qv,t−1‖2] + 2ηEt0 [〈qv,t−1, qm,t−1 + qq,t−1 + qw,t−1 + qξ,t−1〉]. (35)

heading: E PROOF OF LEMMA 5
text: Lemma 5 Let Fthred = O( 4) and η2Tthred ≤ r2. By following the conditions and notations in Theorem 1, Lemma 1 and Lemma 2, we conclude that if SGD with momentum (Algorithm 2) has the APCG property, then we have that Clower := Et0 [‖qv,Tthred−1‖2]+2ηEt0 [〈qv,Tthred−1, qm,Tthred−1 + qq,Tthred−1 + qw,Tthred−1 + qξ,Tthred−1〉] > Cupper.
W.l.o.g, we assume that cm, L, σ2, c′, ch, and ρ are not less than one and that ≤ 1.
E.1 SOME CONSTRAINTS ON β .
We require that parameter β is not too close to 1 so that the following holds,
• 1) L(1− β)3 > 1. • 2) σ2(1− β)3 > 1. • 3) c′(1− β)2 > 1. • 4) η ≤ 1−βL . • 5) η ≤ 1−β .
• 6) Tthred ≥ c(1−β)η log( Lcmσ 2ρc′ch (1−β)δγ ) ≥ 1 + 2β 1−β .
The constraints upper-bound the value of β. That is, β cannot be too close to 1. We note that the β dependence on L, σ, and c′ are only artificial. We use these constraints in our proofs but they are mostly artefacts of the analysis. For example, if a function is L-smooth, and L < 1, then it is also 1-smooth, so we can assume without loss of generality that L > 1. Similarly, the dependence on σ is not highly relevant, since we can always increase the variance of the stochastic gradient, for example by adding an O(1) gaussian perturbation.

heading: E.2 SOME LEMMAS
text: To prove Lemma 5, we need a series of lemmas with the choices of parameters on Table 3.
Upper bounding Et0 [‖qq,t−1‖]:
Lemma 9. Following the conditions in Lemma 1 and Lemma 2, we have
Et0 [‖qq,t−1‖] ≤ ( Πt−1j=1(1 + ηθjλ) ) βLcm (1− β)2
+
( Πt−1j=1(1 + ηθjλ) ) 1− β ρ η 2 8 ( Fthred + 2r2ch + ρ3r 3c3m ) (1− β)2
+
( Πt−1j=1(1 + ηθjλ) ) 1− β ρ ( 8 r 2σ2 (1−β)2 + 4η 2 ( β 1−β )2 c2m + 2r 2c2m ) 2η .
(36)

