ON THE DEPENDENCE OF THE

BERRY–ESSEEN BOUND ON DIMENSION

V. Bentkus^{1}

June 2001

Abstract. Let X be a random vector with values in ℝ^{d}. Assume that X has mean zero and identity covariance. Write β = E |X|^{3} and a_{d} = sup_{ε>0} sup_{A∈C} Φ((∂A)^{ε})/(2ε), where C is the class of convex subsets of ℝ^{d}, the measure Φ is the standard d-dimensional normal distribution, and (∂A)^{ε} is ε-neighborhood of the boundary ∂A of the set A. For Δ_{n} = sup_{A∈C} |ℙ\{S_{n} ∈ A\} − Φ(A)|, we prove a Berry–Esseen bound Δ_{n} ≤ 100b_{d}β/\sqrt{n} with b_{d} = max{1, a_{d}}, where S_{n} is a normalized sum of n independent copies of X. This result reduces estimation of Δ_{n} to estimation of the constants a_{d}. The best known to the date estimate of a_{d} is a_{d} ≤ 4d^{1/4}, which yields Δ_{n} ≤ 400d^{1/4}β/\sqrt{n}.

1. Introduction and results

We provide a Berry–Esseen type bound with apparently optimal dependence on dimension. In essence, the question is reduced to a question about optimal dependence on dimension of standard normal d-dimensional measures of convex shells. Our proof seems to be less technically involved compared to the classical proofs using characteristic functions (see Bhattacharya and Rao (1986). It seems to be a difficult problem to prove the result using the Fourier method.

Let ℝ^{d} be a real Euclidean d-dimensional space of vectors x = (x_{1}, …, x_{d}) with the norm |x|^{2} = x_{1}^{2} + ⋯ + x_{d}^{2} and the scalar product ⟨x, x⟩ = |x|^{2}.

Let X, X_{1}, …, X_{n} be i.i.d. random vectors with common mean E X_{j} = 0 and the identity covariance, so that E ⟨X, x⟩^{2} = |x|^{2}. Write β = E |X|^{3}. By Y, Y_{1}, …, Y_{n} we denote i.i.d. standard normal random vectors, that is Gaussian random vectors with mean zero and identity covariance operator. We assume throughout that X’s, Y ’s and

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1Vilnius Institute of Mathematics and Informatics, Lithuania. Research supported by the Max Planck Institute for Mathematics, Bonn.

2000 Mathematics Subject Classification. 60F05.

Key words and phrases. multidimensional, Berry–Esseen bound, dependence on dimension.

other random vectors and variables are independent if the contrary is not clear from the context.

Let A be a class of (measurable) subsets A ⊂ ℝ^{d} (below we assume that all sets and functions are measurable). We shall obtain upper bounds for the quantity

Δ_{n} ≡ Δ_{n}(A) = sup_{A∈A} |ℙ{S_{n} ∈ A} − ℙ{Y ∈ A}|,

where S_{n} = (X_{1} + ⋯ + X_{n})/\sqrt{n}. The bounds have the shape

Δ_{n} ≤ c_{d}(A)β/\sqrt{n},  β = E |X|^{3},  (1.1)

with some constant c_{d}(A) which depends only on the dimension and the class A. For example, for the class B of (Euclidean) balls we have Δ_{n}(B) ≤ cβ/\sqrt{n}, where c is an absolute constant. For the class C of convex subsets of ℝ^{d}, it holds Δ_{n}(C) ≤ 400d^{1/4}β/\sqrt{n}. Whether the dependence on d in the last bound is optimal, remains an open question. One can show (see Nagaev (1976)) that for the class of convex sets Δ_{n}(C) ≥ c_{0}β/\sqrt{n}, where c_{0} > 0 is an absolute positive constant. Hence, the upper and lower bounds for Δ_{n}(C) differ by the factor d^{1/4}. We conjecture that an optimal bound for Δ_{n}(C) has to depend on d as ln^{δ} d with some δ > 0. In view of our result (see Theorem 1.1 below), to prove this it suffices to improve the bound (1.4) for standard Gaussian measures of convex shells, which seems to be a difficult problem.

We need some definitions and conditions related to A and the standard normal distribution Φ(A) = ℙ{Y ∈ A}.

For a class, say A, of subsets A ⊂ ℝ^{d}, consider the following conditions:

i) A is invariant under rescaling, that is, aA ∈ A, if A ∈ A and a number a ∈ ℝ is positive; ii) A is shift invariant, that is, x + A ∈ A, if A ∈ A and x ∈ ℝ^{d}; iii) A is invariant under taking of ε-neighborhoods, for ε > 0. More precisely, A^{ε}, A^{−ε} ∈ A if A ∈ A. Here

A^{ε} = {x ∈ ℝ^{d} : ρ_{A}(x) ≤ ε}  and  A^{−ε} = {x ∈ A : B_{ε}(x) ⊂ A},

where ρ_{A}(x) = inf_{y∈A} |x − y| is the distance between A ⊂ ℝ^{d} and x ∈ ℝ^{d}, and B_{ε}(x) = {y ∈ ℝ^{d} : |x − y| ≤ ε}. For example, the class C of convex subsets of ℝ^{d} satisfies the conditions i)–iii). An-

other example provides the class B of Euclidean balls B_{r}(x), r ≥ 0, x ∈ ℝ^{d}. The class of rectangles does not satisfy the condition iii) since an ε-neighborhood of a rectangle is not a rectangle.

We consider the following condition on a class A of subsets of ℝ^{d} and the standard normal distribution Φ(A) = ℙ{Y ∈ A}: there exist constants, say a_{d} = a_{d}(A), depending only on d and A such that

ℙ{Y ∈ A^{ε} \ A} ≤ a_{d}ε, ℙ{Y ∈ A \ A^{−ε}} ≤ a_{d}ε,  for all A ∈ A and ε > 0. (1.2)

For the class B of balls we have sup_{d≥1} a_{d}(B) < ∞. Ball (1993) has proved that

e^{−1}\sqrt{ln d} ≤ sup_{A∈C} ∫_{∂A} p(x) ds ≤ 4d^{1/4},  (1.3)

where p(x) is the standard normal d-dimensional density and ds is the surface measure on the boundary ∂A of A. Using (1.3) one can show that for the class of convex sets

e^{−1}\sqrt{ln d} ≤ a_{d}(C) ≤ 4d^{1/4}.  (1.4)

To find precise asymptotics of a_{d} as d → ∞ remains an open problem; in view of the next Theorem 1.1, a solution of this problem leads to improvements of bounds of type (1.1).

