Estimation with separated error terms

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Qifan
Qifan on 4 Aug 2015
Edited: John D'Errico on 7 Aug 2015
Dear all,
I meet a estimation problem and wonder if anyone has some creative ideas.
y=X*beta+exp(X*theta+err2)+err1
The parameters vectors need to be estimated are denoted by beta, and theta, y is the dependent variable, X is the independent variable matrix. The strange thing is that there are two error terms in this equation which are err1 and err2 with each of them assumed to be normally distributed. The exp(.) means an exponential term. Can any one give me some hint about how to estimate beta and theta?
Many thanks, Qifan

Answers (2)

Ghada Saleh
Ghada Saleh on 7 Aug 2015
Hi Qifan,
I understand you want to estimate 'beta' and 'theta'. You can check the curve fitting toolbox. You can use the fitting tool for parametric fitting and supply it with a custom function. You can also use least square fitting, where the error is assumed to be from Normal distribution as in your case.
Also, a possible workaround for the issue of having 'err2' in the exponent is to use series expansion for the exponential term and ignore the nonlinear terms of the error. However, the practicality of this solution depends on your use case.
I hope this helps,
Ghada

John D'Errico
John D'Errico on 7 Aug 2015
Edited: John D'Errico on 7 Aug 2015
This now becomes effectively an "errors in variables" problem. Nothing stops you from trying to estimate the parameters using traditional methods. Essentially, the curve fitting tools or optimization TB tools, stats tools, etc., do not explicitly force you to say where the error comes in. (There are implicit assumptions with all of these tools.)
You will merely fit the model as:
y = X*beta+exp(X*theta)
The only problem that will arise is that you may get somewhat biased estimates for the parameters. That is at least the classical problem when you have errors in the independent variable. Your ability to estimate the parameters well will depend on the magnitude of that noise, compared to the variations in X.

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