Data-Driven Bandwidth Selection for Recursive Kernel Density Estimators Under Double Truncation

Abstract

In this paper we proposed a data-driven bandwidth selection procedure of the recursive kernel density estimators under double truncation. We showed that, using the selected bandwidth and a special stepsize, the proposed recursive estimators outperform the nonrecursive one in terms of estimation error in many situations. We corroborated these theoretical results through simulation study. The proposed estimators are then applied to data on the luminosity of quasars in astronomy. We corroborated these theoretical results through simulation study, then, we applied the proposed estimators to data on the luminosity of quasars in astronomy.

Appendices

Appendix A: Proofs

First, we introduce the following asymptotically equivalent version of the proposed recursive estimators (2.3),

$$\begin{array}{@{}rcl@{}} f_{n}\left( x\right)=\left( 1-\gamma_{n}\right)f_{n-1}\left( x\right)+\gamma_{n}\alpha h_{n}^{-1}\frac{K\left( h_{n}^{-1}\left[x-X_{n}\right]\right)}{G\left( X_{n}\right)}, \end{array} $$

(5.1)

and the asymptotically equivalent version of the non-recursive estimator (2.4),

$$\begin{array}{@{}rcl@{}} \widetilde{f}_{n}\left( x\right)=\frac{\alpha}{nh_{n}}\sum\limits_{k = 1}^{n}\frac{K\left( \frac{x-X_{k}}{h_{n}}\right)}{G_{n}\left( X_{k}\right)}. \end{array} $$

(5.2)

Remark 1.

The consistency results of $\frac {\alpha _{n}}{G_{n}\left (.\right )}$ can be obtained from Shen (2010) and Moreira and de Uña-Àlvarez (2010).

Throughout this section we use the following notation:

$$\begin{array}{@{}rcl@{}} {\Pi}_{n}&=&\prod\limits_{j = 1}^{n}\left( 1-\gamma_{j}\right),\\ Z_{n}\left( x\right)&=&h_{n}^{-1}\alpha \frac{K\left( \frac{x-X_{n}}{h_{n}}\right)}{G\left( X_{n}\right)}. \end{array} $$

(5.3)

Let us first state the following technical lemma.

Lemma 1.

Let $\left (v_{n}\right )\in \mathcal {GS}\left (v^{*}\right )$, $\left (\eta _{n}\right )\in \mathcal {GS}\left (-\eta \right )$, and $m>0$ such that $m-v^{*}\xi >0$ where $\xi$ is defined inEq. 3.2. We have

$$\begin{array}{@{}rcl@{}} \lim_{n \to +\infty}v_{n}{{\Pi}_{n}^{m}}\sum\limits_{k = 1}^{n}{\Pi}_{k}^{-m}\gamma_{k}v_{k}^{-1} =\left( m-v^{*}\xi\right)^{-1}. \end{array} $$

Moreover, for all positive sequence $\left (\beta _{n}\right )$ such that $\lim _{n \to +\infty }\beta _{n}= 0$, and all $C \in \mathbb {R}$,

$$\begin{array}{@{}rcl@{}} \lim_{n \to +\infty}v_{n}{{\Pi}_{n}^{m}}\left[\sum\limits_{k = 1}^{n} {\Pi}_{k}^{-m} \eta_{k}v_{k}^{-1}\beta_{k}+C\right]= 0. \end{array} $$

Lemma 1 is widely applied throughout the proofs. Let us underline that it is its application, which requires Assumption $(A2)(iii)$ on the limit of $(n\gamma _{n})$ as n goes to infinity.

Our proofs are organized as follows. Propositions 1 and 2 in Sections 5.1 and 5.2 respectively, Theorem 1 in Section 5.3.

