10/36-702: Minimax Theory

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1 0/36-70: Minimax Theory Introduction When solving a statistical learning problem, there are often many procedures to choose from. This leads to the following question: how can we tell if one statistical learning procedure is better than another? One answer is provided by minimax theory which is a set of techniques for finding the minimum, worst case behavior of a procedure. Definitions and Notation Let P be a set of distributions and let X,..., X n be a sample from some distribution P P. Let θp be some function of P. For example, θp could be the mean of P, the variance of P or the density of P. Let θ = θx,..., X n denote an estimator. Given a metric d, the minimax risk is R n R n P = inf θ sup E P [d θ, θp ] where the infimum is over all estimators. The sample complexity is } nɛ, P = min n : R n P ɛ. Example Suppose that P = Nθ, : θ R} where Nθ, denotes a Gaussian with mean θ and variance. Consider estimating θ with the metric da, b = a b. The minimax risk is R n = inf sup E P [ θ θ ]. θ 3 In this example, θ is a scalar. Example Let X, Y,..., X n, Y n be a sample from a distribution P. Let mx = E P Y X = x = y dp y X = x be the regression function. In this case, we might use the metric dm, m = m x m x dx in which case the minimax risk is In this example, θ is a function. [ R n = inf sup E P m mx mx ]. 4

2 Notation. Recall that the Kullback-Leibler distance between two distributions P 0 and P with densities p 0 and p is defined to be dp0 p0 x KLP 0, P = log dp 0 log p 0 xdx. dp p x The appendix defines several other distances between probability distributions and explains how these distances are related. We write a b = mina, b} and a b = maxa, b}. If P is a distribution with density p, the product distribution for n iid observations is P n with density p n x = n i= px i. It is easy to check that KLP0 n, P n = nklp 0, P. For positive sequences a n and b n we write a n = Ωb n to mean that there exists C > 0 such that a n Cb n for all large n. a n b n if a n /b n is strictly bounded away from zero and infinity for all large n; that is, a n = Ob n and b n = Oa n. 3 Bounding the Minimax Risk Usually, we do not find R n directly. Instead, we find an upper bound U n and a lower bound L n on R n. To find an upper bound, let θ be any estimator. Then R n = inf θ sup E P [d θ, θp ] sup E P [d θ, θp ] U n. 5 So the maximum risk of any estimator provides an upper bound U n. Finding a lower bound L n is harder. We will consider three methods: the Le Cam method, the Fano method and Tsybakov s bound. If the lower and upper bound are close, then we have succeeded. For example, if L n = cn α and U n = Cn α for some positive constants c, C and α, then we have established that the minimax rate of convergence is n α. All the lower bound methods involve a the following trick: we reduce the problem to a hypothesis testing problem. It works like this. First, we will choose a finite set of distributions M = P,..., P N } P. Then R n = inf θ sup E P [d θ, θp ] inf max E j[d θ, θ j ] 6 θ P j M where θ j = θp j and E j is the expectation under P j. Let s = min j k dθ j, θ k. By Markov s inequality, P d θ, θ > t E[d θ, θ] t and so E[d θ, θ] tp d θ, θ > t. Setting t = s/, and using 6, we have R n s inf θ max P jd θ, θ j > s/. 7 P j M

3 Given any estimator θ, define Now, if ψ j then, letting k = ψ, s dθ j, θ k dθ j, θ + dθ k, θ ψ = argmin d θ, θ j. j dθ j, θ + dθ j, θ since ψ j implies that d θ, θ k d θ, θ j = dθ j, θ. So ψ j implies that dθ j, θ s/. Thus P j d θ, θ j > s/ P j ψ j inf ψ P jψ j where the infimum is over all maps ψ form the data to,..., N}. We can think of ψ is a multiple hypothesis test. Thus we have We can summarize this as a theorem: R n s inf ψ max P jψ j. P j M Theorem 3 Let M = P,..., P N } P and let s = min j k dθ j, θ k. Then R n = inf θ sup E P [d θ, θp ] s inf max P jψ j. 8 ψ P j M Getting a good lower bound involves carefully selecting M = P,..., P N }. If M is too big, s will be small. If M is too small, then max Pj M P j ψ j will be small. 4 Distances We will need some distances between distributions. Specifically, Total Variation TVP,Q = sup A P A QA L P Q = p q Kullback-Leibler KLP,Q = p logp/q χ χ P, Q = p q dq = p q Hellinger HP,Q = p q. We also define the affinity between P and q by ap, q = p q. 3

4 We then have TVP, Q = P Q = ap, q and H P, Q TVP, Q KLP, Q χ P, Q. The appendix contains more information about these distances. 5 Lower Bound Method : Le Cam Theorem 4 Let P be a set of distributions. For any pair P 0, P P, inf sup E P [d θ, θp ] s [p n θ 4 0x p n x]dx = s [ ] TVP0 n, P n 4 9 where s = dθp 0, θp. We also have: inf θ sup E P [d θ, θp ] s 8 e nklp 0,P s P 0,P 8 e nχ 0 and inf θ sup E P [d θ, θp ] s 8 p 0 p n. Corollary 5 Suppose there exist P 0, P P such that KLP 0, P log /n. Then inf θ sup E P [d θ, θp ] s 6 where s = dθp 0, θp. Proof. Let θ 0 = θp 0, θ = θp and s = dθ 0, θ. First suppose that n = so that we have a single observation X. From Theorem 3, inf θ sup E P [d θ, θp ] s π where π = inf ψ max P jψ j. 3 j=0, 4

5 Since a maximum is larger than an average, π = inf ψ Define the Neyman-Pearson test max P jψ j inf j=0, ψ ψ x = P 0 ψ 0 + P ψ. 0 if p0 x p x if p 0 x < p x. In Lemma 7 below, we show that the sum of the errors P 0 ψ 0 + P ψ is minimized by ψ. Now P 0 ψ 0 + P ψ = p 0 xdx + p xdx p >p 0 p 0 >p = [p 0 x p x]dx + [p 0 x p x]dx = [p 0 x p x]dx. p >p 0 p 0 >p Thus, Thus we have shown that P 0 ψ 0 + P ψ inf θ = sup E P [d θ, θp ] s 4 [p 0 x p x]dx. [p 0 x p x]dx. Now suppose we have n observations. Then, replacing p 0 and p with p n 0x = n i= p 0x i and p n x = n i= p x i, we have inf sup E P [d θ, θp ] s [p n θ 4 0x p n x]dx. In Lemma 7 below, we show that p q e KLP,Q. Since KLP0 n, P n = nklp 0, P, we have inf sup E P [d θ, θp ] s θ 8 e nklp 0,P. The other results follow from the inequalities on the distances. Lemma 6 Let ψ be the Neyman-Pearson test. For any test ψ, P 0 ψ = + P ψ = 0 P 0 ψ = + P ψ = 0. 5

