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The difference between an estimator's expected value and the true value of the parameter being estimated is known as bias in statistics. Unbiased refers to an estimator or decision rule that has no bias. "Bias" is an objective feature of an estimator in statistics.

(a)

As mentioned in the hint, the proof follows simply

\(\begin{aligned}V(\bar X) &= V\left( {\frac{{\rm{1}}}{{\rm{n}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{X}}_{\rm{i}}}} } \right)\\ &= \frac{{\rm{1}}}{{{{\rm{n}}^{\rm{2}}}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{V}} \left( {{{\rm{X}}_{\rm{i}}}} \right)\\\ &= {{{{\rm{n}}^{\rm{2}}}}}{\rm{ \times n \times V}}\left( {{{\rm{X}}_{\rm{1}}}} \right)\\ &= \frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}\end{aligned}\)

2. Each \({X_i}\)has the same distribution and is independent of the others. Because of this, the estimate is skewed. \(E\left( {{{\bar X}^2}} \right) \ne {\mu ^2}\) and the bias is

Hence, the estimator is biased

(b)

The following holds:

\(\begin{aligned}{\rm{E}}\left( {{{{\rm{\bar X}}}^{\rm{2}}}{\rm{ - k}}{{\rm{S}}^{\rm{2}}}} \right) &= E \left( {{{{\rm{\bar X}}}^{\rm{2}}}} \right){\rm{ - kE}}\left( {{{\rm{S}}^{\rm{2}}}} \right)\\\ &= \frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}{\rm{ + }}{{\rm{\mu }}^{\rm{2}}}{\rm{ - k}}{{\rm{\sigma }}^{\rm{2}}}\end{aligned}\)

(3): The estimator \({S^2}\)is an unbiased estimator of the \({\sigma ^2}.\)parameter.

The question is whether the following holds for\(k \in \mathbb{R}\).

\(\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}{\rm{ + }}{{\rm{\mu }}^{\rm{2}}}{\rm{ - k}}{{\rm{\sigma }}^{\rm{2}}}{\rm{ = }}{{\rm{\mu }}^{\rm{2}}}\)

From this, obviously,

\({\rm{k = }}\frac{{\rm{1}}}{{\rm{n}}}\)

Hence, the estimator \({S^2}\)is an unbiased estimator of the \({\sigma ^2}.\)parameter, and the value is \({\rm{k = }}\frac{{\rm{1}}}{{\rm{n}}}\).