1.1 A.1 Proof of Proposition 1

In view of Eqs. 5.1 and 5.3, we have

$$\begin{array}{@{}rcl@{}} f_{n}\left( x\right) - f\left( x\right)\!\!\!&=&\!\!\!\left( 1-\gamma_{n}\right)\left( f_{n-1}\left( x\right)-f\left( x\right)\right)+\gamma_{n}\left( Z_{n}\left( x\right)-f\left( x\right)\right)\\ \!\!\!&=&\!\!\!\sum\limits_{k = 1}^{n-1}\left[\prod\limits_{j=k + 1}^{n}\left( 1-\gamma_{j}\right)\right]\gamma_{k}\left( Z_{k}\left( x\right) - f\left( x\right)\right)+\gamma_{n}\left( Z_{n}\left( x\right)-f\left( x\right)\right) \\ &&\!\!\!+\left[\prod\limits_{j = 1}^{n}\left( 1-\gamma_{j}\right)\right]\left( f_{0}\left( x\right)-f\left( x\right)\right)\\ \!\!\!&=&\!\!\!{\Pi}_{n}\sum\limits_{k = 1}^{n}{\Pi}_{k}^{-1}\gamma_{k}\left( Z_{k}\left( x\right)-f\left( x\right)\right)+{\Pi}_{n}\left( f_{0}\left( x\right)-f\left( x\right)\right). \end{array} $$

It follows that

$$\begin{array}{@{}rcl@{}} \mathbb{E}\left( f_{n}\left( x\right)\right)-f\left( x\right) \!\!\!&=&\!\!\!{\Pi}_{n}{\sum}_{k = 1}^{n}{\Pi}_{k}^{-1}\gamma_{k}\left( \mathbb{E}\left( Z_{k}\left( x\right)\right) - f\left( x\right)\right)+{\Pi}_{n}\left( f_{0}\left( x\right)-f\left( x\right)\right). \end{array} $$

Moreover, for simplicity, we let $H\left (x\right )=\frac {f_{X}\left (x\right )}{G\left (x\right )}$, and in view of Eq. ??, we have

$$\begin{array}{@{}rcl@{}} \mathbb{E}\left[Z_{k}^{p}\left( x\right)\right] \!\!\!&=&\!\!\!h_{k}^{-p}\alpha^{p}\mathbb{E}\left[\frac{K^{p}\left( \frac{x-X_{k}}{h_{k}}\right)}{G^{p}\left( X_{k}\right)}\right]\\ \!\!\!&=&\!\!\!h_{k}^{-p}\alpha^{p}{\int}_{\mathbb{R}}K^{p}\left( \frac{x-y}{h_{k}}\right)\frac{f_{X}\left( y\right)}{G^{p}\left( y\right)}dy\\ \!\!\!&=&\!\!\!h_{k}^{-p + 1}\alpha^{p}{\int}_{\mathbb{R}}K^{p}\left( z\right)\frac{f_{X}\left( x-zh_{k}\right)}{G^{p}\left( x-zh_{k}\right)}dy\\ \!\!\!&=&\!\!\!h_{k}^{-p + 1}\alpha^{p}{\int}_{\mathbb{R}}K^{p}\left( z\right)\frac{H\left( x-zh_{k}\right)}{G^{p-1}\left( x-zh_{k}\right)}dz\\ \!\!\!&=&\!\!\!h_{k}^{-p + 1}\alpha^{p-1}{\int}_{\mathbb{R}}K^{p}\left( z\right)\frac{f\left( x-zh_{k}\right)}{G^{p-1}\left( x-zh_{k}\right)}dz. \end{array} $$

(4)

Then, it follows from Eq. 5.4, for $p = 1$, that

$$\begin{array}{@{}rcl@{}} \mathbb{E}\left[Z_{k}\left( x\right)\right]-f\left( x\right)\!\!\!&=&\!\!\!{\int}_{\mathbb{R}}K\left( z\right)\left[f\left( x-zh_{k}\right)-f\left( x\right)\right]dz\\ \!\!\!&=&\!\!\!\frac{{h_{k}^{2}}}{2}f^{\left( 2\right)}\left( x\right)\mu_{2}\left( K\right)+\eta_{k}\left( x\right), \end{array} $$

