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Asked by Julio
on 24 Jul 2012

I am currently writing code for the project that my professor is thinking of, and this involves writing code for local linear regression and Nadaraya-Watson regression. The first code that I wrote uses a loop for the NW estimator and the local linear regression estimator.

% This code simulates the local linear estimator and NW kernel.

% Generating noisy data. (I use the same as the example in the Lo paper.) x=linspace(0,4*pi,100); y=sin(x)+0.5*randn(size(x));

% Set up the space to store the estimates. yhatnw=zeros(size(x)); yhatll=zeros(size(x)); n=length(x);

% Set up the bandwidth. hx=median(abs(x-median(x)))/0.6745*(4/3/n)^0.2; hy=median(abs(y-median(y)))/0.6745*(4/3/n)^0.2; h=sqrt(hy*hx);

% Find smooth at each value of x. for i=1:n w=wfun(h,x(i),x); xc=x-x(i); s2=mean(xc.^2.*w); s1=mean(xc.*w); s0=mean(w); yhatnw(i)=sum(w.*y)/sum(w); % Nadaraya-Watson kernel yhatll(i)=sum(((s2-s1*xc).*w.*y)/(s2*s0-s1^2))/n; % local linear estimator end

plot(x,y,'.',x,yhatnw,'-',x,yhatll,':') The corresponding weighting function is:

function w=wfun(h,mu,x); w=exp((-1/2)*(((x-mu)/h).^2))/sqrt(2*pi*h^2);

I tried to vectorize the code, and it works for the NW estimator (since I am getting the same graph), but not for the local linear estimator (the graph with the loop is better than that of the matrix). The relevant code (for the local linear estimator) is here:

% Evaluating the weighting matrix. xi=ones(n,1)*x; data=x'*ones(1,n); w=wfun(h,xi,data);

% % Local linear regression xc=data-xi; s2=mean(xc.^2.*w); s1=mean(xc.*w); s0=mean(w); s2i=ones(n,1)*s2; s1i=ones(n,1)*s1; s0i=ones(n,1)*s0; yhatll=sum(((s2i-s1i*xc).*w.*yi)'./(s2i*s0i-(s1i^2)))/n;

I think there is something wrong with the way I am vectorizing the local linear regression, but I am unsure, so I'd like to ask all of you. Thanks!

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Answer by Andrei Bobrov
on 25 Jul 2012

Edited by Andrei Bobrov
on 25 Jul 2012

xc = bsxfun(@minus,x',x); w = exp(-.5*(xc/h).^2)/sqrt(2*pi*h^2); xw = xc.*w; s0 = mean(w); s1 = mean(xw); s2 = mean(xw.*xc); yhatnw = sum(bsxfun(@times,w,y'))./sum(w); yhatll = sum(... bsxfun(@rdivide,... bsxfun(@times,... bsxfun(@minus,s2,... bsxfun(@times,xc,s1)).*w,y'),s2.*s0 - s1.^2))/n;

or your variant

[ii,jj] = meshgrid(1:n); xc = x(jj)-x(ii); w=exp( -1/2*(xc/h).^2 )/sqrt(2*pi*h^2); s2=mean(xc.^2.*w); s1=mean(xc.*w); s0=mean(w); yhatll3=sum(((s2(ii)-s1(ii).*xc).*w.*y(jj))./(s2(ii).*s0(ii)-s1(ii).^2))/n;

## 2 Comments

## bym (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/44433#comment_91426

I think your post is missing the definition of yi, which sidetracked me for awhile. Please confirm that yi is defined something like

## Julio (view profile)

Direct link to this comment:http://www.mathworks.com/matlabcentral/answers/44433#comment_91478

Hi, I was looking at it, and you are right. I was missing my definition of yi. I'll put it right here: