^{1}

^{2}

^{*}

^{3}

^{4}

This paper is to investigate the convergence rate of asymptotic normality of frequency polygon estimation for density function under mixing random fields, which include strongly mixing condition and some weaker mixing conditions. A Berry-Esseen bound of frequency polygon is established and the convergence rates of asymptotic normality are derived. In particularly, for the optimal bin width , it is showed that the convergence rate of asymptotic normality reaches to when mixing coefficient tends to zero exponentially fast.

Denote the integer lattice points in the N-dimensional Euclidean space by Z N for N ≥ 1 . Let { X i : i ∈ Z N } be a strictly stationary random field with common density f ( x ) on the real line R. Throughout this paper, let ‖ i ‖ = ( i 1 2 + i 2 2 + ⋯ + i N 2 ) 1 / 2 , i ^ = i 1 i 2 ⋯ i N , i ≼ j denote i k ≤ j k ( 1 ≤ k ≤ N ) for i = ( i 1 , i 2 , ⋯ , i N ) ∈ Z N and j = ( j 1 , j 2 , ⋯ , j N ) ∈ Z N , and 1 = ( 1,1, ⋯ ,1 ) ∈ Z N . The limit process n → ∞ denotes

min { n i ; 1 ≤ i ≤ N } → ∞ and n i / n j ≤ C ( 1 ≤ i , j ≤ N )

for some constant C > 0 .

For a set of sites S ⊂ Z N , B ( S ) = B ( X i ; i ∈ S ) denotes the σ-field generated by the random variables ( X i ; i ∈ S ) . Card ( S ) denotes the cardinality of S, and dist ( S , S ′ ) denotes the Euclidean distance between S and S ′ , that is dist ( S , S ′ ) = min { ‖ i − j ‖ ; i ∈ S , j ∈ S ′ } . We will use the following mixing coefficient

α ( B ( S ) , B ( S ′ ) ) = sup { | P ( A B ) − P ( A ) P ( B ) | , A ∈ B ( S ) , B ∈ B ( S ′ ) } ≤ C h ( Card ( S ) , Card ( S ′ ) ) φ ( dist ( S , S ′ ) ) , (1)

where C is some positive constant, φ ( u ) ↓ 0 as u → ∞ , and h ( n , m ) is a symmetric positive function nondecreasing in each variable.

If h ≡ 1 , then { X i : i ∈ Z N } is called strongly mixing. In Carbon et al. [

h ( n , m ) ≤ min { n , m } , (2)

or

h ( n , m ) ≤ ( n + m + 1 ) k ˜ (3)

where k ˜ ≥ 1 . Conditions (2) and (3) are also used by Neaderhouser [

In recent years, there is a growing interest in statistical problem for random fields, because spatial data are modeled as finite observations of random fields. For asymptotic properties of kernel density estimators for spatial random fields, one can refer to Tran [

The purpose of this paper is going to investigate the convergence rate of asymptotic normality of frequency polygon estimation of density function for mixing random fields. The frequency polygon has the advantage to be conceptually and computationally simple. Furthermore, Scott [

This paper is organized as follows: Next section presents the main results. Section 3 gives some lemmas, which will be used later. Section 4 provides the proofs of theorems. Throughout this paper, the letter C will be used to denote positive constants whose values are unimportant and may vary, but not dependent on n .

Suppose that we observe { X n } on a rectangular region { i : 1 ≼ i ≼ n } . Consider a partition ⋯ < x − 2 < x − 1 < x 0 < x 1 < x 2 ⋯ of the real line into equal intervals I k = [ ( k − 1 ) b n , k b n ) of length b n , where b n is the bin width and k = 0 , ± 1 , ± 2 , ⋯ . For x ∈ [ ( k 0 − 1 / 2 ) b n , ( k 0 + 1 / 2 ) b n ) , consider the two adjacent histogram bins I k 0 and I k 0 + 1 . Denote the number of observations falling in these intervals respectively by v k 0 and v k 0 + 1 . Then the values of the histogram in these previous bins are given by

f k 0 = v k 0 / ( n ^ b n ) , f k 0 + 1 = v k 0 + 1 / ( n ^ b n ) . (4)

Thus the frequency polygon estimation of the density function f ( x ) is defined as follows

f n ( x ) = ( 1 2 + k 0 − x b n ) f k 0 + ( 1 2 − k 0 + x b n ) f k 0 + 1 (5)

for x ∈ [ ( k 0 − 1 / 2 ) b n , ( k 0 + 1 / 2 ) b n ) .

