Help with Linear Regression

Question

liu James 2017-6-14

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/344761-help-with-linear-regression

评论： Star Strider 2017-6-15

I'm trying to conduct a simple linear regression fitting using fitlm, but the results that is provided when I plot the curve with the original data and the linear equation line found through fitlm is really off. I'm just using the first 20 data points from the list

    mdl = fitlm(time(1:20),price(1:20))
  mdl = 
Linear regression model:
    y ~ 1 + x1
Estimated Coefficients:
                    Estimate       SE        tStat      pValue  
                   __________    _______    _______    _________
      (Intercept)    2.4736e+05      42743     5.7871    1.748e-05
      x1                -5.7654    0.99663    -5.7849    1.756e-05
Number of observations: 20, Error degrees of freedom: 18
Root Mean Squared Error: 0.0178
R-squared: 0.65,  Adjusted R-Squared 0.631
F-statistic vs. constant model: 33.5, p-value = 1.76e-05

Then I prepped for plot using.

   x=time(1:20)
   y1=2.4736e+05+(x*-5.7654)
   plot(time(1:20),price(1:20),x,y1)

I would get this plot attached. I've also attached the data. Please help with a better fit or explanation as to why it is so far off. I'm not sure what I'm doing incorrectly.

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Answer 1

Star Strider 2017-6-14

0
链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/344761-help-with-linear-regression#answer_270710

在 MATLAB Online 中打开

This works:

d = load('liu James Linear .mat');
Price = d.Price;
time = d.time;
mdl = fitlm(time, Price, 'linear')
ypred = predict(mdl, [min(time) max(time)]');
figure(1)
plot(time, Price, '+')
hold on
plot([min(time) max(time)]', ypred, '-r')
hold off
grid

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liu James 2017-6-15

在 MATLAB Online 中打开

by the way, I changed the upper code for range of 1:20.

mdl = fitlm(time(1:20), Price(1:20), 'linear')
ypred = predict(mdl, time(1:20));
figure(1)
plot(time(1:20), Price(1:20), '+')
hold on
plot([min(time) max(time)]', ypred, '-r')
hold off
grid
mdl = 
Linear regression model:
    y ~ 1 + x1
Estimated Coefficients:
                    Estimate       SE        tStat      pValue  
                   __________    _______    _______    _________
      (Intercept)    2.4736e+05      42743     5.7871    1.748e-05
      x1                -5.7654    0.99663    -5.7849    1.756e-05
Number of observations: 20, Error degrees of freedom: 18
Root Mean Squared Error: 0.0178
R-squared: 0.65,  Adjusted R-Squared 0.631
F-statistic vs. constant model: 33.5, p-value = 1.76e-05

The mdl gives the same as above.

Which I wrote the slope formula as

y1=2.4736e+05+(x*-5.7654)

Is this incorrect?

Star Strider 2017-6-15

在 MATLAB Online 中打开

That appears to be correct for the first 20 values. (I get the same result, not surprisingly.)

I would use the predict function rather than writing your own function to calculate the fit (that you then use to plot the line). The predict function uses full internal precision of the slope and intercept, while your equation uses only the precision that fitlm reports in its results. That is likely the reason your results seem to be in error. For example, for ‘time(1)’ and ‘time(20)’, predict gives [94.5918, 94.5157], and your equation gives [97.0681, 96.9920]. The difference will be noticeable on the plot.

So your equation is mathematically correct, but computationally incorrect, in that it does not use the full precision that fitlm calculates.

You have to ask for the full precision parameter estimates with:

coefs = mdl.Coefficients.Estimate;

Your equation rewritten as:

y1 = coefs(1) + coefs(2)*x;

then gives results identical to those predict produces.

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