Integration by parts with empirical measure

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I'm currently reading through the paper Asymptotic normality of nearest neighbor regression function estimates and am struggling to understand the asymptotic equalities that were shown in the proof of Lemma 4 (p. 923) where the last one was proven by using integration by parts:

$beginalign*&qquad fracm(x_0)sqrta_n^3int left[alpha_n(x_0)-alpha_n(x)right]K'left(fracF(x_0)-F(x)a_nright)F(d x)\ &= -fracm(x_0)sqrta_n^3int alpha_n(x)K'left(fracF(x_0)-F(x)a_nright)F(d x)\
&= -fracm(x_0)sqrta_nint Kleft(fracF(x_0)-F(x)a_nright)alpha_n(dx). endalign*$

Here $m(x) = E[Y|X=x]$, $alpha_n(x) = sqrtn [F_n(x)-F(x)]$ denotes the empirical process of the random variable $X$ (having continuous distribution function $F$), $K$ being a twice continuously differentiable probability kernel with bounded support and $na_n^3 to infty$, $a_n to 0$ for $n to infty$.

I know that I could expand the second to last term like so:
$beginalign
&quad-fracm(x_0)sqrta_n^3int alpha_n(x)K'left(fracF(x_0)-F(x)a_nright)F(d x)\
&= -fracm(x_0)sqrtnsqrta_n^3int F_n(x)K'left(fracF(x_0)-F(x)a_nright)F(d x) + fracm(x_0)sqrtnsqrta_n^3int F(x)K'left(fracF(x_0)-F(x)a_nright)F(d x)\ endalign$

where the second integral contains transformations with $F$, so a transformation of variables and the integral a la $u = F(x)$ is possible and then application of the "classic" integration by parts yields the first of the following terms and a "constant" term.

$beginalign &= fracm(x_0)sqrtnsqrta_nint Kleft(fracF(x_0)-F(x)a_nright) F(d x) -fracm(x_0)sqrtnsqrta_nint Kleft(fracF(x_0)-F(x)a_nright) F_n(d x)\
&= -frac1sqrta_nm(x_0)int Kleft(fracF(x_0)-F(x)a_nright) alpha_n(d x)endalign$

It would be nice if someone could provide me an explanation on the upper equalities since I'm not even sure if I'm on the right track. I only consider the classic integration by parts, maybe the Lebesgue-Stieltjes variant is more helpful.

integration probability-theory measure-theory regression empirical-processes

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asked Jul 26 at 11:43

PeterMcCoy

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I'm currently reading through the paper Asymptotic normality of nearest neighbor regression function estimates and am struggling to understand the asymptotic equalities that were shown in the proof of Lemma 4 (p. 923) where the last one was proven by using integration by parts:

$beginalign*&qquad fracm(x_0)sqrta_n^3int left[alpha_n(x_0)-alpha_n(x)right]K'left(fracF(x_0)-F(x)a_nright)F(d x)\ &= -fracm(x_0)sqrta_n^3int alpha_n(x)K'left(fracF(x_0)-F(x)a_nright)F(d x)\
&= -fracm(x_0)sqrta_nint Kleft(fracF(x_0)-F(x)a_nright)alpha_n(dx). endalign*$

Here $m(x) = E[Y|X=x]$, $alpha_n(x) = sqrtn [F_n(x)-F(x)]$ denotes the empirical process of the random variable $X$ (having continuous distribution function $F$), $K$ being a twice continuously differentiable probability kernel with bounded support and $na_n^3 to infty$, $a_n to 0$ for $n to infty$.

I know that I could expand the second to last term like so:
$beginalign
&quad-fracm(x_0)sqrta_n^3int alpha_n(x)K'left(fracF(x_0)-F(x)a_nright)F(d x)\
&= -fracm(x_0)sqrtnsqrta_n^3int F_n(x)K'left(fracF(x_0)-F(x)a_nright)F(d x) + fracm(x_0)sqrtnsqrta_n^3int F(x)K'left(fracF(x_0)-F(x)a_nright)F(d x)\ endalign$

where the second integral contains transformations with $F$, so a transformation of variables and the integral a la $u = F(x)$ is possible and then application of the "classic" integration by parts yields the first of the following terms and a "constant" term.

$beginalign &= fracm(x_0)sqrtnsqrta_nint Kleft(fracF(x_0)-F(x)a_nright) F(d x) -fracm(x_0)sqrtnsqrta_nint Kleft(fracF(x_0)-F(x)a_nright) F_n(d x)\
&= -frac1sqrta_nm(x_0)int Kleft(fracF(x_0)-F(x)a_nright) alpha_n(d x)endalign$

It would be nice if someone could provide me an explanation on the upper equalities since I'm not even sure if I'm on the right track. I only consider the classic integration by parts, maybe the Lebesgue-Stieltjes variant is more helpful.

integration probability-theory measure-theory regression empirical-processes

share|cite|improve this question

asked Jul 26 at 11:43

PeterMcCoy

37139

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I'm currently reading through the paper Asymptotic normality of nearest neighbor regression function estimates and am struggling to understand the asymptotic equalities that were shown in the proof of Lemma 4 (p. 923) where the last one was proven by using integration by parts:

