How do I vectorize the loop for the local linear regression?

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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!
  2 Comments
bym
bym on 24 Jul 2012
I think your post is missing the definition of yi, which sidetracked me for awhile. Please confirm that yi is defined something like
yi = repmat(y,100,1);
Julio
Julio on 25 Jul 2012
Hi, I was looking at it, and you are right. I was missing my definition of yi. I'll put it right here:
yi=ones(n,1)*y;

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Answers (1)

Andrei Bobrov
Andrei Bobrov on 25 Jul 2012
Edited: 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;

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