We know that the curve estimated by the frequency polygon is a non-smooth curve, but it tends to be a smooth density curve as the interval length b n of interpolation gradually tends to zero. So we always assume that b n tends to zero as n → ∞ . In addition, we need the following basic assumptions.

Assumption (A1) The density f ( x ) with bounded derivative. For all i , j and some constant M > 0 ,

| f j | i ( y | x ) | ≤ M ,

where f j | i ( y | x ) is the conditional density of X j given X i .

Assumption (A2) The random field { X i : i ∈ Z N } satisfies (1) with φ ( u ) = O ( u − θ ) for some θ > 2 N .

Under Assumption (A2), we can take β such that N θ − 1 < β < 1 / 2 , then ∑ i = 1 ∞ i N − 1 φ β ( i ) < ∞ . Carefully checking the proof of Theorem 3.1 in Carbon et al [

Proposition 1 Suppose that Assumption (A1) and (A2) are satisfied. Then, for x ∈ [ ( k 0 − 1 / 2 ) b n , ( k 0 + 1 / 2 ) b n ) , we have

n b n V a r ( f n ( x ) ) = σ n 2 ( x ) + o ( 1 ) , (6)

where

σ n 2 ( x ) = [ 1 2 + 2 ( k 0 − x b n ) 2 ] f ( x ) . (7)

It should be reminded that, as in Remark 3 in El Machkouri (2013), it should be ( 1 / 2 + 2 ( k 0 − x / b n ) 2 ) f ( x ) instead of ( 1 / 2 + ( 2 k 0 − x / b n ) 2 ) f ( x ) for the asymptotic variance σ n 2 ( x ) .

Let S n = ( n ^ b n ) 1 / 2 [ f n ( x ) − E f n ( x ) ] σ n − 1 ( x ) , F S n ( u ) = P ( S n < u ) and Φ ( u ) denote the distribution function of N ( 0,1 ) . Now we give our main results as follow.

Theorem 1. Suppose that Assumption (A1) and (A2) hold. Assume that there exist integers p = p n → ∞ and q = q n → ∞ such that

τ 1, n → 0, τ 2, n → 0, τ 3, n → 0 (8)

where τ 1, n = q p − 1 , τ 2, n = ( n ^ b n ) − 1 / 2 p N and τ 3, n = ( n ^ b n − 1 ) 1 / 2 q − θ h ( n ^ , p N ) . Then, for x such that f ( x ) > 0 and as n → ∞ , we have

sup u ∈ R | F S n ( u ) − Φ ( u ) | = O ( τ n ) (9)

where τ n = τ 1, n 1 / 3 + τ 2, n + τ 3 , n 1 / 2 + τ 4 , n 1 / 3 and τ 4 , n = b n − 1 p N q − θ .

Remark 1. In the theorem above, it does not need to assume that τ 4 , n → 0 because 0 ≤ τ 4 , n ≤ C τ 2, n τ 3 , n → 0 from (8).

Theorem 1 provides a general result for Berry-Esseen bound of frequency polygon estimation. Some specific bounds can be obtained by choosing different b n , p and q.

Theorem 2. Suppose that Assumption (A1) and (A2) hold. Let b n = C n ^ − ν for some ν ∈ ( 0,1 ) . Denote that η 1 = ( 1 + ν ) N ( 1 − ν ) ε + 2 ( 1 − ε ) N ( 1 + 3 N ) ε , η 2 = η 1 + ( ε + 3 N ) N ( 1 + 3 N ) ε and η 3 = η 1 + 4 N k ˜ ( 1 − ν ) ε for some ε ∈ ( 0,1 ) .

1) If h ≡ 1 and

θ ≥ max { 2 N , η 1 } , (10)

2) or if (2) is satisfied and

θ ≥ max { 2 N , η 2 } , (11)

3) or if (3) is satisfied and

θ ≥ max { 2 N , η 3 } , (12)

then, for x such that f ( x ) > 0 and as n → ∞ , we have

sup u ∈ R | F S n ( u ) − Φ ( u ) | = O ( n ^ − ( 1 − ν ) ( 1 − ε ) 2 ( 1 + 3 N ) ) (13)

Carbon [

b o p t = 2 ( 15 49 R 2 ( f ) ) 1 / 5 n ^ − 1 / 5 (14)

where R 2 ( f ) = ∫ − ∞ ∞ [ f ″ ( x ) ] 2 d x , when θ > 2 N + 3 / 2 . For the optimal bin width, it is ease to get the following result by Theorem 2.