$beginalign*&qquad fracm(x_0)sqrta_n^3int left[alpha_n(x_0)-alpha_n(x)right]K'left(fracF(x_0)-F(x)a_nright)F(d x)\ &= -fracm(x_0)sqrta_n^3int alpha_n(x)K'left(fracF(x_0)-F(x)a_nright)F(d x)\
&= -fracm(x_0)sqrta_nint Kleft(fracF(x_0)-F(x)a_nright)alpha_n(dx). endalign*$

Here $m(x) = E[Y|X=x]$, $alpha_n(x) = sqrtn [F_n(x)-F(x)]$ denotes the empirical process of the random variable $X$ (having continuous distribution function $F$), $K$ being a twice continuously differentiable probability kernel with bounded support and $na_n^3 to infty$, $a_n to 0$ for $n to infty$.

I know that I could expand the second to last term like so:
$beginalign
&quad-fracm(x_0)sqrta_n^3int alpha_n(x)K'left(fracF(x_0)-F(x)a_nright)F(d x)\
&= -fracm(x_0)sqrtnsqrta_n^3int F_n(x)K'left(fracF(x_0)-F(x)a_nright)F(d x) + fracm(x_0)sqrtnsqrta_n^3int F(x)K'left(fracF(x_0)-F(x)a_nright)F(d x)\ endalign$

where the second integral contains transformations with $F$, so a transformation of variables and the integral a la $u = F(x)$ is possible and then application of the "classic" integration by parts yields the first of the following terms and a "constant" term.

$beginalign &= fracm(x_0)sqrtnsqrta_nint Kleft(fracF(x_0)-F(x)a_nright) F(d x) -fracm(x_0)sqrtnsqrta_nint Kleft(fracF(x_0)-F(x)a_nright) F_n(d x)\
&= -frac1sqrta_nm(x_0)int Kleft(fracF(x_0)-F(x)a_nright) alpha_n(d x)endalign$

It would be nice if someone could provide me an explanation on the upper equalities since I'm not even sure if I'm on the right track. I only consider the classic integration by parts, maybe the Lebesgue-Stieltjes variant is more helpful.

integration probability-theory measure-theory regression empirical-processes

share|cite|improve this question

asked Jul 26 at 11:43

PeterMcCoy

37139

I'm currently reading through the paper Asymptotic normality of nearest neighbor regression function estimates and am struggling to understand the asymptotic equalities that were shown in the proof of Lemma 4 (p. 923) where the last one was proven by using integration by parts:

$beginalign*&qquad fracm(x_0)sqrta_n^3int left[alpha_n(x_0)-alpha_n(x)right]K'left(fracF(x_0)-F(x)a_nright)F(d x)\ &= -fracm(x_0)sqrta_n^3int alpha_n(x)K'left(fracF(x_0)-F(x)a_nright)F(d x)\
&= -fracm(x_0)sqrta_nint Kleft(fracF(x_0)-F(x)a_nright)alpha_n(dx). endalign*$

Here $m(x) = E[Y|X=x]$, $alpha_n(x) = sqrtn [F_n(x)-F(x)]$ denotes the empirical process of the random variable $X$ (having continuous distribution function $F$), $K$ being a twice continuously differentiable probability kernel with bounded support and $na_n^3 to infty$, $a_n to 0$ for $n to infty$.

I know that I could expand the second to last term like so:
$beginalign
&quad-fracm(x_0)sqrta_n^3int alpha_n(x)K'left(fracF(x_0)-F(x)a_nright)F(d x)\
&= -fracm(x_0)sqrtnsqrta_n^3int F_n(x)K'left(fracF(x_0)-F(x)a_nright)F(d x) + fracm(x_0)sqrtnsqrta_n^3int F(x)K'left(fracF(x_0)-F(x)a_nright)F(d x)\ endalign$

where the second integral contains transformations with $F$, so a transformation of variables and the integral a la $u = F(x)$ is possible and then application of the "classic" integration by parts yields the first of the following terms and a "constant" term.

$beginalign &= fracm(x_0)sqrtnsqrta_nint Kleft(fracF(x_0)-F(x)a_nright) F(d x) -fracm(x_0)sqrtnsqrta_nint Kleft(fracF(x_0)-F(x)a_nright) F_n(d x)\
&= -frac1sqrta_nm(x_0)int Kleft(fracF(x_0)-F(x)a_nright) alpha_n(d x)endalign$

It would be nice if someone could provide me an explanation on the upper equalities since I'm not even sure if I'm on the right track. I only consider the classic integration by parts, maybe the Lebesgue-Stieltjes variant is more helpful.

integration probability-theory measure-theory regression empirical-processes

share|cite|improve this question

asked Jul 26 at 11:43

PeterMcCoy

37139

share|cite|improve this question

share|cite|improve this question

asked Jul 26 at 11:43

PeterMcCoy

37139

asked Jul 26 at 11:43

PeterMcCoy

37139

asked Jul 26 at 11:43

PeterMcCoy

37139

37139

add a comment | 

add a comment | 

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