# Regression Proof that the point of averages (x,y) lies on the estimated regression line

How do you show that the point of averages (x,y) lies on the estimated regression line?

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What have you done so far? – user28 Nov 2 '10 at 1:30
I don't know where to start. – Justin Meltzer Nov 2 '10 at 1:38
I know the simple regression line is y=B0 + B1x – Justin Meltzer Nov 2 '10 at 1:39
And I guess by the average x and y is E(x) and E(y), not sure how to link that to the regression line though. – Justin Meltzer Nov 2 '10 at 1:40
The regression line is the line that minimizes the sum of squared errors. Knowing that, and a basic knowledge of calculus, find the values of B0 and B1 that minimize that sum of squared errors. The rest requires a little bit of high school level algebra. – Christopher Aden Nov 2 '10 at 2:06
To get you started: $\bar y = 1/n \sum y_i = 1/n \sum (\hat y_i + \hat \epsilon_i)$ then plug in, how the $\hat y_i$ are estimated by the $x_i$ and you're almost done.
EDIT: since no one replied, here the rest for sake of completeness: the $\hat y_i$ are estimated by $\hat y_i=\hat \beta_0 + \hat \beta_1 x_{i1} + \ldots + \hat \beta_n x_{in} + \hat \epsilon_i$, so you get $\bar y = 1/n \sum \hat \beta_0 + \hat \beta_1 x_{i1} + \ldots + \hat \beta_n x_{in}$ (the $\hat \epsilon_i$ sum to zero) and finally:
$\bar y = \hat \beta_0 + \hat \beta_1 \bar x_{1} + \ldots + \hat \beta_n \bar x_{n}$. And that's it: The regression line goes through the point $(\bar x, \bar y)$