# For the model $y_i=\beta_0+\beta_1x_{1i}+e_i,\quad i=1,\ldots,n$ , does $e_1=e_2$ imply $y_1=y_2$?

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 $$x_1=x_2$$.

Model is

$y=\beta_0+\beta_1x+ϵ \ \ \text{where} \ \ \varepsilon \sim N(0,\sigma^2)$

or

$E(y) = \beta_0 + \beta_1x$

You are estimating $\beta_0$, $\beta_1$ and $\sigma$.

After you get all the estimates, you calculate $\varepsilon_i$'s to check if it's a good fit using

$y_i = \beta_0 + \beta_1x_i+\varepsilon_i$

or

$\varepsilon_i= y_i - \beta_0 + \beta_1x_i$

You'd better to get logic right before you start to validate your results.

$\varepsilon$ is random variable with known mean zero and unknown variance.

$\varepsilon_i$'s are the residuals of your model.

So $\varepsilon_i$'s should follow a nice normal distribution with your estimated variance if your model is good.

• Your answer is likely to be misinterpreted unless you clearly distinguish parameters from their estimates in your notation. Use $\TeX$ markup (enclose it between dollar signs \\$).
– whuber
Commented Apr 28, 2015 at 15:25