6 Proof. Recall that p 0 > p when ψ = 0 and that p 0 < p when ψ =. So P 0 ψ = + P ψ = 0 = p 0 xdx + p xdx ψ= ψ=0 = p 0 xdx + p 0 xdx + p xdx + p xdx ψ=,ψ = ψ=,ψ =0 ψ=0,ψ =0 ψ=0,ψ = p 0 xdx + p xdx + p xdx + p 0 xdx ψ=,ψ = ψ=,ψ =0 ψ=0,ψ =0 ψ=0,ψ = = p 0 xdx + p xdx ψ = ψ =0 = P 0 ψ = + P ψ = 0. Lemma 7 For any P and Q, p q e KLP,Q. Proof. First note that, since a b + a b = a + b, we have p q + p q =. 4 Hence p q p q p q = p q [ = p q p q from 4 ] p q p q p q Cauchy Schwartz pq = = exp log = exp log p q/p pq exp where we used Jensen s inequality in the last inequality. p log q = e KLP,Q p Example 8 Consider data X, Y,..., X n, Y n where X i Uniform0,, Y i = mx i +ɛ i and ɛ i N0,. Assume that } m M = m : my mx L x y, for all x, y [0, ]. 6

7 So P is the set of distributions of the form px, y = pxpy x = φy mx where m M. How well can we estimate mx at some point x? Without loss of generality, let s take x = 0 so the parameter of interest is θ = m0. Let dθ 0, θ = θ 0 θ. Let m 0 x = 0 for all x. Let 0 ɛ and define Lɛ x 0 x ɛ m x = 0 x ɛ. Then m 0, m M and s = m 0 m 0 0 = Lɛ. The KL distance is KLP 0, P = = = = ɛ 0 ɛ Now, KLNµ,, Nµ, = µ µ /. So = 0 KLP 0, P = L p0 x, y p 0 x, y log dydx p x, y p0 xp 0 y x p 0 xp 0 y x log p xp y x φy φy log dydx φy m x φy φy log dydx φy m x KLN0,, Nm x, dx. ɛ 0 ɛ x dx = L ɛ 3 6. dydx Let ɛ = 6 log /L n /3. Then, KLP 0, P = log /n and hence, by Corollary 5, inf θ sup E P [d θ, θp ] s 6 = Lɛ 6 = L 6 /3 6 log c /3 =. 5 L n n It is easy to show that the regressogram regression histogram θ = m0 has risk E P [d θ, θp ] /3 C. n Thus we have proved that inf θ sup E P [d θ, θp ] n 3. 6 The same calculations in d dimensions yield inf θ sup E P [d θ, θp ] n d+. 7 7

8 On the squared scale we have inf θ Similar rates hold in density estimation. sup E P [d θ, θp ] n d+. 8 There is a more general version of Le Cam s lemma that is sometimes useful. Lemma 9 Let P, Q,..., Q N be distributions such that dθp, θq j s for all j. Then inf sup E P [d θ, θp ] s p n q n θ 4 where q = N j q j. Example 0 Let Y i = θ i + n Z i, i =,..., d where Z, Z,..., Z d N0, and θ = θ,..., θ d Θ where Θ = θ R d : θ 0 }. Let P = N0, n I. Let Q j have mean 0 expect that j th coordinate has mean a log d/n where 0 < a <. Let q = N j q j. Some algebra good homework question! shows that χ q, p 0 as d. By the generalized Le Cam lemma, R n a log d/n using squared error loss. We can estimate θ by thresholding Bonferroni. This gives a matching upper bound. 6 Lower Bound Method II: Fano For metrics like df, g = f g, Le Cam s method will usually not give a tight bound. Instead, we use Fano s method. Instead of choosing two distributions P 0, P, we choose a finite set of distributions P,..., P N P. We start with Fano s lemma. Lemma Fano Let X,..., X n P where P P,..., P N }. Let ψ be any function of X,..., X n taking values in,..., N}. Let β = max j k KLP j, P k. Then N P j ψ j nβ + log. log N 8

9 Proof: See Lemma 8 in the appendix.. Now we can state and prove the Fano minimax bound. Theorem Let F = P,..., P N } P. Let θp be a parameter taking values in a metric space with metric d. Then E P d θ, θp s nβ + log 9 log N inf θ sup where and s = min j k d θp j, θp k, 0 β = max j k KLP j, P k. Corollary 3 Fano Minimax Bound Suppose there exists F = P,..., P N } P such that N 6 and β = max KLP j, P k log N j k 4n. Then Proof. From Theorem 3, inf θ max E P [ ] d θ, θp s 4. 3 R n s inf ψ max P jψ j s P j F N P j ψ j where the latter is due to the fact that a max is larger than an average. By Fano s lemma, N P j ψ j nβ + log. log N Thus, inf θ sup E P d θ, θp inf max θ P F E P d θ, θp s nβ + log. 4 log N 9

10 7 Lower Bound Method III: Tsybakov s Bound This approach is due to Tsybakov 009. appendix. The proof of the following theorem is in the Theorem 4 Tsybakov 009 Let X,..., X n P P. Let P 0, P,..., P N } P where N 3. Assume that P 0 is absolutely continuous with respect to each P j. Suppose that Then where N inf θ s = KLP j, P 0 log N 6. sup E P [d θ, θp ] s 6 max dθp j, θp k. 0 j<k N 8 Hypercubes To use Fano s method or Tsybakov s method, we need to construct a finite class of distributions F. Sometimes we use a set of the form } F = P ω : ω Ω where Ω = } ω = ω,..., ω m : ω i 0, }, i =,..., m which is called a hypercube. There are N = m distributions in F. For ω, ν Ω, define the Hamming distance Hω, ν = m Iω k ν k. One problem with a hypercube is that some pairs P, Q F might be very close together which will make s = min j k d θp j, θp k small. This will result in a poor lower bound. We can fix this problem by pruning the hypercube. That is, we can find a subset Ω Ω which has nearly the same number of elements as Ω but such that each pair P, Q F = P ω : ω Ω } is far apart. We call Ω a pruned hypercube. The technique for constructing Ω is the Varshamov-Gilbert lemma. Lemma 5 Varshamov-Gilbert Let Ω = } ω = ω,..., ω N : ω j 0, }. Suppose that N 8. There exists ω 0, ω,..., ω M Ω such that i ω 0 = 0,..., 0, ii M N/8 and iii Hω j, ω k N/8 for 0 j < k M. We call Ω = ω 0, ω,..., ω M } a pruned hypercube. 0