heading: Proof.
text: Et0 [‖qq,t−1‖] = Et0 [‖ − t−1∑ s=1 ( Πt−1j=s+1Gj ) s∑ k=1 βs−k ( ∇f(wt0+k)−∇Q(wt0+s) ) ‖]
(a) ≤ Et0 [ t−1∑ s=1 ‖ ( Πt−1j=s+1Gj ) s∑ k=1 βs−k ( ∇f(wt0+k)−∇Q(wt0+s) ) ‖]
(b) ≤ Et0 [ t−1∑ s=1 ‖ ( Πt−1j=s+1Gj ) ‖2‖ s∑ k=1 βs−k ( ∇f(wt0+k)−∇Q(wt0+s) ) ‖]
(c) ≤ Et0 [ t−1∑ s=1 ‖ ( Πt−1j=s+1Gj ) ‖2 s∑ k=1 βs−k‖ ( ∇f(wt0+k)−∇Q(wt0+s) ) ‖]
(d) ≤ Et0 [ t−1∑ s=1 ‖ ( Πt−1j=s+1Gj ) ‖2 s∑ k=1 βs−k ( ‖∇f(wt0+k)−∇f(wt0+s)‖+ ‖∇f(wt0+s)−∇Q(wt0+s)‖ ) ]
(37) where (a), (c), (d) is by triangle inequality, (b) is by the fact that ‖Ax‖2 ≤ ‖A‖2‖x‖2 for any matrix A and vector x. Now that we have an upper bound of ‖∇f(wt0+k)−∇f(wt0+s)‖,
‖∇f(wt0+k)−∇f(wt0+s)‖ (a) ≤ L‖wt0+k − wt0+s‖ (b) ≤ Lη(s− k)cm. (38)
where (a) is by the assumption of L-Lipschitz gradient and (b) is by applying the triangle inequality (s − k) times and that ‖wt − wt−1‖ ≤ η‖mt−1‖ ≤ ηcm, for any t. We can also derive an upper bound of Et0 [‖∇f(wt0+s)−∇Q(wt0+s)‖],
Et0 [‖∇f(wt0+s)−∇Q(wt0+s)‖] (a)
≤ Et0 [ ρ
2 ‖wt0+s − wt0‖2]
(b) ≤ ρ 2 (8ηs(Fthred + 2r2ch + ρ3r3c3m) (1− β)2 + 8 r2σ2 (1− β)2 + 4η2 ( β 1− β )2 c2m + 2r 2c2m. )
(39) Above, (a) is by the fact that if a function f(·) has ρ Lipschitz Hessian, then
‖∇f(y)−∇f(x)−∇2f(x)(y − x)‖ ≤ ρ 2 ‖y − x‖2 (40)
(c.f. Lemma 1.2.4 in (Nesterov (2013))) and using the definition that
Q(w) := f(wt0) + 〈w − wt0 ,∇f(wt0)〉+ 1
2 (w − wt0)>H(w − wt0),
(b) is by Lemma 1 and η2t ≤ r2 for 0 ≤ t ≤ Tthred Et0 [‖wt0+t − wt0‖2] ≤ 8ηt ( Fthred + 2r2ch + ρ3r 3c3m )
(1− β)2 + 8η2
tσ2
(1− β)2 + 4η2 ( β 1− β )2 c2m + 2r 2c2m
≤ 8ηt ( Fthred + 2r2ch + ρ3r 3c3m )
(1− β)2 + 8
r2σ2
(1− β)2 + 4η2 ( β 1− β )2 c2m + 2r 2c2m.
(41) Combing (37), (38), (39), we have that
Et0 [‖qq,t−1‖] (37) ≤ Et0 [ t−1∑ s=1 ‖ ( Πt−1j=s+1Gj ) ‖2 s∑ k=1 βs−k ( ‖∇f(wt0+k)−∇f(wt0+s)‖+ ‖∇f(wt0+s)−∇Q(wt0+s)‖ ) ]
(38),(39) ≤ t−1∑ s=1 ‖ ( Πt−1j=s+1Gj ) ‖2 s∑ k=1 βs−kLη(s− k)cm
+ t−1∑ s=1 ‖ ( Πt−1j=s+1Gj ) ‖2 s∑ k=1 βs−k ρ 2 (8ηs(Fthred + 2r2ch + ρ3r3c3m) (1− β)2 + 8 r2σ2 (1− β)2 + 4η2 ( β 1− β )2 c2m + 2r 2c2m )
:= t−1∑ s=1 ‖ ( Πt−1j=s+1Gj ) ‖2 s∑ k=1 βs−kLη(s− k)cm + t−1∑ s=1 ‖ ( Πt−1j=s+1Gj ) ‖2 s∑ k=1 βs−k ρ 2 (νs + ν),
(42) where on the last line we use the notation that
νs := 8ηs ( Fthred + 2r2ch + ρ3r 3c3m )
(1− β)2
ν := 8 r2σ2
(1− β)2 + 4η2 ( β 1− β )2 c2m + 2r 2c2m.
(43)
To continue, let us analyze ‖ ( Πt−1j=s+1Gj ) ‖2 first.
‖ ( Πt−1j=s+1Gj ) ‖2 = ‖Πt−1j=s+1(I − η j∑ k=1 βj−kH)‖2
(a) ≤ Πt−1j=s+1(1 + ηθjλ) = Πt−1j=1(1 + ηθjλ)
Πsj=1(1 + ηθjλ)
(b) ≤ Πt−1j=1(1 + ηθjλ)
(1 + η )s . (44)
Above, we use the notation that θj := ∑j k=1 β
j−k. For (a), it is due to that λ := −λmin(H), λmax(H) ≤ L, and the choice of η so that 1 ≥ ηL1−β , or equivalently,
η ≤ 1− β L . (45)
For (b), it is due to that θj ≥ 1 for any j and λ ≥ . Therefore, we can upper-bound the first term on r.h.s of (42) as
t−1∑ s=1 ‖ ( Πt−1j=s+1Gj ) ‖2 s∑ k=1 βs−kLη(s− k)cm = t−1∑ s=1 ‖ ( Πt−1j=s+1Gj ) ‖2 s−1∑ k=1 βkkLηcm
(a) ≤ t−1∑ s=1 ‖ ( Πt−1j=s+1Gj ) ‖2
β
(1− β)2 Lηcm
(b) ≤ ( Πt−1j=1(1 + ηθjλ) ) βLηcm (1− β)2 t−1∑ s=1
1
(1 + η )s
(c) ≤ ( Πt−1j=1(1 + ηθjλ) ) βLηcm (1− β)2 1 η = ( Πt−1j=1(1 + ηθjλ) ) βLcm (1− β)2 , (46)
where (a) is by that fact that ∑∞ k=1 β
kk ≤ β(1−β)2 for any 0 ≤ β < 1, (b) is by using (44), and (c) is by using that ∑∞ s=1( 1 1+η )
s ≤ 1η . Now let us switch to bound∑t−1 s=1 ‖ ( Πt−1j=s+1Gj ) ‖2 ∑s k=1 β s−k ρ 2 (νs + ν) on (42). We have that t−1∑ s=1 ‖ ( Πt−1j=s+1Gj ) ‖2 s∑ k=1 βs−k ρ 2 (νs + ν) (a) ≤ 1 1− β t−1∑ s=1 ‖ ( Πt−1j=s+1Gj ) ‖2 ρ 2 (νs + ν)
(b) ≤ ( Πt−1j=1(1 + ηθjλ) ) 1− β t−1∑ s=1
1 (1 + η )s ρ 2 νs +
( Πt−1j=1(1 + ηθjλ) ) 1− β t−1∑ s=1
1 (1 + η )s ρ 2 ν
(c) ≤ ( Πt−1j=1(1 + ηθjλ) ) 1− β t−1∑ s=1
1 (1 + η )s ρ 2 νs +
( Πt−1j=1(1 + ηθjλ) ) 1− β ρν 2η
=
( Πt−1j=1(1 + ηθjλ) ) 1− β t−1∑ s=1
1 (1 + η )s ρ 2 νs +
( Πt−1j=1(1 + ηθjλ) ) 1− β ρ ( 8 r 2σ2 (1−β)2 + 4η 2 ( β 1−β )2 c2m + 2r 2c2m ) 2η
(d) ≤ ( Πt−1j=1(1 + ηθjλ) ) 1− β ρ (η )2 8η ( Fthred + 2r2ch + ρ3r 3c3m ) (1− β)2
+
( Πt−1j=1(1 + ηθjλ) ) 1− β ρ ( 8 r 2σ2 (1−β)2 + 4η 2 ( β 1−β )2 c2m + 2r 2c2m )
2η (47) where (a) is by the fact that ∑s k=1 β
s−k ≤ 1/(1 − β), (b) is by (44), (c) is by using that∑∞ s=1( 1 1+η ) s ≤ 1η , (d) is by ∑∞ k=1 z kk ≤ z(1−z)2 for any |z| ≤ 1 and substituting z = 1 1+η ,
which leads to ∑∞ k=1 z kk ≤ z(1−z)2 = 1/(1+η ) (1−1/(1+η ))2 = 1+η (η )2 ≤ 2 (η )2 in which the last inequality is by chosen the step size η so that η ≤ 1. By combining (42), (46), and (47), we have that
Et0 [‖qq,t−1‖ (42) ≤ t−1∑ s=1 ‖ ( Πt−1j=s+1Gj ) ‖2 s∑ k=1 βs−kLη(s− k)cm + t−1∑ s=1 ‖ ( Πt−1j=s+1Gj ) ‖2 s∑ k=1 βs−k ρ 2 (νs + ν)
(46),(47) ≤ ( Πt−1j=1(1 + ηθjλ) ) βLcm (1− β)2
+
( Πt−1j=1(1 + ηθjλ) ) 1− β ρ η 2 8 ( Fthred + 2r2ch + ρ3r 3c3m ) (1− β)2
+
( Πt−1j=1(1 + ηθjλ) ) 1− β ρ ( 8 r 2σ2 (1−β)2 + 4η 2 ( β 1−β )2 c2m + 2r 2c2m ) 2η ,
(48) which completes the proof.

