I am reading Simple Regression Model from this book, Section 6.5 (page 267 in downloaded pdf, 276 if viewed online).

The author starts with below equation for a simple linear regression model,

$$ Y_i = \alpha_1 + \beta x_i + \varepsilon_i $$

And then after few lines, he lets for conveience that, $\alpha_1 = \alpha - \beta\overline{x}$ so that,

$$ Y_i = \alpha + \beta(x_i - \overline{x}) + \varepsilon_i $$

where $\overline{x} = \dfrac{1}{n}\sum\limits_{i=1}^nx_i$

My questions:
1. It is not convincing to bring in $\overline{x}$ just for convenience sake in the equation. Can any one please explain the logic behind bring that in the equation?
2. After above equation, the author says, $Y_i$ is equal to a nonrandom quantity, $\alpha + \beta(x_i - \overline{x})$, plus a mean zero normal random variable $\varepsilon_i$. Does that mean, $\alpha + \beta(x_i - \overline{x})$ has no randomness involved in that?

Kindly help.


1 Answer 1

  1. $\alpha_1$s in two equations are different. Let $\alpha_2$ be the $\alpha$ in the second equation, then $$ \alpha_1 = \alpha_2 + \beta \bar x$$

At the time that the computer was not popular or had no computer, the line was fit by using calculators. Bringing in $\bar x$ is really simplified the computation.

  1. From the first equation, $\epsilon$ is the only random component. So source of randomness of $Y$ is $\epsilon$, the other parts $\alpha + \beta x$ are known or unknown constant.
  • $\begingroup$ I just corrected $\alpha_1$ to $\alpha$ in 2nd equation. Still the reason is not convincing that it simplified the computation. Can you kindly elaborate further? How could $\overline{x}$ suddenly enter the equation without an associated mathematical logic. $\endgroup$ Oct 20, 2018 at 18:32
  • $\begingroup$ Let $z_i=x_i-\bar x$, then (1) $\sum z_i = 0$ vs calculating $\sum x_i$, (2) $\sum z_i^2$ is easier easier than $\sum x_i^2$, and (3) $\sum z_iY_i$ is easier easier than $\sum x_iY_i$. introducing $\bar x$ into equation does not change anything in equation, similar to $ + a - a$ , which we used to proof something in math. $\endgroup$
    – user158565
    Oct 20, 2018 at 18:43
  • $\begingroup$ $Y_i = \alpha_1 + \beta x_i + \varepsilon_i$ ==> $Y_i = \alpha_1 + \beta x_i + \varepsilon_i - \beta \bar x + \beta \bar x $ ==> $Y_i = (\alpha_1 +\beta \bar x) + \beta (x_i - \bar x) + \varepsilon_i$ ==> $Y_i = \alpha + \beta (x_i - \bar x) + \varepsilon_i$ $\endgroup$
    – user158565
    Oct 20, 2018 at 18:56

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