Which one notation is correct and why ?
$y_1=\beta_0+\beta_1x_{11}+\epsilon_1$
or,
$y_1=\beta_0+\beta_1x_{11}+e_1$
or,
$Y_1=\beta_0+\beta_1x_{11}+\epsilon_1$
or,
$Y_1=\beta_0+\beta_1x_{11}+e_1$
Where $\epsilon_1$ is the statistical error (disturbance) and $e_1$ is the residual(fitting error) .
Also, I am confused with the notation $Y_1$ and $y_1$ .
Now my second question is : When $e_1=e_2$ , does it imply $y_1=y_2$ ?
To check it , i suppose , the sample size is $n$ . And the model is :
$$y_i=\beta_0+\beta_1x_{1i}+e_i,\quad i=1,\ldots,n$$
say, $\beta_0=.5$ and $\beta_1=2.1$ and $x_{11}=2$ , $x_{12}=2.2$
If $$e_1=e_2$$ $$\Rightarrow y_1-(\beta_0+\beta_1x_{11})=y_2-(\beta_0+\beta_1x_{12})$$ $$\Rightarrow y_1-[.5+(2.1)(2)]=y_2-[.5+(2.1)(2.2)]$$ $$\Rightarrow y_1-4.7=y_2-5.12$$ $$\Rightarrow y_1=y_2-5.12+4.7$$ $$\Rightarrow y_1=y_2-.42$$ $$\Rightarrow y_1\ne y_2$$
So , $e_1=e_2$ does not imply $y_1=y_2$ unless $x_{11}=x_{12}$ .
But when i started to cross-check , that is ,
Holding $\beta_0=.5$ and $\beta_1=2.1$ and $x_{11}=2$ , $x_{12}=2.2$ and the relationship $y_1=y_2-.42$ , when $y_1=5$ , $y_2=4.58$ , then
$$e_1=y_1-(\beta_0+\beta_1x_{11})=5-[.5+(2.1)(2)]=5-4.7=0.3$$
and
$$e_2=y_2-(\beta_0+\beta_1x_{12})=4.58-[.5+(2.1)(2.2)]=4.58-5.12=-.54$$
That is , $$e_1\ne e_2\quad\text{!!! }$$
But for this example I established the relationship if $$e_1=e_2\quad\text{then}\quad y_1=y_2-.42$$ Then why is the converse not true , i.e., if $$y_1=y_2-.42\quad\text{then}\quad e_1=e_2$$
???
EDIT (for my second question) :
If $y_1=5$ , then $y_2=5.42$ . I did wrong to calculate $y_2$ . Now it comes for the example if $y_1=y_2-.42$ , then $e_1=e_2$ .
So my conclusion is If $e_1=e_2$ that doesn't imply $y_1=y_2$ unless $x_1=x_2$.