Corollary 1. Suppose that Assumption (A1) and (A2) hold and h ≡ 1 . Let b n = C n ^ − 1 / 5 . 1) If

θ ≥ max { 2 N , 3 N 2 ε + 2 ( 1 − ε ) N ( 1 + 3 N ) ε } (15)

for some ε ∈ ( 0,1 ) , then, for x such that f ( x ) > 0 ,

sup u ∈ R | F S n ( u ) − Φ ( u ) | = O ( n ^ − 2 ( 1 − ε ) 5 ( 1 + 3 N ) ) . (16)

2) If φ ( u ) tends to zero exponentially fast as u tends to infinity, then, for x such that f ( x ) > 0 ,

sup u ∈ R | F S n ( u ) − Φ ( u ) | = O ( n ^ − 2 5 ( 1 + 3 N ) ) . (17)

Remark 2. The asymptotic normality of frequency polygon under the strongly mixing conditions established by Carbon [

In the later proof, we need to estimate the upper bounds of covariance and variance of dependent variables. The following two lemmas give the upper bounds of covariance and variance respectively.

Lemma 1. Roussas and Ioannides [

| E ( ξ η ) − ( E ξ ) ( E η ) | ≤ 4 C 1 C 2 α ( d i s t ( S , S ′ ) ) . (18)

Let

Y i , k = I ( ( k − 1 ) b n ≤ X i < k b n ) , Y ˜ i , k = Y i , k − E Y i , k . (19)

Lemma 2. Gao et al. [

E ( ∑ a ≼ i ≼ a + m Y ˜ i , k ) 2 ≤ C m ^ b n . (20)

Lemma 3. Lemma 3.7 in Yang [

sup u | F ζ n ( u ) − Φ ( u ) | = O ( γ n ) , (21)

then for any ε > 0 ,

sup u | F ζ n + η n ( u ) − Φ ( u ) | = O ( γ n + ε + P ( | η n | ≥ ε ) ) . (22)

Proof of Theorem 1 We will use the methodology of using “small” and “big” blocks which is similar to that of Carbon et al. [

Z i , k 0 = b n − 1 / 2 { ( 1 2 + k 0 − x b n ) Y i , k 0 + ( 1 2 − k 0 + x b n ) Y i , k 0 + 1 } . (23)

and Z i , k 0 = Z i , k 0 − E Z i , k 0 . Then

S n ( x ) = n ^ − 1 / 2 ∑ 1 ≼ i ≼ n Z ˜ i , k 0 . (24)

Now we divide S n ( x ) into the sum of large blocks and the sum of small blocks. According to the block size method, we assume q < p and p , q satisfy (8). Assume for some integer vector r = ( r 1 , r 2 , ⋯ , r N ) , we have n 1 = r 1 ( p + q ) , ⋯ , n N = r N ( p + q ) . If it is not this case, there will be a remainder term in the splitting block, but it will not change the proof much. For 1 ≼ j ≼ r , let

U ( 1 , n , j ) = n ^ − 1 / 2 ∑ i k = ( j k − 1 ) ( p + q ) + 1 ; 1 ≤ k ≤ N ( j k − 1 ) ( p + q ) + p Z ˜ i , k 0

U ( 2 , n , j ) = n ^ − 1 / 2 ∑ i k = ( j k − 1 ) ( p + q ) + 1 ; 1 ≤ k ≤ N − 1 ( j k − 1 ) ( p + q ) + p ∑ i N = ( j N − 1 ) ( p + q ) + p + 1 j N ( p + q ) Z ˜ i , k 0

U ( 3 , n , j ) = n ^ − 1 / 2 ∑ i k = ( j k − 1 ) ( p + q ) + 1 ; 1 ≤ k ≤ N − 2 ( j k − 1 ) ( p + q ) + p ∑ i N − 1 = ( j N − 1 − 1 ) ( p + q ) + p + 1 j N − 1 ( p + q ) ∑ i N = ( j N − 1 ) ( p + q ) + 1 ( j N − 1 ) ( p + q ) + p Z ˜ i , k 0

U ( 4 , n , j ) = n ^ − 1 / 2 ∑ i k = ( j k − 1 ) ( p + q ) + 1 ; 1 ≤ k ≤ N − 2 ( j k − 1 ) ( p + q ) + p ∑ i N − 1 = ( j N − 1 − 1 ) ( p + q ) + p + 1 j N − 1 ( p + q ) ∑ i N = ( j N − 1 ) ( p + q ) + p + 1 j N ( p + q ) Z ˜ i , k 0

an so on. Note that

U ( 2 N − 1, n , j ) = n ^ − 1 / 2 ∑ i k = ( j k − 1 ) ( p + q ) + p + 1 ; 1 ≤ k ≤ N − 1 j k ( p + q ) ∑ i N = ( j N − 1 ) ( p + q ) + 1 ( j N − 1 ) ( p + q ) + p Z ˜ i , k 0 .

