Question

Consider again the conditions of Exercise 12, and suppose that it is desired to estimate the value of \(\theta = 5 - 4{\beta _0} + {\beta _1}\). Find an unbiased estimator \(\hat \theta \) of

\(\theta \). Determine the value of \(\hat \theta \)and the M.S.E. of\(\hat \theta \).

Short answer

The unbiased estimator of \(\hat \theta \) is \(\theta \). The value of \(\hat \theta \) is \(\hat \theta = - 158.020\).

MSE of \(\hat \theta \) is \({\mathop{\rm MSE}\nolimits} (\hat \theta ) = 11.519{\sigma ^2}\)

Step-by-step solution

Given Information

Define a variable\(\theta = 5 - 4{\beta _0} + {\beta _1}\)where \({\beta _0}\) and \({\beta _1}\) are the slope and intercept of linear regression.

Determine the unbiased estimator of \(\hat \theta \)

\({\hat \beta _0}\)and \({\hat \beta _1}\) are unbiased estimators of \({\beta _0}\) and \({\beta _1}\), respectively.

Thus,

\(\begin{array}{l}E\left( {{{\hat \beta }_0}} \right) = {\beta _0}\\E\left( {{{\hat \beta }_1}} \right) = {\beta _1}\end{array}\)

Further, define,

\(\hat \theta = 5 - 4{\hat \beta _0} + {\hat \beta _1},\)

To show:

\({\bf{E}}\left( {{\bf{\hat \theta }}} \right){\bf{ = \theta }}\)

Thus,

\(\begin{array}{c}E(\hat \theta ) = E\left( {5 - 4{{\hat \beta }_0} + {{\hat \beta }_1}} \right)\\ = 5 - 4E\left( {{{\hat \beta }_0}} \right) + E\left( {{{\hat \beta }_1}} \right)\\ = 5 - 4{\beta _0} + {\beta _1}\\ = \theta \end{array}\)

Thus, the unbiased estimator of \(\hat \theta \) is \(\theta \).

From the exercise 12,

\({\hat \beta _0} = 40.892\)and \({\hat \beta _1} = 0.548\)

Thus, the value of unbiased estimate of \(\theta \) is,

\(\begin{array}{c}\hat \theta = 5 - 4 \times 40.892 + 0.548\\ = - 158.020\end{array}\)

Therefore, the value of \(\hat \theta \) is \( - 158.020\).

Determine the M.S.E of \(\hat \theta \)

The MSE of an estimator\(\hat \theta \)can be written as,

\({\bf{MSE(\hat \theta ) = Var(\hat \theta ) + Bia}}{{\bf{s}}^{\bf{2}}}\)

As the parameter is proved to be an unbiased estimated, the bias would be 0.

Therefore,

\(\begin{array}{c}MSE(\hat \theta ) = Var(\hat \theta )\\ = Var\left( {5 - 4{{\hat \beta }_0} + {{\hat \beta }_1}} \right)\\ = Var\left( {4{{\hat \beta }_0}} \right) + Var\left( {{{\hat \beta }_1}} \right) - 2Cov\left( {4{{\hat \beta }_0},{{\hat \beta }_1}} \right)\\ = {4^2}Var\left( {{{\hat \beta }_0}} \right) + Var\left( {{{\hat \beta }_1}} \right) - 2Cov\left( {4{{\hat \beta }_0},{{\hat \beta }_1}} \right)\\ = 16Var\left( {{{\hat \beta }_0}} \right) + Var\left( {{{\hat \beta }_1}} \right) - 8Cov\left( {{{\hat \beta }_0},{{\hat \beta }_1}} \right)\end{array}\)

The variances of the estimated parameters \({\hat \beta _0}\) and \({\hat \beta _1}\) are,

\(\begin{array}{l}Var\left( {{{\hat \beta }_0}} \right) = {\sigma ^2}\left( {\frac{1}{n} + \frac{{{{\bar x}^2}}}{{s_x^2}}} \right)\quad {\rm{ }}\\Var\left( {{{\hat \beta }_1}} \right) = \frac{{{\sigma ^2}}}{{s_x^2}}.\end{array}\)

Also, the covariance of \({\hat \beta _0}\) and \({\hat \beta _1}\) is given as,

\(Cov\left( {{{\hat \beta }_0},{{\hat \beta }_1}} \right) = - \frac{{\bar x{\sigma ^2}}}{{s_x^2}}\)

From exercise 12 and 13,

\(\begin{array}{l}Var\left( {{{\hat \beta }_0}} \right) = 0.607{\sigma ^2}\\Var\left( {{{\hat \beta }_1}} \right) = 0.095{\sigma ^2}\\Cov\left( {{{\hat \beta }_0},{{\hat \beta }_1}} \right) = - 0.214{\sigma ^2}\end{array}\)

Substitute the values as follows,

\(\begin{array}{l}MSE(\hat \theta ) = 16\left( {0.607{\sigma ^2}} \right) + 0.095{\sigma ^2} + 8\left( {0.214{\sigma ^2}} \right)\\MSE(\hat \theta ) = 11.519{\sigma ^2}\end{array}\)

Hence, the MSE of \(\hat \theta \)is \({\mathop{\rm MSE}\nolimits} (\hat \theta ) = 11.519{\sigma ^2}\)