11 Proof. Let D = N/8. Set ω 0 = 0,..., 0. Define Ω 0 = Ω and Ω = ω Ω : Hω, ω 0 > D}. Let ω be any element in Ω. Thus we have eliminated ω Ω : Hω, ω 0 D}. Continue this way recursively and at the j th step define Ω j = ω Ω j : Hω, ω j > D} where j =,..., M. Let n j be the number of elements eliminated at step j, that is, the number of elements in A j = ω Ω j : Hω, ω j D}. It follows that n j D i=0 N i The sets A 0,..., A M partition Ω and so n 0 + n + + n M = N. Thus, D N M + N. i Thus M + D i=0 N N i i=0 =. N P i= Z i m/8 where Z,..., Z N are iid Bernoulli / random variables. By Hoeffding s inequaity, N P Z i m/8 e 9N/3 < N/4. i= Therefore, M N/8 as long as N 8. Finally, note that, by construction, Hω j, ω k D + N/8. Example 6 Consider data X, Y,..., X n, Y n where X i Uniform0,, Y i = fx i + ɛ i and ɛ i N0,. The assumption that X is uniform is not crucial. Assume that f is in the Holder class F defined by } F = f : f l y f l x L x y β l, for all x, y [0, ] where l = β. P is the set of distributions of the form px, y = pxpy x = φy mx where f F. Let Ω be a pruned hypercube and let } m F = f ω x = ω j φ j x : ω Ω where m = cn β+, φj x = Lh β Kx X j /h, and h = /m. Here, K is any sufficiently smooth function supported on /, /. Let d f, g = f g. Some calculations show that, for ω, ν Ω, df ω, f ν = Hω, νlh β+ K m 8 Lhβ+ K c h β.

12 We used the Varshamov-Gilbert result which implies that Hω, ν m/8. Furthermore, To apply Corollary 3, we need to have Now log N 4n KLP ω, P ν c h β. KLP ω, P ν log N 4n. = log m/8 4n = m 3n = 3nh. So we set h = c/n /β+. In that case, df ω, f ν c h β = c c/n β/β+. Corollary 3 implies that inf sup E P [d f, f] n β β+. f It follows that inf f sup E P f f n β β+. It can be shown that there are kernel estimators that achieve this rate of convergence. The kernel has to be chosen carefully to take advantage of the degree of smoothness β. A similar calculation in d dimensions shows that inf f sup E P f f n β β+d. 9 Further Examples 9. Parametric Maximum Likelihood For parametric models that satisfy weak regularity conditions, the maximum likelihood estimator is approximately minimax. Consider squared error loss which is squared bias plus variance. In parametric models with large samples, it can be shown that the variance term dominates the bias so the risk of the mle θ roughly equals the variance: Rθ, θ = Var θ θ + bias Var θ θ. 5 The variance of the mle is approximately Var θ where Iθ is the Fisher information. niθ Hence, nrθ, θ Iθ. 6 Typically, the squared bias is order On while the variance is of order On.

13 For any other estimator θ, it can be shown that for large n, Rθ, θ Rθ, θ. For d- dimensional vectors we have Rθ, θ Iθ /n = Od/n. Here is a more precise statement, due to Hájek and Le Cam. The family of distributions P θ : θ Θ with densities P θ : θ Θ is differentiable in quadratic mean if there exists l θ such that pθ+h p θ ht l θ pθ dµ = o h. 7 Theorem 7 Hájek and Le Cam Suppose that P θ : θ Θ is differentiable in quadratic mean where Θ R k and that the Fisher information I θ is nonsingular. Let ψ be differentiable. Then ψ θ n, where θ n is the mle, is asymptotically, locally, uniformly minimax in the sense that, for any estimator T n, and any bowl-shaped l, sup lim inf I I n sup E θ+h/ n l h I n T n ψ θ + h n ElU 8 where I is the class of all finite subsets of R k and U N0, ψ θ I θ ψ θ T. For a proof, see van der Vaart 998. Note that the right hand side of the displayed formula is the risk of the mle. In summary: in well-behaved parametric models, with large samples, the mle is approximately minimax. 9. Estimating a Smooth Density Here we use the general strategy to derive the minimax rate of convergence for estimating a smooth density. See Yu 008 for more details. Let F be all probability densities f on [0, ] such that 0 < c 0 fx c <, f x c <. We observe X,..., X n P where P has density f F. We will use the squared Hellinger distance d f, g = 0 fx gx dx as a loss function. Upper Bound. Let f n be the kernel estimator with bandwidth h = n /5. Then, using bias-variance calculations, we have that sup E f fx fx dx f F Cn 4/5 3

14 for some C. But fx gx fx gx dx = dx C fx + gx fx gx dx 9 for some C. Hence sup f E f d f, f n Cn 4/5 which gives us an upper bound. Lower Bound. For the lower bound we use Fano s inequality. Let g be a bounded, twice differentiable function on [ /, /] such that / / / gxdx = 0, g xdx = a > 0, / / / g x dx = b > 0. Fix an integer m and for j =,..., m define x j = j //m and g j x = c m gmx x j for x [0, ] where c is a small positive constant. Let M denote the Varshamov-Gilbert pruned version of the set m f τ = + τ j g j x : τ = τ,..., τ m, +} }. m For f τ M, let fτ n denote the product density for n observations and let M n = } M. Some calculations show that, for all τ, τ, f n τ : f τ KLfτ n, fτ n = nklf τ, f τ C n β. 30 m4 By Lemma 5, we can choose a subset F of M with N = e c 0m elements where c 0 is a constant and such that d f τ, f τ C m 4 α 3 for all pairs in F. Choosing m = cn /5 gives β log N/4 and d f τ, f τ C. Fano s n 4/5 lemma implies that max E j d f, f j j C n 4/5. Hence the minimax rate is n 4/5 which is achieved by the kernel estimator. Thus we have shown that R n P n 4/5. This result can be generalized to higher dimensions and to more general measures of smoothness. Since the proof is similar to the one dimensional case, we state the result without proof. 4

