Residual values for a linear regression fit

Question

NA 2020-10-16

0
链接

此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/616413-residual-values-for-a-linear-regression-fit

评论： Star Strider 2020-10-17

采纳的回答： Star Strider

Aa.JPG

在 MATLAB Online 中打开

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.

0 个评论
显示 -2更早的评论隐藏 -2更早的评论

请先登录，再进行评论。

请先登录，再回答此问题。

Answer 1

Star Strider 2020-10-16

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/616413-residual-values-for-a-linear-regression-fit#answer_516208

在 MATLAB Online 中打开

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 个评论
显示 4更早的评论隐藏 4更早的评论

NA 2020-10-17

在 MATLAB Online 中打开

aaJPG.JPG

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 2020-10-17

在 MATLAB Online 中打开

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.

请先登录，再进行评论。