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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 $x_1=x_2$.

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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.

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  • $\begingroup$ 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 \$). $\endgroup$ – whuber Apr 28 '15 at 15:25

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