15 Theorem 8 Let Z be a compact subset of R d. Let Fp, C denote all probability density functions on Z such that p z p z p fz d d dz C where the sum is over al p,..., p d such that j p j = p. Then there exists a constant D > 0 such that p inf sup E f f n z fz p+ dz D. 3 f f Fp,C n The kernel estimator with an appropriate kernel with bandwidth h n = n /p+d achieves this rate of convergence. 9.3 Minimax Classification Let us now turn to classification. We focus on some results of Yang 999, Tsybakov 004, Mammen and Tsybakov 999, Audibert and Tsybakov 005 and Tsybakov and van de Geer 005. The data are Z = X, Y,..., X n, Y n where Y i 0, }. Recall that a classifier is a function of the form hx = Ix G for some set G. The classification risk is RG = PY hx = PY IX G = EY IX G. 33 The optimal classifier is h x = Ix G where G = x : mx /} and mx = EY X = x. We are interested in how close RG is to RG. Following Tsybakov 004 we define dg, G = RG RG = mx dp Xx 34 G G where A B = A B c A c B and P X is the marginal distribution of X. There are two common types of classifiers. The first type are plug-in classifiers of the form ĥx = I mx / where m is an estimate of the regression function. The second type are empirical risk minimizers where ĥ is taken to be the h that minimizes the observed error rate n n i= Y i hx i as h varies over a set of classifiers H. Sometimes one minimizes the error rate plus a penalty term. According to Yang 999, the classification problem has, under weak conditions, the same order of difficulty in terms of minimax rates as estimating the regression function mx. Therefore the rates are given in Example 39. According to Tsybakov 004 and Mammen and Tsybakov 999, classification is easier than regression. The apparent discrepancy is due to differing assumptions. 5

16 To see that classification error cannot be harder than regression, note that for any m and corresponding Ĝ dg, Ĝ = mx dp X x 35 G Ĝ mx mx dp X x mx mx dp X x 36 so the rate of convergence of dg, G is at least as fast as the regression function. Instead of putting assumptions on the regression function m, Mammen and Tsybakov 999 put an entropy assumption on the set of decision sets G. They assume log Nɛ, G, d Aɛ ρ 37 where Nɛ, G, d is the smallest number of balls of radius ɛ required to cover G. They show that, if 0 < ρ <, then there are classifiers with rate sup EdĜ, G = On / 38 P independent of dimension d. Moreover, if we add the margin or low noise assumption P X 0 < mx t Ct α for all t > 0 39 we get sup EdĜ, G = O n +α/+α+αρ 40 P which can be nearly /n for large α and small ρ. The classifiers can be taken to be plug-in estimators using local polynomial regression. Moreover, they show that this rate is minimax. We will discuss classification in the low noise setting in more detail in another chapter. 9.4 Estimating a Large Covariance Matrix Let X,..., X n be iid Gaussian vectors of dimension d. Let Σ = σ ij i,j d be the d d covariance matrix for X i. Estimating Σ when d is large is very challenging. Sometimes we can take advantage of special structure. Bickel and Levina 008 considered the class of covariance matrices Σ whose entries have polynomial decay. Specifically, Θ = Θα, ɛ, M is all covariance matrices Σ such that 0 < ɛ λ min Σ λ max Σ /ɛ and such that } max j i σ ij : i j > k Mk α 6

17 for all k. The loss function is Σ Σ where is the operator norm A = sup Ax. x: x = Bickel and Levina 008 constructed an estimator that that converges at rate log d/n α/α+. Cai, Zhang and Zhou 009 showed that the minimax rate is min n α/α+ + log d n, d } n so the Bickel-Levina estimator is not rate minimax. Cai, Zhang and Zhou then constructed an estimator that is rate minimax. 9.5 Semisupervised Prediction Suppose we have data X, Y,..., X n, Y n for a classification or regression problem. In addition, suppose we have extra unlabelled data X n+,..., X N. Methods that make use of the unlabeled are called semisupervised methods. We discuss semisupervised methods in another Chapter. When do the unlabeled data help? Two minimax analyses have been carried out to answer that question, namely, Lafferty and Wasserman 007 and Singh, Nowak and Zhu 008. Here we briefly summarize the results of the latter. Suppose we want to estimate mx = EY X = x where x R d and y R. Let p be the density of X. To use the unlabelled data we need to link m and p in some way. A common assumption is the cluster assumption: m is smooth over clusters of the marginal px. Suppose that p has clusters separated by a amount γ and that m is α smooth over each cluster. Singh, Nowak and Zhu 008 obtained the following upper and lower minimax bounds as γ varies in 6 zones which we label I to VI. These zones relate the size of γ and the number of unlabeled points: γ semisupervised supervised unlabelled data help? upper bound lower bound I n α/α+d n α/α+d NO II n α/α+d n α/α+d NO III n α/α+d n /d YES IV n /d n /d NO V n α/α+d n /d YES VI n α/α+d n /d YES The important message is that there are precise conditions when the unlabeled data help and conditions when the unlabeled data do not help. These conditions arise from computing the minimax bounds. 7

18 9.6 Graphical Models Elsewhere in the book, we discuss the problem of estimating graphical models. Here, we shall briefly mention some minimax results for this problem. Let X be a random vector from a multivariate Normal distribution P with mean vector µ and covariance matrix Σ. Note that X is a random vector of length d, that is, X = X,..., X d T. The d d matrix Ω = Σ is called the precision matrix. There is one node for each component of X. The undirected graph associated with P has no edge between X j and X j if and only if Ω jk = 0. The edge set is E = j, k : Ω jk 0}. The graph is G = V, E where V =,..., d} and E is the edge set. Given a random sample of vectors X,..., X n P we want to estimate G. Only the edge set needs to be estimated; the nodes are known. Wang, Wainwright and Ramchandran 00 found the minimax risk for estimating G under zero-one loss. Let G d,r λ denote all the multivariate Normals whose graphs have edge sets with degree at most r and such that min i,j E Ω jk Ωjj Ω kk λ. The sample complexity nd, r, λ is the smallest sample size n needed to recover the true graph with high probability. They show that for any λ [0, /], log d r log d nd, r, λ > max, r 4λ. 4 log + rλ λ Thus, assuming λ /r, we get that n Cr logd r. rλ +r λ 9.7 Deconvolution and Measurement Error A problem has seems to have received little attention in the machine learning literature is deconvolution. Suppose that X,..., X n P where P has density p. We have seen that the minimax rate for estimating p in squared error loss is n β β+ where β is the assumed amount of smoothness. Suppose we cannot observe X i directly but instead we observe X i with error. Thus, we observe Y,..., Y n where Y i = X i + ɛ i, i =,..., n. 4 The minimax rates for estimating p change drastically. A good account is given in Fan 99. As an example, if the noise ɛ i is Gaussian, then Fan shows that the minimax risk satisfies β R n C log n 8