Finally

U ( 2 N , n , j ) = n ^ − 1 / 2 ∑ i k = ( j k − 1 ) ( p + q ) + p + 1 ; 1 ≤ k ≤ N j k ( p + q ) Z ˜ i , k 0 .

For each integer i ∈ [ 1,2 N ] , define

T ( i , n ) = ∑ 1 ≼ j ≼ r U ( i , n , j ) (25)

and B n = ∑ i = 2 2 N T ( i , n ) . Then

S n ( x ) = T ( 1, n ) + B n . (26)

Enumerate the random variables { U ( 1 , n , j ) : 1 ≼ j ≼ r } in an arbitrary manner and refer to them as V 1 , V 2 , ⋯ , V r ^ . Note that | V i | ≤ C n ^ − 1 / 2 p N b n − 1 / 2 . Using Theorem 4 in Rio [

E | V i − V ˜ i | ≤ C n ^ − 1 / 2 p N b n − 1 / 2 h ( ( r ^ − 1 ) p N , p N ) φ ( q ) ≤ C n ^ − 1 / 2 p N b n − 1 / 2 h ( n ^ , p N ) φ ( q ) . (27)

Let T ˜ ( 1, n ) = ∑ i = 1 r ^ V ˜ i and A n = T ( 1, n ) − T ˜ ( 1, n ) . Thus

S n ( x ) = T ˜ ( 1, n ) + A n + B n . (28)

By Lemma 3, it is sufficient to show that

P ( | A n | > τ 3 , n 1 / 2 ) = O ( τ 3 , n 1 / 2 ) , (29)

P ( | B n | > τ 1 , n 1 / 3 + τ 4 , n 1 / 3 ) = O ( τ 1 , n 1 / 3 + τ 4 , n 1 / 3 ) , (30)

and

sup u ∈ R | F T ˜ ( 1, n ) ( u ) − Φ ( u ) | = O ( τ 2, n ) . (31)

Obviously, from (27)

P ( | A n | > τ 3 , n 1 / 3 ) ≤ C τ 3 , n − 1 / 2 ∑ i = 1 r ^ E | V i − V ˜ i | ≤ C τ 3 , n − 1 / 2 r ^ n ^ − 1 / 2 p N b n − 1 / 2 h ( n ^ , p N ) φ ( q ) ≤ C τ 3 , n − 1 / 2 ( n ^ b n − 1 ) 1 / 2 q − θ h ( n ^ , p N ) = C τ 3 , n 1 / 2 , (32)

it follows (29). Now consider that

P ( | B n | > τ 1 , n 1 / 3 + τ 4 , n 1 / 3 ) ≤ ∑ i = 2 2 N P ( | T ( i , n ) | > τ 1 , n 1 / 3 + τ 4 , n 1 / 3 ) ≤ C ( τ 1 , n 1 / 3 + τ 4 , n 1 / 3 ) − 2 ∑ i = 2 2 N E T 2 ( i , n ) . (33)

Note that

E T 2 ( 2 , n ) = r ^ E U 2 ( 2 , n , j ) + ∑ 1 ≼ j , j ′ ≼ r , j ≠ j ′ Cov ( U ( 2 , n , j ) , U ( 2 , n , j ′ ) ) = : Λ 1 + Λ 2 (34)

By Lemma 2,

Λ 1 ≤ C r ^ n ^ − 1 b n − 1 p N − 1 q b n ≤ C p − 1 q = C τ 1, n . (35)