(5)

with

$$\begin{array}{@{}rcl@{}} \eta_{k}\left( x\right)={\int}_{\mathbb{R}}K\left( z\right)\left[f\left( x-zh_{k}\right)-f\left( x\right)-\frac{1}{2}z^{2}{h_{k}^{2}}f^{\left( 2\right)}\left( x\right)\right]dz, \end{array} $$

and, since f is bounded and continuous at x, we have $\lim _{k\to \infty }\eta _{k}\left (x\right )= 0$. In the case $a\leq \gamma /5$, we have $\lim _{n\to \infty }\left (n\gamma _{n}\right )>2a$; the application of Lemma 1 then gives

$$\begin{array}{@{}rcl@{}} \mathbb{E}\left[f_{n}\left( x\right)\right]-f\left( x\right)\!\!\!&=&\!\!\!\frac{1}{2}f^{\left( 2\right)}\left( x\right){\int}_{\mathbb{R}}z^{2}K\left( z\right)dz{\Pi}_{n}{\sum}_{k = 1}^{n}{\Pi}_{k}^{-1}\gamma_{k}{h_{k}^{2}}\left[1+o\left( 1\right)\right]\\ &&+{\Pi}_{n}\left( f_{0}\left( x\right)-f\left( x\right)\right)\\ \!\!\!&=&\!\!\!\frac{1}{2\left( 1-2a\xi\right)}f^{\left( 2\right)}\left( x\right)\mu_{2}\left( K\right)\left[h_{n}^{2}+o\left( 1\right)\right], \end{array} $$

and Eq. ?? follows. In the case $a>\gamma /5$, we have ${h_{n}^{2}}=o\left (\sqrt {\gamma _{n}h_{n}^{-1}}\right )$, and $\lim _{n\to \infty }\left (n\gamma _{n}\right )>\left (\gamma -a\right )/2$, then Lemma 1 ensures that

$$\begin{array}{@{}rcl@{}} \mathbb{E}\left[f_{n}\left( x\right)\right]-f\left( x\right)\!\!\!&=&\!\!\!{\Pi}_{n}\sum\limits_{k = 1}^{n}{\Pi}_{k}^{-1}\gamma_{k}o\left( \sqrt{\gamma_{k}h_{k}^{-1}}\right)+O\left( {\Pi}_{n}\right)\\ \!\!\!&=&\!\!\!o\left( \sqrt{\gamma_{n}h_{n}^{-1}}\right). \end{array} $$

which gives (3.4). Further, we have

$$\begin{array}{@{}rcl@{}} Var\left[f_{n}\left( x\right)\right]\!\!\!&=&\!\!\!{{\Pi}_{n}^{2}}\sum\limits_{k = 1}^{n}{\Pi}_{k}^{-2}{\gamma_{k}^{2}}Var\left[Z_{k}\left( x\right)\right]\\ \!\!\!&=&\!\!\!{{\Pi}_{n}^{2}}\sum\limits_{k = 1}^{n}{\Pi}_{k}^{-2}{\gamma_{k}^{2}}\left( \mathbb{E}\left( {Z_{k}^{2}}\left( x\right)\right)-\left( \mathbb{E}\left( Z_{k}\left( x\right)\right)\right)^{2}\right). \end{array} $$

(6)

Moreover, in view of Eq. 5.4, for $p = 2$, that

$$\begin{array}{@{}rcl@{}} \mathbb{E}\left( {Z_{k}^{2}}\left( x\right)\right)\!\!\!&=&\!\!\!h_{k}^{-1}\alpha{\int}_{\mathbb{R}}\frac{f\left( x-zh_{k}\right)}{G\left( x-zh_{k}\right)}K^{2}\left( z\right)dz\\ \!\!\!&=&\!\!\!h_{k}^{-1}\alpha \frac{f\left( x\right)}{G\left( x\right)}{\int}_{\mathbb{R}}K^{2}\left( z\right)dz+\nu_{k}\left( x\right), \end{array} $$

(7)

with

$$\begin{array}{@{}rcl@{}} \nu_{k}\left( x\right)=h_{k}^{-1}\alpha{\int}_{\mathbb{R}}K^{2}\left( z\right)\left[\frac{f\left( x-zh_{k}\right)}{G\left( x-zh_{k}\right)}-\frac{f\left( x\right)}{G\left( x\right)}\right]dz. \end{array} $$