heading: Upper bounding ‖qv,t−1‖:
text: Lemma 10. Following the conditions in Lemma 1 and Lemma 2, we have
‖qv,t−1‖ ≤ ( Πt−1j=1(1 + ηθjλ) ) rcm. (49)
Proof. ‖qv,t−1‖ ≤ ‖ ( Πt−1j=1Gj )( − rmt0 ) ‖ ≤ ‖ ( Πt−1j=1Gj ) ‖2‖ − rmt0‖ ≤ ( Πt−1j=1(1 + ηθjλ) ) rcm, (50) where the last inequality is because η is chosen so that 1 ≥ ηL1−β and the fact that λmax(H) ≤ L.
Lower bounding Et0 [2η〈qv,t−1, qq,t−1〉]: Lemma 11. Following the conditions in Lemma 1 and Lemma 2, we have
Et0 [2η〈qv,t−1, qq,t−1〉] ≥ −2η ( Πt−1j=1(1 + ηθjλ) )2 rcm×[ βLcm
(1− β)2 +
ρ η 2 8 ( Fthred + 2r2ch + ρ3r 3c3m ) (1− β)3 + ρ ( 8 r 2σ2 (1−β)2 + 4η 2 ( β 1−β )2 c2m + 2r 2c2m ) 2η (1− β) ].
(51)
Proof. By the results of Lemma 9 and Lemma 10
Et0 [2η〈qv,t−1, qq,t−1〉] ≥ −Et0 [2η‖qv,t−1‖‖qq,t−1‖] Lemma 10 ≥ −Et0 [2η ( Πt−1j=1(1 + ηθjλ) ) rcm‖qq,t−1‖]
Lemma 9 ≥ −2η ( Πt−1j=1(1 + ηθjλ) )2 rcm×[ βLcm
(1− β)2 +
ρ η 2 8 ( Fthred + 2r2ch + ρ3r 3c3m ) (1− β)3 + ρ ( 8 r 2σ2 (1−β)2 + 4η 2 ( β 1−β )2 c2m + 2r 2c2m ) 2η (1− β) ].
(52)
Lower bounding Et0 [2η〈qv,t−1, qξ,t−1〉]: Lemma 12. Following the conditions in Lemma 1 and Lemma 2, we have
Et0 [2η〈qv,t−1, qξ,t−1〉] = 0. (53)

