The model is given by $E(Y_i)=\gamma+\beta (x_i-\bar x)$

Show $\mathrm{cov}(E_i,E_j)=\sigma^2p_{ij}$ where $$p_{ij}=1/n+\frac{(x_i-\bar x)(x_j-\bar x)}{S_{xx}}$$

I already showed that $E(E_i)=0$ and $\mathrm{cov}(Y_i,\bar Y)=\sigma^2/n$

I think I'm missing some easy properties that I can use.

I can see that somewhat related (not equal) $\sigma^2 p_{ij}=\mathrm{var}(\hat Y)$

I tried:

$$\mathrm{cov}(E_i,E_j)=\mathrm{cov}(Y_i-\bar y - \hat\beta(x_i-\bar x),Y_j-\bar y - \hat\beta(x_j-\bar x))$$

But then I don't know how to procceed because I have $3$ random variables. I can find $E(Y_i)=\gamma+\beta (x_i-\bar x),E(\bar y)=\gamma$ and $E(\hat\beta)=\beta$

  • $\begingroup$ It seems that you confused some notations. What is $\bar{Y}_i$? Perhaps you meant $cov(\hat{Y}_i - \bar{Y}_n, \hat{Y}_j - \bar{Y}_n)$? $\endgroup$ – V. Vancak Aug 19 '16 at 9:41
  • $\begingroup$ Sorry, it was late when I typed it up. $E_i=Y_i-\hat Y_i=Y_i-\bar y -\hat\beta(x_i-\bar x)$ $\endgroup$ – GRS Aug 19 '16 at 10:31

It is not a complete answer, i.e., it seems that there is no way to avoid some tedious calculations, but may be it can help you:

1) $ cov(\hat{\beta}(x_i - \bar{x}), \hat{\beta}(x_j - \bar{x}) ) = \sigma^2\frac{(x_i-\bar{x})(x_j-\bar{x})}{\sum (x_i - \bar{x})^n}$.

2) $cov(\bar{Y}_n, \bar{Y}_n) = var(\bar{Y}_n)=\sigma^2/n $

3) If the error terms are i.i.d then $cov(Y_i, \bar{Y}_n) = cov(Y_i, Y_i/n)=\sigma^2/n $

  • $\begingroup$ Thanks a lot, I'll work through it $\endgroup$ – GRS Aug 19 '16 at 11:30

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