Residual values for a linear regression fit

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NA
NA el 16 de Oct. de 2020
Comentada: Star Strider el 17 de Oct. de 2020
I have these points
x = [1,1,2,2,3,4,4,6]';
y = [8,1,1,2,2,3,4,1]';
I want to remove the point from above set that makes the residual largest.
This is the code I use
d=zeros(length(x),1);
for i=1:length(x)
x_bk = x;
y_bk = y;
x(i) = [];
y(i) = [];
X = [ones(length(x),1) x];
b = X\y;
yhat = X*b;
d(i) = abs(sum(y - yhat));
x = x_bk;
y = y_bk;
end
index = find(min(d)==d);
x(index) = [];
y(index) = [];
X = [ones(length(x),1) x];
b = X\y;
yhat_r = X*b;
plot(x,y,'o')
hold on
plot(x,yhat_r,'--')
I think the result should be black line (attached file), but I get red dashed line.

Respuesta aceptada

Star Strider
Star Strider el 16 de Oct. de 2020
I would do something like this:
x = [1,1,2,2,3,4,4,6]';
y = [8,1,1,2,2,3,4,1]';
xv = x;
yv = y;
for k = 1:numel(x)
X = [xv(:), ones(size(xv(:)))];
b = X \ yv(:);
yhat = X*b;
rsdn(k) = norm(yv - X*b);
xv = x;
yv = y;
xv(k) = [];
yv(k) = [];
end
figure
plot((1:numel(x)), rsdn)
grid
[rsdnmin,idxn] = min(rsdn(2:end));
[rsdnmax,idxx] = max(rsdn(2:end));
lowest = idxn+1
hihest = idxx+1
idxv = [lowest; hihest];
figure
for k = 1:2
subplot(2,1,k)
xv = x;
yv = y;
xv(idxv(k)) = [];
yv(idxv(k)) = [];
plot(xv,yv,'ob')
yhat = [xv(:), ones(size(xv(:)))]*bmtx(:,idxv(k));
hold on
plot(xv, yhat, '--r')
hold off
title(sprintf('Eliminating Set %d', idxv(k)))
end
Here, the norm of residuals (the usual metric) is least when eliminating ‘row=2’, and greatest when eliminating ‘row=6’.
Experiment to get the result you want.
  6 comentarios
NA
NA el 17 de Oct. de 2020
I want to show that if I remove only one set of data the regression line changes a lot. (But I do not know, this is practically true or not).
For this reason, I make this set:
a0 = 4.5882;
a1 = 0.2353;
x = (0:1:8)';
y = a0+a1*x+randn(size(x));
But, it does not show any difference (please see the attachment). I think the way of producing data set is not correct.
Star Strider
Star Strider el 17 de Oct. de 2020
In that simulation, you are defining a particular slope and intercept and adding a normally-distributed random vector to it. The slopes and intercepts of the fitted lines will not change much.
You can see that most easily if you add this text call to each plot (in the loop):
text(1.1*min(xlim),0.9*max(ylim), sprintf('Y = %.3f\\cdotX%+.3f',bmtx(:,k)), 'HorizontalAlignment','left')
That will print the regression equation in the upper-left corner of each one. You can then compare them.
Note that the residual norms do not change much, either. In the original data set, they varied between 2.73 and 5.97. In this data set, they are within about ±0.5 of each other.

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