Moreover, it follows from Eq. 5.5, that

$$\begin{array}{@{}rcl@{}} \mathbb{E}\left[Z_{k}\left( x\right)\right]&=&f\left( x\right)+\widetilde{\nu}_{k}\left( x\right), \end{array} $$

(8)

with

$$\begin{array}{@{}rcl@{}} \widetilde{\nu}_{k}\left( x\right)={\int}_{\mathbb{R}}K\left( z\right)\left[f\left( x-zh_{k}\right)-f\left( x\right)\right]dz. \end{array} $$

Then, it follows from Eqs. 5.6, 5.7 and 5.8, that

$$\begin{array}{@{}rcl@{}} Var\left[f_{n}\left( x\right)\right]&=&\alpha \frac{f\left( x\right)}{G\left( x\right)}R\left( K\right){{\Pi}_{n}^{2}}\sum\limits_{k = 1}^{n}{\Pi}_{k}^{-2}{\gamma_{k}^{2}}h_{k}^{-1}+{{\Pi}_{n}^{2}}\sum\limits_{k = 1}^{n}{\Pi}_{k}^{-2}{\gamma_{k}^{2}}\nu_{k}\left( x\right)\\ &&-f^{2}\left( x\right){{\Pi}_{n}^{2}}\sum\limits_{k = 1}^{n}{\Pi}_{k}^{-2}{\gamma_{k}^{2}}-2f\left( x\right){{\Pi}_{n}^{2}}\sum\limits_{k = 1}^{n}{\Pi}_{k}^{-2}{\gamma_{k}^{2}}\widetilde{\nu}_{k}\left( x\right)\\ &&-{{\Pi}_{n}^{2}}\sum\limits_{k = 1}^{n}{\Pi}_{k}^{-2}{\gamma_{k}^{2}}\widetilde{\nu}_{k}^{2}\left( x\right). \end{array} $$

Since f and $fG^{-1}$ are bounded continuous, we have $\lim _{k\to \infty }\nu _{k}\left (x\right )= 0$ and $\lim _{k\to \infty }\widetilde {\nu _{k}}\left (x\right )= 0$. In the case $a\geq \gamma /5$, we have $\lim _{n\to \infty }\left (n\gamma _{n}\right )>\left (\gamma -a\right )/2$, and the application of Lemma 1 gives

$$\begin{array}{@{}rcl@{}} Var\left[f_{n}\left( x\right)\right]&=&\frac{\gamma_{n}}{h_{n}}\left( 2-\left( \gamma-a\right)\xi\right)^{-1}\alpha \frac{f\left( x\right)}{G\left( x\right)}R\left( K\right) +o\left( \frac{\gamma_{n}}{h_{n}}\right), \end{array} $$

which proves (3.5). Now, in the case $a<\gamma /5$, we have $\gamma _{n}h_{n}^{-1}=o\left ({h_{n}^{4}}\right )$, and $\lim _{n\to \infty }\left (n\gamma _{n}\right )>2a$, then the application of Lemma 1 gives

$$\begin{array}{@{}rcl@{}} Var\left[f_{n}\left( x\right)\right]&=&{{\Pi}_{n}^{2}}\sum\limits_{k = 1}^{n}{\Pi}_{k}^{-2}\gamma_{k}o\left( {h_{k}^{4}}\right)\\ &=&o\left( {h_{n}^{4}}\right), \end{array} $$

which proves (3.6).

1.2 A.2 Proof of Proposition 2

Following similar steps as the proof of the Proposition 2 of Mokkadem et al. (2009), we proof the Proposition 2.

1.3 A.3 Proof of Theorem 1

Let us at first assume that, if $a\geq \gamma /5$, then

$$\begin{array}{@{}rcl@{}} \sqrt{\gamma_{n}^{-1} h_{n}}\left( f_{n}\left( x\right) - \mathbb{E}\left[f_{n}\left( x\right)\right]\right)\stackrel{\mathcal{D}}{\rightarrow}\mathcal{N}\left( 0,\left( 2 - \left( \gamma - a\right)\xi\right)^{-1}\alpha \frac{f\left( x\right)}{G\left( x\right)}R\left( K\right)\right). \end{array} $$

(9)

