How to obtain exponential equation parameter from probplot function?
1 visualización (últimos 30 días)
Mostrar comentarios más antiguos
fransec
el 28 de Mayo de 2019
Comentada: fransec
el 30 de Mayo de 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.
0 comentarios
Respuesta aceptada
Akira Agata
el 29 de Mayo de 2019
2 comentarios
Akira Agata
el 30 de Mayo de 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;
Más respuestas (1)
Jeff Miller
el 30 de Mayo de 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.
Ver también
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!