Ich stimme Glen_b zu. Bei Regressionsproblemen liegt der Schwerpunkt auf den Parametern und nicht auf der unabhängigen Variablen oder dem Prädiktor x. Und dann kann man entscheiden, ob man das Problem durch einfache Transformationen linearisieren oder als solches vorgehen will.
y=ax+bx2+cx3+dx2/3+e/x+fx−4/7. This function is nonlinear in x. But for regression problems, the nonlinearity in x is not an issue. One has to check whether the parameters are linear or linear. In this case, a, b, c,.. f all have power 1. So, they are linear.
Remark that, in y=exp(ax), though a looks like it has power 1, but when expanded
exp(ax)=1+ax/1!+(ax)2/2!+…. You can clearly see that it is a nonlinear parameter since a has a power more than 1. But, this problem can be linearised by invoking a logarithmic transformation. That is, a nonlinear regression problem is converted to a linear regression problem.
Similarly, y=a/(1+bexp(cx) is a logistic function. It has three parameters, namely a, b and c. The parameters b and c have power more than 1, and when expanded they multiply with each other bringing nonlinearity. So, they are not linear. But, they can be also linearised using a proper substitution by setting first (a/y)−1=Y and then invoking a logarithmic function on both the sides to linearise.
Now suppose y=a1/(1+b1exp(c1x))+a2/(1+b2exp(c2x)). This is once again nonlinear with respect to the parameters. But, it cannot be linearised. One needs to use a nonlinear regression.
In principle, using a linear strategy to solve a nonlinear regression problem is not a good idea. So, tackle linear problems (when all the parameters have power 1) using linear regression and adopt nonlinear regression if your parameters are nonlinear.
In your case, substitute the weighting function back in the main function. The parameter β0 would be the only parameter with power 1. All the other parameters are nonlinear (β1 eventually multiplies with θ1 and θ2 (these two are nonlinear parameters) making it also nonlinear. Therefore, it is a nonlinear regression problem.
Adopt a nonlinear least squares technique to solve it. Choose initial values cleverly and use a multistart approach to find the global minima.
This vide will be helpful (though it does not talk about global solution): http://www.youtube.com/watch?v=3Fd4ukzkxps
Using GRG nonlinear solver in the Excel spreadsheet (install the solver toolpack by going to options - Add-Ins - Excel Add-Ins and then choosing Solver Add-In)and invoking the multistart in the options list by prescribing intervals to the parameters and demanding the constraint precision and the convergence to be small, a global solution can be obtained.
If you are using Matlab, use the global optimisation toolbox. It has multistart and globalsearch options. Certain codes are available here for a global solution, here
and
here.
If you are using Mathematica, look here.
If you are using R, try here.
The main function is linear.
It does not matter if nonlinear known functions==>B(L;θ) <== appear in the equations.
I would proceed with a linear least squares if I were you.
This is how you confirm or deny linearity:
https://en.wikipedia.org/wiki/Non-linear#Definition
You might also like:
https://en.wikipedia.org/wiki/Linear_combination
https://en.wikipedia.org/wiki/Least_squares
http://en.m.wikipedia.org/wiki/Linear_least_squares_(mathematics)
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It will be easy to understand, if I explain it in the context of functions.
Linear: A function which has a constant slope. Algebraically,a polynomial with highest exponent equal to 1. It's a function whose graph is a line. For example,
y=2x+3
Non-Linear: A function which has opposite properties of a linear function. A function which has a varying slope. It's a polynomial with exponent equal to 2 or more. It's graph is not a line. For example,
y=x^2
[http://study.com/academy/lesson/nonlinear-function-definition-examples.html][1]
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