Let F_n(x) be empirical function of the distribution constructed by simple sample and S_N(x) be cubic spline interpolating the values F_n(x_k) in the nodes of the mesh (1). If we choose the boundary conditions for S_N(x) in the form

then the derivative of the spline function is defined by the equality

Here, for, 0

and

C_i,j(x) are defined by the following relations:

where

, for the other i and j.

Let F_n(x) be empirical function of the distribution constructed by simple sample and S_N(x) be cubic spline interpolating the values F_n(x_k). in the mesh (1). If we choose the boundary conditions for S_N(x) in the form

Then the derivative of the spline function is defined by the equality

where, for, ,

,

, ,

, ,

, ,

,

and C_i,j are defined by formula (2).

We introduce the following denotations:

is the simple sample from the general population

;

is empirical function of distribution of the sample;

is the empirical process;

is the sequence of wiener processes;

is the brownian bridge.

We give the auxiliary lemmas.

There exists a probability space (Ω, F, P).

On which it can be defined version and the sequence of Brownian bridges B_n(t) such that for all x > 0

where a = 3.26, b = 4.86, с = 2.70.

Let be modulus of continuity of the brownian bridge B_n(t),

and. Then with probability 1 does not exceed the quantity.

Here is the random variable which is not less than 1 almost everywhere and.

Let be the spline estimation.

1) If and are defined as in Theorem 2, then for

.

2) If and are defined as in Theorem 1, then

where 0 < x < 1,

[y] is the integer part of the number y.

3) Suppose, , , d = i ? j, and, then for

Proof. By virtue of, Theorems 9, 11, 12 from Stechkin and Subbotin [1] and Theorems 1 from Lii [5] follows the first statement of Theorem 3. The second and the third statement of Theorem 3 are proved in Lii [5] .

Suppose as. Then in order with probability 1

it is necessary and sufficient that the function g(x) is the density of the distribution F(x) concentrated and continuous on the interval [0,1] with respect to Lebesgue measure.

Proof. Sufficiency. It is clear that

where

First we estimate the term in the right hand part of (3). We have

From Lemma 1 it follows that with probability 1 for

If we denote the modulus of continuity by then from

Lemma 2

where

with probability and

This, combining (3)-(6) and using Theorem 3 we get the sufficiency condition of Theorem 4.

Necessity. Let with probability 1

Hence, from continuity of it follows continuity of g(x) on the interval [0, 1].

Therefore, the sequence random variables

are uniformly integrable. Therefore according to Theorem 5 from Shiryaev [8] and the inequalities

it follows that for

By virtue of (7) it is easy to see that the sequence of functions

uniformly converges to some continuous function g₀(x), i.e. for

We show now continuity of F(x) on the interval [0, 1].

We assume the inverse that there exists a point x₀, such that. Then by virtue of (8) and

it follows continuity of F(x) on the interval [0, 1].

By (8) for all

From another side, according to Theorem 11 from Stechkin and Subbotin (1976)

By virtue of (9)-(11)

Theorem 4 is proved.