Our result is the following theorem.

Theorem 1.1. Let a class A of convex sets satisfy the conditions i)–iii). Furthermore, assume that A and the standard normal distribution satisfy the condition (1.2). Then

Δ_{n} ≤ 100b_{d}β/\sqrt{n}, b_{d} = max\{1, a_{d}\}.

In particular, for the class of convex sets we have Δ_{n} ≤ 400d^{1/4}β/\sqrt{n}. In the case of the class of all Euclidean balls the bound improves to Δ_{n} ≤ cβ/\sqrt{n}, where c is an absolute constant.

The constants in Theorem 1.1 can be considerably improved, especially for larger d. Our aim was just to provide explicit constants keeping the proof and notation as simple as possible. The conditions i)–iii) on the class A can be relaxed using a little more refined techniques.

The question of the dependence of the Berry–Esseen bound on dimension has drawn attention of many authors, let us mention here only Nagaev (1976), Senatov (1980), Sazonov (1981), Götze (1991), where the bounds depend on the dimension as d. Our approach is related to that of Bentkus (1986a), where the bound depends on d as \sqrt{d}.

The proof of Theorem 1.1 is based on induction in n and Taylor expansions. Usually such proofs are very robust against extensions to general spaces and abstract settings, and can be used as well to prove large deviation theorems (see Bentkus (1986b)). Direct inductional proofs is a commonly used tool to obtain convergence rates in the CLT, see, for example, books Sazonov (1981), Paulauskas and Račkauskas (1989), Senatov (1998). A feature of our proof is that we apply a special smoothing with differentiable functions approximating indicator functions of convex sets. The smoothness properties of these functions rely on nice differentiability properties of Euclidean distances. In the context of limit theorems smoothness of Euclidean distances were earlier applied by Senatov (1984) to estimate Prokhorov’s metric.

2. Auxiliary Lemmas

We provide the proof of Theorem 1.1 a little later since first we would like to formulate and prove several auxiliary Lemmas.

Results like the next (very simple technically) Lemma 2.1 usually are called smoothing inequalities or lemmas. Smoothing inequalities allow to replace probabilities like ℙ{S_{n} ∈ A} = E 𝕀{S_{n} ∈ A}, which are expectations of non-differentiable indicator functions x ∣→ 𝕀{x ∈ A}, by expectations of smooth functions, say φ. Such a replacement allows to apply the differential calculus. In Lemma 2.1 we use smooth functions x ∣→ φ(x) which approximate indicator functions. Another often used way to make smoothing is via convolutions with certain measures and Fourier transforms (or characteristic functions), that is, via the family of exponential functions φ_{t}(x) = exp{i⟨t, x⟩}, t ∈ ℝ^{d}, see Bhattacharya and Rao (1986).

Lemma 2.1. Let ε > 0 be a positive number. Let a set A ⊂ ℝ^{d} and functions 0 ≤ φ_{1}, φ_{2} ≤ 1 satisfy

φ_{1}(x) = 1, for x ∈ A, φ_{1}(x) = 0, for x ∉ A^{ε},
φ_{2}(x) = 1, for x ∈ A^{−ε}, φ_{2}(x) = 0, for x ∉ A.

Then, for arbitrary r. v. X and Y, the difference γ(A) = ℙ{X ∈ A} − ℙ{Y ∈ A} satisfies

|γ(A)| ≤ max_{k=1,2} |γ(φ_{k})| + max {ℙ{Y ∈ A^{ε} \ A}, ℙ{Y ∈ A \ A^{−ε}}},  (2.1)

where γ(φ_{k}) = E φ_{k}(X) − E φ_{k}(Y).

Let a class A of sets satisfy the condition iii). Assume that we have a family, say {φ_{ε,A} : A ∈ A}, of functions φ_{ε,A} : ℝ^{d} → ℝ such that 0 ≤ φ_{ε,A} ≤ 1 and φ_{ε,A}(x) = 1 for x ∈ A, φ_{ε,A}(x) = 0, for x ∉ A^{ε}. Then

sup_{A∈A} |γ(A)| ≤ sup_{A∈A} |γ(φ_{ε,A})| + max { sup_{A∈A} ℙ{Y ∈ A^{ε} \ A}, sup_{A∈A} ℙ{Y ∈ A \ A^{−ε}}} (2.2)

with γ(φ_{ε,A}) = E φ_{ε,A}(X) − E φ_{ε,A}(Y).

Proof. We prove only (2.2) since the proof of (2.1) is simpler. We consider first the case γ(A) ≥ 0. Writing for brevity φ = φ_{ε,A}, we have

|γ(A)| = γ(A) ≤ E φ(X) − E φ(Y) + E φ(Y) − ℙ{Y ∈ A}
≤ | E φ(X) − E φ(Y)| + ℙ{Y ∈ A^{ε} \ A}  (2.3)

since the function φ satisfies 0 ≤ φ ≤ 1 and equals 1 on the set A. Taking sup_{A∈A} in (2.3)

leads to (2.2).

Now consider the case γ(A) < 0. We have to estimate −γ(A) = ℙ{Y ∈ A} − ℙ{X ∈ A}. The proof depends on whether A^{−ε} = ∅ or A^{−ε} ≠ ∅.

If A^{−ε} = ∅ then ℙ{Y ∈ A^{−ε}} = 0 and

−γ(A) ≤ ℙ{Y ∈ A} = ℙ{Y ∈ A} − ℙ{Y ∈ A^{−ε}} ≤ ℙ{Y ∈ A \ A^{−ε}}.  (2.4)

Taking sup_{A∈A} in (2.4) we get (2.2).