Define J ( 2, n , j ) = { i : ( j k − 1 ) ( p + q ) + 1 ≤ i k ≤ ( j k − 1 ) ( p + q ) + p , 1 ≤ k ≤ N − 1, ( j N − 1 ) ( p + q ) + p + 1 ≤ i N ≤ j N ( p + q ) } . By Lemma 1,

| Λ 2 | ≤ n ^ − 1 ∑ 1 ≼ j , j ′ ≼ r , j ≠ j ′ ∑ i ∈ J ( 2, n , j ) , i ′ ∈ J ( 2, n , j ′ ) | C o v ( Z ˜ i , k 0 , Z ˜ i ′ , k 0 ) | ≤ C n ^ − 1 b n − 1 ∑ 1 ≼ j , j ′ ≼ r , j ≠ j ′ ∑ i ∈ J ( 2, n , j ) , i ′ ∈ J ( 2, n , j ′ ) φ ( ‖ i − i ′ ‖ ) ≤ C n ^ − 1 b n − 1 ∑ 1 ≼ j , j ′ ≼ r , j ≠ j ′ ∑ i ∈ J ( 2, n , j ) , i ′ ∈ J ( 2, n , j ′ ) φ ( ‖ j − j ′ ‖ q ) ≤ C n ^ − 1 b n − 1 p 2 N − 2 q 2 ∑ 1 ≼ j , j ′ ≼ r , j ≠ j ′ ( ‖ j − j ′ ‖ q ) − θ

≤ C n ^ − 1 b n − 1 p 2 N − 2 q 2 r ^ q − θ ∑ 1 ≼ j ≼ r ‖ j ‖ − θ ≤ C n ^ − 1 b n − 1 p 2 N r ^ q − θ ( q p − 1 ) 2 ≤ C b n − 1 p N q − θ = C τ 4, n . (36)

Combining (34)-(36), we have

E T 2 ( 2, n ) ≤ C ( τ 1 , n + τ 4 , n ) . (37)

similarly, E T 2 ( i , n ) ≤ C ( τ 1 , n + τ 4 , n ) for 3 ≤ i ≤ 2 N . Thus, we obtain (30) from (33).

Finely, to show that (31). Clearly,

Var ( T ( 1, n ) ) = r ^ Var ( V 1 ) + ∑ 1 ≤ t , t ′ ≤ r ^ , t ≠ t ′ Cov ( V t , V t ′ ) (38)

Define J ( 1, n , j ) = { i : ( j k − 1 ) ( p + q ) + 1 ≤ i k ≤ ( j k − 1 ) ( p + q ) + p ,1 ≤ k ≤ N } . Recalling (36), we have

| ∑ 1 ≤ t , t ′ ≤ r ^ , t ≠ t ′ Cov ( V t , V t ′ ) | ≤ ∑ 1 ≼ j , j ′ ≼ r , j ≠ j ′ | Cov ( U ( 1, n , j ) , U ( 1, n , j ′ ) ) | = n ^ − 1 ∑ 1 ≼ j , j ′ ≼ r , j ≠ j ′ ∑ i ∈ J ( 2, n , j ) , i ′ ∈ J ( 2, n , j ′ ) Cov ( Z ˜ i , k 0 , Z ˜ i ′ , k 0 ) ≤ C n ^ − 1 b n − 1 ∑ 1 ≼ j , j ′ ≼ r , j ≠ j ′ ∑ i ∈ J ( 2, n , j ) , i ′ ∈ J ( 2, n , j ′ ) φ ( ‖ i − i ′ ‖ )

≤ C n ^ − 1 b n − 1 ∑ 1 ≼ j , j ′ ≼ r , j ≠ j ′ ∑ i ∈ J ( 2, n , j ) , i ′ ∈ J ( 2, n , j ′ ) φ ( ‖ j − j ′ ‖ q ) ≤ C n ^ − 1 b n − 1 p 2 N ∑ 1 ≼ j , j ′ ≼ r , j ≠ j ′ ( ‖ j − j ′ ‖ q ) − θ ≤ C n ^ − 1 b n − 1 p 2 N r ^ q − θ ∑ 1 ≼ j ≼ r ‖ j ‖ − θ ≤ C n ^ − 1 b n − 1 p 2 N r ^ q − θ (39)

and by Lemma 2

r ^ Var ( V 1 ) = r ^ Var ( U ( 1, n , 1 ) ) ≤ C r ^ n ^ − 1 p N ≤ C . (40)

Combining (38)-(40) yields that Var ( T ( 1, n ) ) = r ^ Var ( V 1 ) + o ( 1 ) and Var ( T ( 1, n ) ) ≤ C , so that Var ( S n ) = Var ( T ( 1, n ) + B n ) = Var ( T ( 1, n ) ) + o ( 1 ) from E B n 2 → 0 . Hence

