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Positive or zero intercept for linear function

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Zhannur Issayev
Zhannur Issayev 2019 年 11 月 30 日
Answered: Star Strider 2019 年 11 月 30 日
Trying to fit a linear function to a set of points with intercept either 0 or positive.
Code:
yld = [0.016848189
0.01682922
0.019936858
0.027905862
0.03179837
0.028801833
0.028801884
0.044003395
0.048011902
0.048013812
0.043509457
0.043510379
0.059143077
0.044998549
0.04499765
0.044998484
0.044898415
0.04499845
0.044408223
0.044222909
0.049398674
0.052317402
0.054167487
0.054265443
0.048420988
0.04656471
0.04998282];
t = [0.320547945
0.320547945
0.550684932
1.849315068
2.471232877
2.542465753
2.542465753
2.55890411
2.909589041
2.909589041
3.030136986
3.030136986
4.063013699
4.975342466
4.975342466
4.975342466
4.975342466
4.975342466
5.663013699
5.663013699
5.663013699
9.512328767
15.01643836
16.02465753
19.61917808
19.61917808
19.61917808];
x = [ones(length(t), 1) log(t) log(t).^2];
y = log(yld + 1);
betas = x\y;
tau = [0.083 0.25 0.5 1 2 3 4 5 7 8 9 10 11 12 13 14 15 16 17 18 19 20];
yfit = betas(1) + betas(2)*log(tau) + betas(3)*(log(tau).^2);
plot(yfit);
I realize than as t goes to 0 log(t) goes to -inf, however, could there be a workaround, perhaps with a worse fit, but with nonnegative intercept?

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Star Strider
Star Strider 2019 年 11 月 30 日
Linearising the exponiential function may not be the best option. I am not certain what you are modeling, however using a nonlinear objective function with fminsearch (or any other appropriate optimiser) may be worth considering.
Try this to fit :
objfcn = @(b,t) b(1) + b(2).*(1 - exp(b(3).*t));
[B,rsdnorm] = fminsearch(@(b) norm(yld - objfcn(b,t)), randn(3,1))
yfit = objfcn(B,tau);
I was able to get an acceptable fit with the parameter set:
B =
0.010429559813784
0.040470293461631
-0.456482676457038

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