If A^{−ε} ≠ ∅, we can in essence repeat the proof of the case γ(A) ≥ 0. Writing for brevity φ = φ_{ε,A^{−ε}}, we have

−γ(A) ≤ ℙ{Y ∈ A} − E φ(Y) + E φ(Y) − E φ(X)
≤ ℙ{Y ∈ A \ A^{−ε}} + | E φ(X) − E φ(Y)|  (2.5)

since the function φ satisfies 0 ≤ φ ≤ 1 and equals 1 on the set A^{−ε}. Noting that by the condition iii) A^{−ε} ∈ A and taking in (2.5) sup_{A^{−ε}} ≤ sup_{A}, we obtain (2.2). □

Lemma 2.2. Let A ⊂ ℝ^{d} be a closed convex set. Then the distance function

ρ(x) = min{|x − y| : y ∈ A}

satisfies |ρ′(x)| ≤ 1, for x ∈ A^{c} = ℝ^{d} \ A. Furthermore, its derivative satisfies the following Lipschitz type condition

|ρ′(x) − ρ′(y)| ≤ |x − y|/r,  for all x, y ∈ A^{c},  (2.6)

with r = min{ρ(x), ρ(y)}.

Proof. Without loss of generality we assume that r = ρ(x) ≥ s = ρ(y).

For x ∈ A^{c}, by x_{0} we denote a unique point x_{0} ∈ A closest to x. Consider the unit vector e_{1} = (x − x_{0})/|x − x_{0}| and the ray R(x_{0}) = {x_{0} + te_{1} : t > 0}. Then ρ′(z) = e_{1}, for all z ∈ R_{1}. Let e_{2} and R(y_{0}) be a unit vector and ray corresponding to y_{0}. It is clear that x = x_{0} +re_{1} ∈ ℝ(x_{0}) and y = y_{0} +se_{2} ∈ ℝ(y_{0}). Introduce the hyper-plane P_{1} (resp. P_{2}) which is a hyper-plane orthogonal to the ray R_{1} (resp. R_{2}) such that x_{0} ∈ P_{1} (resp. y_{0} ∈ P_{2}). Analytically P_{1} is given by P_{1} = {x_{0} + u ∈ ℝ^{d} : ⟨u, e_{1}⟩ = 0}. It is clear that the set A is a subset of the half-space P_{1}^{−} = {x_{0} + u ∈ ℝ^{d} : ⟨u, e_{1}⟩ ≤ 0}.

We consider separately the following to cases:

i) P_{1} ∩ P_{2} ≠ ∅;

ii) P_{1} ∩ P_{2} = ∅.

i) There exists z ∈ P_{1} ∩ P_{2}. Consider the rays

R_{3} = {z + te_{1} : t > 0}  and  R_{4} = {z + te_{2} : t > 0}

and points \overline{x} = z + re_{1} ∈ R_{3} and \overline{y} = z + se_{2} ∈ R_{4}. It is clear that

|\overline{x} − \overline{y}| ≤ |x − y|.  (2.7)

Indeed, A ⊂ P_{1}^{−} ∩ P_{2}^{−} since A is a convex set. The angle between planes P_{1} and P_{2} is bounded from above by π since A is convex. Therefore, if we have two rays, say R(u) and R(v), then the distance between points u + re_{1} ∈ R(u) and v + se_{2} ∈ R(v) becomes minimal when u = v ∈ P_{1}∩P_{2}, which proves (2.7) (this fact is obvious geometrically and it is not difficult to provide an analytical proof of it). The inequality (2.7) and ρ′(x) = e_{1}, ρ′(y) = e_{2} show that (2.6) is implied by the following inequality

|e_{1} − e_{2}| ≤ |\overline{x} − \overline{y}|/r = |e_{1} − se_{2}/r|.  (2.8)

The inequality (2.8) is just a simple fact from the planar geometry. Nonetheless, let us provide an analytical proof of it. Taking the square, (2.8) reduces to

2 − 2⟨e_{1}, e_{2}⟩ ≤ 1 + \frac{s^{2}}{r^{2}} − 2 \frac{s}{r} ⟨e_{1}, e_{2}⟩.  (2.9)

Note that 0 ≤ ⟨e_{1}, e_{2}⟩ ≤ 1 since the angle between planes P_{1} and P_{2} is bounded from above by π. Hence, for s ≥ r, the right hand side of (2.9) attains its minimal value 2 − 2⟨e_{1}, e_{2}⟩ when s = r, which proves (2.9) and (2.6) in the case i).

ii) Now the planes P_{1} and P_{2} are parallel and e_{2} = −e_{1}. Therefore ρ′(x) = e_{1} and ρ′(y) = −e_{1}. The inequality (2.6) takes the form 2 ≤ |x_{0} − y_{0} + (r + s)e_{1}|/r. It is clear geometrically that |x_{0} − y_{0} + (r + s)e_{1}| assumes its minimal value when P_{1} = P_{2}, x_{0} = y_{0} and s = r. This value is 2, which proves ii) and the lemma. □

Lemma 2.3. Let a set A ⊂ ℝ^{d} be convex. Then for any ε > 0 there exist a function φ (which depends only on ε and A) such that

φ(x) = 1, for x ∈ A,  φ(x) = 0, for x ∈ ℝ^{d} \ A^{ε},  0 ≤ φ ≤ 1,  (2.10)

and

|φ′(x)| ≤ \frac{2}{ε} ,  |φ′(x) − φ′(y)| ≤ \frac{8|x − y|}{ε^{2}} .  (2.11)

Furthermore, we can choose φ to have the form φ(x) = ψ(ρ(x)/ε), where ρ(x) is the distance between x and A, and ψ : ℝ → ℝ is a continuously differentiable non-negative non-increasing function such that ∫_{ℝ} |ψ′(t)| dt = 1.

Proof. Without loss of generality we assume that the set A is closed.

Let ψ : ℝ → [0, 1] be a function such that ψ′′(x) = 0, for x ≤ 0 or x ≥ 1, and ψ′′(x) = −4, for 0 < x ≤ 1/2, ψ′′(x) = 4, for 1/2 < x < 1. We define

ψ′(x) = ∫_{−∞}^{x} ψ′′(t)dt  and  ψ(x) = 1 + ∫_{−∞}^{x} ψ′(t)dt.