19 which means that the problem is essentially hopeless. Similar results hold for nonparametric regression. In the usual nonparametric regression problem we observe Y i = mx i + ɛ i and we want to estimate the function m. If we observe Xi = X i + δ i instead of X i then again the minimax rates change drastically and are logarithmic of the δ i s are Normal Fan and Truong 993. This is known as measurement error or errors in variables. This is an interesting example where minimax theory reveals surprising and important insight. 9.8 Normal Means Perhaps the best understood cases in minimax theory involve normal means. First suppose that X,..., X n Nθ, σ where σ is known. A function g is bowl-shaped if the sets x : gx c} are convex and symmetric about the origin. We will say that a loss function l is bowl-shaped if lθ, θ = gθ θ for some bowl-shaped function g. Theorem 9 The unique estimator that is minimax for every bowl-shaped loss function is the sample mean X n. For a proof, see Wolfowitz 950. Now consider estimating several normal means. Let X j = θ j + ɛ j / n for j =,..., n and suppose we and to estimate θ = θ,..., θ n with loss function l θ, θ = n θ j θ j. Here, ɛ,..., ɛ n N0, σ. This is called the normal means problem. There are strong connections between the normal means problem and nonparametric learning. For example, suppose we want to estimate a regression function fx and we observe data Z i = fi/n +δ i where δ i N0, σ. Expand f in an othonormal basis: fx = j θ jψ j x. An estimate of θ j is X j = n n i= Z i ψ j i/n. It follows that X j Nθ j, σ /n. This connection can be made very rigorous; see Brown and Low 996. The minimax risk depends on the assumptions about θ. Theorem 0 Pinsker. If Θ n = R n then R n = σ and θ = X = X,..., X n is minimax.. If Θ n = θ : n j θ j C } then Up to sets of measure 0. lim inf n inf θ sup R θ, θ = σ C θ Θ n σ + C. 43 9

20 Define the James-Stein estimator Then θ JS = Hence, θ JS is asymptotically minimax. n σ n X. 44 n X j lim sup R θ JS, θ = σ C n θ Θ n σ + C Let X j = θ j + ɛ j for j =,,..., where ɛ j N0, σ /n. } Θ = θ : θj a j C 46 where a j = πj p. Let R n denote the minimax risk. Then min n p p+ σ p p p+ Rn = C p+ p p+ p + p+. 47 n π p + Hence, R n n p p+. An asymptotically minimax estimator is the Pinsker estimator defined by θ = w X, w X,..., where w j = [ a j /µ] + and µ is determined by the equation σ a j µ a j + = C. n j The set Θ in 46 is called a Sobolev ellipsoid. This set corresponds to smooth functions in the function estimation problem. The Pinsker estimator corresponds to estimating a function by smoothing. The main message to take away from all of this is that minimax estimation under smoothness assumptions requires shrinking the data appropriately. 0 Adaptation The results in this chapter provide minimax rates of convergence and estimators that achieve these rates. However, the estimators depend on the assumed parameter space. For example, estimating a β-times differential regression function requires using an estimator tailored to the assumed amount of smoothness to achieve the minimax rate n β β+. There are estimators that are adaptive, meaning that they achieve the minimax rate without the user having to know the amount of smoothness. See, for example, Chapter 9 of Wasserman 006 and the references therein. 0

21 Minimax Hypothesis Testing Let Y,..., Y n P where P P and let P 0 P. We want to test H 0 : P = P 0 versus H : P P 0. Recall that a size α test is a function φ of y,..., y n such that φy,..., y n 0, } and P n 0 φ = α. Let Φ n be the set level α tests based on n observations where 0 < α < is fixed. We want to find the minimax type II error β n ɛ = inf sup φ Φ n ɛ P n φ = 0 48 where Pɛ = } P P : dp 0, Q > ɛ and d is some metric. The minimax testing rate is ɛ n = inf ɛ : β n ɛ δ }. Lower Bound. Define Q by QA = P n AdµP 49 where µ is any distribution whose support is contained in Pɛ. In particular, if µ is uniform on a finite set P,..., P N then QA = Pj n A. 50 N j Define the likelihood ratio L n = dq dp n 0 = py n p 0 y n dµp = py j dµp. 5 p 0 y j j Lemma Let 0 < δ < α. If then β n ɛ δ. E 0 [L n] + 4 α δ 5

22 Proof. Since P n 0 φ = α for each φ, we have β n ɛ = inf φ sup P n φ = 0 inf ɛ φ Qφ = 0 inf P 0 n φ = 0 + [Qφ = 0 P0 n φ = 0] φ α + [Qφ = 0 P0 n φ = 0] α sup QA P0 n A A = α Q P n 0. Now, Q P0 n = L n y n dp 0 y n = E 0 L n y n E 0 [L n]. The result follows from 5. Upper Bound. Let φ be any size α test. Then gives an upper bound. β n ɛ sup P n φ = 0 ɛ. Multinomials Let Y,..., Y n P where Y i,..., d} S. The minimax estimation rate under L loss for this problem is O d/n. We will show that the minimax testing rate is d /4 / n. We will focus on the uniform distribution p 0. So p 0 y = /d for all y. Let } Pɛ = p : p p 0 > ɛ. We want to find ɛ n such that β n Pɛ n = δ. Theorem Let where ɛ Cd/4 n C = [log + 4 α δ ] /4. Assume that ɛ. Then β n ɛ δ.

23 Proof. Let γ = ɛ/d. Let η = η,..., η d where η j, +} and j η j = 0. Let R be all such sequences. This is like a Radamacher sequence except that they are not quite independent. Define p η by p η y = p 0 y + γ j η j ψ j y where ψ j y = if y = j and ψ j y = 0 otherwise. Then p η j 0 for all j and j p ηj =. Also p 0 p η = γη j = γd = ɛ. j Let N be the number of such probability functions. Now and, for any η, ν R, L n = N η ν i L n = N p η Y i p ν Y i p 0 Y i p 0 Y i p η Y i p 0 Y i η = p 0 Y i + γ j η jψ j Y i p 0 Y i + γ j ν jψ j Y i N p η ν i 0 Y i p 0 Y i = + dγ η N j ψ j Y i + dγ ν j ψ j Y i. η ν i j j Taking the expected value over Y,..., Y n, and using the fact that ψ j Y i ψ k Y i = 0 for j k, and that E 0 [ψ j Y ] = P 0 Y = j = /d, we have E 0 [L n] = + dγ n η N j ν j exp ndγ η, ν N η ν j η ν = E η,ν e n dγ η,ν where E η,ν denotes expectation over random draws of η and ν. The average over η and ν can thought of as averages with respect to random assignmetn of the s and s. Hence, because these are negatively associated, i E 0 [L n] E η,ν [e ndγ η,ν ] j Ee ndγ η j ν j = j coshndγ j + n d γ 4 j e n d γ 4 = e n d 3 γ 4 = e n d 3 ɛ 4 /d α δ 3

