# Fit parameter to Weibull distribution

12 vues (au cours des 30 derniers jours)
Kalle le 4 Mai 2023
Commenté : Jeff Miller le 4 Juin 2023
Hi, i wounder why my customized 3-parametric Weibull distribution function can't fit this specific data but the 2-parametric Weibull distribution function does.
%% INPUT DATA
A = [5.05002673467000 4.03148539861043 5.39815092086792 5.80624039967855 6.40216302871704 5.76023737589518 4.58155600229899 5.00490283966065 6.05405394236247 3.56419595082601 4.50625197092692 6.45612986882528 5.02723805109660 5.93072589238485 5.20658095677694 6.97059233983358 6.17937358220418 4.19038120905558 5.45702393849691 4.59787972768148 2.83733654022217 5.06135972340902 4.45322751998902 3.77954316139221 6.04727951685588 5.98482131958008 4.78870089848836 5.37800931930542 5.78453779220581];
A = A';
%% WEIBULL SETUP
custompdf = @(x,a,b,c) (x>c).*(b/a).*(((x-c)/a).^(b-1)).*exp(-((x-c)/a).^b);
customcdf = @(x,a,b,c) (x>c).*1-exp(-((x-c)/a).^b);
opt = statset('MaxIter',1e5,'MaxFunEvals',1e5,'FunValCheck','off');
params = mle(A,'pdf',custompdf,'start',[mean(A) std(A) min(A)],'Options',opt,'LowerBound',[0 0 -Inf],'UpperBound',[Inf Inf min(A)]);
a = params(1,1);
b = params(1,2);
c = params(1,3);
x = [c+eps(c):0.1:max(A)*4.5];
%% QUANTILES
p = wblfit(A);
[nlogl,pcov] = wbllike(p,A);
[quantile_Wbl2,q95lo,q95up] = wblinv(0.9991666,p(1),p(2),pcov);
Q = quantile(customcdf(x, a, b, c),0.9991666);
quantile_Wbl3 = fsolve(@(x) customcdf(x, a, b, c) - 0.9991666, quantile_Wbl2);
%% FIGURES
plot(x,customcdf(x, a, b, c),'LineWidth',2);
hold on
plot(x,wblcdf(x,p(1),p(2)));
hold on
cdfplot(A);
legend('WBL-3-Parameter','WBL-2-Parameter','daten')
hold off
%% Anpassungstest
% kolmogorov-smirnov Anpassungstest
[h_wbl3,p_wbl3] = kstest(A,[A customcdf(A,a,b,c)],0.05,0);
[h_wbl2,p_wbl2] = kstest(A,[A wblcdf(A,p(1),p(2))],0.05,0);
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### Réponse acceptée

Jeff Miller le 3 Juin 2023
The problem is a bad choice of starting values. You will get much better parameter estimates with
params = mle(A,'pdf',custompdf,'start',[1 1 1],'Options',opt,'LowerBound',[0 0 -Inf],'UpperBound',[Inf Inf min(A)]);
The 2- and 3-parameter fits are almost identical because the mle estimate of 'c' is pretty close to zero (-0.1464).
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Kalle le 4 Juin 2023
Thank you. That helps alot. :)
Jeff Miller le 4 Juin 2023
You are welcome. If the answer resolves the issue, please 'Accept' it.

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### Plus de réponses (1)

Sandeep le 23 Mai 2023
Hi Kalle,
There can be more than one reason as to why a customized 3-parametric Weibull distribution function may not adequately fit a specific data set while the standard 2-parametric Weibull distribution function does. Firstly, including more parameters in the model can lead to overfitting, which negatively affects the generalization performance of the model. The customized 3-parameter Weibull distribution function may overfit the data, leading to poor predictions in new instances. The customized 3-parameter Weibull distribution function may require more data points to estimate the parameters accurately, leading to poor parameter estimation and sub-optimal results.
In summary, the customized 3-parameter Weibull distribution function may not fit your specific data because it is too complex or you do not have enough data to estimate the parameters accurately, overfitting to random errors, or sub-optimal parameter estimation.
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Kalle le 2 Juin 2023
Thx

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