The function ψ is non-increasing and has the following properties: ψ(x) = 1, for x ≤ 0, ψ(x) = 0, for x ≥ 1, 0 ≤ ψ ≤ 1,

|ψ′(t)| ≤ 4|t|,  |ψ′(t)| ≤ 2,  |ψ′(t) − ψ′(s)| ≤ 4|t − s|.  (2.12)

We define φ(x) = ψ(ρ(x)/ε). The function φ clearly satisfies the relations (2.10). Using Lemma 2.2 and the properties of ψ, we have

|φ′(x)| = ε^{−1}|ψ′(ρ(x)/ε)ρ′(x)| ≤ 2/ε,

which proves the first inequality in (2.11).

Let us prove the second inequality in (2.11). Let x, y ∈ ℝ^{d}.

If both x, y ∈ A, then φ′(x) = φ′(y) = 0 and the second inequality in (2.11) is obviously fulfilled. To check that φ′(x) = 0, for x ∈ A, we consider two cases: i) x is in the interrior A \ ∂A of A; ii) x ∈ ∂A. In the case i) φ′(x) = 0 since φ is a constant function in the open set A \ ∂A. In the case ii), using |ψ′(t)| ≤ 4|t|, we have

|φ(x + h) − φ(x)| ≤ ∫_{0}^{ρ(x+h)}|ψ′(t)| dt ≤ 4ρ^{2}(x + h) ≤ 4|h|^{2},

which implies φ′(x) = 0.

Now let both x, y ∈ A^{c}. We can assume that r = ρ(x) ≤ s = ρ(y). Lemma 2.2 and properties of ψ yield

|φ′(x) − φ′(y)| = ε^{−1}|ψ′(r/ε)ρ′(x) − ψ′(s/ε)ρ′(y)|
≤ ε^{−1}|ψ′(r/ε)| |ρ′(x) − ρ′(y)| + ε^{−1}|ψ′(r/ε) − ψ′(s/ε)| |ρ′(y)|
≤ \frac{4r}{ε^{2}} \frac{|x − y|}{r} + \frac{4}{ε^{2}} |r − s| ≤ \frac{8}{ε^{2}} |x − y|,

which proves the second inequality in (2.11).

It remains to consider the case x ∈ A and y ∈ A^{c}. Now φ′(x) = 0 and the second inequality in (2.11) takes the form |φ′(y)| ≤ 8|x − y|/ε^{2}. Write s = ρ(y). Using |ψ′(t)| ≤ 4|t|, we have

|φ′(y)| = ε^{−1} |ψ′(s/ε)| ≤ \frac{4s}{ε^{2}} ≤ \frac{4|x − y|}{ε^{2}}

since s = ρ(y) ≤ |x − y|. This completes the proof of (2.11) and of the lemma. □

The next two simple technical Lemmas will be often used in the forthcoming proof of Theorem 1.1. To prove the bound Δ_{n} ≤ cb_{d}β/\sqrt{n} with an inexplicit absolute constant c, all we need is that expectations and integrals in Lemmas 2.4–2.5 are bounded from above by c_{0}β (with an absolute constant c_{0}), which is almost obvious.

Lemma 2.4. Write σ = 2/\sqrt{2π}. We have

1 ≤ d^{−3/2}β, E |X| ≤ β, E |Y | ≤ σβ ≤ β, E |Y |^{3} ≤ 2σβ ≤ 2β  (2.13)
E |X|^{2}|⟨X, Y ⟩| ≤ σβ ≤ β, E |Y_{1}|^{2}|⟨Y_{1}, Y ⟩| ≤ 2σ^{2}β ≤ 2β,  (2.14)

E |⟨X, Y ⟩| ≤ σβ ≤ β, E |⟨Y_{1}, Y ⟩| ≤ σ^{2}β ≤ β,
E |⟨X, Y ⟩|^{3} ≤ 2σβ ≤ 2β, E |⟨Y_{1}, Y ⟩|^{3} ≤ 4σ^{2}β ≤ 3β.  (2.15)

Proof. To prove that β ≥ d^{3}/2, it suffices to notice that d = E |X|^{2} ≤ β^{2/3}. The inequality E |X| ≤ β is implied by an application of Hölder inequality and β ≥ 1.

To prove other inequalities, let us consider first the case d = 1. Then ⟨X, Y ⟩ = XY, the random variable Y is a standard normal variable, and (2.13)–(2.15) are implied by an application of E |Y | = σ and E |Y |^{3} = 2σ.

Henceforth we assume that d ≥ 2. Let η, η_{j} be independent real standard normal variables. Using that coordinates η_{i} of Y = (η_{1}, …, η_{d}) are independent and identically distributed, E η^{4} = 3 and E η^{2} = 1, we have

E |Y |^{4} = d E η^{4} + (d^{2} − d)( E η^{2})^{2} = d^{2} + 2d.

Therefore, for d ≥ 2, we have

E |Y |^{3} ≤ ( E |Y |^{2})^{1/2}( E |Y |^{4})^{1/2} = d^{3/2}\sqrt{1 + 2/d} ≤ \sqrt{2}β ≤ 2σβ.

Using polar coordinates and integrating by parts we have

E |Y |^{3} = s_{d−1} ∫_{0}^{∞}r^{d+2} exp{−r^{2}/2}dr = (d + 1)s_{d−1} ∫_{0}^{∞}r^{d}{−r^{2}/2}dr = (d + 1) E |Y |,

where s_{d−1} is the area of the unit sphere {x ∈ ℝ^{d} : |x| = 1}. Hence, E |Y |^{3} ≥ 2 E |Y |, for d ≥ 2. Combining this inequality with the already proved E |Y |^{3} ≤ 2σβ, we conclude the proof of (2.13).

Let us prove (2.14) and (2.15). Given X, the random variable ⟨X, Y ⟩ is a mean zero normal random variable with variance |X|^{2}. Therefore, conditioning on X, we have E |X|^{2}|⟨X, Y ⟩| = σ E |X|^{3} = σβ. The proof of the second inequality in (2.14) is similar. We have E |⟨X, Y ⟩| = σ E |X| ≤ σβ and E |⟨X, Y ⟩|^{3} = 2σ E |X|^{3} = 2σβ. The proof of the last inequality in (2.15) is similar. □

Lemma 2.5. Let

p(y) = (2π)^{−d/2} exp{−|y|^{2}/2},  y ∈ ℝ^{d},  (2.16)

be the standard normal d-dimensional density. Then we have

A = E ∫_{ℝ^{d}} |p′(y)X| dy ≤ σβ ≤ β,  B = E ∫_{ℝ^{d}} |p′(y)Y | dy ≤ σ^{2}β ≤ β, (2.17)

and

C = E ∫_{ℝ^{d}} |p′′′(y)X^{3}| dy ≤ ζβ ≤ 2β,  D = E ∫_{ℝ^{d}} |p′′′(y)Y^{3}| dy ≤ 2σζβ ≤ 3β (2.18)

with σ = 2/\sqrt{2π} and ζ = 2(1 + 4 e^{−3/2})/\sqrt{2π}.

