How to obtain the standard deviation of the fitting parameters?

Question

Qili Hu 2024-3-21

0
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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2097006-how-to-obtain-the-standard-deviation-of-the-fitting-parameters

评论： Star Strider 2024-3-22

I have a question about how to obtain the standard deviation of the fitting parameters according to the following program codes.

x=[0 5 10 15 20 30 45 60 75 90 105 120];

y=[0 3.87 4.62 4.98 5.21 5.40 5.45 5.50 5.51 5.52 5.54 5.53];

plot(x,y,'bo');

hold on

pause(0.1);

beta0=[39,0.002];

% syms n t

% fun=@(beta,t) beta(1)*(1-6/(pi^2)*symsum((1./n.^2).*exp(-beta(2)*(n.^2).*t),n,1,Inf));

% betafit = nlinfit(x,y,fun,beta0);

beta1=beta0;

delta = 1e-8; % desired objective accuracy

R0=Inf; % initial objective function

for K=1:10000

fun=@(beta,t) beta(1)*(1-6/(pi^2)*sum((1./(1:K)'.^2).*exp(-beta(2)*((1:K)'.^2).*t),1));

[betafit,R] = nlinfit(x,y,fun,beta1);

R = sum(R.^2);

if abs(R0-R)<delta

break;

end

beta1=betafit;

R0 = R;

end

plot(x,fun(betafit,x),'.-r');

xlabel('x');

ylabel('y');

legend('experiment','model');

title(strcat('\beta=[',num2str(betafit),'];----stopped at--','K=',num2str(K)));

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Answer 1

Star Strider 2024-3-21

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https://ww2.mathworks.cn/matlabcentral/answers/2097006-how-to-obtain-the-standard-deviation-of-the-fitting-parameters#answer_1428576

在 MATLAB Online 中打开

The most meaningful measure of the parameters is the confidence interval. You could possibly recover some information from the covariance matrix ‘CovB’, however that would take some effort.

x=[0 5 10 15 20 30 45 60 75 90 105 120];

y=[0 3.87 4.62 4.98 5.21 5.40 5.45 5.50 5.51 5.52 5.54 5.53];

plot(x,y,'bo');

hold on

pause(0.1);

beta0=[39,0.002];

% syms n t

% fun=@(beta,t) beta(1)*(1-6/(pi^2)*symsum((1./n.^2).*exp(-beta(2)*(n.^2).*t),n,1,Inf));

% betafit = nlinfit(x,y,fun,beta0);

beta1=beta0;

delta = 1e-8; % desired objective accuracy

R0=Inf; % initial objective function

% K=1:10000;

for K=1:10000

fun=@(beta,t) beta(1)*(1-6/(pi^2)*sum((1./(1:K)'.^2).*exp(-beta(2)*((1:K)'.^2).*t),1));

[betafit,Rv,J,CovB,MSE] = nlinfit(x,y,fun,beta1);

R = sum(Rv.^2);

if abs(R0-R)<delta

break;

end

beta1=betafit;

R0 = R;

end

fprintf('\nbeta = %8.4f\nbeta = %8.4f\n\n',beta1)

beta = 5.4900 beta = 0.1374

CovMtx = CovB

CovMtx = 2×2

1.0e-03 * 0.4686 -0.0534 -0.0534 0.0206

ci = nlparci(beta1,Rv,'Covar',CovB)

ci = 2×2

5.4418 5.5382 0.1273 0.1475

plot(x,fun(betafit,x),'.-r');

xlabel('x');

ylabel('y');

legend('experiment','model', 'Location','best');

title(strcat('\beta=[',num2str(betafit),'];----stopped at--','K=',num2str(K)));

.

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Star Strider 2024-3-21

在 MATLAB Online 中打开

This calculation is not usually performed, since the standard deviation of a parameter in a multi-parametric system is not important.

