Question

Let\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)be a random sample from a pdf\({\rm{f(x)}}\)that is symmetric about\({\rm{\mu }}\), so that\({\rm{\backslash widetildeX}}\)is an unbiased estimator of\({\rm{\mu }}\). If\({\rm{n}}\)is large, it can be shown that\({\rm{V (\tilde X)\gg 1/}}\left( {{\rm{4n(f(\mu )}}{{\rm{)}}^{\rm{2}}}} \right)\).

a. Compare\({\rm{V(\backslash widetildeX)}}\)to\({\rm{V(\bar X)}}\)when the underlying distribution is normal.

b. When the underlying pdf is Cauchy (see Example 6.7),\({\rm{V(\bar X) = \yen}}\), so\({\rm{\bar X}}\)is a terrible estimator. What is\({\rm{V(\tilde X)}}\)in this case when\({\rm{n}}\)is large?

Short answer

a) The comparison is\({\rm{V(\tilde X) > V(\bar X)}}\)

b) The variance now becomes\({\rm{V(\tilde X) = }}\frac{{{{\rm{\pi }}^{\rm{2}}}{{\rm{\beta }}^{\rm{2}}}}}{{{\rm{4n}}}}{\rm{.}}\)

Step-by-step solution

Introduction

An estimator is a rule for computing an estimate of a given quantity based on observable data: the rule (estimator), the quantity of interest (estimate), and the output (estimate) are all distinct.

Explanation

The pdf \({\rm{f(x)}}\)of normally distributed random variable with parameters \({\rm{\mu }}\)and \({\rm{\sigma }}\) is

\({\rm{f(x) = }}\frac{{\rm{1}}}{{\sqrt {{\rm{2\pi }}} {\rm{ \times \sigma }}}}{\rm{exp}}\left\{ {{\rm{ - }}\frac{{{{{\rm{(x - \mu )}}}^{\rm{2}}}}}{{{\rm{2}}{{\rm{\sigma }}^{\rm{2}}}}}} \right\}{\rm{,}}\quad {\rm{x\hat I R}}{\rm{.}}\)

As given in the exercise, it can be shown that

\({\rm{V(\tilde X)\gg }}\frac{{\rm{1}}}{{{\rm{4n \times (f(\mu )}}{{\rm{)}}^{\rm{2}}}}}\)

where \({\rm{f(\mu )}}\)is

\(\begin{aligned}f(\mu ) = \frac{{\rm{1}}}{{\sqrt {{\rm{2\pi }}} {\rm{ \times \sigma }}}}{\rm{exp}}\left\{ {{\rm{ - }}\frac{{{{{\rm{(\mu - \mu )}}}^{\rm{2}}}}}{{{\rm{2}}{{\rm{\sigma }}^{\rm{2}}}}}} \right\}\\ &= \frac{{\rm{1}}}{{\sqrt {{\rm{2\pi }}} {\rm{ \times \sigma }}}}{\rm{.}}\end{aligned}\)

Then, the variance is

\({\rm{V(\tilde X)\gg }}\frac{{\rm{1}}}{{{\rm{4n \times (f(\mu )}}{{\rm{)}}^{\rm{2}}}}}{\rm{ = }}\frac{{{\rm{2\pi }}{{\rm{\sigma }}^{\rm{2}}}}}{{{\rm{4n}}}}\)

Also, the variance of random variable \({\rm{\bar X}}\)is

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

(1): all \({{\rm{X}}_{\rm{i}}}\)have the same distribution and are independent.

It is obvious that

\({\rm{V(\tilde X)\gg }}\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ \times }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}{\rm{ > }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}{\rm{ = V(\bar X)}}\)

because

\(\frac{{\rm{\pi }}}{{\rm{2}}}{\rm{ > 1}}\)

Therefore, the comparison is\({\rm{V(\tilde X) > V(\bar X)}}\)

Explanation

The pdf \({\rm{f(x)}}\)of Cauchy distributed random variable with parameter \({\rm{\beta }}\)is

\({\rm{f(x) = }}\frac{{\rm{1}}}{{{\rm{\pi \beta }}\left( {{\rm{1 + ((x - \mu )/\beta }}{{\rm{)}}^{\rm{2}}}} \right)}}{\rm{,}}\quad {\rm{x\hat I R}}{\rm{.}}\)

As given in the exercise, it can be shown that

\({\rm{V(\tilde X)\gg }}\frac{{\rm{1}}}{{{\rm{4n \times (f(\mu )}}{{\rm{)}}^{\rm{2}}}}}{\rm{,}}\)

where \({\rm{f(\mu )}}\)is

\(\begin{aligned}f(\mu ) &= \frac{{\rm{1}}}{{{\rm{\pi \beta }}\left( {{\rm{1 + }}{{\left( {{{{\rm{(\mu - \mu )}}}^{\rm{2}}}{\rm{/\beta }}} \right)}^{\rm{2}}}} \right)}}\\&= \frac{{\rm{1}}}{{{\rm{\pi \beta }}}}{\rm{.}}\end{aligned}\)

Therefore, the variance now becomes\({\rm{V(\tilde X)\gg }}\frac{{\rm{1}}}{{{\rm{4n \times (f(\mu )}}{{\rm{)}}^{\rm{2}}}}}{\rm{ = }}\frac{{{{\rm{\pi }}^{\rm{2}}}{{\rm{\beta }}^{\rm{2}}}}}{{{\rm{4n}}}}{\rm{.}}\)