heading: Proof.
text: Et0 [2η〈qv,t−1, qξ,t−1〉] = Et0 [2η〈qv,t−1,− t−1∑ s=1 ( Πt−1j=s+1Gj ) s∑ k=1 βs−kξt0+k〉]
(a) = Et0 [2η〈qv,t−1, s∑ k=1 αkξt0+k〉]
(b) = Et0 [2η s∑ k=1 Et0+k−1[〈qv,t−1, αkξt0+k〉]]
(c) = Et0 [2η s∑ k=1 〈qv,t−1,Et0+k−1[αkξt0+k]〉]
= Et0 [2η s∑
k=1
αk〈qv,t−1,Et0+k−1[ξt0+k]〉]
(d) = 0,
(54)
where (a) holds for some coefficients αk, (b) is by the tower rule, (c) is because qv,t−1 is measureable with t0, and (d) is by the zero mean assumption of ξ’s.
Lower bounding Et0 [2η〈qv,t−1, qm,t−1〉]: Lemma 13. Following the conditions in Lemma 1 and Lemma 2, we have
Et0 [2η〈qv,t−1, qm,t−1〉] ≥ 0. (55)
Proof. Et0 [2η〈qv,t−1, qm,t−1〉]
= 2ηrEt0 [〈 ( Πt−1j=1Gj ) mt0 , t−1∑ s=1 ( Πt−1j=s+1Gj ) βsmt0〉]
(a) = 2ηrEt0 [〈mt0 , Bmt0〉]
(b) ≥ 0,
(56)
where (a) is by defining the matrix B := ( Πt−1j=1Gj )>(∑t−1 s=1 ( Πt−1j=s+1Gj ) βs ) . For (b), notice that the matrix B is symmetric positive semidefinite. To see that the matrix B is symmetric positive semidefinite, observe that each Gj := (I − η ∑j k=1 β
j−kH) can be written in the form of Gj = UDjU
> for some orthonormal matrix U and a diagonal matrix Dj . Therefore, the matrix product( Πt−1j=1Gj )>( Πt−1j=s+1Gj ) = U(Πt−1j=1Dj)(Π t−1 j=s+1Dj)U
> is symmetric positive semidefinite as long as each Gj is. So, (b) is by the property of a matrix being symmetric positive semidefinite.
Lower bounding 2ηEt0 [〈qv,t−1, qw,t−1〉]: Lemma 14. Following the conditions in Lemma 1 and Lemma 2, if SGD with momentum has the APCG property, then
2ηEt0 [〈qv,t−1, qw,t−1〉] ≥ − 2ηrc′
(1− β) (Πt−1j=1(1 + ηθjλ)) 2 . (57)
Proof. Define Ds := Πt−1j=1GjΠ t−1 j=s+1Gj .
2ηEt0 [〈qv,t−1, qw,t−1〉] = 2ηEt0 [〈 ( Πt−1j=1Gj )( rmt0 ) , t−1∑ s=1 ( Πt−1j=s+1Gj ) s∑ k=1 βs−k∇f(wt0)〉]
= 2ηEt0 [〈rmt0 , t−1∑ s=1 ( Πt−1j=1GjΠ t−1 j=s+1Gj ) s∑ k=1 βs−k∇f(wt0)〉]
= 2ηr t−1∑ s=1 s∑ k=1 βs−kEt0 [〈mt0 , Ds∇f(wt0)〉]
(a) ≥ −2η2rc′ t−1∑ s=1 s∑ k=1 βs−k‖Ds‖2‖∇f(wt0)‖2
≥ −2η 2rc′
1− β t−1∑ s=1 ‖Ds‖2‖∇f(wt0)‖2, (58)
where (a) is by the APCG property. We also have that ‖Ds‖2 = ‖Πt−1j=1GjΠ t−1 j=s+1Gj‖2 ≤ ‖Π t−1 j=1Gj‖2‖Π t−1 j=s+1Gj‖2
(a) ≤ ‖Πt−1j=1Gj‖2 Πt−1j=1(1 + ηθjλ)
(1 + η )s
(b) ≤ ( Πt−1j=1(1 + ηθjλ) )2 (1 + η )s
(59)
where (a) and (b) is by (44). Substituting the result back to (58), we get
2ηEt0 [〈qv,t−1, qw,t−1〉] ≥ − 2η2rc′
1− β t−1∑ s=1 ‖Ds‖2‖∇f(wt0)‖2
≥ −2η 2rc′
1− β t−1∑ s=1
( Πt−1j=1(1 + ηθjλ) )2 (1 + η )s ‖∇f(wt0)‖2 ≥ − 2η2rc′ (1− β)η ( Πt−1j=1(1 + ηθjλ) )2‖∇f(wt0)‖2 (60)
Using the fact that ‖∇f(wt0)‖ ≤ completes the proof.

