MATLAB: How to fix one point with no error while curve fitting

curve fittingfittingMATLABplotStatistics and Machine Learning Toolbox

I have a data to fit linear plot. But I want to fit the first point with no error. I mean, the fitted curve should fix on the first point and then fit to the rest of the point in 'least square fit' method. How can I do that?

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[Edited] removed least squares comment

You could give more importance to the first observation using weights, I do this using the fitlm function in the statistics toolbox:

x = 1:10; y = randn(10,1);
lm1 = fitlm(x,y);
lm2 = fitlm(x,y,'weights',[100, ones(1,length(y)-1)]);
scatter(x,y), hold on
plot(x, lm1.Fitted,'r')
plot(x, lm2.Fitted,'b')

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MATLAB: Multivariate fit with two unknown coefficients and known intercepts

Using the modeling tools you know, this is correct:

% some random data
x1 = randn(100,1);
x2 = randn(100,1);
Y = x1 + 2*x2 + 1 +randn(size(x1))/100;
Y1 = Y-1;
T = table(Y,x1,x2);
lm = fitlm(T,'Y ~ x1 + x2 - 1')
lm = 
Linear regression model:
    Y ~ x1 + x2
Estimated Coefficients:
          Estimate       SE       tStat       pValue  
          ________    ________    ______    __________
    x1     0.799       0.10355    7.7158    1.0195e-11
    x2    1.8409      0.095852    19.206    5.4359e-35
Number of observations: 100, Error degrees of freedom: 98
Root Mean Squared Error: 0.977

Effectively, subtract 1 from Y, then fit a model with no intercept at all. To use the model, just add 1 back into your prediction at the end. So you create a model with no constant term at all, and then put it back in when you use the model.

If you have other tools available, for example, LSQLIN from the optimization toolbox, there are other ways you can do it. Or you can download my own LSE from the file exchange . It can constrain a parameter to a specific given value.

coefs = lse([ones(size(x1)),x1,x2],Y,[1 0 0],1)
coefs =
          1
     1.0009
      1.998

Using lsqlin, I would do this:

lsqlin([ones(size(x1)),x1,x2],Y,[],[],[1 0 0],1)
Minimum found that satisfies the constraints.
Optimization completed because the objective function is non-decreasing in 
feasible directions, to within the default value of the optimality tolerance,
and constraints are satisfied to within the default value of the constraint tolerance.
<stopping criteria details>
ans =
            1
       1.0009
        1.998

In either of the latter two cases, you get the coefficients as just a vector, with less additional information that the stats toolbox gives you.

MATLAB: Confusion about the representation of Root Mean Square, R Squared …

Residuals and measures related to them are not a percentage. In the context of a one-dimensional situation, residuals are analogous to deviations from the mean, and measures derived from them are roughly analogous to the variance or standard deviation. (With heavy emphasis on ‘roughly’.)

The Coefficient of Determination (R-Squared) value could be thought of as a decimal fraction (though not a percentage), in a very loose sense. From the documentation:

Coefficient of determination (R-squared) indicates the proportionate amount of variation in the response variable y explained by the independent variables X in the linear regression model. The larger the R-squared is, the more variability is explained by the linear regression model.

So the higher the R-Squared value, the better the fit of the model to the data.

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Related Solutions

MATLAB: Multivariate fit with two unknown coefficients and known intercepts

MATLAB: Confusion about the representation of Root Mean Square, R Squared …

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