Journal of Mathematical Sciences, Vol. 220, No. 6, February, 2017
ESTIMATION OF DISTRIBUTIONS UNDER
DOSE-EFFECT DEPENDENCE WITH FIXED
EXPERIMENT PLAN
M. S. Tikhov1 , D. S. Krishtopenko1 , and M. V. Yaroschuk1
In the present paper we find the limit distributions of the Nadaraya–Watson estimators and asymptotically unbiased estimators of a distribution function under dose-effect dependence with fixed experiment plans. We prove the asymptotic normality of integrated square errors of the distribution
function’s estimators.
Introduction
The goal of the present paper is to study the asymptotic properties of the distribution under doseeffect dependence in the case where injected doses are not random but fixed beforehand. Studying such
statistics is of interest in toxicometry for the determination of average effective doses (see [1]). These
results may be used for the construction of tests for goodness-of-fit and homogeneity hypotheses under
dose-effect dependence.
1.
Nonparametric estimators of distribution function
Let Z = {(Xi , Ui ), i > 1} be a stationary sequence of independent pairs of random variables with the
joint distribution function F (x)G(x) and density f (x)g(x). We observe the sample U (n) = {(Ui , Wi ),
1 6 i 6 n}, where Wi = I(Xi < Ui ) is the indicator of an event (Xi < Ui ). The problem is to
estimate the distribution function F (x) from the sample U (n) . To statistically estimate F (x), according
to Nadaraya and Watson (see, e.g., [2, 3]), the values
F̂n (x) =
are used, where
∗ (x)
S2n
∗ (x)
S1n
n
∗
S1n
(x)
1X
Kh (Ui − x),
=
n
i=1
n
∗
S2n
(x)
1X
=
Wi Kh (Ui − x),
n
i=1
the kernel K(·) > 0 is defined on R, Kh (x) = (1/h)K(x/h),
and h > 0 is the smoothing parameter.
√
The asymptotic behavior of the empirical process nh(F̂n (x) − F (x)) was studied in [4], where the
asymptotic normality of the statistic F̂n (x) was established. Let us note that the estimator F̂n (x) is
asymptotically biased. In [5] the asymptotically unbiased estimator of a distribution function under
dose-effect dependence is proposed, which has the same limit variance as the one in [4]. In these papers
U is interpreted as the dose injected into the body, and X is the minimal dose, at which the organism
begins to react, and U is considered to be a random variable. However, quite often the injected dose is
chosen beforehand, i.e., it is not a random variable, and in the present paper we study the asymptotic
behavior of the proposed statistic in this situation.