heading: E.3 PROOF OF LEMMA 5
text: Recall that the strategy is proving by contradiction. Assume that the function value does not decrease at least Fthred in Tthred iterations on expectation. Then, we can get an upper bound of the expected distance Et0 [‖wt0+Tthred − wt0‖2] ≤ Cupper but, by leveraging the negative curvature, we can also show a lower bound of the form Et0 [‖wt0+Tthred − wt0‖2] ≥ Clower. The strategy is showing that the lower bound is larger than the upper bound, which leads to the contradiction and concludes that the function value must decrease at least Fthred in Tthred iterations on expectation. To get the contradiction, according to Lemma 1 and Lemma 3, we need to show that Et0 [‖qv,Tthred−1‖2] + 2ηEt0 [〈qv,Tthred−1, qm,Tthred−1 + qq,Tthred−1 + qw,Tthred−1 + qξ,Tthred−1〉] > Cupper. (61) Yet, by Lemma 13 and Lemma 12, we have that ηEt0 [〈qv,Tthred−1, qm,Tthred−1〉] ≥ 0 and ηEt0 [〈qv,Tthred−1, qξ,Tthred−1〉] = 0. So, it suffices to prove that Et0 [‖qv,Tthred−1‖2] + 2ηEt0 [〈qv,Tthred−1, qq,Tthred−1 + qw,Tthred−1〉] > Cupper, (62) and it suffices to show that
• 14Et0 [‖qv,Tthred−1‖ 2] + 2ηEt0 [〈qv,Tthred−1, qq,Tthred−1〉] ≥ 0.
• 14Et0 [‖qv,Tthred−1‖ 2] + 2ηEt0 [〈qv,Tthred−1, qw,Tthred−1〉] ≥ 0.
• 14Et0 [‖qv,Tthred−1‖ 2] ≥ Cupper.
E.3.1 PROVING THAT 14Et0 [‖qv,Tthred−1‖ 2] + 2ηEt0 [〈qv,Tthred−1, qq,Tthred−1〉] ≥ 0:
By Lemma 4 and Lemma 11, we have that
1 4 Et0 [‖qv,Tthred−1‖2] + Et0 [2η〈qv,Tthred−1, qq,Tthred−1〉] ≥ 1 4 ( ΠTthred−1j=1 (1 + ηθjλ) )2 r2γ − 2η ( ΠTthred−1j=1 (1 + ηθjλ) )2 rcm × [ βLcm (1− β)2 + ρ η 2 8 ( Fthred + 2r2ch + ρ3r 3c3m ) (1− β)3 + ρ ( 8 r 2σ2 (1−β)2 + 4η 2 ( β 1−β )2 c2m + 2r 2c2m ) 2η (1− β) ].
(63)
To show that the above is nonnegative, it suffices to show that
r2γ ≥ 24ηrβLc 2 m
(1− β)2 , (64)
and
r2γ ≥ 24ηrcmρ (1− β)η 2
8 ( Fthred + 2r2ch + ρ3r 3c3m )
(1− β)2 , (65)
and
r2γ ≥ 24ηrcm 1− β
ρ ( 8 r 2σ2 (1−β)2 + 4η 2 ( β 1−β )2 c2m + 2r 2c2m )
2η . (66)
Now w.l.o.g, we assume that cm, L, σ2, c′, and ρ are not less than one and that ≤ 1. By using the values of parameters on Table 3, we have the following results; a sufficient condition of (64) is that
cr cη ≥ 24Lc
2 m 2
(1− β)2 . (67)
A sufficient condition of (65) is that
cr cF ≥ 576cmρ (1− β)3 , (68)
and 1 ≥ 1152cmρchcr
(1− β)3 , (69)
and
1 ≥ 192c 4 mρ 2c2r (1− β)3 . (70)
A sufficient condition of (66) is that
1 ≥ 96cmρ(σ 2 + 3c2m)cr
(1− β)3 , (71)
and a sufficient condition for the above (71), by the assumption that both σ2 ≥ 1 and cm ≥ 1, is
1 ≥ 576c 3 mρσ 2cr
(1− β)3 . (72)
Now let us verify if (67), (68), (69), (70), (72) are satisfied. For (67), using the constraint of cη on Table 3, we have that 1cη ≥ c5mρL 2σ2c′ch c1
. Using this inequality, it suffices to let cr ≥ c0 2
c3mρLσ 2c′ch(1−β)2 for getting (67), which holds by using the constraint that c
′(1−β)2 > 1 and ≤ 1.
For (68), using the constraint of cF on Table 3, we have that 1cF ≥ c4mρ 2Lσ4ch c2
. Using this inequality, it suffices to let cr ≥ c0c3mρLσ4(1−β)3 , which holds by using the constraint that σ 2(1− β)3 > 1. For (69), it needs (1−β) 3
1152cmρch ≥ c0c3mρLσ2ch ≥ cr, which hold by using the constraint that σ 2(1− β)3 > 1.
For (70), it suffices to let (1−β) 2
14c2mρ ≥ c0c3mρLσ2ch ≥ cr which holds by using the constraint that
σ2(1− β)3 > 1. For (72), it suffices to let (1−β) 3
576c3mρσ 2 ≥ c0 c3mρLσ 2ch ≥ cr, which holds by using the
constraint that L(1− β)3 > 1 and ≤ 1. Therefore, by choosing the parameter values as Table 3, we can guarantee that 14Et0 [‖qv,Tthred−1‖ 2] + 2ηEt0 [〈qv,Tthred−1, qq,Tthred−1〉] ≥ 0.
E.3.2 PROVING THAT 14Et0 [‖qv,Tthred−1‖ 2] + 2ηEt0 [〈qv,Tthred−1, qw,Tthred−1〉] ≥ 0:
By Lemma 4 and Lemma 14, we have that
1 4 Et0 [‖qv,Tthred−1‖2] + 2ηEt0 [〈qv,Tthred−1, qw,Tthred−1〉]
≥ 1 4
( ΠTthred−1j=1 (1 + ηθjλ) )2 r2γ − 2ηrc ′
(1− β) (ΠTthred−1j=1 (1 + ηθjλ))
2 . (73)
To show that the above is nonnegative, it suffices to show that
r2γ ≥ 8ηrc ′
(1− β) . (74)
A sufficient condition is crcη ≥ 8 4c′ 1−β . Using the constraint of cη on Table 3, we have that 1 cη ≥ c5mρL 2σ2c′ch c1 . So, it suffices to let cr ≥ c0 4 3c5mρL 2σ2ch(1−β) , which holds by using the constraint that L(1− β)3 > 1 (so that L(1− β) > 1) and ≤ 1.
E.3.3 PROVING THAT 14Et0 [‖qv,Tthred−1‖ 2] ≥ Cupper :
From Lemma 4 and Lemma 1, we need to show that
1
4
( ΠTthred−1j=1 (1 + ηθjλ) )2 r2γ
≥ 8ηt ( Fthred + 2r2ch + ρ3r 3c3m )
(1− β)2 + 8
r2σ2
(1− β)2 + 4η2 ( β 1− β )2 c2m + 2r 2c2m. (75)
We know that 14 ( ΠTthred−1j=1 (1 + ηθjλ) )2 r2γ ≥ 14 ( ΠTthred−1j=1 (1 + ηθj ) )2 r2γ. It suffices to show that 1
4
( ΠTthred−1j=1 (1 + ηθj ) )2 r2γ
≥ 8ηt ( Fthred + 2r2ch + ρ3r 3c3m )
(1− β)2 + 8
r2σ2
(1− β)2 + 4η2 ( β 1− β )2 c2m + 2r 2c2m. (76)
Note that the left hand side is exponentially growing in Tthred. We can choose the number of iterations Tthred large enough to get the desired result. Specifically, we claim that Tthred ≥ c(1−β) η log( Lcmσ 2ρc′ch
(1−β)δγ ) for some constant c > 0. To see this, let us first apply log on both sides of (76),
2 ( Tthred−1∑
j=1
log(1 + ηθj ) ) + log(r2γ) ≥ log(8aTthred + 8b) (77)
where we denote a := 4η ( Fthred+2r2ch+ ρ3 r 3c3m ) (1−β)2 and b := 4 r2σ2 (1−β)2 + 2η 2 ( β 1−β )2 c2m + r
2c2m. To proceed, we are going to use the inequality log(1 + x) ≥ x2 , for x ∈ [0,∼ 2.51]. We have that
1 ≥ η (1− β)
(78)
as guaranteed by the constraint of η. So,
2 ( Tthred−1∑
j=1 log(1 + ηθj ) ) (a) ≥ Tthred−1∑ j=1 ηθj = Tthred−1∑ j=1 j−1∑ k=0 βkη
= Tthred−1∑ j=1 1− βj 1− β η ≥ 1 1− β (Tthred − 1− β 1− β )η .
(b) ≥ Tthred − 1 2(1− β) η , (79)
where (a) is by using the inequality log(1 + x) ≥ x2 with x = ηθj ≤ 1 and (b) is by making Tthred−1 2(1−β) ≥ β (1−β)2 , which is equivalent to the condition that
Tthred ≥ 1 + 2β
1− β (80)
Now let us substitute the result of (79) back to (77). We have that
Tthred ≥ 1 + 2(1− β)
η log( 8aTthred + 8b γr2 ), (81)
which is what we need to show. By choosing Tthred large enough,
Tthred ≥ c(1− β)
η log(
Lcmσ 2ρc′ch
(1− β)δγ ) = O((1− β) log( 1 (1− β) ) −6) (82)
for some constant c > 0, we can guarantee that the above inequality (81) holds.