24 where we used the fact that, γ = ɛ n /d and for small x, coshx + x. From Lemma, it follows that that β n Pɛ, L δ. For the upper bound we use a test due to Paninski 008. Let T n be the number of bins with example one observation. Define t n by Let φ = IT n < t n. Thus φ Φ n. P 0 T n < t n = α. Lemma 3 There exists C such that, if p p 0 > C d /4 / n then P φ = 0 < δ.. Testing Densities Now consider Y,..., Y n P where Y i [0, ] d. Let p 0 be the uniform density. We want to test H 0 : P = P 0. Define Pɛ, s, L = f : f 0 f ɛ} Hs L 53 where f H s L if, for all x, y, and f t L for t =,..., s. f t y f t x L x y t Theorem 4 Fix δ > 0. Define ɛ n by βɛ n = δ. Then, there exist c, c > 0 such that c n s 4s+d ɛn c n s 4s+d. Proof. Here is a proof outline. Divide the spce into k = n /4s+d equal size bins. Let η be a Radamacher sequence. Let ψ be a smooth function such that ψ = 0 and ψ =. For the j th bin B j, let ψ j be the function ψ rescaled and recentered to be supported on B j with ψ j =. Define p η y = p 0 y + γ j η j ψ j y where γ = cn s+d/4s+d. It may be verified p η is a density in H s L and that p 0 p η ɛ. Let N be the number of Rademacher sequences. Then L n = N p η Y i p 0 Y i η i 4

25 and L n = N η ν = N η ν i i p η Y i p ν Y i p 0 Y i p 0 Y i = p 0 Y i + j γη jψ j Y i p 0 Y i + j γν jψ j Y i N p η ν i 0 Y i p 0 Y i + +. j γη jψ j Y i p 0 Y i j γν jψ j Y i p 0 Y i Taking the expcted value over Y,..., Y n, and using the fact that the ψ j s are orthogonal, E 0 [L n] = + n γ η N j ν j exp n γ η N j ν j. η ν j η ν j Thus E 0 [L n] E η,ν e n η,ν where η, ν = γ j η jν j. Hence, E 0 [L n] E η,ν e n η,ν = j Ee nη jν j = j coshnρ j j + n ρ 4 j j e n γ 4 = e kn γ 4 C 0. From Lemma, it follows that that β n Pɛ, L δ. The upper bound can be obtained by using a χ test on the bins. Surprisingly, the story gets much more complicated for non-uniform densities. Summary Minimax theory allows us to state precisely the best possible performance of any procedure under given conditions. The key tool for finding lower bounds on the minimax risk is Fano s inequality. Finding an upper bound usually involves finding a specific estimator and computing its risk. 3 Bibliographic remarks There is a vast literature on minimax theory however much of it is scattered in various journal articles. Some texts that contain minimax theory include Tsybakov 009, van de Geer 000, van der Vaart 998 and Wasserman 006. References: Arias-Castro Ingster xxxx, Yu 008, Tsybakov 009, van der Vaart 998, Wasserman 04. 5

26 4 Appendix 4. Metrics For Probability Distributions Minimax theory often makes use of various metrics for probability distributions. Here we summarize some of these metrics and their properties. Let P and Q be two distributions with densities p and q. We write the distance between P and Q as either dp, Q or dp, q whichever is convenient. We define the following distances and related quantities. Total variation TVP, Q = sup A P A QA L distance d P, Q = p q L distance d P, Q = p q Hellinger distance hp, Q = p q Kullback-Leibler distance KLP, Q = p logp/q χ χ P, Q = p q /p Affinity P Q = p q = minpx, qx}dx Hellinger affinity AP, Q = pq There are many relationships between these quantities. These are summarized in the next two theorems. We leave the proofs as exercises. Theorem 5 The following relationships hold:. TVP, Q = d P, Q = p q. Scheffés Theorem.. TVP, Q = P A QA where A = x : px > qx} hp, Q. 4. h P, Q = AP, Q. 5. P Q = d P, Q. h P,Q 6. P Q A P, Q = 7. h P, Q TVP, Q = d P, Q hp, Q 8. TVP, Q KLP, Q/. Pinsker s inequality. 9. log dp/dq + dp KLP, Q + KLP, Q/. 0. P Q e KLP,Q.. TVP, Q hp, Q KLP, Q χ P, Q.. Le Cam s inequalities. h P,Q 4. 6

27 Let P n denote the product measure based on n independent samples from P. Theorem 6 The following relationships hold:. h P n, Q n = n h P,Q.. P n Q n A P n, Q n = 3. P n Q n d P, Q n. 4. KLP n, Q n = nklp, Q. h P, Q n. 4. Fano s Lemma For 0 < p < define the entropy hp = p log p p log p and note that 0 hp log. Let Y, Z be a pair of random variables each taking values in,..., N} with joint distribution P Y,Z. Then the mutual information is defined to be IY ; Z = KLP Y,Z, P Y P Z = HY HY Z 54 where HY = j PY = j log PY = j is the entropy of Y and HY Z is the entropy of Y given Z. We will use the fact that IY ; hz IY ; Z for an function h. Lemma 7 Let Y be a random variable taking values in,..., N}. Let P,..., P N } be a set of distributions. Let X be drawn from P j for some j,..., N}. Thus P X A Y = j = P j A. Let Z = gx be an estimate of Y taking values in,..., N}. Then, HY X PZ Y logn + hpz = Y. 55 We follow the proof from Cover and Thomas 99. Proof. Let E = IZ Y. Then HE, Y X = HY X + HE X, Y = HY X since HE X, Y = 0. Also, HE, Y X = HE X + HY E, X. But HE X HE = hp Z = Y. Also, HY E, X = P E = 0HY X, E = 0 + P E = HY X, E = P E = hp Z = Y logn 7

