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The question is:

"Show how the nonlinear regression equation $y=a(x-b)+c(x-b)^2$ can be converted to a linear regression equation solvable by the method of least squares."

if we take that \begin{align} a(x-b) - c(x-b)^2 &= ax - ab + cx^2 -2cbx - cb^2\\ & = \underbrace{-ab-cb^2}_{\beta_0} + x\underbrace{(a-2cb)}_{\beta_1} + x^2\underbrace{c}_{\beta_2}\\ \end{align} thus, $$ y_i = \beta_0+\beta_1 x_i + \beta_2 x_i^2 + \epsilon_i, \quad i=1,...,n. $$

$ \beta_0$ and $\beta_1 $ are dependent each other ,so if you take the partial derivative of $y_i$ for $\beta_0$ you got dependence on $\beta_1$ ,so the gradient of Y by given X dependent on unknown parameters and we got non linear model.

or if you take the next model

\begin{align} xb + (xb)^2 +(xb)^3 &= \underbrace{xb}_{\beta_0} + x^2\underbrace{(b^2)}_{\beta_1} + x^3\underbrace{b^3}_{\beta_2}\\ \end{align} thus, $$ y_i = \beta_0x+\beta_1 x_i^2 + \beta_2 x_i^3 + \epsilon_i, \quad i=1,...,n. $$

and you got a linear model but of course is not a linear model because $\beta_0$ and $\beta_1$ are dependent.

Any help explaining this would be greatly appreciated!

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Note that \begin{align} a(x-b) - c(x-b)^2 &= ax - ab + cx^2 -2cbx - cb^2\\ & = \underbrace{-ab-cb^2}_{\beta_0} + x\underbrace{(a-2cb)}_{\beta_1} + x^2\underbrace{c}_{\beta_2}\\ \end{align} thus, $$ y_i = \beta_0+\beta_1 x_i + \beta_2 x_i^2 + \epsilon_i, \quad i=1,...,n. $$

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  • $\begingroup$ but betha_0 and betha_1 are dependent each other and in linear model the parameters are independent each other? $\endgroup$ – Tzur Roy Feb 7 '18 at 17:53
  • $\begingroup$ Nope, in general $cov(\hat{\beta}_0, \hat{\beta}_1) \neq 0$. $\endgroup$ – V. Vancak Feb 7 '18 at 17:56
  • $\begingroup$ so what defines a linear regression? $\endgroup$ – Tzur Roy Feb 7 '18 at 18:29
  • $\begingroup$ Whether the partial derivatives w.r.t. $\beta_j$ is independent of the other unknown $\beta$s. $\endgroup$ – V. Vancak Feb 7 '18 at 18:31
  • $\begingroup$ two questions:1)if we take y =β0+ β1^2*x is non linear model because the β1^2 but if we take the partial derivative for β1 we got 2*β1*x its not dependent on β0 and you wrote "independent of the other unknown βs"(mark on "other" I think you mean other betha not include betha1) another question if we take this model y=β0+β1x+β2x^2 $\endgroup$ – Tzur Roy Feb 7 '18 at 18:52

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