How to fit a gaussian to unnormalized data
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Niklas Kurz
el 3 de Sept. de 2021
Editada: Niklas Kurz
el 4 de Sept. de 2021
I do know this question has been asked in several kinds plus it's rather a mathematical question for mathstack like sites.
But here I am, bothering you with my data-points.
I've got the X-values
X = -6:1:6;
and Y values, corresponding to how often each X value was hit.
Y = [1 3 1 8 5 16 18 10 6 2 1 1 0];
For later on I calculated Mean and standard deviation as followed:
mean = sum(X.*Y)/(sum(Y));
std = 0;
for i =1:1:size(Y,2)
std = std+ Y(i).*(X(i)-m).^2;
end
std = sqrt(std/(n-1));
Now to the crucial part: fitting the data to a gaussian curve.
First of I normalized the data: Heres probably my problem located:
Yn = Y/max(Y)
Actually the normalization should lead to a total area of one but
trapz(X,Yn)
is not equal to one. I use it anyways.
In cftool I rigorously typed in the gaussian distribution equation for fitting:
1/(sqrt(2*pi)*s)*exp(-(x-m)^2/(2*s^2)) % alias: s/std m/mean
It doesn't happen to fit the data points quite well.
Also it's deviating from plotting the eqatuion above with mean and std calculated
I still believe something with the normalization turned out wrong.
You can name what?
2 comentarios
Jeff Miller
el 4 de Sept. de 2021
Why are you using cftool at all? The maximum likelihood estimates of the gaussian mu and sigma can be computed directly from the data, something like this (unchecked):
mu_est = sum(X.*Y)/(sum(Y));
sigma_est = sqrt( sum(Y.*((X-mu_est).^2)) / n); % note division by n rather than n-1 for maximum likelihood
Respuesta aceptada
Paul
el 4 de Sept. de 2021
Editada: Paul
el 4 de Sept. de 2021
I think you need to normalize Y by it's sum (given the unit spacing of X), not its max
X = -6:1:6;
Y = [1 3 1 8 5 16 18 10 6 2 1 1 0];
Yn = Y/sum(Y);
Xvals = repelem(X,Y);
histogram(Xvals,'Normalization','pdf');
hold on
plot(X,Yn,'-o','LineWidth',1);
mu = mean(Xvals);
sigma = std(Xvals);
plot(X,normpdf(X,mu,sigma),'g-x','LineWidth',1)
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