How to find a solution using vpasolve or maybe other?
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Hi, Can anyone please help me how to do this? I want to solve these 3 equations but MATLAB didn't give me any solutions sometimes when the data is simulated.
N = 100;
n=5;
lambda = 1;
for i = 1:N
U = normrnd(0,1,n,1);
V = normrnd(0,1,n,1);
W = (lambda/sqrt(1+lambda^2))*abs(U) + (1/sqrt(1+lambda^2))*V;
m1 = sum(W)/n;
m2 = sum((W-m1).^2)/n;
m3 = sum((W-m1).^3)/n;
A = (m2^3)/(m3^2);
if A > ((pi-2)^3)/(2*(4-pi)^2)
theta_Hat = (2/pi) + (A*(2/pi)*((4/pi)-1)^2)^(1/3);
lambda_Hatt = sqrt(1./(theta_Hat-1));
else
lambda_Hatt = 0;
end
lambdaMM = sign(m3).*abs(lambda_Hatt);
sigmaMM = sqrt(m2./(1-((2/pi).*(lambdaMM.^2./(1+lambdaMM.^2)))));
muMM = m1 - (sigmaMM.*sqrt(2./pi).*lambdaMM./sqrt(1+lambdaMM.^2));
eqn_mu = ((m1 - mu_Hat)./sigma_Hat) - (lambda_Hat./n).*sum(normpdf((lambda_Hat.*(W-mu_Hat)./sigma_Hat))./normcdf((lambda_Hat.*(W-mu_Hat)./sigma_Hat)));
eqn_sigma = (-n./sigma_Hat) + (1/sigma_Hat.^3).*sum((W-mu_Hat).^2) - (lambda_Hat./sigma_Hat.^2).*sum((W-mu_Hat).*normpdf((lambda_Hat.*(W-mu_Hat)./sigma_Hat))./normcdf((lambda_Hat.*(W-mu_Hat)./sigma_Hat)));
eqn_lambda = ((-3*pi^2.*lambda_Hat)./(2.*(pi.^2+(8.*lambda_Hat.^2)))) + sum(((W-mu_Hat)./sigma_Hat).*normpdf((lambda_Hat.*(W-mu_Hat)./sigma_Hat))./normcdf((lambda_Hat.*(W-mu_Hat)./sigma_Hat)));
equations = [eqn_mu,eqn_sigma,eqn_lambda];
vars = [mu_Hat,sigma_Hat,lambda_Hat];
initials = [muMM,sigmaMM,lambdaMM];
[muML,sigmaML,lambdaML] = vpasolve(equations==0, vars, initials);
end
Please help. Thank you so much.
14 Comments
Walter Roberson
on 10 Feb 2018
Unrecognized function or variable 'mu_Hat'.
Mikie
on 11 Feb 2018
Walter Roberson
on 11 Feb 2018
Not sure why you are overwriting your entire output variables each time through the loop?
Walter Roberson
on 11 Feb 2018
I find that the values of muML and lambdaML jump around a lot.
I put in a breakpoint and tested the equations at the first point that vpasolve() failed to return a value. It turns at that in that case, there is a solution (at least in limit form) of sigma_Hat = +/- infinity and lambda_Hat = 0 and mu_Hat = anything. If the other equations have the same structure then this would explain why there is so much jumping around with large values of sigma_Hat and why the mu_Hat value is pretty much meaningless: you are getting whichever side of 0 happens to be tried first.
Mikie
on 14 Feb 2018
Walter Roberson
on 14 Feb 2018
Your equations have theoretical solutions with one of the parameters being either -inf or +inf, and no theoretical based solver can choose between those two. When you add limited precision of calculation, you add a large number of solutions where that variable merely assumes large magnitude leading to division of the numerators by values large enough that the sub expressions become negligible. vpasolve does not have a preference for perusing a positive value instead of a negative value: it uses a Newton type method that gives it projected values based upon the local conditions near where it starts, and if round off error happens to favour one side of 0 over the other then it is happy to persue that side. Therefore for nearly identical starting equations, it might run off towards negative infinity or positive infinity so if you get 9e7 for one iteration and -9e7 for the next it doesn't mean oscillating, it just means that there are multiple solutions with one being chosen by random factors.
I would suggest that you need to think more about constraints for your values.
Mikie
on 26 Feb 2018
Walter Roberson
on 26 Feb 2018
"Why does [...] solving by giving an initial value and no giving any intial value come out different?"
vpasolve() is not a global minimizer. It uses a quasi-Newton approach, so it is prone to getting stuck in local minima.
vpasolve() is explicitly a solver that uses a limited number of significant digits, software floating point, so if two different inputs give an output that is the same to within those limited number of digits, then it cannot tell the two apart.
Your equations involve exp(-expression / sigma_Hat^2) . For finite expression and large enough sigma_Hat, the exp() is going to underflow to 0, and once that starts happening, the system effectively cannot tell apart different values of sigma_Hat
Mikie
on 27 Feb 2018
Walter Roberson
on 28 Feb 2018
My tests hint that the solution for the equations has mu_Hat = min(W) - 5*eps, with all mu_Hat greater than min(W)+17*eps leading to NaN in the calculations, and with a quite steep gradient near min(W) - 5*eps
The values of the other variables have much much less influence on the calculation.
The tolerances are so tight that it must be assumed that a minor rearrangement of the calculation could give a quite different calculation for any one position.
Mikie
on 28 Feb 2018
Walter Roberson
on 28 Feb 2018
You are using fminunc for Lpmle, which is an unconstrained minimization. The Lpmle function perhaps has several minima, but one of them is p(2) -> +inf and p(3) = 0, which is a combination which gives -inf for Lpmle.
Mikie
on 5 Mar 2018
Walter Roberson
on 10 Jun 2018
No. You can test the exit flag and do something different if those cases are detected, but you cannot tell fminunc to detect the situations internally and try somewhere else.
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