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I know that

$\widehat{\beta}_1 = \frac{\sum_{i=1}^n \left( x_i-\bar{x} \right) \left( y_i-\bar{y} \right)}{\sum_{i=1}^n \left(x_i-\bar{x} \right)^2} =\frac{\sum_{i=1}^n \left( x_i-\bar{x} \right)y_i}{\sum_{i=1}^n \left(x_i-\bar{x} \right)^2}$

$\widehat{\beta}_0 = \bar{y}-\widehat{\beta}_1\bar{x}$

and I am trying to prove that

$\sqrt{Var\left[\widehat{\beta}_1\right]} = \frac{\displaystyle s_{y|x}}{\displaystyle s_x\sqrt{N-1}}$

$\sqrt{Var\left[\widehat{\beta}_0\right]} = s_{y|x} \sqrt{\frac{1}{N}+\frac{\bar{x}^2}{s_x^2(N-1)}}$

I found some hints here but I am still stuck in the derivations.

  • Is $Var[\widehat{\beta}_1] = \frac{Var[\sum_{i=1}^n \left( x_i-\bar{x} \right)y_i]}{Var\left[\sum_{i=1}^n \left(x_i-\bar{x} \right)^2\right]} = \frac{Var[\sum_{i=1}^n \left( x_i-\bar{x} \right)y_i]}{s_x^2(N-1)}$ ? Because the covariance of denominator and numerator is zero?

  • Is $Var[\sum_{i=1}^n \left( x_i-\bar{x} \right)y_i] = s_{y|x}^2$ ? Why?

  • Why $Cov\left[\bar{y}. \widehat{\beta}_1\right]=Cov\left[\frac{1}{n}\sum_{i=1}^n y_i, \frac{\sum_{i=1}^n \left( x_i-\bar{x} \right)y_i}{\sum_{i=1}^n \left(x_i-\bar{x} \right)^2} \right]=0$?

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user2513881
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  • Sorry that I have no basic knowledge of matrices in statistics. – user2513881 Mar 19 '16 at 07:21
  • Try it now -- it should work this time – Glen_b Mar 19 '16 at 08:19
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    Based on your comment: You might want to ditch the question on standard errors and focus on the Linear Algebra a bit. Even if someone explains your question wonderfully, you will still lack a major asset in terms of Statistics education; check this [thread](http://stats.stackexchange.com/questions/21346) for some resources. Even some basic knowledge will get you huge mileage compared to no knowledge of matrices in Statistics. (repost.) – usεr11852 Mar 19 '16 at 08:22

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