How to linearize a non linear function So, consider a function $F(x,\theta)$ that needs to be linear in relation to the parameters $\theta$. If
$y_i = \alpha\beta + \beta^2x + \epsilon$
Then, it is possible to linearize it by defining $\alpha'=\alpha\beta$ and $\beta' = \beta^2$. However if we consider another function:
$y_i = \alpha + \alpha^2x + \epsilon$
It is not possible to linearize it (in relation to parameters $\theta$). Why is that? Can't I just define $\alpha' = \alpha^2$? How can I prove that a function is not linearizable?
 A: The following argument indicates how to address such questions generally.
Let's suppose there is a vector parameter $\theta\in\Theta\subset\mathbb{R}^p$ and a one-to-one differentiable reparameterization $\alpha = h(\theta)$ (where $h:\mathbb{R}^p\to\mathbb{R}$) and, if necessary, a re-expression of the variable $x$ in terms of a vector variable $y\in \mathbb{R}^p$ for which $x=g(y)$, so that the formula becomes a bilinear function of $y,\theta;$ that is,
$$\sum_{j=1}^p y_j\,\theta_j = \alpha(\theta) + \alpha(\theta)^2 x(y).$$
Differentiating both sides with respect to a fixed $\theta_i$ gives
$$y_i = \left(1 + 2\alpha(\theta)\, x(y)\right)\frac{\partial\alpha}{\partial \theta_i}.$$
The left hand side does not vary with $\theta_i,$ but the right hand side does unless $\alpha$ is (piecewise) constant as a function of $\theta_i.$  But in that case, unless the set in which $\alpha$ is assumed to lie contains no intervals at all, $\alpha$ cannot everywhere be a one-to-one function, showing it is not a reparameterization, QED.

It is instructive to see what happens when taking this approach with a linearizable formula. When we analyze the first example, we obtain $p$ equations
$$y_i = \frac{\partial (\alpha\beta)}{\partial \theta_i}+\frac{\partial (\beta^2)}{\partial \theta_i} x.$$
Because the left side does not vary with $\theta,$ we conclude that either $\beta^2$ is a (positive) linear function of $\theta_i$ and $\alpha\beta$ is a linear function of $\theta_i$ (in which case $y_i$ is an affine function of $x$) or $\beta^2$ does not depend on $\theta_i$ and $\alpha\beta$ is a linear function of $\theta_i.$  The simplest solution requires $p=2$ and takes $\beta^2 = \theta_1 \ge 0$ and $\alpha\beta=\theta_2,$ as shown in the question, but there are plenty of others.  


*

*One, with $p=3$ parameters, is $y_1=-3x,$ $y_2=1,$ $y_3=-4,$ $\beta^2 = -3\theta_1 \le 0,$ and $\alpha\beta = \theta_2 - 4\theta_3.$  You can check that $$\alpha\beta + \beta^2 x = (\theta_2 - 4\theta_3) + (-3\theta_1)(x) = \theta_1 y_1 + \theta_2 y_2 + \theta_3 y_3.$$  (The parameters in this solution are not identifiable, however.)  

*Another two-parameter solution is $y_1=1-x,$ $y_2=1,$ $\beta^2=-\theta_1,$ and $\alpha\beta=\theta_1 + \theta_2.$  You can check that $$\alpha \beta + \beta^2 x = (\theta_1 + \theta_2) + (-\theta_1)(1-y_1) = \theta_1 y_1 + \theta_2 y_2.$$
A: Your first example is a model with two effective parameters:$$y=\beta_0+\beta_x x+\varepsilon$$
You have two degrees of freedom $\alpha,\beta$ so you were able to linearize the model. Having the same degrees of freedom is not a sufficient condition but it's necessary. I show thesufficient conditions further in answer.
Your second example has two parameters too, the same as the first, but due to the constraint $\beta_x=\beta_0^2$ you have only one degree of freedom $\alpha$. Hence, you can't linearize this model.
It must be necessary that whether it is a linear or nonlinear transformation the degrees of freedom must be preserved! You can't expand one dimensional space $\alpha$ into two dimensional space $\beta_0,\beta_x$.
Sufficient condition and psudo-proof
You have a model $y=g(a,b)+h(a,b)x+\varepsilon$. It may be possible to transform it into $y=c+dx+\varepsilon$. Subtract one from another:
$$0=g(a,b)-c+(h(a,b)-d)x$$
Since, estimates $(\hat c,\hat d)$ are random variables that are a function of a random sample $(x,y)$ this can only work when:
$$g(\hat a,\hat b)=\hat c\\h(\hat a,\hat b)=\hat d$$
It's a system of algebraic equations that may have a unique solution. So, you might be able to map $(\hat c,\hat d)$ back to $(\hat a,\hat b)$. The sufficient condition is that the system has a solution.
Example: imagine that your equation was $y=g(a)+h(a)x+\varepsilon$. There's only one degree of freedom $a$ despite two functions $g,h$, so we have the following two to satisfy simultaneously:
$$g(\hat a)=\hat c\\h(\hat a)=\hat d$$
This is generally impossible to satisfy, unless somehow $\hat c=\hat d$ always. In other words you'd have to deal with truly one-dimensional system: $y=c(1+x)+\varepsilon$, essentially a regression through origin $y=x'+\varepsilon$
