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How to obtain exponential equation parameter from probplot function?

4 vues (au cours des 30 derniers jours)
fransec
fransec le 28 Mai 2019
Commenté : fransec le 30 Mai 2019
Hello everybody,
I have my dataset that here we name data. It is a vector.
I apply the probplot function from an exponential distribution
h=probplot('exponential',data)
I would like to obtain the exponential parameters to build an exp equation such as
f(x) = a*exp(b*x)
which represents the exponential fit for my data.
Thank you.

Réponse acceptée

Akira Agata
Akira Agata le 29 Mai 2019
I believe fitdist function should be some help.
  2 commentaires
fransec
fransec le 29 Mai 2019
Unfortunately, fitdist doesn't solve my request.
I explain better my issue.
This is my exponential probabilistic plot.
figure,probplot('exponential',data)
exp probplot.jpg
I would like to find the function (or equation) of the reference black dashed line that represents the theoretical exponential distribution, in order to work on data that diverge from that line.
Akira Agata
Akira Agata le 30 Mai 2019
If my uderstanding is correct, you have data vector (e.g N-by-1 array) data, and want to estimate exponential distribution parameter a and b in the equation f(x) = a*exp(b*x).
It seems that the following is one possible step to estimate a and b.
% Sample data (e.g 1000-by-1 array)
pd = makedist('Exponential','mu',2);
data = random(pd,1000,1);
% Estimate parameter mu in f(x) = (1/mu)*exp(-x/mu)
pd2 = fitdist(data,'Exponential');
mu = pd2.mu;
% Since a = (1/mu) and b = (-1/mu),
a = 1/mu;
b = -1/mu;

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Plus de réponses (1)

Jeff Miller
Jeff Miller le 30 Mai 2019
For a standard exponential distribution,
probability = 1 - exp(-lambda*x)
where x is the data value and lambda is the parameter of the exponential. I think this is the equation of the black dashed line that you show. (But maybe you some other distribution in mind since you show a function with two parameters, a and b).
The value of lambda used in the previous equation would usually be the maximum likelihood estimate:
lambda_est = 1 / mean(data);
So, your graph makes it look like your data set is more compressed at the high end than would be expected from a true exponential.
  1 commentaire
fransec
fransec le 30 Mai 2019
Yeah, I think the equation that you proposed represents the black dashed line.
I think that lambda could be 1/mu.
The problem is that I don't know how to plot the probability equation (in plot) over the probability plot, in order to find the intrsection between the two functions...

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