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In a simple linear regression model

$$E(Y|X=x)=\beta_0+\beta_1x,$$

where the parameters $\beta_0, \beta_1$ are estimated via OLS as

$$\hat{\beta}_1=\frac{\mathrm{Cov}(X,Y)}{\mathrm{Var}(X)}, \text{and}$$ $$\hat{\beta}_0=\mathrm{E}(Y)-\hat{\beta}_1\mathrm{E}(X),$$

it is well known that the variance of the residuals is $$\mathrm{Var}(y-\hat{\beta}_0-\hat{\beta}_1x|X=x)=\left(1-\frac{\mathrm{Cov}^2(X,Y)}{\mathrm{Var}(X)\mathrm{Var}(Y)}\right)\mathrm{Var}(Y)=(1-r^2)\mathrm{Var}(Y).$$

What can be said about:

  1. $\mathrm{E}(|y-\hat{\beta}_0-\hat{\beta}_1x|\,|X=x)$? Note: not $\mathrm{E}(y-\hat{\beta}_0-\hat{\beta}_1x|X=x)$, which would be equal to 0.
  2. $\mathrm{E}[(y-\hat{\beta}_0-\hat{\beta}_1x)^2\,|X=x]$?
  3. $\mathrm{Var}(|y-\hat{\beta}_0-\hat{\beta}_1x|\,|X=x)$?
user255695
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  • This seems like a homework problem - if that is the case, you are supposed to explain what you have tried to solve it and where you got stuck. If other people on this forum solve the problem for you, you are not going to learn very much. – Isabella Ghement Aug 06 '19 at 13:26
  • @IsabellaGhement It is a self-study question. The problem with $\mathrm{E}(|y-\hat{\beta}_0-\hat{\beta}_1x|\,|X=x)$ is that I could apply the triangle inequality and use linearity of expectation to work on each term separately, but that would give me an upper bound. – user255695 Aug 06 '19 at 14:04
  • You can imagine the distribution of the residuals. One of the assumptions of the residuals is that there is no real pattern to the residuals and if you were to graph a density to them, you would end up with something approaching a uniform distribution. – M Waz Aug 06 '19 at 17:45

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