heading: F PROOF OF LEMMA 15
text: Lemma 15 (Daneshmand et al. (2018)) Let us define the event Υk := {‖∇f(wkTthred)‖ ≥ or λmin(∇2f(wkTthred)) ≤ − }. The complement is Υck := {‖∇f(wkTthred)‖ ≤ and λmin(∇2f(wkTthred)) ≥ − }, which suggests that wkTthred is an ( , )-second order stationary points. Suppose that
E[f(w(k+1)Tthred)− f(wkTthred)|Υk] ≤ −∆
E[f(w(k+1)Tthred)− f(wkTthred)|Υ c k] ≤ δ
∆ 2 .
(83)
Set T = 2Tthred ( f(w0) − minw f(w) ) /(δ∆). We return w uniformly randomly from w0, wTthred , w2Tthred , . . . , wkTthred , . . . , wKTthred , where K := bT/Tthredc. Then, with probability at least 1− δ, we will have chosen a wk where Υk did not occur.
Proof. Let Pk be the probability that Υk occurs. E[f(w(k+1)Tthred)− f(wkTthred)] = E[f(w(k+1)Tthred)− f(wkTthred)|Υk]Pk + E[f(w(k+1)Tthred)− f(wkTthred)|Υ c k](1− Pk)
≤ −∆Pk + δ∆/2(1− Pk) = δ∆/2− (1 + δ/2)∆Pk ≤ δ∆/2−∆Pk.
(84) Summing over all K, we have
1
K + 1 K∑ k=0 E[f(w(k+1)Tthred)− f(wkTthred)] ≤ ∆ 1 K + 1 K∑ k=0 (δ/2− Pk)
⇒ 1 K + 1 K∑ k=0 Pk ≤ δ/2 + f(w0)−minw f(w) (K + 1)∆ ≤ δ
⇒ 1 K + 1 K∑ k=0 (1− Pk) ≥ 1− δ.
(85)