28 since Y = gx when E = 0 and, when E =, HY X, E = by the log number of remaining outcomes. Combining these gives HY X PZ Y logn +hpz = Y. Lemma 8 Fano s Inequality Let P = P,..., P N } and β = max j k KLP j, P k. For any random variable Z taking values on,..., N}, N P j Z j nβ + log. log N Proof. For simplicity, assume that n =. The general case follows since KLP n, Q n = nklp, Q. Let Y have a uniform distribution on,..., N}. Given Y = j, let X have distribution P j. This defines a joint distribution P for X, Y given by P X A, Y = j = P X A Y = jp Y = j = N P ja. Hence, From 55, N P Z j Y = j = P Z Y. HY Z P Z Y logn + hp Z = Y P Z Y logn + h/ = P Z Y logn + log. Therefore, P Z Y logn HY Z log = HY IY ; Z log = log N IY ; Z log log N β log. 56 The last inequality follows since IY ; Z IY ; X = N KLP j, P N KLP j, P k β 57 j,k where P = N N P j and we used the convexity of K. Equation 56 shows that and the result follows. P Z Y logn log N β log 8

29 4.3 Tsybakov s Method Proof of Theorem 4. Let X = X,..., X n and let ψ ψx 0,,..., N}. Fix τ > 0 and define } dp 0 A j = τ. dp j Then P 0 ψ 0 = = τ τ P 0 ψ = j P 0 ψ = j A j P 0 ψ = j A j P j ψ = j P j ψ = j A j A j = τn N P j ψ = j A j P j ψ = j τ P j A c j P j ψ = j τn N = τnp 0 a P j A c j where p 0 = N P j ψ = j, a = N dp0 P j < τ. dp j 9

30 Hence, max P jψ j = max P 0 ψ 0, max P jψ j 0 j N j N max τnp 0 a, max P jψ j j N } max τnp 0 a, P j ψ j N } τnp 0 a, p 0 = max min 0 p max = [ τn + τn N τnp a, p } = } } ] dp0 P j τ. dp j τn a + τn In Lemma 9 below we show that N dp0 P j τ dp j a + a / log/τ where α = N j KP j, P 0. Choosing τ = / N we get N max P jψ j 0 j N + a + a / N log/τ N = + a + a / N log N = By Theorem 3, R n s 3 8 s 8. Lemma 9 Let a = N j KP j, P 0. Then, N dp0 P j τ dp j a + a /. log/τ 30

31 Proof. Note that dp0 dpj P j τ = P j dp j dp 0 τ = P j log dp j log dp 0. τ By Markov s inequality, P j log dp [ j log P j log dp ] j dp 0 τ dp 0 + log τ [ log dp ] j dp j. log/τ dp 0 + According to Pinsker s second inequality see Thheorem 5 in the appendix and Tsyabakov Lemma.5, [ log dp ] j dp j KP j, P 0 + KP j, K 0 /. dp 0 + So P j log dp [ ] j log KP j, P 0 + KP j, K 0 /. dp 0 τ log/τ Using Jensen s inequality, N j KP j, P 0 N KP j, P 0 = a. j So N dp0 P j τ KP j, P 0 dp j log/τ N log/τ N j a log/τ a / log/τ. j KP j, P 0 / 4.4 Assouad s Lemma Assouad s Lemma is another way to get a lower bound using hypercubes. Let } Ω = ω = ω,..., ω N : ω j 0, } be the set of binary sequences of length N. Let P = P ω : ω Ω} be a set of N distributions indexed by the elements of Ω. Let hω, ν = N Iω j ν j be the Hamming distance between ω, ν Ω. 3

32 Lemma 30 Let P ω : ω Ω} be a set of distributions indexed by ω and let θp be a parameter. For any p > 0 and any metric d, max E ω d p θ, θp ω N d p θp ω, θp ν min min P ω Ω p+ ω,ν hω, ν ω,ν ω P ν. 58 hω,ν 0 hω,ν= For a proof, see van der Vaart 998 or Tsybakov The Bayesian Connection Another way to find the minimax risk and to find a minimax estimator is to use a carefully constructed Bayes estimator. In this section we assume we have a parametric family of densities px; θ : θ Θ} and that our goal is to estimate the parameter θ. Since the distributions are indexed by θ, we can write the risk as Rθ, θ n and the maximum risk as sup θ Θ Rθ, θ n. Let Q be a prior distribution for θ. The Bayes risk with respect to Q is defined to be B Q θ n = Rθ, θ n dqθ 59 and the Bayes estimator with respect to Q is the estimator θ n that minimizes B Q θ n. For simplicity, assume that Q has a density q. The posterior density is then qθ X n = px..., X n ; θqθ mx,..., X n where mx,..., x n = px,..., x n ; θqθdθ. Lemma 3 The Bayes risk can be written as Lθ, θ n qθ x,..., x n dθ mx,..., x n dx dx n. It follows from this lemma that the Bayes estimator can be obtained by finding θ n = θ n x..., x n to minimize the inner integral Lθ, θ n qθ x,..., x n dθ. Often, this is an easy calculation. Example 3 Suppose that Lθ, θ n = θ θ n. Then the Bayes estimator is the posterior mean θ Q = θ qθ x,..., x n dθ. 3

33 Now we link Bayes estimators to minimax estimators. Theorem 33 Let θ n be an estimator. Suppose that i the risk function Rθ, θ n is constant as a function of θ and ii θ n is the Bayes estimator for some prior Q. Then θ n is minimax. Proof. We will prove this by contradiction. Suppose that θ n is not minimax. Then there is some other estimator θ such that sup θ Θ Rθ, θ < sup Rθ, θ n. 60 θ Θ Now, B Q θ = Rθ, θ dqθ definition of Bayes risk sup Rθ, θ θ Θ average is less than sup < sup Rθ, θ n from 60 = θ Θ Rθ, θ n dqθ since risk is constant = B Q θ n definition of Bayes risk. So B Q θ < B Q θ n. This is a contradiction because θ n is the Bayes estimator for Q so it must minimize B Q. Example 34 Let X Binomialn, θ. The mle is X/n. Let Lθ, θ n = θ θ n. Define X θ + n 4n n =. + Some calculations show that this is the posterior mean under a Betaα, β prior with α = β = n/4. By computing the bias and variance of θ n it can be seen that Rθ, θ n is constant. Since θ n is Bayes and has constant risk, it is minimax. n Example 35 Let us now show that the sample mean is minimax for the Normal model. Let X N p θ, I be multivariate Normal with mean vector θ = θ,..., θ p. We will prove that θ n = X is minimax when Lθ, θ n = θ n θ. Assign the prior Q = N0, c I. Then the posterior is c x N + c, c + c I. 6 33

