How can I get the 95% confidence interval for the 3 parameters of a Weibull distribution?

Hello everyone,

I am having problems with the mle function. I need to estimate the 3 parameters of a Weibull distribution for a set of data (aproximately 30 values), with the 95% confidence interval for the parameters.

As Matlab's built in Weibull pdf is defined for 2 parameters, I created a custom pdf which takes into account the 3rd scale parameter c, with the following code:

custpdf = @(x,a,b,c) (x>c).*(b/a).*(((x-c)/a).^(b-1)).*exp(-((x-c)/a).^b);

If I then use the mle function to estimate the three parameters, this line of code works fine and returns plausible values for the parameters:

opt = statset('MaxIter',1e5,'MaxFunEvals',1e5,'FunValCheck','off');
Weibull = mle(data,'pdf',custpdf,'start',[Wa_ini Wb_ini Wc_ini],'Options',opt,...
    'LowerBound',[0 0 0],'UpperBound',[Inf Inf min(sigma_i)])

(data is the set of values I am evaluating, and [Wa_ini Wb_ini Wc_ini] are the values that should be taken as start values by the algorithm)

Now when I use the form of the mle function which also returns the 95% confidence interval (code below), Matlab still returns the correct values for the 3 parameters, but the lower and upper limits of the confidence interval are completely incoherent : for example for the parameter a=107.3528, the confidence interval is [-450.0639;+664.7696].

[Weibull, pci] = mle(sigma_i,'pdf',custpdf,'start',[Wa_ini Wb_ini Wc_ini],'Options',opt,...
    'LowerBound',[0 0 0],'UpperBound',[Inf Inf min(sigma_i)])

Can anyone help me with this issue?

I am using Matlab2017a.

Thank you!

[Ben M]

Hello,

i know your post is a few years old, but i am facing the same problem for myself. Have you been able to solve your problem concerning the confidence interval? If so, would you share your solution with me?

Many greetings

[Anas Boukili Makhoukhi]

Hi there,

Have you been able to solve your problem concerning the confidence interval? If so, would you share your solution with me ?

Answer 1

[laurent jalabert]

Dear Anas, I am not specialist for this question. I found this function for Wiebull + 3 parameters

https://www.real-statistics.com/other-key-distributions/weibull-distribution/three-parameter-weibull-distribution/

I thought that you can define this function and use non linear regression fit to get those 3 parameters. Personnaly, I started to use non linear regression fit for my research, it worked very well, and gave results very closed to Mathematica (you can also try it if you have).

Or you can try this :

https://fr.mathworks.com/matlabcentral/answers/405367-how-to-calculate-standard-errors-for-estimated-parameters-for-a-3-parameter-weibull-distribution

[Anas Boukili Makhoukhi]

Many thanks for your answer :)

It seems really helpful

Answer 2

[laurent jalabert]

I am not sure but : if you define the Weibull function and use the Non Linear Regression fit (fitnlm) in Statistics/Machine Learning, you will get the standard deviation on the fit parameters.

modelfun1 = @(a,x) a(1).*(1+a(2).*x(:,1)); --> replace by your function

beta1 are initial range of parameters, tbl is your data arranged in a table

mdl = fitnlm(tbl,modelfun1,beta1,'CoefficientNames',{'a1';'a2'});

[Anas Boukili Makhoukhi]

Thank you very much for the quick response.

Honestly, I want to estimate the quaniles and their Confidence Intervals by a Weibull distribution with three parameters. Can you help me at this point please?

For Weibull with two parameters, it works like this: [X, XLO, XUP] = wblinv (P, A, B, PCOV, alpha)

But I can't do it for Weibull with 3 parameters

[Jeff Miller]

I think the third parameter of the Weibull is just a constant shift value, so you can get whatever you want from the 2-parameter Weibull and at the shift at the end. For example, if your shift is 71, then just use wblinv with your two parameters and then add 71 to the resulting X, XLO, and XUP

[Anas Boukili Makhoukhi]

Many thanks for your answer :)

Exactly, we have the following result :

If X follows a 2-parameter Weibull distribution, then Y = X + c follows a 3-parameter Weibull distribution with c is the location parameter.

Is that means, i should estimate separately the mean of my dataset and add it to my estimates ? I dont think its logic because tha gonna overestimate the quantiles !

[laurent jalabert]

Maybe you can can upload a graph of your distribution, it might be helpful to figure out the data you are talking about

[laurent jalabert]

http://ipl.physics.harvard.edu/wp-uploads/2013/03/PS3_Error_Propagation_sp13.pdf

[Mohammed Benali]

Hi guys,

i‘m facing the exact same problem as mentioned by Louise above. If someone has a solution to estimate the confidence bounds for a 3p-weibull in matlab, please share your code. Thanks!

[Jeff Miller]

@Anas Boukili Makhoukhi I think c is estimated by the minimum rather than the mean of your dataset.