heading: G PROOF OF THEOREM 1
text: Theorem 1 Assume that the stochastic momentum satisfies CNC. Set r = O( 2), η = O( 5), and Tthred = c(1−β)η log( Lcmσ 2ρc′ch (1−β)δγ ) = O((1 − β) log( Lcmσ 2ρc′ch (1−β)δγ )
−6) for some constant c > 0. If SGD with momentum (Algorithm 2) has APAG property when gradient is large (‖∇f(w)‖ ≥ ), APCGTthred property when it enters a region of saddle points that exhibits a negative curvature (‖∇f(w)‖ ≤ and λmin(∇2f(w)) ≤ − ), and GrACE property throughout the iterations, then it reaches an ( , ) second order stationary point in T = 2Tthred(f(w0)−minw f(w))/(δFthred) = O((1− β) log(Lcmσ
2ρc′ch (1−β)δγ ) −10) iterations with high probability 1− δ, where Fthred = O( 4).

heading: G.1 PROOF SKETCH OF THEOREM 1
text: In this subsection, we provide a sketch of the proof of Theorem 1. The complete proof is available in Appendix G. Our proof uses a lemma in (Daneshmand et al. (2018)), which is Lemma 15 below. The lemma guarantees that uniformly sampling a w from {wkTthred}, k = 0, 1, 2, . . . , bT/Tthredc gives an ( , )-second order stationary point with high probability. We replicate the proof of Lemma 15 in Appendix F.
Lemma 15. (Daneshmand et al. (2018)) Let us define the event Υk := {‖∇f(wkTthred)‖ ≥ or λmin(∇2f(wkTthred)) ≤ − }. The complement is Υck := {‖∇f(wkTthred)‖ ≤ and λmin(∇2f(wkTthred)) ≥ − }, which suggests that wkTthred is an ( , )-second order stationary points. Suppose that
E[f(w(k+1)Tthred)− f(wkTthred)|Υk] ≤ −∆ & E[f(w(k+1)Tthred)− f(wkTthred)|Υ c k] ≤ δ
∆ 2 . (86) Set T = 2Tthred ( f(w0) − minw f(w) ) /(δ∆). 8 We return w uniformly randomly from w0, wTthred , w2Tthred , . . . , wkTthred , . . . , wKTthred , where K := bT/Tthredc. Then, with probability at least 1− δ, we will have chosen a wk where Υk did not occur.
To use the result of Lemma 15, we need to let the conditions in (86) be satisfied. We can bound E[f(w(k+1)Tthred)− f(wkTthred)|Υk] ≤ −Fthred, based on the analysis of the large gradient norm regime (Lemma 7) and the analysis for the scenario when the update is with small gradient norm but a large negative curvature is available (Subsection 3.2.1). For the other condition, E[f(w(k+1)Tthred)− f(wkTthred)|Υck] ≤ δ Fthred 2 , it requires that the expected amortized increase of function value due to taking the large step size r is limited (i.e. bounded by δFthred2 ) when wkTthred is a second order stationary point. By having the conditions satisfied, we can apply Lemma 15 and finish the proof of the theorem.

heading: G.2 FULL PROOF OF THEOREM 1
text: Proof. Our proof is based on Lemma 15. So, let us consider the events in Lemma 15, Υk := {‖∇f(wkTthred)‖ ≥ or λmin(∇2f(wkTthred)) ≤ − }. We first show that E[f(w(k+1)Tthred)− f(wkTthred)|Υk] ≤ Fthred.

heading: When ‖∇f(wkTthred)‖ ≥ :
text: 8One can use any upper bound of f(w0)−minw f(w) as f(w0)−minw f(w) in the expression of T .
Consider that Υk is the case that ‖∇f(wkTthred)‖ ≥ . Denote t0 := kTthred in the following. We have that
Et0 [f(wt0+Tthred)− f(wt0)] = Tthred−1∑ t=0 Et0 [E[f(wt0+t+1)− f(wt0+t)|w0:t0+t]]
= Et0 [f(wt0+1)− f(wt0)] + Tthred−1∑ t=1 Et0 [E[f(wt0+t+1)− f(wt0+t)|w0:t0+t]]
(a) ≤ −r 2 ‖∇f(wt0)‖2 + Lr2c2m 2 + Tthred−1∑ t=1 Et0 [E[f(wt0+t+1)− f(wt0+t)|w0:t0+t]]
(b) ≤ −r 2 ‖∇f(wt0)‖2 + Lr2c2m 2 + Tthred−1∑ t=1 ( η2ch + ρ 6 η3c3m ) (c) ≤ −r 2 ‖∇f(wt0)‖2 + Lr2c2m 2 + r2ch + ρ 6 r3c3m
(d) ≤ −r 2 ‖∇f(wt0)‖2 + Lr2c2m + r2ch
(e) ≤ −r 2 2 + Lr2c2m + r 2ch (f) ≤ −r 4 2 (g) ≤ −Fthred,
(87)
where (a) is by using Lemma 6 with step size r, (b) is by using Lemma 8, (c) is due to the constraint that η2Tthred ≤ r2, (d) is by the choice of r, (e) is by ‖∇f(wt)‖ ≥ , (f) is by the choice of r so that r ≤ 2
4(Lc2m+ch) , and (g) is by
r 4 2 ≥ Fthred. (88)

