0
$\begingroup$

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!

|cite|improve this question|||||
$\endgroup$
0
$\begingroup$

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, $$ y_i = \beta_0+\beta_1 x_i + \beta_2 x_i^2 + \epsilon_i, \quad i=1,...,n. $$

|cite|improve this answer|||||
$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.