Proof. We have

p′(x)h = −⟨x, h⟩p(x),  p′′′(x)h^{3} = (3|h|^{2}⟨x, h⟩ − ⟨x, h⟩^{3})p(x).  (2.19)

Let us prove (2.17). Using (2.19) we can write A = E |⟨Y, X⟩|. Applying (2.15) we get A ≤ σβ. Similarly B ≤ σ^{2}β.

To prove (2.18) we notice that

E |3|h|^{2}⟨Y, h⟩ − ⟨Y, h⟩^{3} | = |h|^{3} E |3η − η^{3}| = 2(1 + 4 e^{−3/2})/\sqrt{2π},

where η is a standard real normal random variable. Applying (2.13), we derive (2.18). □

For a sufficiently smooth function f : ℝ^{d} → ℝ, we shall apply the Taylor formula

f(x+h) = f(x)+f′(x)h+⋯+ \frac{1}{s!} f^{(s)}(x)h^{s} + \frac{1}{s!} E (1−τ)^{s}f^{(s+1)}(x+τh)h^{s+1}, (2.20)

where τ is a random variable uniformly distributed on the interval [0, 1].

3. Proof of Theorem 1.1

The proof is based on induction in n. We shall prove that

Δ_{n} ≤ Mb_{d}δ,  δ =^{def} β/\sqrt{n},  (3.1)

holds for all n provided that an absolute constant M ≥ 10 is sufficiently large (recall that b_{d} = max{1, a_{d}}).

Since β ≥ 1 and Δ_{n} ≤ 1, for n ≤ M^{2}β^{2} we have

Δ_{n} ≤ 1 ≤ b_{d}\sqrt{M^{2}β^{2}/n} = Mb_{d}δ.  (3.2)

Hence, (3.1) holds for n ≤ 100 ≤ M^{2}β^{2}. In particular, (3.1) holds for n = 1.

Now assume that (3.1) holds for 1, …, n − 1. We shall prove that then (3.1) holds for n as well. In the proof we can assume that n ≥ M^{2}β^{2} ≥ 100, cf. (3.2).

Let ε > 0 be a positive number (later we shall choose ε = aδ with a sufficiently large absolute constant a > 0). Applying the smoothing Lemma 2.1, the condition (1.2) and using a_{d} ≤ b_{d}, we have

Δ_{n} ≤ sup_{A∈A} | E φ(S_{n}) − E φ(Y)| + b_{d}ε,  (3.3)

where a function φ = φ_{ε,A} satisfies the conditions of Lemma 2.1. Fix A ∈ A. By Lemma 2.3 we can choose φ(x) = ψ(ρ(x)/ε), where ρ(x) ≡ ρ_{A}(x) is the distance between x and A, and φ and ψ satisfy the conditions of Lemma 2.3.

To simplify the notation, we write

U = X/\sqrt{n},  U_{k} = X_{k}/\sqrt{n},  V = Y/\sqrt{n},  V_{k} = Y_{k}/\sqrt{n}.

Write

W_{k} = V_{1} + ⋯ + V_{k−1} + U_{k+1} + ⋯ + U_{n},  Z_{n} = V_{1} + ⋯ + V_{n}.  (3.4)

Due to the i.i.d. assumption, S_{n} is distributed as W_{1} +U. Moreover, Z_{n}, W_{n} +V and Y are i.i.d. standard normal random vectors. We have

| E φ(S_{n}) − E φ(Y)| = | E φ(W_{1} + U) − E φ(W_{n} + V)|
≤ | E φ(W_{1} + U) − E φ(W_{1} + V)| + | E φ(W_{1} + V) − E φ(W_{n} + V)|
= | E φ(W_{1} + U) − E φ(W_{1} + V)| + | E φ(W_{2} + U) − E φ(W_{n} + V)| (3.5)

since by the i.i.d. assumption E φ(W_{1} + V) = E φ(W_{2} + U). Repetitions of arguments used to obtain (3.5), lead to

| E φ(S_{n}) − E φ(Y)| ≤ γ_{1} + ⋯ + γ_{n},  (3.6)

where

γ_{k} = | E φ(W_{k} + U) − E φ(W_{k} + V)|.

Introduce

θ^{2} = n/(k − 1),  α^{2} = n/(n − k),  Z = U_{k+1} + ⋯ + U_{n}.  (3.7)

Then, for 2 ≤ k ≤ n, we can write

γ_{k} = | E φ(Y_{1}/θ + U + Z) − E φ(Y_{1}/θ + V + Z)|.  (3.8)

Write n_{0} = [n/2], where [z] stands for the integer part of z. Below we shall prove that, for any m = 2, …, n_{0},

γ_{1} ≤ b_{d}(c_{1}δ + c_{2}Mδ^{2}/ε),  (3.9)
γ_{2} + ⋯ + γ_{m} ≤ ^{\frac{βb_{d}\sqrt{m − 1}}{ε^{2}n}} (c_{3}Mδ + c_{4}ε),  (3.10)
γ_{m+1} + ⋯ + γ_{n_{0}} ≤ \frac{c_{5}βb_{d}Mδ}{\sqrt{m − 1}} + c_{6}b_{d}δ,  (3.11)
γ_{n_{0}+1} + ⋯ + γ_{n} ≤ c_{7}b_{d}δ  (3.12)

with some absolute constants c_{1}, …, c_{7}. These constants can be chosen to be

c_{1} = (2 + 2σ)\sqrt{100/99}, c_{2} = 2c_{1}, c_{3} = 2c_{4}, c_{4} = 16\sqrt{2}(σ + 2σ^{2})/3,
c_{5} = ζ(1 + 2σ)\sqrt{2}/3, c_{6} = c_{5}\sqrt{100/98}, c_{7} = c_{5}\sqrt{100/97}  (3.13)

with σ = 2/\sqrt{2π} and ζ = 2(1 + 4 e^{−3/2})/\sqrt{2π}.