heading: When ‖∇f(wkTthred)‖ ≤ and λmin(∇2f(wkTthred)) ≤ − :
text: The scenario that Υk is the case that ‖∇f(wkTthred)‖ ≤ and λmin(∇2f(wkTthred)) ≤ − has been analyzed in Appendix E, which guarantees that E[f(wt0+Tthred) − f(wt0)] ≤ −Fthred under the setting.

heading: When ‖∇f(wkTthred)‖ ≤ and λmin(∇2f(wkTthred)) ≥ − :
text: Now let us switch to show that E[f(w(k+1)Tthred) − f(wkTthred)|Υck] ≤ δ Fthred 2 . Recall that Υ c k means that ‖∇f(wkTthred)‖ ≤ and λmin(∇2f(wkTthred)) ≥ − . Denote t0 := kTthred in the following. We have that
Et0 [f(wt0+Tthred)− f(wt0)] = Tthred−1∑ t=0 Et0 [E[f(wt0+t+1)− f(wt0+t)|w0:t0+t]]
= Et0 [f(wt0+1)− f(wt0)] + Tthred−1∑ t=1 Et0 [E[f(wt0+t+1)− f(wt0+t)|w0:t0+t]]
(a) ≤ r2ch + ρ
6 r3c3m + Tthred−1∑ t=1 Et0 [E[f(wt0+t+1)− f(wt0+t)|w0:t0+t]]
(b) ≤ r2ch + ρ
6 r3c3m + Tthred−1∑ t=1 ( η2ch + ρ 6 η3c3m ) (c) ≤ 2r2ch + ρ
3 r3c3m ≤ 4r2ch
(d) ≤ δFthred 2 . (89)
where (a) is by using Lemma 8 with step size r, (b) is by using Lemma 8 with step step size η, (c) is by setting η2Tthred ≤ r2 and η ≤ r, (d) is by the choice of r so that 8r2ch ≤ δFthred.
Now we are ready to use Lemma 15, since both the conditions are satisfied. According to the lemma and the choices of parameters value on Table 3, we can set T = 2Tthred ( f(w0) −
minw f(w) ) /(δFthred) = O((1− β) log(Lcmσ 2ρc′ch (1−β)δγ )
−10), which will return a w that is an ( , ) second order stationary point. Thus, we have completed the proof.

heading: G.3 COMPARISON TO DANESHMAND ET AL. (2018)
text: Theorem 2 in Daneshmand et al. (2018) states that, for CNC-SGD to find an ( , ρ1/2 ) stationary point, the total number of iterations is T = O( ` 10L3
δ4γ4 log 2( `L δγ ) −10), where ` is the bound of the stochastic gradient norm ‖gt‖ ≤ `which can be viewed as the counterpart of cm in our paper. By translating their result for finding an ( , ) stationary point, it is T = O( ` 10L3ρ5
δ4γ4 log 2(ρ`L δγ ) −10). On the other hand, using the parameters value on Table 3, we have that T = 2Tthred ( f(w0)−minw f(w) ) /(δFthred) = O( (1−β)c9mL 3ρ3(σ2)3c2hc ′
δ4γ4 log( Lcmσ 2c′ch (1−β)δγ ) −10) for Algorithm 2.
Before making a comparison, we note that their result does not have a dependency on the variance of stochastic gradient (i.e. σ2), which is because they assume that the variance is also bounded by the constant ` (can be seen from (86) in the supplementary of their paper where the variance terms ‖ζi‖ are bounded by `). Following their treatment, if we assume that σ2 ≤ cm, then on (71) we can instead replace (σ2 + 3c2m) with 4c 2 m and on (72) it becomes 1 ≥ 576c3mρcr (1−β)3 . This will remove all the parameters’ dependency on σ2. Now by comparing Õ((1− β)c9mc2hc′ · ρ3L3 δ4γ4 −10) of ours and T = Õ(ρ2`10 · ρ 3L3
δ4γ4 −10) of Daneshmand et al. (2018), we see that in the high momentum regime
where (1−β) << ρ 2`10
c9mc 2 hc ′ , Algorithm 2 is strictly better than that of Daneshmand et al. (2018), which
means that a higher momentum can help to find a second order stationary point faster.

Review summaries:
This paper makes an interesting theoretical contribution; namely, that SGD with momentum (and with a slight modification to the step-size rule) is guaranteed to quickly converge to a second-order stationary point, implying it quickly escapes saddle points. SGD with momentum is widely used in the practice of deep learning, but a theoretical analysis has remained largely elusive. This paper sheds light theoretical properties justifying its use for deep learning.