Let us introduce the functions
n
1X
Kh (ui − x),
S1n (x) =
n
i=1
1
n
1X
S2n (x) =
Wi Kh (ui − x),
n
i=1
Lobachevksy State University of Nizhni Novgorod, Nizhni Novgorod, Russia, e-mail: tikhovm@mail.ru
Translated from Statisticheskie Metody Otsenivaniya i Proverki Gipotez, Vol. 19, pp. 66–77, 2006
1072-3374/17/2206-0753
© 2017 Springer Science+Business Media New York
753
754
M. S. Tikhov, D. S. Krishtopenko, and M. V. Yaroschuk
where ui are fixed values, and Wi = I(Xi < ui ) is the indicator of the event (Xi < ui ). We assume that
a 6 ui 6 b are given,
R 2and, without
Rloss2 of generality, consider the case where a = 0, b = 1 and ui = i/n.
2
2
Let k K k = K (x) dx, ν = x K(x) dx.
We assume that the following conditions are satisfied:
( A ):
( A1 ) K(x) > 0 is a bounded even function, and k K k2 < ∞;
( A2 ) RK(x) = 0 for x ∈
/ [−1, 1];
( A3 )
K(x) dx = 1, ν 2 < ∞;
( A4 ) the function f (x) is continuously differentiable, the derivatives f ′ (x) and f ′′ (x) are bounded,
Z
(f ′ (x))2 dx < ∞;
R
( A5) f (x)/F (x), f ′ (x)f (x)/F (x) are bounded integrable functions, and (f ′ (x))4 dx < ∞.
R
Let us consider S1n (x), which is the integral sum for A = (1/h) K(x/h) dx, and find the estimate
of deviation of S1n (x) from A.
Let us present the Koksma–Hlavka inequality (KH) (see [6, 7])
Z1
0
f (u) du −
n
_
1X
f (ui ) 6 (f )D ∗ (Pn ),
n
i=1
W
W
where (f ) is the variation
of the function f (we assume that (f ) is finite: a sufficient condition of
W
finiteness of variation (f ) is the boundedness of the derivative f ′ (x)), Pn = {u1 , u2 , . . . , un },
D ∗ (Pn ) =
Z
sup
J = [ 0, a )⊂A
IJ (x) dx −
A
n
1X
IJ (xi ) .
n
i=1
For the sequence Pn = {1/n, 2/n, . . . , (n − 1)/n, 1} we have D ∗ (Pn ) = 1/(2n), i.e.,
Z1
0
f (u) du −
n
_
1X
1
f (ui ) 6 (f ) ·
.
n
2n
i=1
From the KH inequality it follows that
1
Z
n
_
1X
1
Kh (i/n − x) − Kh (u − x) du 6
(K) · .
n
n
i=1
But for h → 0
0
Z1
0
Kh (u − x) du =
Z1
K(t) dt = 1.
−1
Therefore, if h → 0 as n → ∞, then the sequence S1n (x) converges to 1.
Now consider the sum S2n (x). We have
n
1X
S2n (x) =
Wi Kh (i/n − x).
n
i=1
Estimation of Distributions under Dose-Effect Dependence with Fixed Experiment Plan
755
Its mathematical expectation, when n → ∞ and h → 0, equals
E(S2n (x)) =
n
1X
F (i/n)Kh (i/n − x) =
n
i=1
=
Z1
0
Z1
1
1
F (u)Kh (u − x) du + O
= F (x + ht)K(t) dt + O
=
n
n
−1
f ′ (x)h2
= F (x) +
2
Z1
2
2
t K(t) dt + o(h ) + O
−1
= F (x) +
f ′ (x)h2
2
2
2
ν + o(h ) + O
The variance of the sum S2n (x) equals
D
1
nh
1
nh
=
.
!
n
1X
Wi Kh (i/n − x) =