The bounds (3.9)–(3.12) imply the desired bound for Δ_{n}. Indeed, summing the inequalities (3.9)–(3.12), using (3.3) and setting ε = aδ with some 1 ≤ a ≤ M to be chosen later, we have (below c_{8} = c_{8}(c_{1}, …, c_{7}) and c_{9} = c_{9}(c_{8}) are sufficiently large absolute constants)

Δ_{n} ≤ b_{d}δ(a + c_{8} + c_{8} ^{\frac{\sqrt{m − 1}}{aβ}} + c_{8}M( \frac{1}{a} + ^{\frac{\sqrt{m − 1}}{a^{2}β}} + \frac{β}{\sqrt{m − 1}} ))
≤ b_{d}δ(a + c_{9}M( \frac{1}{a} + ^{\frac{\sqrt{m − 1}}{a^{2}β}} + \frac{β}{\sqrt{m − 1}} ))

using \frac{M}{a} ≥ 1. Therefore, the inequality Δ_{n} ≤ Mb_{d}δ reduces to checking that

a + c_{9}M( \frac{1}{a} + ^{\frac{\sqrt{m − 1}}{a^{2}β}} + \frac{β}{\sqrt{m − 1}} ) ≤ M.  (3.14)

We choose m − 1 such that a^{2}β^{2} ≤ m − 1 ≤ 2a^{2}β^{2}. Then (3.14) is implied by

a + c_{10}M/a ≤ M,  (3.15)

where c_{10} ≥ 1 is an absolute constant. Let us take a = 2c_{10} and M = 10c_{10}. Then (3.15) holds. Notice, that we have to check the condition 1 ≤ m ≤ n/2. Since m − 1 ≤ 2a^{2}β^{2}, it suffices to check that 3a^{2}β^{2} ≤ n/4 or, equivalently, that 12c_{10}^{2}β^{2} ≤ n. This inequality holds since we assume n ≥ M^{2}β^{2} = 100c_{10}^{2}β^{2}. A little more careful choice of the parameters using numerical values (3.13) of constants shows that it suffices to take M ≤ 100, which proves the theorem.

It remains to prove (3.9)–(3.12). We shall prove (3.9)–(3.12) with

c_{1} ≤ 8, c_{2} ≤ 16, c_{3} ≤ 192, c_{4} ≤ 48, c_{5} ≤ 20, c_{6} ≤ 10, c_{7} ≤ 20.

Better constants (3.13) can be obtained using refined versions of inequalities of Lemmas 2.4–2.5, applying n ≥ 100 to estimate better θ and α, and using Eτ = 1/2 and E τ^{2} = 1/3. In order to keep the notation as simple as possible, we omit related details.

Let us prove (3.9). A little later we prove that

γ_{1} ≤ \frac{4β}{ε\sqrt{n}} (2Δ_{n−1} + αb_{d}ε),  α^{2} = n/(n − 1).  (3.16)

Using the induction assumption Δ_{n−1} ≤ Mb_{d}β/\sqrt{(n − 1)} = Mb_{d}δα and estimating α ≤ 2, for n ≥ 2, the inequality (3.16) yields (3.9).

Let us prove the inequality (3.16). Using the Taylor formula (2.20) with s = 0, noting that φ′(x) = φ′(x)𝕀{A^{ε} \ A} since φ is a constant function outside the set A^{ε} \ A^{−ε}, and estimating the derivative φ′ by (2.11) of Lemma 2.3, we have

γ_{1} = | E φ′(W_{1} + τU)U − E φ′(W_{1} + τV)V | ≤ 2 E I(U)/ε + 2 E I(V)/ε  (3.17)

with

I(U) = |U|𝕀{W_{1} + τU ∈ (A^{ε} \ A}, I(V) = |V |𝕀{W_{1} + τV ∈ (A^{ε} \ A}.

We shall prove that

E I(U) ≤ E |U|(2Δ_{n−1} + αb_{d}ε)  E I(V) ≤ E |V |(2Δ_{n−1} + αb_{d}ε).  (3.18)

By (2.13) of Lemma 2.4, we have E |V |, E |U| ≤ β/\sqrt{n}, and (3.17)–(3.18) imply (3.16).

Let us prove (3.18). We shall prove only the bound for E I(U) since a proof of the bound for E I(V) is similar. Conditioning on τ and U, we have

E I(U) = E |U|ℙ{W_{1} ∈ A^{ε} \ A − τU | τ, U } ≤ E |U| sup_{z∈ℝ^{d}} ℙ{W_{1} ∈ A^{ε} \ A + z }

≤ E |U| sup_{A∈A} ℙ{W_{1} ∈ A^{ε} \ A} = E |U| sup_{A∈A} ℙ{S_{n−1}/α ∈ A^{ε} \ A}

= E |U| sup_{A∈A} ℙ{S_{n−1} ∈ A^{αε} \ A}.  (3.19)

We estimate ℙ{S_{n−1} ∈ A^{αε} \ A} = ℙ{S_{n−1} ∈ A^{αε} } − ℙ{S_{n−1} ∈ A} applying the induction assumption. Using in addition the condition (1.2) to estimate ℙ{Y ∈ A^{αε} \ A}, we get

ℙ{S_{n−1} ∈ A^{αε} \ A} ≤ 2Δ_{n−1} + ℙ{Y ∈ A^{αε} \ A} ≤ 2Δ_{n−1} + αb_{d}ε.  (3.20)

Combining (3.19) and (3.20), we obtain (3.18), which concludes the proof of (3.16) and of (3.9).

