Question

Using a long rod that has length \({\rm{\mu }}\)you are going to lay out a square plot in which the length of each side is \({\rm{\mu }}\).Thus the area of the plot will be \({{\rm{\mu }}^{\rm{2}}}\)However, you do not know the value of \({\rm{\mu }}\), so you decide to make \({\rm{n}}\)independent measurements \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)of the length. Assume that each \({{\rm{X}}_{\rm{i}}}\)has mean \({\rm{\mu }}\) (unbiased measurements) and variance \({{\rm{\sigma }}^{\rm{2}}}\).

a. Show that \({{\rm{\bar X}}^{\rm{2}}}\)is not an unbiased estimator for \({{\rm{\mu }}^{\rm{2}}}\). (Hint: For any \({\rm{Y,E}}\left( {{{\rm{Y}}^{\rm{2}}}} \right){\rm{ = V(Y) + (E(Y)}}{{\rm{)}}^{\rm{2}}}\)Apply this with \({\rm{Y = \bar X}}\))

b. For what value of \({\rm{k}}\)is the estimator \({{\rm{\bar X}}^{\rm{2}}}{\rm{ - k}}{{\rm{S}}^{\rm{2}}}\)unbiased for \({{\rm{\mu }}^{\rm{2}}}\)? (Hint: Compute \({\rm{E}}\left( {{{{\rm{\bar X}}}^{\rm{2}}}{\rm{ - k\;}}{{\rm{S}}^{\rm{2}}}} \right)\))

Short answer

a.The estimator is biased.

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

Step-by-step solution

Concept introduction

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.

Finding whether it’s biased or unbiased

(a)

As mentioned in the hint, the proof follows simply

1. : the following holds for the variance:

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

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

Finding whether it’s biased or unbiased

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