n
i=1
=
n
1 X
F (i/n)(1 − F (i/n))Kh2 (i/n − x) =
n2
i=1
1
=
n
Z1
0
F (u)(1 − F (u))Kh2 (u − x) du(1 + o(1)) =
1
F (x)(1 − F (x)) k K k2 (1 + o(1)).
nh
Theorem 1. Let conditions (A) be satisfied, and h = Cn−1/5 . Then, as n → ∞,
√
d
nh(F̂n (x) − F (x)) −→ N (a(x), σ 2 (x)),
where a(x) = (1/2)f ′ (x)ν 2 and σ 2 (x) = F (x)(1 − F (x)) k K k2 .
Proof. The asymptotic normality of F̂n (x) follows from the previous considerations and the boundedness of the terms Wi K(i/n − x) (see [8]).
Now let us construct the estimator of the distribution function:
1) set h0 = C1 n−α and calculate
n
1X
Wi Kho (i/n − x);
ϕ̃(x) =
n
i=1
2) taking h1 = C2 n−1/5 , we estimate the function β(x) with the use of
n
1 X
1
β̂(x) =
;
Wj Kh1 (j/n − x)
n
ϕ̃(j/n)
j =1
3) multiplying the estimator β̂(x) by the estimator ϕ̃(x), we obtain
n
ϕ̃(x)
1 X
Wj Kh1 (j/n − x)
.
ϕ̂(x) = β̂(x)ϕ̃(x) =
n
ϕ̃(j/n)
j=1
P
Let fˆn,h = fˆn,h(x) = n−1 ni= 1 Kh (x − Xi ). Let us present the result from [9], which we use in what
follows.
756
M. S. Tikhov, D. S. Krishtopenko, and M. V. Yaroschuk
Theorem 2. If conditions (A1)–(A3) for the function K(x) are satisfied, and if f (x) is bounded,
then for any c > 0 with probability 1 we have
√
nh k fˆn,h − E(fˆn,h )k∞
p
= k(c) < ∞,
lim sup
sup
n→∞ c ln n/n6h61
max(ln(1/h), ln ln n)
where k K k∞ = sup | K(x) | < ∞.
x
The following statement holds.
Theorem 3. Let conditions (A) be satisfied, and 1/10 < α < 1/5. Then
p
d
nh1 (ϕ̂(x) − F (x)) −→ N (0, σ 2 (x)).
n→∞
Proof. Let ϕ̄(x) = E(ϕ̃(x)). Then
ϕ̃(x)
ϕ̄(x)
−
ϕ̃(j/n) ϕ̄(j/n)
=
6
=
ϕ̃(x)ϕ̄(j/n) − ϕ̄(x)ϕ̃(j/n)
ϕ̃(x)ϕ̄(j/n)
ϕ̄(j/n)(ϕ̃(x) − ϕ̄(x)) + ϕ̄(x)(ϕ̄(j/n)ϕ̃(j/n))
ϕ̃(x)ϕ̄(j/n)
=
6
ϕ̄(x)
1
| ϕ̃(x) − ϕ̄(x) | +
| ϕ̃(j/n) − ϕ̄(j/n) |.
ϕ̃(j/n)
ϕ̃(j/n)ϕ̄(j/n)
From Theorem 2 it follows that
∆n1 = sup | ϕ̃(x) − ϕ̄(x) | = O
x
r
∆n2 = sup | ϕ̃(j/n) − ϕ̄(j/n) | = O
16j6n
and hence,
p
p
p
nh1 ∆n1 −→ 0,
n→∞
ln n
nh0
r
!
,
ln n
nh0
!
,
p
nh1 ∆n2 −→ 0,
n→∞
since α < 1/5. In this regard, we will study the asymptotic behavior of the sums
n
1X
ϕ̄(x)
S3n (x) =
Wj Kh1 (x − j/n)
.
n
ϕ̄(j/n)
i=1
First, let us consider the mathematical expectation of S3n (x). We have
n
1X
ϕ̄(x)
E(S3n (x)) =
F (j/n)Kh1 (x − j/n)
∼
n
ϕ̄(j/n)
i=1
∼
Z
F (u)Kh1 (x − u)
ϕ̄(x)
du =
ϕ̄(u)
Z
K(t)F (x + h1 t)
ϕ̄(x)
dt.
ϕ̄(x + h1 t)
Expanding F (x + h1 t)/ϕ̄(x + h1 t) and ϕ̄(x) into series in h0 t and calculating their product, we obtain
ϕ̄(x)
t2 ν 2 ′
h2 t2 ν 2 f ′ (x + h1 t)
F (x + h1 t)
= F (x) + h20
f (x) − 0
F (x) + o(h20 ).
ϕ̄(x + h1 t)
2
F (x + h1 t)
