Center of gaussian mixed distribution area

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Aishwarya Rao
Aishwarya Rao el 27 de Mzo. de 2023
Comentada: Aishwarya Rao el 11 de Abr. de 2023
Hi! I want to calculate the center of a gaussian mixed distribution. The code goes as follows:
dd = importdata('gmdata.txt'); %%%% import data
xx = dd(:,1);
yy = dd(:,2);
obj = gmdistribution.fit([xx,yy],2); %%%% fit a distribution
figure;
scatter(xx,yy,10,'.') %%%% Scatter plot with points of size 10
hold on
gmPDF = @(x,y) arrayfun(@(x0,y0) pdf(obj,[x0 y0]),x,y);
hh = fcontour(gmPDF,[-8 6]);
I am getting the above distribution as output. I want to recreate the value of GM distribution centers as given in "Tanabe, Hiroko, Keisuke Fujii, and Motoki Kouzaki. "Intermittent muscle activity in the feedback loop of postural control system during natural quiet standing." Scientific reports 7.1 (2017): 10631." The sentences read as:
Kindly help me with the code. I guess the center they are mentioning as the center of the fcontour rings, but not sure.
  1 comentario
Torsten
Torsten el 27 de Mzo. de 2023
Editada: Torsten el 27 de Mzo. de 2023
Your data ressemble a usual multivariate Gaussian, don't they ?

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Respuestas (1)

the cyclist
the cyclist el 27 de Mzo. de 2023
Editada: the cyclist el 27 de Mzo. de 2023
I downloaded and looked over the paper. I am also not certain what they mean by the "center" of the mixed gaussian model.
But, I think a pretty reasonable guess is they mean the peak of the joint PDF. (I think is also what you mean by the "center of the contour rings").
You can use the fminsearch function to find this peak:
dd = importdata('gmdata.txt'); %%%% import data
xx = dd(:,1);
yy = dd(:,2);
obj = gmdistribution.fit([xx,yy],2); %%%% fit a distribution
figure;
scatter(xx,yy,10,'.') %%%% Scatter plot with points of size 10
hold on
gmPDF = @(x,y) arrayfun(@(x0,y0) pdf(obj,[x0 y0]),x,y);
hh = fcontour(gmPDF,[-8 6]);
% Define the function to be minimized (using the negative because we actually want the maximum)
f = @(x) -obj.pdf(x);
% Initial guess as to where the maximum is.
% (This is a relatively poor guess, but the result looks OK anyway)
xy_init = [0 0];
% Use fminsearch to find the maximum
xy_max = fminsearch(f, xy_init)
xy_max = 1×2
-1.9393 -5.2685
Eye-balling the plot, this looks accurate.
  2 comentarios
Torsten
Torsten el 27 de Mzo. de 2023
Yes, it's one of the obj.mu (where both of them look pretty much the same).
Aishwarya Rao
Aishwarya Rao el 11 de Abr. de 2023
Thank you @the cyclist. The code worked!

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