Let us prove (3.10). We shall show that (now α^{2} = n/(n − k))

γ_{k} ≤ \frac{24β}{ε^{2}n\sqrt{k − 1}} (2Δ_{n−k} + b_{d}αε),  for k = 2, …, n − 1.  (3.21)

This inequality implies (3.10). Indeed, using the induction assumption we can estimate Δ_{n−k} ≤ Mb_{d}αδ. Furthermore, α ≤ 2 since 1 ≤ k ≤ n_{0} = [n/2], and

∑_{k=2}^{m} \frac{1}{\sqrt{k − 1}} ≤ 1 + ∫_{1}^{m} \frac{dt}{\sqrt{t}} = 2\sqrt{m} − 1 ≤ 2\sqrt{m − 1}, for m ≥ 2.

We note that the Lipschitz condition (2.11) implies

|φ′(x) − φ′(y)| ≤ \frac{8|x − y|}{ε^{2}} (𝕀\{x ∈ A^{ε} \ A\} + 𝕀\{y ∈ A^{ε} \ A\}).  (3.22)

Indeed, if at least one of x and y belong to A^{ε}\A, then 𝕀{x ∈ A^{ε}\A}+𝕀{x ∈ A^{ε}\A} ≥ 1, and (3.22) becomes (2.11). If neither x nor y belongs to A^{ε} \ A, then φ′(x) = φ′(y) = 0 since φ is a constant function outside the set A^{ε} \A, and (3.22) is just the identity 0 = 0.

Let us return to the proof of (3.21). Using the notation (3.7)–(3.8), we can write

γ_{k} = |I − J| I =^{def} E φ(Y_{1}/θ + U + Z), J =^{def} E φ(Y_{1}/θ + V + Z).

Let us show that

I = I_{0} + I_{1},  J = I_{0} + J_{1}  (3.23)

with I_{0} = E φ(Y_{1}/θ + Z) and

I_{1} = θ E τ⟨Y_{1}, U⟩φ′(Y_{1}/θ + τ_{1}τU + Z)U,
J_{1} = θ E τ⟨Y_{1}, V ⟩φ′(Y_{1}/θ + τ_{1}τV + Z)V,  (3.24)

where τ and τ_{1} are independent random variables uniformly distributed in the interval [0, 1]. We have

I = E ∫_{ℝ^{d}} φ(x/θ + U + Z)p(x) dx,  (3.25)

where p is standard normal density (2.16). Changing the variable x as x/θ + U = y, we can rewrite (3.25) as

I = E ∫_{ℝ^{d}} φ(y + Z)p(θy − θU)θ^{d} dy.  (3.26)

Given U, expanding the density

p(θy − θU) = p(θy) − θ E p′(θy − τθU)U = p(θy) + θ E ⟨θy − τθU, U⟩p(θy − θτU)

(cf. (2.19)) and changing the variable y in integrals as θy = x and θy − τθU = x, the relation (3.25) yields

I = I_{0} + I_{1}, I_{1} =^{def} θ E ⟨Y_{1}, U⟩φ(Y_{1}/θ + τU + Z).  (3.27)

Expanding the function φ by (2.20) with s = 0 and using E U = 0, we have

I_{1} = θ E τ⟨Y_{1}, U⟩φ′(Y_{1}/θ + τ_{1}τU + Z)U.  (3.28)

The relations (3.25)–(3.28) yield (3.23) for I. The proof for J is similar, we have just to replace everywhere U by V.

Write

I_{2} = θ E τ⟨Y_{1}, U⟩φ′(Y_{1}/θ + Z)U, J_{2} = θ E τ⟨Y_{1}, V ⟩φ′(Y_{1}/θ + Z)V

and notice that I_{2} = J_{2} since the covariances of U and V are equal. Therefore (3.23) implies

γ_{k} = |I − J| = |I_{1} − J_{1}| ≤ |I_{1} − I_{2}| + |J_{1} − J_{2}|.  (3.29)

Applying to the function φ′ the Lipschitz condition (3.22), writing for a while

ξ = \frac{8θ}{ε^{2}} τ_{1}τ^{2}|U|^{2} |⟨Y_{1}, U⟩|

and ℙ_{∗} for the conditional probability given all random variables except Z, we have

|I_{1} − I_{2}| ≤ E ξ(ℙ_{∗}{Y_{1}/θ + τ_{1}τU + Z ∈ A^{ε} \ A} + ℙ_{∗}{Y_{1}/θ + Z ∈ A^{ε} \ A})
≤ 2 E ξ sup_{x∈ℝ^{d}} ℙ{x + Z ∈ A^{ε} \ A}.  (3.30)

Note that Z is a normalized sum of n − k independent copies of X. Therefore, using the induction assumption and arguing similarly to (3.19)–(3.3), we get

ℙ{x + Z ∈ A^{ε} \ A} ≤ 2Δ_{n−k} + b_{d}αε.  (3.31)

In order to estimate E ξ we notice that τ, τ_{1} ≤ 1 and E |U|^{2} |⟨Y_{1}, U⟩| ≤ βn^{−3/2} (cf. (2.14) of Lemma 2.4). We get

E ξ ≤ 8θβ/(ε^{2}n^{3/2}).  (3.32)

Collecting the bounds (3.30)–(3.32), we obtain

|I_{1} − I_{2}| ≤ 8θβ(2Δ_{n−k} + b_{d}αε)/(ε^{2}n^{3/2}).  (3.33)

Similarly to the proof of (3.33 ) we have

|J_{1} − J_{2}| ≤ 16θβ(2Δ_{n−k} + b_{d}αε)/(ε^{2}n^{3/2}).  (3.34)

Now (3.29) combined with (3.33) and (3.34) yields (3.21) since θ = \sqrt{n/(k − 1)}.

Let us prove (3.11) and (3.12). Below we show that

γ_{k} ≤ 5βΔ_{n−k}(k − 1)^{−3/2} + 5β(n − k)^{−3/2},  for k = 2, …, n.  (3.35)
γ_{k} ≤ 5β(k − 1)^{−3/2}  for k = 2, …, n.  (3.36)

These bounds imply (3.11) and (3.12). To prove (3.11) we use (3.35) and the induction assumption Δ_{n−k} ≤ 2b_{d}Mδ, for k ≤ n_{0}. Using

∑_{k=m+1}^{∞} \frac{1}{(k − 1)^{3/2}} ≤ ∫_{m−1}^{∞} \frac{dt}{t^{3/2}} = \frac{2}{(m − 1)^{1/2}} ,

estimating n − k ≥ n/2, for k ≤ n_{0}, we have

γ_{m+1} + ⋯ + γ_{n_{0}} ≤ \frac{20b_{d}βMδ}{\sqrt{m − 1}} + \frac{10β}{\sqrt{n}} ,

which proves (3.11) since b_{d} ≥ 1. To prove (3.12), we use (3.36). Summing the bound (3.36) over k > n_{0} = [n/2] and using

∑_{k=n_{0}+1}^{∞} \frac{1}{(k − 1)^{3/2}} ≤ ∫_{n_{0}−1}^{∞} \frac{dt}{t^{3/2}} = \frac{2}{\sqrt{n_{0} − 1}} ≤ \frac{4}{\sqrt{n}} ,

we obtain (3.12).