34 The Bayes risk B Q θ n = Rθ, θ n dqθ is minimized by the posterior mean θ = c X/ + c. Direct computation shows that B Q θ = pc /+c. Hence, if θ is any estimator, then pc + c = B Q θ B Q θ = Rθ, θdqθ sup Rθ, θ. θ This shows that RΘ pc / + c for every c > 0 and hence But the risk of θ n = X is p. So, θ n = X is minimax. RΘ p Nonparametric Maximum Likelihood and the Le Cam Equation In some cases, the minimax rate can be found by finding ɛ to solve the equation Hɛ n = nɛ n where Hɛ = log Nɛ and Nɛ is the smallest number of balls of size ɛ in the Hellinger metric needed to cover P. Hɛ is called the Hellinger entropy of P. The equation Hɛ = nɛ is known as the Le Cam equation. In this section we consider one case where this is true. For more general versions of this argument, see Shen and Wong 995, Barron and Yang 999 and Birgé and Massart 993. Our goal is to estimate the density function using maximum likelihood. The loss function is Hellinger distance. Let P be a set of probability density functions. We have in mind the nonparametric situation where P does not correspond to some finite dimensional parametric family. Let Nɛ denote the Hellinger covering number of P. We will make the following assumptions: A We assume that there exist 0 < c < c < such that c px c for all x and all p P. A We assume that there exists a > 0 such that where Bp, δ = q : hp, q δ}. Haɛ, P, h sup Hɛ, Bp, 4ɛ, h p P A3 We assume nɛ n as n where Hɛ n nɛ n. 34

35 Assumption A is a very strong and is made only to make the proofs simpler. Assumption A says that the local entropy and global entropy are of the same order. This is typically true in nonparametric models. Assumption A3 says that the rate of convergence is slower than O/ n which again is typical of nonparametric problems. An example of a class P that satisfies these conditions is P = p : [0, ] [c, c ] : 0 pxdx =, 0 p x dx C }. Thanks to A we have, p q KLp, q χ p, q = p q p c = p q p + q c where C = 4c /c. 4c c p q = Ch p, q 63 Let ɛ n solve the Le Cam equation. More precisely, let } ɛ ɛ n = min ɛ : H nɛ. 64 C 6C We will show that ɛ n is the minimax rate. Upper Bound. To show the upper bound, we will find an estimator that achieves the rate. Let P n = p,..., p N } be an ɛ n / C covering set where N = Nɛ n / C. The set P n is an approximation to P that grows with sample size n. Such a set is called sieve. Let p be the mle over P n, that is, p = argmax p Pn Lp where Lp = n i= px i is the likelihood function. We call p, a sieve maximum likelihood estimator. It is crucial that the estimator is computed over P n rather than over P to prevent overfitting. Using a sieve is a type of regularization. We need the following lemma. Lemma 36 Wong and Shen Let p 0 and p be two densities and let δ = hp 0, p. Let Z,..., Z n be a sample from p 0. Then Lp P Lp 0 > / e e nδ nδ /4. 35

36 Proof. Lp P Lp 0 > / e nδ n n pz i = P p i= 0 Z i > /4 e nδ e nδ /4 pz i E p i= 0 Z i n n = e nδ /4 pz i E = e nδ /4 p0 p p 0 Z i n = e nδ /4 h p 0, p = e nδ /4 exp n log h p 0, p e nδ /4 e nh p 0,p/ = e nδ /4. In what follows, we use c, c, c,..., to denote various positive constants. Theorem 37 sup E p hp, p = Oɛ n. Proof. Let p 0 denote the true density. Let p be the element of P n that minimizes KLp 0, p j. Hence, KLp 0, p Cd p 0, p Cɛ n/c = ɛ n/. Let where A = / C. Then Now B = p P n : dp, p > Aɛ n } Ph p, p 0 > Dɛ n Ph p, p + hp 0, p > Dɛ n Ph p, p + ɛ > Dɛ n C = Ph p, p > Aɛ n = P p B P sup P sup p B P sup p B P + P. Lp Lp > e nɛ n A /+ Lp Lp > e nɛ n A / p B Lp Lp > Lp 0 + P sup p B Lp > enɛ n P Lp P Lp > e nɛ n A / Nɛ/ Ce nɛ n A /4 e nɛ n /6C p B where we used Lemma 36 and the definition of ɛ n. To bound P, define K n = n Hence, EK n = KLp 0, p ɛ /. Also, σ Var log p 0Z E log p 0Z c log p Z p Z c c c ɛ = log KLp 0, p log c 3ɛ n c c 36 E log p 0Z p Z n i= log p 0Z i p Z i.

37 where we used 63. So, by Bernstein s inequality, P = PK n > ɛ n = PK n KLp 0, p > ɛ n KLp 0, p P K n KLp 0, p > ɛ n nɛ 4 n exp 8σ + c 4 ɛ n exp c 5 nɛn. Thus, P + P exp c 6 nɛ n. Now, Eh p, p 0 = Ph p, p 0 > tdt 0 Dɛn = Ph p, p 0 > tdt + Ph p, p 0 > tdt 0 Dɛ n Dɛ n + exp c 6 nɛn c7 ɛ n. Lower Bound. Now we derive the lower bound. Theorem 38 Let ɛ n be the smallest ɛ such that Haɛ 64C nɛ. Then inf p Proof. Pick any p P. Let B = q : packing set for B. Then Hence, for any P j, P k F, sup E p hp, p = Ωɛ n. hp, q 4ɛ n }. Let F = p,..., p N } be an ɛ n N = log P ɛ n, B, h log Hɛ n, B, h log Haɛ n 64C nɛ. KLP n j, P n k = nklp j, P k Cnh P j, P k 6Cnɛ n N 4. It follows from Fano s inequality that as claimed. inf p sup E p hp, p p P 4 min hp j, p k ɛ n j k 4 In summary, we get the minimax rate by solving Hɛ n nɛ n. Now we can use the Le Cam equation to compute some rates: 37

38 Example 39 Here are some standard examples: Space Entropy Rate Sobolev α ɛ /α n α/α+ Sobolev α dimension d ɛ d/α n α/α+d Lipschitz α ɛ d/α n α/α+d Monotone /ɛ n /3 Besov Bp,q α ɛ d/α n α/α+d Neural Nets see below see below m-dimensional m log/ɛ m/n parametric model In the neural net case we have fx = c 0 + i c iσv T i x + b i where c C, v i = and σ is a step function or a Lipschitz sigmoidal function. Then and hence /+/d Hɛ ɛ /+/d log/ɛ 65 ɛ n +/d/+/d log n +/d+/d/+/d ɛ n n/ log n +/d/+/d