The standard error of the estimate is

with σ denoting the standard deviation, and the confidence intervals multiply it by the critical values of the t distribution for the problem being considered,

. The confidence intervals use the inverse t distribution with n-2 degrees -of-freedom (since this estimates 2 parameters), so one way of calculating it could be —

tci = tinv([0.025 0.975], 12-2)
tci = 1×2
   -2.2281    2.2281
beta =   [5.4900
          0.1374];                              % Parameter Estimates
ci = [5.4418    5.5382
      0.1273    0.1475];                        % Confidence Intervals
sigma = ((ci-beta)./tci) * sqrt(12)             % Estimated Standard Deviations
sigma = 2×2
    0.0749    0.0749
    0.0157    0.0157

This is the best that I can do to estimate the standard deviations, based on assumptions regarding confidence intervals using the standard errors of the mean for vectors. I cannot claim that it is accurate, since I have never done this calculation before, nor have I ever seen it done.

.

Qili Hu 2024-3-22

Dear Strider

According to the codes you write, I add them to my codes. But, it do not work during running. When I change sigma=((ci-beta)./tci)*sqrt(12) to sigma=((ci-beta1)./tci)*sqrt(12), the running result is not consistent with you. It seems that beta do not appear in the codes. So, I hope you can help me to check the following codes. Thank you very much.

x=[0 5 10 15 20 30 45 60 75 90 105 120];

y=[0 3.87 4.62 4.98 5.21 5.40 5.45 5.50 5.51 5.52 5.54 5.50];

plot(x,y,'ro','MarkerFaceColor','r');

hold on

beta0=[5,0.1];

beta1=beta0;

delta=1e-8;

R0=Inf;

for n=1:15000

fun=@(beta,t) beta(1)*(1-6/(pi^2)*sum((1./(1:n)'.^2).*exp(-beta(2)*((1:n)'.^2).*t),1));

[betafit,Rv,J,CovB,MSE]=nlinfit(x,y,fun,beta1);

R=sum(Rv.^2);

if abs(R0-R)<delta

break;

end

beta1=betafit;

R0=R;

end

CovMtx=CovB;

ci=nlparci(beta1,Rv,'Covar',CovB);

tci=tinv([0.025 0.975],12-2);

sigma=((ci-beta)./tci)*sqrt(12);

plot(x,fun(betafit,x),'-r');

xlabel('t (min)');

ylabel('qt (mg/g)');

title(strcat('\beta=[',num2str(betafit),'];----stopped at--','n=',num2str(n)));

Star Strider 2024-3-22

在 MATLAB Online 中打开

It seems that you only copied part of my code.

When I provided the rest of it (slightly modified from the earlier version), it works as expected —

x=[0 5 10 15 20 30 45 60 75 90 105 120];

y=[0 3.87 4.62 4.98 5.21 5.40 5.45 5.50 5.51 5.52 5.54 5.50];

plot(x,y,'ro','MarkerFaceColor','r');

hold on

beta0=[5,0.1];

beta1=beta0;

delta=1e-8;

R0=Inf;

for n=1:15000

fun=@(beta,t) beta(1)*(1-6/(pi^2)*sum((1./(1:n)'.^2).*exp(-beta(2)*((1:n)'.^2).*t),1));

[betafit,Rv,J,CovB,MSE]=nlinfit(x,y,fun,beta1);

R=sum(Rv.^2);

if abs(R0-R)<delta

break;

end

beta1=betafit;

R0=R;

end

% CovMtx=CovB;

betav = beta1(:) % Convert 'beta1' To The column Vector 'betav'

betav = 2x1

5.4856 0.1379

ci=nlparci(beta1,Rv,'Covar',CovB) % PArameter Confidence Intervals

ci = 2x2

5.4387 5.5326 0.1280 0.1478

tci=tinv([0.025 0.975],12-2); % Inverse 't' Distribution

sigma=((ci-betav)./tci)*sqrt(12) % Calculate Standard Deviations

sigma = 2x2

0.0731 0.0731 0.0154 0.0154

fprintf('Beta(1) = %8.4f\t\tStandatrd Deviation = %8.4f\nBeta(2) = %8.4f\t\tStandatrd Deviation = %8.4f\n',[betav sigma(:,1)].')

Beta(1) = 5.4856 Standatrd Deviation = 0.0731 Beta(2) = 0.1379 Standatrd Deviation = 0.0154

plot(x,fun(betafit,x),'-r');

xlabel('t (min)');

ylabel('qt (mg/g)');

title(strcat('\beta=[',num2str(betafit),'];----stopped at--','n=',num2str(n)));

I added the fprintf call to be certain that there are no ambiguities.

.

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