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Consider the standard form of linear regression:

$$Y_i = \beta_0+\beta_{1}x_{i} +\epsilon_i \ \text{for} \ i = 1 \dots n$$

We can rewrite this as $$Y_i = (\beta_0+\beta_{1}\bar{x})+\beta_{1}(x_i-\bar{x}) + \epsilon_i \ \text{for} \ i = 1 \dots n$$

If we let $\alpha = \beta_0+\beta_{1}\bar{x}$, then $$Y_i = \alpha+\beta_{1}(x_i-\bar{x}) + \epsilon_i \ \text{for} \ i = 1 \dots n$$

What is the interpretation of $\alpha$? Would it be the value of $Y_i$ at $x_i = \bar{x}$?

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    $\begingroup$ but you've got the equation for alpha written right there... what more can you want? you could think of it as kind of the centre point of the regression line. $\endgroup$ – Miss Palmer Apr 5 '17 at 13:55
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You don't need all this, just take the expectation of left and right sides of main model equation: $$E[Y]=\beta_0+\beta_1E[X]\equiv\alpha$$

In fact, this shows a very important role of the intercept in OLS: it allows for errors to have mean zero. Think of this if there was no intercept, then you can't have a balance because $$E[Y]\ne \beta_1 E[X]$$

So, you'll be forced to have $$E[Y]=\beta_1E[X]+E[e],$$ where $E[e]=E[Y]-\beta_1E[X]\ne 0$

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