Question

Consider Model \((9.6 .1) .\) Let \(\eta_{0}=E\left(Y \mid x=x_{0}-\bar{x}\right) .\) The least squares estimator of \(\eta_{0}\) is \(\hat{\eta}_{0}=\hat{\alpha}+\hat{\beta}\left(x_{0}-\bar{x}\righ(a) Using \)(9.6 .9)\(, show that \)\hat{\eta}_{0}\( is an unbiased estimator and show that its variance is given by $$ V\left(\hat{\eta}_{0}\right)=\sigma^{2}\left[\frac{1}{n}+\frac{\left(x_{0}-\bar{x}\right)^{2}}{\sum_{i=1}^{n}\left(x_{1}-\bar{x}\right)^{2}}\right] $$ (b) Obtain the distribution of \)\hat{\eta}_{0}\( and use it to determine a \)(1-\alpha) 100 \%\( confidence interval for \)\eta_{0}\(.t)\).

Short answer

It could be shown that the estimator is unbiased, its variance was calculated and it follows a normal distribution. A confidence interval for the parameter \(\eta_{0}\) was also computed.

Step-by-step solution

Unbiasedness of the estimator

In order to check unbiasedness of the estimator \(\hat{\eta}_{0}\), the expected value of the estimator needs to be equal to the parameter it is estimating, \(\eta_{0}\). Hence the requirement is that \(E(\hat{\eta}_{0}) = \eta_{0}\). Using equation (9.6.9), this becomes \(E(\hat{\alpha} + \hat{\beta}(x_{0}-\bar{x})) = \alpha + \beta(x_{0}- \bar{x})\), which, since \(\hat{\alpha}\) and \(\hat{\beta}\) are unbiased estimators, is true.

Variance of the estimator

Next, the variance of the estimator is computed. The variance of \(\hat{\eta}_{0}\) is \(Var(\hat{\alpha} + \hat{\beta}(x_{0}- \bar{x}))\). Using the properties of variance, this becomes \(Var(\hat{\alpha}) + (x_{0}- \bar{x})^2Var(\hat{\beta}) + 2(x_{0}- \bar{x}) Cov(\hat{\alpha}, \hat{\beta})\). The covariance between \(\hat{\alpha}\) and \(\hat{\beta}\) is 0, and using the formulas for \(Var(\hat{\alpha})\) and \(Var(\hat{\beta})\) from equation (9.6.1), the variance of \(\hat{\eta}_{0}\) is shown to be \(\sigma^{2}\left[\frac{1}{n}+\frac{(x_{0}-\bar{x})^{2}}{\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}}\right]\) .

Distribution of the estimator

The estimator \(\hat{\eta}_{0}\) is a linear combination of \(\hat{\alpha}\) and \(\hat{\beta}\), whose distributions are both normal, hence \(\hat{\eta}_{0}\) follows a normal distribution with mean \(\eta_{0}\) and the calculated variance from step 2.

Confidence interval for \(\eta_{0}\)

The confidence interval for \(\eta_{0}\) can now be obtained. From the normal distribution of \(\hat{\eta}_{0}\), a \(1-\alpha\) confidence interval for \(\eta_{0}\) is \(\hat{\eta}_{0} \pm z_{\frac{\alpha}{2}}\sqrt{Var(\hat{\eta}_{0})}\), where \(z_{\frac{\alpha}{2}}\) is the value from the standard normal distribution which leaves a probability of \(\alpha / 2\) in each tail.