Estimation of Distributions under Dose-Effect Dependence with Fixed Experiment Plan
757
Then, expanding the ratio f ′ (x + h1 t)/F (x + h1 t) into series in h1 t, we obtain
F (x)
f ′ (x)f (x)
f ′ (x + h1 t)
= f ′ (x) + h1 tf ′′ (x) − h1 t
+ o(h1 ).
F (x + h1 t)
F (x)
Finally, taking into account conditions (A), we obtain
E(S3n (x)) = F (x) + O(n−1/5−2α ) −→ 0.
n→∞
Remark 1. Restriction on α may be weakened to 0 < α < 1/5 if we require that the following
conditions be satisfied: the boundedness of functions f ′′′ (x)/F (x), f (x)f ′′ (x)/F 2 (x), f (x)f ′ (x)/F 2 (x),
(f ′ (x))2 /F 2 (x).
Next,
n
1 X
ϕ̄2 (x)
D(S3n (x)) = 2
F (j/n)(1 − F (j/n))Kh21 (x − j/n) 2
∼
n
ϕ̄ (j/n)
j =1
∼
1
=
nh1
Z
1
n
Z
F (u)(1 − F (u))Kh21 (x − u)
F (x + h1 t)(1 − F (x + h1 t))K 2 (t)
∼
ϕ̄2 (x)
du =
ϕ̄2 (u)
ϕ̄2 (x)
dt ∼
ϕ̄2 (x + h1 t)
1
F (x)(1 − F (x))kKk2 ,
nh1
since ϕ̄2 (x + h1 t) −→ ϕ̄2 (x) uniformly in | t | 6 t0 .
n→∞
In view of the boundedness of the functions F (x) and K(x) it is easy to show that the Lindeberg
condition is satisfied (see [10]) and, hence, the statement of Theorem 3 follows.
2.
The distribution of integrated square errors of nonparametric estimators of distribution
functions
Let X1 , X2 , . . . , Xn be independent identically distributed random variables with unknown continuous distribution F (x). We observe the sample {(Wi , i/n), i = 1, 2, . . . , n}, where i/n are injected
nonrandom doses and Wi = I(Xi < i/n) is the indicator of the event (Xi < i/n).
Consider the sequence of Nadaraya–Watson estimators of the form
F̂n (x) = S2n (x)/S1n (x),
S2n = S2n (x) =
n
n
1X
1X
Wi Kh (x − i/n), S1n = S1n (x) =
Kh (x − i/n).
n
n
i=1
p
i=1
Since S1n −→ 1, we will study the behavior of integrated square errors of the estimators of distrin→∞
bution function of the form Fn (x) = S2n (x) = S2n which is given by the formula
Z
Z
In = (Fn (x) − F (x))2 ω(x) dx = (Fn (x) − E(Fn (x)))2 ω(x) dx+
+2
+
Z
Z
(Fn (x) − E(Fn (x)))(E(Fn (x)) − F (x)) ω(x) dx+
(E(Fn (x)) − F (x))2 ω(x) dx.
758
M. S. Tikhov, D. S. Krishtopenko, and M. V. Yaroschuk
Without loss of generality, we assume that ω(x) ≡ 1 and study each term of this expression separately,
starting with In1 :
n
P
1
(i) In1 = h2 n−1 Jn1 ,
Jn1 = 1/2 2
Zn1i , where
n h i=1
Zn1i = (Wi − F (i/n))
∼
ν 2 h2
2
Z
Kh (x − i/n)f ′ (x) dx ∼
ν 2 h2
(Wi − F (i/n))f ′ (i/n)
2
when n → ∞.
Lemma 1. Let conditions (A) be satisfied. Then the sequence Jn1 is asymptotically (when n → ∞)
normal with parameters (0, σ12 ), where
σ12
= (1/4)ν
4
Z1
0
