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For the simple linear regression model

$$y_i=\beta_0 +\beta_1 x_i$$

I have derived the estimates of $\beta_0=\bar y -\beta_1\bar x$ and $\beta_1=\frac{\Sigma (y_i-\bar y)(x_i-\bar x)}{\Sigma (x_i-\bar x)^2}$ but my book says that it is possible to express $\beta_0=\Sigma a_i Y_i$ and $\beta_1 =\Sigma c_i Y_i$.

How ever I can't seem to factor out $Y_i$ from either expression. For $\beta_0$ all I can get is $\beta_0=\Sigma \frac{y_i}{n}-\beta_1\bar x$ but this has constant term $\beta_1 \bar x$.

For $\beta_1$ I get $\beta_1=\Sigma y_i\frac{x_i-\bar x}{\Sigma (x_i-\bar x)^2}-\frac{\bar y(x_i-\bar x)}{\Sigma(x_i- \bar x)^2}$ again with constant term that I can't get rid of. Does anyone know how to solve it, why these constants should go or how to get rid of them?

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  • $\begingroup$ What is $\bar{x}_0$ in the expression of $\beta_1$? $\endgroup$ – Shanks May 20 '18 at 13:01
  • $\begingroup$ Should be written $\hat{\beta}_0$ and $\hat{\beta}_1$ instead of just $\beta_0$ and $\beta_1$ while writing the estimates. $\endgroup$ – Shanks May 20 '18 at 13:03
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Hint:

\begin{align} \hat{\beta}_0 &= \bar{y} - \hat{\beta}_1 .\bar{x}\\ & = \left(\frac{1}{n} \sum \limits_{i=1}^n y_i\right) - \frac{\sum \limits_{i=1}^n (y_i-\bar y)(x_i-\bar x)}{\sum \limits_{i=1}^n (x_i-\bar x)^2} \bar{x}\\ &= \left(\frac{1}{n} \sum \limits_{i=1}^n y_i\right) - \frac{\sum \limits_{i=1}^n (y_i-\bar y)x_i}{\sum \limits_{i=1}^n (x_i-\bar x)^2} \bar{x}\\ &= \left(\frac{1}{n} \sum \limits_{i=1}^n y_i\right) - \sum \limits_{i=1}^n \left(\frac{x_i\cdot \bar{x}}{\sum \limits_{i=1}^n (x_i-\bar x)^2}\right)y_i + \bar{y} \frac{\bar{x}\sum\limits _{i=1}^n x_i}{\sum \limits_{i=1}^n (x_i-\bar x)^2}\\ &= \sum \limits_{i=1}^n \left(\frac{1}{n}\right) y_i - \sum \limits_{i=1}^n \left(\frac{x_i\cdot \bar{x}}{\sum \limits_{i=1}^n (x_i-\bar x)^2}\right)y_i + \sum \limits_{i=1}^n \left(\frac{\bar{x}\sum\limits _{i=1}^n x_i}{n\sum \limits_{i=1}^n (x_i-\bar x)^2}\right)y_i \end{align}

Now the rest should be done easily.

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