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I have a $y$ variable and 2 $x$'s ($x_0$ and $x_1$). I am told that the $y$ is a function of the 2 $x$'s. I know the functional form of this relationship, but want to calculate the weights/values of constants that go into the transformation. The functional form is somewhat complicated. The following is not what I have, but to illustrate the point:

$$y = B_0 - \frac{(x_1 + B_1)B_2-B_3}{B_4} + (B_5X_2)^{B_6}$$

Is it possible to determine estimates of the $B_j$'s using regression or another method?

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  • $\begingroup$ To what extent does your example resemble the function you want to estimate? I ask because the one you provided does not seem to be identifiable (e.g. $B_0$ and $B_3$ cancel out each other). $\endgroup$ – Tim Apr 13 '15 at 16:05
  • $\begingroup$ It is actually a little simpler in that there are no exponents. It is fine is only a couple things are not identified. I should have explained a little more - I have data and was provided with the weights that were supposedly used, but they don't seem to be quite rights. So I wanted to estimate them and compare how closely they match up. $\endgroup$ – bill999 Apr 13 '15 at 17:32
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Yes. For further searching, this is often called non-linear regression or curve fitting. For simpler functional forms, such as power functions, log functions, polynomials, etc. it may be easier to transform your dependent variable or your independent variables. However, for something complex like what you have, you'll need a more general approach. If you use R you may want to look into the nls() function. This may help you on your journey.

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  • $\begingroup$ Thank you. I do know R and I'll look into the nls() function. $\endgroup$ – bill999 Apr 13 '15 at 17:32
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    $\begingroup$ From looking at your reply to @Tim's comment, your function appears to not be non-linear at all actually. Without exponents that equation simplifies to $y=B_0+B_1x_1+B_2x_2$, which is a basic linear regression. You will not be able to identify separately $B_2$ and $B_1$ from your equation, because an infinite number of solutions will exist. Instead, the $B_0$ of your regression will equal $B_0 - (B_1B_2 - B_3)/B_4$ from your equation (not exactly, I just don't want to put the full conversion) $\endgroup$ – le_andrew Apr 13 '15 at 17:51

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