0

Provided I have experimental couples $(x,y)$, is the adjustment of $\beta_i$ parameters of the following function:

$$y = f(x) = \beta_0 \cdot \exp (\beta_1\cdot x) + \beta_2$$

A Non Linear Least Squares problem because of the $\beta_2$ parameter?

I know that I can first estimate $\beta_2$ and then apply the following transformation:

$$y' = y - \beta_2$$

This will reduce the problem to a linear least square in term of parameters because the transformation $\log y'$ is possible.

I am just surprised that a simple offset factor could make a Least Square problem Non Linear. Am I right?

jlandercy
  • 510
  • A function is not a problem. What is the actual problem related to your function? – Gerry Myerson Sep 08 '17 at 07:42
  • 1
    @GerryMyerson If I want to estimate $\beta_i$ provided I have experimental data couples $(x,y)$ does it fall into LS problem or NLLS? – jlandercy Sep 08 '17 at 07:51
  • How do you mean you can estimate $\beta_2$ first? Estimating a constant function, and then substracting this constant function and estimating a model for the residuals using a logarithmic transform, does not necessarily lead to an optimal solution to the original problem. – Johan Löfberg Sep 08 '17 at 08:05

1 Answers1

2

You are correct. The model being $$y = \beta_0 \, \exp (\beta_1\, x) + \beta_2$$ assuming that you know more or less the value of $\beta_2$, you can rewrite $$\log(y-\beta_2)=\log(\beta_0)+\beta_1x=\alpha+\beta_1\,x$$ and then linear regression will give you $\alpha$ and $\beta_1$ with $\beta_0=e^{\alpha}$.

But, take care : these are only estimates of the real values since, in this preliminary good step, you minimized $$SSQ_1=\sum_{i=1}^n\left(\alpha+\beta_1\,x_i-\log(y_i-\beta_2) \right)^2$$ which has little to do with $$SSQ_2=\sum_{i=1}^n\left(\beta_0 \, \exp (\beta_1\, x_i) + \beta_2-y_i\right)^2$$ You need to continue with the nonlinear regression.

Claude Leibovici
  • 260,315