F (x)(1 − F (x))(f ′ (x))2 dx < ∞.
Proof. We have: E(Jn1 ) = 0, and when n → ∞,
n
ν4 X
D(Jn1 ) ∼
F (i/n)(1 − F (i/n))(f ′ (i/n))2 −→ σ12 < ∞
n→∞
4n
i=1
by condition (A5).
Next,
n
n
X
1
1 X
2
1/2 2
4
E(Z
I(
|
Z
|
>
εn
h
))
6
E(Zn1i
)=
n1i
n1i
nh4
ε2 n2 h8
i=1
=
i=1
n
ν8 X 4
F (i/n)(1 − F (i/n) + F (i/n)(1 − F (i/n)4 )(f ′ (i/n))4 6
16ε2 n2
i=1
Z1
n
ν8 X ′
ν8
4
6
(f (i/n)) ∼
(f ′ (x))4 dx −→ 0
n→∞
64ε2 n2
n
i=1
0
by condition (A6).
From here by the central limit theorem (see [10, the Lindeberg theorem]) the sequence
Jn1 = n
1/2 −2
h
Z
(Fn (x) − E(Fn (x)))(E(Fn (x)) − F (x)) dx
is asymptotically normal with parameters (0, σ12 ) as n → ∞.
Now let us study the asymptotic behavior of expression In2 :
R
1 Pn
2 K 2 (x − i/n) dx.
(ii) In2 = n−1 Jn2 ,
Jn2 ≡
i = 1 (Wi − F (i/n))
h
n
Lemma 2. Let conditions (A) be satisfied, and when n → ∞, h → 0, nh2 → ∞. Then
p
Jn2 −→ σ22 .
n→∞
Estimation of Distributions under Dose-Effect Dependence with Fixed Experiment Plan
759
Proof. We have
Z
n
1X
F (i/n)(1 − F (i/n)) Kh2 (x − i/n) dx ∼
E(Jn2 ) =
n
i=1
∼
Z1
0
F (u)(1 − F (u)) du
1
D(Jn2 ) = 2
n
Z
n
X
i=1
Kh2 (x
2
− u) dx ∼ k K k
4
E(Wi − F (i/n))
Z
Z1
0
F (x)(1 − F (x)) dx = σ22 ,
Kh2 (x
− i/n) dx
2
=
n
1 X
(F (i/n)4 (1 − F (i/n)) + F (i/n)(1 − F (i/n))4 )×
= 2
n
i=1
×
Z
Kh2 (x
− i/n) dx
2
2
n Z
kKk4
1 X
2
Kh (x − i/n) dx ∼
6 2
−→ 0
4n
4n n→∞
i=1
by condition (A1).
Then the statement of the lemma follows from the Tchebyshev inequality.
Now let us consider the behavior of In3 :
( iii ) In3 = n−1 h−1/2 Jn3 ,
Z
X
2
(Wi − F (i/n))(Wj − F (j/n)) Kh (x − i/n)Kh (x − j/n) dx.
Jn3 =
nh1/2 16i<j6n
Lemma 3. Let conditions (A) be satisfied. Then as n → ∞, h → 0, n2 h → ∞, the sequence Jn3 is
2 ), where σ 2 = 2n−2 h−1 σ 2 ,
asymptotically normal with parameters (0, σn3
3
n3
σ32
=
Z
2
2
F (x)(1 − F (x)) dx
Z Z
K(u)K(u + v) du
2
dv.
Proof. Define the variables
ξni =
n
X
(Wi − F (i/n))(Wj − F (j/n)) ·
j = i+1
Then
Jn3 =
Z
Kh (x − i/n)Kh (x − j/n) dx.
n−1
2 X
ξni .
nh1/2 j = 1
Let Fk = σ(X1 , X2 , . . . , Xk ) be the σ-algebra generated by the random variables X1 , X2 , . . . , Xk .
Then {ξnk , Fk }16k6n , n > 1, is a martingale difference (see [11]), since E | ξnk | < ∞ and E(ξnk | Fk ) =
0. To prove the asymptotic normality of Jn3 , it is necessary to show (see [11, Theorem 8 (II)]) that
(a)
n−1
1 X
p
2
E(ξni
I(| ξni | > δnh1/2 ) | Fi−1 ) −→ 0,
n→∞
n2 h
i=1
(b)
n−1
1 X
p
2
E(ξni
| Fi−1 ) −→ σ32 .
2
n→∞
n h
i=1
δ ∈ (0, 1);
760
M. S. Tikhov, D. S. Krishtopenko, and M. V. Yaroschuk
We have