In the case when $a>\gamma /5$, Part 1 of Theorem 1 follows from the combination of Eqs. 3.4 and 5.9. In the case when $a=\gamma /5$, Parts 1 and 2 of Theorem ?? follow from the combination of Eqs. 3.3 and 5.9. In the case $a<\gamma /5$, Eq. 3.6 implies that

$$\begin{array}{@{}rcl@{}} h_{n}^{-2}\left( f_{n}\left( x\right)-\mathbb{E}\left( f_{n}\left( x\right)\right)\right)\stackrel{\mathbb{P}}{\rightarrow}0, \end{array} $$

and the application of Eq. 3.3 gives Part 2 of Theorem 1.

We now prove (5.9). In view of Eq. 2.3, we have

$$\begin{array}{@{}rcl@{}} f_{n}\left( x\right)-\mathbb{E}\left[f_{n}\left( x\right)\right]={\Pi}_{n}\sum\limits_{k = 1}^{n}{\Pi}_{k}^{-1}\gamma_{k}\left( Z_{k}\left( x\right)-\mathbb{E}\left[Z_{k}\left( x\right)\right]\right). \end{array} $$

Set

$$\begin{array}{@{}rcl@{}} Y_{k}\left( x\right)&=&{\Pi}_{k}^{-1}\gamma_{k}\left( Z_{k}\left( x\right)-\mathbb{E}\left[Z_{k}\left( x\right)\right]\right). \end{array} $$

The application of Lemma 1 ensures that

$$\begin{array}{@{}rcl@{}} {v_{n}^{2}}&=&{\sum}_{k = 1}^{n}Var\left( Y_{k}\left( x\right)\right)\\ &=&\sum\limits_{k = 1}^{n}{\Pi}_{k}^{-2}{\gamma_{k}^{2}}Var\left( Z_{k}\left( x\right)\right)\\ &=&\sum\limits_{k = 1}^{n}{\Pi}_{k}^{-2}{\gamma_{k}^{2}}h_{k}^{-1}\left[\alpha \frac{f\left( x\right)}{G\left( x\right)}R\left( K\right)+o\left( 1\right)\right]\\ &=&{\Pi}_{n}^{-2}\gamma_{n}h_{n}^{-1}\left[\left( 2-\left( 2-\left( \gamma-a\right)\xi\right)\right)^{-1}\alpha \frac{f\left( x\right)}{G\left( x\right)}R\left( K\right)+o\left( 1\right)\right]. \end{array} $$

On the other hand, we have, for all $p>0$,

$$\begin{array}{@{}rcl@{}} \mathbb{E}\left[\left|Z_{k}\left( x\right)\right|^{2+p}\right] &=& O\left( \frac{1}{h_{k}^{1+p}}\right), \end{array} $$

and, since $\lim _{n\to \infty }\left (n\gamma _{n}\right )>\left (\gamma -a\right )/2$, there exists $p>0$ such that $\lim _{n\to \infty }$ $\left (n\gamma _{n}\right )>\frac {1+p}{2+p}\left (\gamma -a\right )$. Applying Lemma 1, we get

$$\begin{array}{@{}rcl@{}} {\sum}_{k = 1}^{n}\mathbb{E}\left[\left|Y_{k}\left( x\right)\right|^{2+p}\right]&=&O\left( \sum\limits_{k = 1}^{n} {\Pi}_{k}^{-2-p}\gamma_{k}^{2+p}\mathbb{E}\left[\left|Y_{k}\left( x\right)\right|^{2+p}\right]\right)\\ &=&O\left( \sum\limits_{k = 1}^{n} \frac{{\Pi}_{k}^{-2-p}\gamma_{k}^{2+p}}{h_{k}^{1+p}}\right)\\ &=&O\left( \frac{\gamma_{n}^{1+p}}{{\Pi}_{n}^{2+p}h_{n}^{1+p}}\right), \end{array} $$

and we thus obtain

$$\begin{array}{@{}rcl@{}} \frac{1}{v_{n}^{2+p}}\sum\limits_{k = 1}^{n}\mathbb{E}\left[\left|Y_{k}\left( x\right)\right|^{2+p}\right]& = & O\left( {\left[\gamma_{n}h_{n}^{-1}\right]}^{p/2}\right)=o\left( 1\right). \end{array} $$

The convergence in Eq. 5.9 then follows from the application of Lyapounov’s Theorem.

