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Show that $Cov(\bar{y},\hat{\beta_1})=0$

For those unfamiliar with statistics, Cov(A,B) refers to the covariance function. $\bar{y}$ refers to the average of the response (dependent variable). $\hat{\beta_1}$ refers to the estimator of the slope.

The solution goes as follows:

$Cov(\bar{y},\hat{\beta_1}) = Cov(\frac{\sum{y_i}}{n},\sum{c_iy_i}) $

Where $c_i = \frac{(x_i-\bar{x})}{S_{xx}} $

And $S_{xx} = \sum{(x_i-\bar{x})^2}$

$Cov(\frac{\sum{y_i}}{n},\sum{c_iy_i}) = \frac{1}{n}Cov(\sum{y_i},\sum{c_iy_i}) $

Can we bring the $\sum{c_i}$ out of the covariance? If so, we would simply be left with $var(y_i)$.

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    $\begingroup$ Indeed, we can use the bilinearity of covariance to write \begin{align} \operatorname{Cov}(\bar y,\hat{\beta_1})&=\operatorname{Cov}\left(\frac{1}{n}\sum_{i=1}^n y_i,\sum_{i=1}^nc_iy_i\right) \\&=\frac{1}{n}\sum_{i=1}^nc_i\operatorname{Var}(y_i) \\&=\frac{\sigma^2}{n}\sum_{i=1}^nc_i=0 \end{align} $\endgroup$
    – StubbornAtom
    Commented Sep 15, 2018 at 7:52
  • $\begingroup$ @StubbornAtom How to show they are independently distributed Normal variables? Can I find such a proof or discussion in George Casella and Berger textbook? $\endgroup$
    – Mariana
    Commented Sep 4, 2021 at 23:23
  • $\begingroup$ @Mariana stats.stackexchange.com/a/362660/119261. $\endgroup$
    – StubbornAtom
    Commented Sep 5, 2021 at 5:48
  • $\begingroup$ @StubbornAtom is the following also correct: \begin{equation} \begin{aligned} \operatorname{Cov}\left(\hat{\beta}_1, \overline{\mathrm{Y}}\right) & =\operatorname{cov}\left(\overline{\mathrm{Y}}, \frac{\sum \mathrm{X}_{\mathrm{i}}-\overline{\mathrm{X}} \mathrm{Y}_{\mathrm{i}}}{\mathrm{s}_{\mathrm{XX}}}\right) \\ & =\frac{\mathrm{X}_{\mathrm{i}}-\overline{\mathrm{X}}}{\mathrm{S}_{\mathrm{XX}}} \operatorname{Var}(\overline{\mathrm{Y}}) \\ & =0 \end{aligned} \end{equation} $\endgroup$
    – bruno
    Commented May 7 at 19:54

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