Let us prove (3.35) and (3.36). We have φ(x) = ψ(ρ_{A}(x)/ε) and ∫_{ℝ} |ψ′(t)| dt = 1 (cf. Lemma 2.3 ). Integrating by parts and using the property iii) of the class A, we obtain

γ_{k} = | ∫_{ℝ} ψ(t/ε)d(ℙ{ρ_{A}(W_{k} + U) ≤ t} − ℙ{ρ_{A}(W_{k} + V) ≤ t})|

≤ ∫_{ℝ} |ψ′(t/ε)| |ℙ{ρ_{A}(W_{k} + U) ≤ t} − ℙ{ρ_{A}(W_{k} + V) ≤ t})|dt/ε

≤ sup_{B∈A} γ_{k}(B)

with

γ_{k}(B) = |P {W_{k} + U ∈ B} − ℙ{W_{k} + V ∈ B})|.

Hence, it suffices to prove that instead of γ_{k} the quantity γ_{k}(B) is bounded as γ_{k} in (3.35) and (3.36). Using the notation (3.7)–(3.8), we have

γ_{k}(B) = |ℙ{Y/θ + U + Z ∈ B} − ℙ{Y/θ + V + Z ∈ B}|.

Changing the variable y as y/θ + U = u, we get

γ_{k}(B) = | E ∫_{ℝ^{d}} 𝕀{y/θ + U + Z ∈ B}p(y)dy − E ∫_{ℝ^{d}} 𝕀{θy + V + Z ∈ B}p(y)dy| = |I_{1} − J_{1}|,

where

I_{1} = E ∫_{ℝ^{d}} 𝕀{u + Z ∈ B}p(θu − θU)θ^{d} du,

and J_{1} is defined as I_{1} just replacing U by V.

Applying the Taylor expansion (2.20) with s = 2 to the density p with h = θU and h = θV, using that U and V have the same means and covariances, changing the variables θu − τθU = y and θu − τθV = y, we obtain

γ_{k}(B) ≤ |I_{0}| + |J_{0}|

with

I_{0} = \frac{θ^{3}}{2} E (1 − τ)^{2} ∫_{ℝ^{d}} 𝕀\{y/θ + τU + Z ∈ B\} p′′′(y)U^{3} dy  (3.37)

and J_{0} defined as I_{0} just replacing U by V.

Let us prove (3.36). Using (3.37) and (2.18) of Lemma 2.4, we have

|I_{0}| ≤ θ^{3} E ∫_{ℝ^{d}} |p′′′(y)U^{3}| dy ≤ \frac{2βθ^{3}}{n^{3/2}} = \frac{2β}{(k − 1)^{3/2}} .

Similarly |J_{0}| ≤ 3β/(k − 1)^{3/2}. These bounds and γ_{k}(B) ≤ |I_{0}| + |J_{0}| prove (3.36).

Let us return to the proof of (3.35). Write α^{2} = n/(n−k). The sum Z is a normalized sum of n − k independent copies of X. Conditioning in (3.37) on all random variables except Z and using the induction assumption, we get

|I_{0}| ≤ Δ_{n−k}I_{2} + |I_{3}|,  (3.38)

where

I_{2} = \frac{θ^{3}}{6} E ∫_{ℝ^{d}} |p′′′(y)U^{3}| dy,  I_{3} = \frac{θ^{3}}{2} E (1 − τ)^{2}I_{4},

I_{4} =^{def} ∫_{ℝ^{d}} 𝕀{y/θ + τU + Y_{1}/α ∈ B} p′′′(y)U^{3} dy.

We can represent I_{4} as

I_{4} = θ^{3} ∫_{ℝ^{d}} ∫_{ℝ^{d}} 𝕀{x + τU + z ∈ B} p′′′(θx)U^{3}p(αz) θ^{d} dxα^{d}dz.

To estimate I_{4}, we integrate by parts. Changing the variable x as x + z = y and writing ∂_{h}[z] for the partial derivative in the direction h with respect to the variable z, we have

I_{4} = − ∫_{ℝ^{d}} ∫_{ℝ^{d}} 𝕀{y + τU ∈ B} (∂_{U}^{3}[z]p(θy − θz))p(αz)θ^{d} dyα^{d}dz

= α^{3} ∫_{ℝ^{d}} ∫_{ℝ^{d}} 𝕀{y + τU ∈ B} p(θy − θz)p′′′(αz)U^{3}θ^{d} dyα^{d}dz.

Therefore

|I_{4}| ≤ α^{3} ∫_{ℝ^{d}} ∫_{ℝ^{d}} p(θy − θz) |p′′′(αz)U^{3}| θ^{d}dy α^{d}dz = α^{3} ∫_{ℝ^{d}} |p′′′(z)U^{3}| dz. (3.39)

Estimating the last integral in (3.39) by (2.18) of Lemma 2.5 and collecting the bounds in (3.38), we get

|I_{0}| ≤ 2β(Δ_{n−k}θ^{3}n^{−3/2} + α^{3}n^{−3/2} ) = 2β(Δ_{n−k}(k − 1)^{−3/2} + (n − k)^{−3/2} ). (3.40)

A similar proof shows that |J_{0}| is bounded as I_{0} in (3.40) but with the factor 3 instead of 2. Therefore the inequality γ_{k}(B) ≤ |I_{0}| + |J_{0}| yields (3.35). □

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Vidmantas Bentkus

Institute of mathematics and informatics Akademijos 4 232600 Vilnius Lithuania

   E-mail address: bentkus@takas.lt