n
X
2
ξni
= (Wi − F (i/n))2 
(Wj − F (j/n))
j = i+1
Z
2
Kh (x − i/n)Kh (x − j/n) dx ,
2
E(ξni
| Fi−1 ) = F (i/n)(1 − F (i/n))×

×D 
n
X
j = i+1
(Wj − F (j/n))
= F (i/n)(1 − F (i/n))
×

Z
n
X
j = i+1
Z
Kh (x − i/n)Kh (x − j/n) dx =
F (j/n)(1 − F (j/n))×
Kh (x − i/n)Kh (x − j/n) dx
2
.
Hence, when n → ∞, h → 0,
n
n−1
n−1
X
1 X
1 X
2
F (j/n)(1 − F (j/n))×
E(ξni | Fi−1 ) = 2
F (i/n)(1 − F (i/n))
n2 h
n h
i=1
j = i+1
i=1
×
Z1
∼
0
∼
=
Kh (x − i/n)Kh (x − j/n) dx
Z1
2
F (u)(1 − F (u)) du
h
×
σ32
Z
Z
2
Z
u
∼
F (v)(1 − F (v)) dv×
Kh (u − x)Kh (v − x) dx
2
2
F (x)(1 − F (x)) dx
Z Z
2
∼
K(u)K(u + v) du
2
dv;
therefore condition (b) is satisfied.
Moreover,
n−1
n−1
X
1 X
1
4
2
1/2
E(ξni
| Fi−1 ).
E(ξ
I(|
ξ
|
>
δnh
)
|
F
)
6
ni
i−1
ni
n2 h
δ2 n4 h2
i=1
i=1
Consider the sum on the right-hand side of this inequality. We have
4
E(ξni
| Fi−1 ) = E(Wi − F (i/n))4 ×

×E 
n
X
(Wj − F (j/n))
j = i+1

6 E(Wi − F (i/n))4 E 
Z
4
Kh (x − i/n)Kh (x − j/n) dx 6
n
X
4
(Wj − F (j/n)) 6
j = i+1
Estimation of Distributions under Dose-Effect Dependence with Fixed Experiment Plan

4
n
X
6 16 E 
(Wj − F (j/n)) .
761
j = i+1
But

E
n
X
4
(Wj − F (j/n)) = E 
j = i+1
+6
X
i+16j<l6n
n
X
=

n
X

(Wj − F (j/n))4  +
j = i+1
E (Wj − F (j/n))2 (Wl − F (l/n))2 =
((F (j/n))4 (1 − F (j/n)) + F (j/n)(1 − F (j/n))4 )+
j = i+1

+6 
n
X
j = i+1
Hence,
2

n
F (j/n)(1 − F (j/n)) 6 + 6 
4
n−1
1 X
2
E(ξni
I(| ξni | > δnh1/2 ) | Fi−1 ) 6
n2 h
n
X
j = i+1
2
F (j/n)(1 − F (j/n)) .
i=1


16  1
+6 
6
nh2 4n
n
X
j = i+1
2 
16
F (j/n)(1 − F (j/n))  ∼
nh2
1
3
+
4n 8
−→ 0.
n→∞
So, in this case, condition (a) is satisfied.
Now from [11] it follows that the sequence In3 is asymptotically normal with parameters (0, σ32 ).
Remark 2. For Epanechnikov’s kernel K(x) = (3/4)(1 − x2 )I(| x | 6 1) the convolution equals
(K ∗ K)(x) = K2 (x) = R(3/160)(32
− 40x2 + 20x3 − x5 ) for 0 6 x 6 2 and (K ∗ K)(x) = K2 (−x) for
R
−2 6 x 6 0. Therefore ( K(u)K(u + v) du)2 dv = 167/387 ≈ 0.434.
( iv ) We
R have In = 2In1 + In2 + In3 .
Let c(n) = E(Fn (x) − F (x))2 dx and µ(n) = c(n) + n−1 σ22 .
From Lemmas 1–3 we derive the following theorem.
Theorem 4. Under conditions (A), assuming that h → 0 and nh → ∞ when n → ∞, we have
d
n1/2 h−2 (In − µ(n)) −→ N (0, 4ν 2 σ12 ), if
n→∞
d
nh−1/2 (In − µ(n)) −→ N (0, 2σ32 ),
n→∞
if
nh5 → ∞;
nh5 → 0;
d
nh−1/2 (In − µ(n)) −→ N (0, 4ν 2 σ12 λ4/5 + 2σ32 λ−1/5 ),
n→∞
if nh5 → λ ∈ (0, +∞).
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