Appendix B: R Source Code

Here we give a source code of the proposed method according to the first model: $U^{*}\sim \mathcal {U}\left (-1,0\right )$, $V^{*}\sim \mathcal {U}\left (0,1\right )$ and $X^{*}\sim \mathcal {N}\left (0,1\right )$.

n=100; #sample size Np=250; #number of discretization points niter=50; #number of iteration ᅟ # initialization of parameters ᅟ FNR=matrix(0,niter,Np); FR=matrix(0,niter,Np); A1=rep(0,niter); A2=rep(0,niter); A=rep(0,niter); Y=matrix(0,n,Np); ᅟ # generation of discretization points ᅟ D=seq(-4,4,8/(Np-1)); ᅟ #Gaussian kernel KG<-function(x){1/sqrt(2*pi)*exp(-x^2/2)} #second derivative of Gaussian kernel Kd<-function(x){y<-(x^2-1)*KG(x)} #computing R(K) K2G<-function(x){1/(2*pi)*exp(-x^2)} IR=integrate(K2G, lower = -Inf, upper = Inf)$value; #computing mu2(K) Kx<-function(x){1/sqrt(2*pi)*x^{2}*exp(-x^{2}/2)} mu=integrate(Kx, lower = -Inf, upper = Inf)$value; ᅟ #start of iterations for (iter in 1:niter) { print(iter) ᅟ #simulation ᅟ Xt=rnorm(n); Ut=runif(n,-1,0); Vt=runif(n,0,1); XX=(Xt>=Ut)*Xt*(Xt<=Vt); X1=(Xt>=Ut)*Xt; X2=Xt*(Xt<=Vt); alpha1=mean(X1!=0); alpha2=mean(X2!=0); alpha=mean(XX!=0); X=Xt;U=Ut;V=Vt; ᅟ for (i in 1:n){ while (X[i]<U[i]|X[i]>V[i]){ U[i]<-runif(1,-1,0) X[i]<-rnorm(1) V[i]<-runif(1,0,1)}} ᅟ #estimation of alpha ᅟ Gn=ecdf(X); G=function(x) {1-x^2} ᅟ alpha=mean(G(XX)); ᅟ #estimation of I1 and I2 ᅟ Q1=quantile(X,0.25) Q3=quantile(X,0.75) d=(Q3-Q1)/1.349; c=min(sd(X),d); ᅟ #estimation of I1 for the non-recursive estimator (see, equation (18)) ᅟ #pilot bandwidth for estimating I1 h=c*n^(-2/5); ᅟ M1=matrix(0,ncol=n,nrow=n); for (i in 1:n){ for (j in 1:n){ M1[i,j]=KG((X[i]-X[j])/h)*(G(X[i])*G(X[j]))^(-1);}} ᅟ I1tilde=(sum(M1)-sum(diag(M1)))/(n*(n-1)*h); ᅟ #estimation of I2 for the non-recursive estimator (see, equation (19)) ᅟ #pilot bandwidth for estimating I2 h=c*n^(-3/14); ᅟ M2 = array (dim = c(n,n,n)); for (i in 1:n){ for (j in 1:n){ for (k in 1:n){ M2[i,j,k]=Kd((X[i]-X[j])/h)*Kd((X[i]-X[k])/h)*(G(X[j]) *G(X[k]))^(-1);}}} L=rep(0,n); for (i in 1:n) { L[i]=sum(diag(M2[i,,]));} I2tilde=(sum(M2)-sum(L))/(n^3*h^6); ᅟ #estimation of I1 for the recursive estimator (see, equation (12)) ᅟ #pilot stepsize for estimating I1 ᅟ gam=1.36/c(2:n); Gl=1-gam; Pn=prod(1-gam); ng=length(Gl); L1=rep(0,ng); for (k in 1:ng) { L1[k]=prod(Gl[1:k]);} P1=L1^(-1); ᅟ #pilot bandwidth