# convert square regression model to linear model

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!

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, $$y_i = \beta_0+\beta_1 x_i + \beta_2 x_i^2 + \epsilon_i, \quad i=1,...,n.$$
• Nope, in general $cov(\hat{\beta}_0, \hat{\beta}_1) \neq 0$. – V. Vancak Feb 7 '18 at 17:56
• Whether the partial derivatives w.r.t. $\beta_j$ is independent of the other unknown $\beta$s. – V. Vancak Feb 7 '18 at 18:31