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Suppose that $\overline{X}=0$ and that linear regression model is valid. Show that $\hat{B_0}$ and $\hat{B_1}$ are independent. What can you conclude about the construction of simultaneous confidence intervals for $B_0$ and $B_1$?.

We have the following model $$Y_i=B_0+B_1XX_i+\epsilon_i$$

and the least squares estimators are given by $$\hat{B_0}=\overline{Y}-\hat{B_1}\overline{X}$$ $$\hat{B_1}=\frac{\sum(Y_i-\overline{Y})(X_i-\overline{X})}{\sum(X_i-\overline{X})^2}$$

since $$\hat{B_0}\sim N(B_0;\frac{\sigma^2\sum X_i^2}{n\sum (X_i-\overline{X})^2})$$ $$\hat{B_1}\sim N(B_1;\frac{\sigma^2}{\sum (X_i-\overline{X})})$$ and $$cov(\hat{B_0};\hat{B_1})=\sigma^2\frac{-\overline{X}^2}{\sum (X_i-\overline{X})^2}=0$$

then $\hat{B_0}$ and $\hat{B_1}$ are independent.

The simultaneous confidence interval with Bonferroni method are given by

$$\hat{B_0}\pm Bs(\hat{B_0})$$ $$\hat{B_1}\pm Bs(\hat{B_1})$$

where $B=t_{1-\alpha/4,n-2}$ and $$s(\hat{B_0})=MSE(\frac{1}{n}+\frac{\overline{X}}{\sum (X_i-\overline{X})^2})$$ $$s(\hat{B_1})=\frac{MSE}{\sum (X_i-\overline{X})^2}$$ then $$IC(\hat{B_1})=\overline{Y}\pm B\frac{MSE}{n}$$ $$IC(\hat{B_1})=\frac{\sum (Y_i-\overline{Y})X_i}{\sum X_i^2}\pm B\frac{MSE}{\sum X_i^2}$$ Is this or is there something special in simultaneous confidence intervals when $\overline{X}=0$?

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  • $\begingroup$ Could you explain what you mean by "something special"? $\endgroup$ – whuber Nov 14 '15 at 21:38
  • $\begingroup$ Well the second term in the standard error of the intercept drops. $\endgroup$ – JohnK Nov 14 '15 at 21:40
  • $\begingroup$ @whuber My question is whether there is some peculiarity in the confidence interval when $\overline{X}=0$. $\endgroup$ – user72621 Nov 14 '15 at 21:41
  • $\begingroup$ @JohnK just that? $\endgroup$ – user72621 Nov 14 '15 at 21:43
  • $\begingroup$ I don't know what precisely you are looking for but it's worth pointing out that this is a nonnegative term so it can only add to the variability of the estimate. By the way, you have forgotten a square in the sample mean. $\endgroup$ – JohnK Nov 14 '15 at 21:46

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