for estimating I1 hk=c*c(1:n)^(-2/5); ᅟ N1=matrix(0,ncol=n,nrow=n); for (i in 1:ng){ for (j in 1:ng){ N1[i,j]=(P1[j]*gam[j]/hk[j]*KG((X[i]-X[j])/hk[j]))*(G(X[i]) *G(X[j]))^(-1);}} ᅟ I1hat=Pn*n^(-1)*(sum(N1)-sum(diag(N1))); ᅟ #estimation of I2 for the recursive estimator (see, equation (13)) ᅟ #pilot stepsize for estimating I2 ᅟ gam=1.48/c(2:n); Gl=1-gam; Pn=prod(1-gam); ng=length(Gl); L1=rep(0,ng); for (k in 1:ng) { L1[k]=prod(Gl[1:k]);} P1=L1^(-1); ᅟ #pilot bandwidth for estimating I2 hk=c*c(1:n)^(-3/14); ᅟ N2=array (dim = c(ng,ng,ng)); for (i in 1:ng){ for (j in 1:ng){ for (k in 1:ng){ N2[i,j,k]=P1[j]*gam[j]*P1[k]*gam[k]*hk[j]^(-3)*hk[k]^(-3) *Kd((X[i]-X[j])/hk[j]) *Kd((X[i]-X[k])/hk[k])*(G(X[j])*G(X[k]))^(-1);}}} Ln=rep(0,ng); for (i in 1:ng) { Ln[i]=sum(diag(N2[i,,]));} ᅟ I2hat=Pn^2*n^(-1)*(sum(N2)-sum(Ln)); ᅟ #Optimale Bandwidth for the non-recursive estimator (see, equation (20)) ᅟ hN=(I1tilde/I2tilde)^(1/5)*(IR/(mu^2))^(1/5)*alpha^(1/5) *n^(-1/5); ᅟ #Optimale Bandwidth for the recursive estimator (see, equation (15)) ᅟ hR=(3/10)^(1/5)*(I1hat/I2hat)^(1/5)*(IR/(mu^2))^(1/5) *alpha^(1/5)*c(2:n)^(-1/5); ᅟ #Non-recursive estimator ᅟ I1=matrix(0,Np,n); for (i in 1:Np){ for (j in 1:n) { I1[i,j]=sum(hN^(-1)*KG(hN^(-1)*(D[i]-X[j]))*G(X[j])^(-1))} } FNR[iter,]=alpha*rowSums(I1)/n; ᅟ #Recursive estimator with stepsize gamma_n=n^(-1) ᅟ I2=matrix(0,Np,(n-1)); for (i in 1:Np){ for (j in 1:(n-1)) { I2[i,j]=sum(hR[j]^(-1)*KG(hR[j]^(-1)*(D[i]-X[j])) *G(X[j])^(-1))} } FR[iter,]=alpha*rowSums(I2)/n; ᅟ }#end of iterations ᅟ #output the result FNR1=FNR[!rowSums(!is.finite(FNR)),] FR1=FR[!rowSums(!is.finite(FR)),] FNR1=colMeans(FNR1); #nonrecursive FR1=colMeans(FR1); #recursive ᅟ #qualitative comparaison between the two estimators (3) and (4) MD=dnorm(D); #Density for the standard normal distribution NR1=c(mean(abs(FNR1-MD)),mean((FNR1-MD)^2), mean(abs(FNR1/MD-1))) NR1=round(NR1,4); print("Non recursive") print(NR1) Rec1=c(mean(abs(FR1-MD)),mean((FR1-MD)^2),mean(abs(FR1/MD-1))) Rec1=round(Rec1,4); print("Recursive") print(Rec1) ᅟ #quantitative comparaison between the two estimators (3) and (4) plot(D,MD,xlab="",ylab="",ylim=c(0,0.65)) Data1=data.frame(D,FNR1) lines(Data1,type="l",lty=1,lwd=2) Data2=data.frame(D,FR1) lines(Data2,type="l",lty=2,lwd=2) legend("topleft",legend=c("Nonrecursive","Recursive"), lty=c(1,2),lwd=2)