Question

Prove that \({{\bf{\sigma }}^{{\bf{'2}}}}\), defined in Eq.\({\bf{(11}}{\bf{.5}}{\bf{.8)}}\)is an unbiased estimator of \({{\bf{\sigma }}^{\bf{2}}}\). You may assume that \({{\bf{S}}^{\bf{2}}}\)has a \({{\bf{X}}^{\bf{2}}}\) distribution with \({\bf{n - p}}\) degrees of freedom.

Short answer

Using the statistic\(\frac{{{S^2}}}{{{\sigma ^2}}}\) follows a chi-square distribution with\(n - p\) degrees of freedom, and the fact that the expected value of a chi-square random variable is the number of its degrees of freedom, it is proved.

Step-by-step solution

Define linear statistical models

A linear model describes the correlation between the dependent and independent variables as a straight line.

\(y = {a_0} + {a_1}{x_1} + {a_2}{x_2} + \tilde A,\hat A1/4 + {a_n}{x_n}\)

Models using only one predictor are simple linear regression models. Multiple predictors are used in multiple linear regression models. For many response variables, multiple regression analysis models are used.

Step 2:Find the proof for unbiased estimator

We need to show that the statistic

\({\sigma ^{2}} = \frac{{{S^2}}}{{n - p}} = \frac{1}{{n - p}} \cdot \mathop \sum \limits_{i = 1}^n {\left( {{Y_i} - {z_{i0}}{{\hat \beta }_0} - \ldots - {z_{i,p - 1}}{{\hat \beta }_{p - 1}}} \right)^2}\)

is an unbiased estimator of \({\sigma ^2}\).

\(\frac{{{S^2}}}{{{\sigma ^2}}}\)follows chi-square distribution with \(n - p\)degrees of freedom.

According to theorem \(8.2.1\),

\(E\left( {\frac{{{S^2}}}{{{\sigma ^2}}}} \right) = n - p\)

since the expected value of a chi-square random variable is actually the number of its degrees of freedom.

Therefore due to the linearity of the expectation,

\(E\left( {{S^2}} \right) = (n - p){\sigma ^2}\)

Dividing by \(n - p\)we obtain,

\(E\left( {\frac{{{S^2}}}{{n - p}}} \right) = {\sigma ^2}\)

\(E\left( {{\sigma ^{2}}} \right) = {\sigma ^2}\)

Hence we proved the given statement.