Loss function and model parameters estimation and link it to optimization Suppose my objective function is to find the minimum value of the following :
$$\min_{\alpha,\beta} \sum_{i=1}^{n} \rho ( e_i ) $$
where $e_i= y_i - ({\alpha+\beta x_i}) $ and $\rho$ is any loss function.
My question is, because the first term $y_i$ does not contains any $\alpha$ or $\beta$ is it correct to first solve the optimization problem which is $$\min_{\alpha,\beta} \alpha+\beta x_i $$ with some constraints and then add $y_i$ term to the results?
I know when we fit a regression line we do it without any complication with some built-in function but I want to understand the intuition behind our calculation and how the optimization process has been done in this kind of problem.
Note: my loss function is not only mean square error it could be any other type.
Any ideas/references will be appreciated.

No, it's not correct. You are assuming $\rho$ is linear, so you can write $\rho(x+y) = \rho(x) + \rho(y)$, and therefore
$$
\min_{\alpha, \beta} \rho(y_i−(\alpha+\beta x_i)) = \min_{\alpha, \beta} \left[\rho(y_i)−\rho(\alpha+\beta x_i) \right]
$$
but that's not always true.
For example, if $\rho(x) = (z - x)^2$ then
$$\rho(x + y) = \\
(z - (x+y))^2 = \\
z^2 + (x+y)^2 - 2z(x+y) = \\
z^2 + x^2 + y^2 +2xy - 2z(x+y)$$
which is not the same as
$$
\rho(x) +\rho(y) = \\
(z - x)^2 + (z-y)^2 = \\
2z^2 + x^2 +y^2 - 2yz - 2xz 
$$
Therefore, you can't find optimal $\alpha$ and $\beta$ using
$$
\min_{\alpha, \beta} \alpha + \beta x_i
$$
Also, if your approach was correct, then the optimal parameters would be $\alpha = -\infty$ and $\beta = -\infty$.