Var ( T ˜ ( 1, n ) ) = r ^ Var ( V ˜ 1 ) = r ^ Var ( V 1 ) = Var ( T ( 1, n ) ) + o ( 1 ) = Var ( S n ) + o ( 1 ) = σ n 2 ( x ) + o ( 1 ) . (41)

Let Δ n ≡ { Var ( T ˜ ( 1, n ) ) } − 3 / 2 ∑ i = 1 r ^ E | V ˜ i | 3 . Note that Var ( T ˜ ( 1, n ) ) ≥ σ n 2 ( x ) / 2 ≥ f ( x ) / 4 for x ∈ ( ( k 0 − 1 / 2 ) b n , ( k 0 + 1 / 2 ) b n ) . From (40), we have

Δ n ≤ C ∑ i = 1 r ^ E | V ˜ i | 3 ≤ C ( n ^ b n ) − 1 / 2 p N r ^ Var ( V ˜ 1 ) ≤ C ( n ^ b n ) − 1 / 2 p N → 0, (42)

yields (31) by Berry-Esseen theorem. Complete the proof.

Proof of Theorem 2 In Theorem 1, take p = [ n ^ ρ ] and q = [ n ^ τ ] where ρ = ( 1 − ν ) ( ε + 3 N ) 2 N ( 1 + 3 N ) and τ = ( 1 − ν ) ε 2 N for 0 < ν < 1 and 0 < ε < 1 . Notes that b n = C n ^ − ν . Then

τ 1, n = q p − 1 = n ^ − 3 ( 1 − ν ) ( 1 − ε ) 2 ( 1 + 3 N ) , (43)

τ 2, n = ( n ^ b n ) − 1 / 2 p N = n ^ − ( 1 − ν ) ( 1 − ε ) 2 ( 1 + 3 N ) , (44)

τ 4, n = b n − 1 p N q − θ = n ^ − [ θ τ − ν − ( 1 − ν ) ( ε + 3 N ) 2 ( 1 + 3 N ) ] . (45)

First consider the case (1), that is that h ≡ 1 and the condition (10) holds. At this time, we have

τ 3, n = ( n ^ b n − 1 ) 1 / 2 q − θ h ( n ^ , p N ) = n ^ − ( 1 − ν ) ε θ − ( 1 + ν ) N 2 N . (46)

The condition (10) implies that θ ≥ ( 1 + ν ) N ( 1 − ν ) ε + 2 ( 1 − ε ) N ( 1 + 3 N ) ε . Combining this with (45) and (46), we can get that

τ 3 , n 1 / 2 = O ( n ^ − ( 1 − ν ) ( 1 − ε ) 2 ( 1 + 3 N ) ) , (47)

τ 4 , n 1 / 3 = O ( n ^ − ( 1 − ν ) ( 1 − ε ) 2 ( 1 + 3 N ) ) . (48)

From (43),(44), (47) and (48), it is ease to know that

τ n = τ 1 , n 1 / 3 + τ 2 , n + τ 3 , n 1 / 2 + τ 4 , n 1 / 3 = O ( n ^ − ( 1 − ν ) ( 1 − ε ) 2 ( 1 + 3 N ) ) . (49)

It follows the desired result (13). For the case (2) and the case (3), the proving methods are similar to the method used to prove the case (1). Complete the proof.

The frequency polygon estimation has the advantage of simple calculation. It can save calculation cost in the face of large data, so it is a valuable and worth studying method. In the existing literature, the asymptotic normality of the frequency polygon estimation has been studied, but its convergence rate has not been established. This paper proves a Berry-Esseen bound of the frequency polygon and derives the convergence rate of asymptotic normality under weaker mixing conditions. In particularly, for the optimal bin width b o p t = C n ^ − 1 / 5 , it is showed that the convergence rate of asymptotic normality reaches to n ^ − 2 / 5 ( 1 + 3 N ) when mixing coefficient tends to zero exponentially fast. These conclusions show that the asymptotic normality of the frequency polygon estimator also has a good convergence rate under the dependent samples. Therefore, when the sample size is large, the normal distribution can be used to give a better confidence interval estimation.

This research was supported by the Natural Science Foundation of China (11461009) and the Scientific Research Project of the Guangxi Colleges and Universities (KY2015YB345).

The authors declare that there are no conflicts of interest regarding the publication of this paper.

Yang, S.C., Yang, X., Xing, G.D. and Li, Y.M. (2018) The Rate of Asymptotic Normality of Frequency Polygon Density Estimation for Spatial Random Fields. Open Journal of Statistics, 8, 962-973. https://doi.org/10.4236/ojs.2018.86064