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Chapter 6: Q18E (page 264)

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

Expert verified

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

01

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.

02

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\({\rm{V(\tilde X) > V(\bar X)}}\)

03

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

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Most popular questions from this chapter

Let\({\rm{X}}\)represent the error in making a measurement of a physical characteristic or property (e.g., the boiling point of a particular liquid). It is often reasonable to assume that\({\rm{E(X) = 0}}\)and that\({\rm{X}}\)has a normal distribution. Thus, the pdf of any particular measurement error is

\({\rm{f(x;\theta ) = }}\frac{{\rm{1}}}{{\sqrt {{\rm{2\pi \theta }}} }}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/2\theta }}}}\quad {\rm{ - \yen < x < \yen}}\)

(Where we have used\({\rm{\theta }}\)in place of\({{\rm{\sigma }}^{\rm{2}}}\)). Now suppose that\({\rm{n}}\)independent measurements are made, resulting in measurement errors\({{\rm{X}}_{\rm{1}}}{\rm{ = }}{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{ = }}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}{\rm{ = }}{{\rm{x}}_{\rm{n}}}{\rm{.}}\)Obtain the mle of\({\rm{\theta }}\).

An estimator \({\rm{\hat \theta }}\) is said to be consistent if for any \( \in {\rm{ > 0}}\), \({\rm{P(|\hat \theta - \theta |}} \ge \in {\rm{)}} \to {\rm{0}}\) as \({\rm{n}} \to \infty \). That is, \({\rm{\hat \theta }}\) is consistent if, as the sample size gets larger, it is less and less likely that \({\rm{\hat \theta }}\) will be further than \( \in \) from the true value of \({\rm{\theta }}\). Show that \({\rm{\bar X}}\) is a consistent estimator of \({\rm{\mu }}\) when \({{\rm{\sigma }}^{\rm{2}}}{\rm{ < }}\infty \) , by using Chebyshev鈥檚 inequality from Exercise \({\rm{44}}\) of Chapter \({\rm{3}}\). (Hint: The inequality can be rewritten in the form \({\rm{P}}\left( {\left| {{\rm{Y - }}{{\rm{\mu }}_{\rm{Y}}}} \right| \ge \in } \right) \le {\rm{\sigma }}_{\rm{Y}}^{\rm{2}}{\rm{/}} \in \). Now identify \({\rm{Y}}\) with \({\rm{\bar X}}\).)

Let\({\rm{X}}\)have a Weibull distribution with parameters\({\rm{\alpha }}\)and\({\rm{\beta }}\), so

\(\begin{array}{l}{\rm{E(X) = \beta \times \Gamma (1 + 1/\alpha )V(X)}}\\{\rm{ = }}{{\rm{\beta }}^{\rm{2}}}\left\{ {{\rm{\Gamma (1 + 2/\alpha ) - (\Gamma (1 + 1/\alpha )}}{{\rm{)}}^{\rm{2}}}} \right\}\end{array}\)

a. Based on a random sample\({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\), write equations for the method of moments estimators of\({\rm{\beta }}\)and\({\rm{\alpha }}\). Show that, once the estimate of\({\rm{\alpha }}\)has been obtained, the estimate of\({\rm{\beta }}\)can be found from a table of the gamma function and that the estimate of\({\rm{\alpha }}\)is the solution to a complicated equation involving the gamma function.

b. If\({\rm{n = 20,\bar x = 28}}{\rm{.0}}\), and\({\rm{\Sigma x}}_{\rm{i}}^{\rm{2}}{\rm{ = 16,500}}\), compute the estimates. (Hint:\(\left. {{{{\rm{(\Gamma (1}}{\rm{.2))}}}^{\rm{2}}}{\rm{/\Gamma (1}}{\rm{.4) = }}{\rm{.95}}{\rm{.}}} \right)\)

Suppose a certain type of fertilizer has an expected yield per acre of \({{\rm{\mu }}_{\rm{2}}}\)with variance \({{\rm{\sigma }}^{\rm{2}}}\)whereas the expected yield for a second type of fertilizer is with the same variance \({{\rm{\sigma }}^{\rm{2}}}\).Let \({\rm{S}}_{\rm{1}}^{\rm{2}}\) and \({\rm{S}}_{\rm{2}}^{\rm{2}}\)denote the sample variances of yields based on sample sizes \({{\rm{n}}_{\rm{1}}}\)and \({{\rm{n}}_{\rm{2}}}\),respectively, of the two fertilizers. Show that the pooled (combined) estimator

\({{\rm{\hat \sigma }}^{\rm{2}}}{\rm{ = }}\frac{{\left( {{{\rm{n}}_{\rm{1}}}{\rm{ - 1}}} \right){\rm{S}}_{\rm{1}}^{\rm{2}}{\rm{ + }}\left( {{{\rm{n}}_{\rm{2}}}{\rm{ - 1}}} \right){\rm{S}}_{\rm{2}}^{\rm{2}}}}{{{{\rm{n}}_{\rm{1}}}{\rm{ + }}{{\rm{n}}_{\rm{2}}}{\rm{ - 2}}}}\)

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

Let\({\rm{X}}\)denote the proportion of allotted time that a randomly selected student spends working on a certain aptitude test. Suppose the pdf of\({\rm{X}}\)is

\({\rm{f(x;\theta ) = }}\left\{ {\begin{array}{*{20}{c}}{{\rm{(\theta + 1)}}{{\rm{x}}^{\rm{\theta }}}}&{{\rm{0拢 x拢 1}}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

where\({\rm{ - 1 < \theta }}\). A random sample of ten students yields data\({{\rm{x}}_{\rm{1}}}{\rm{ = }}{\rm{.92,}}{{\rm{x}}_{\rm{2}}}{\rm{ = }}{\rm{.79,}}{{\rm{x}}_{\rm{3}}}{\rm{ = }}{\rm{.90,}}{{\rm{x}}_{\rm{4}}}{\rm{ = }}{\rm{.65,}}{{\rm{x}}_{\rm{5}}}{\rm{ = }}{\rm{.86}}\),\({{\rm{x}}_{\rm{6}}}{\rm{ = }}{\rm{.47,}}{{\rm{x}}_{\rm{7}}}{\rm{ = }}{\rm{.73,}}{{\rm{x}}_{\rm{8}}}{\rm{ = }}{\rm{.97,}}{{\rm{x}}_{\rm{9}}}{\rm{ = }}{\rm{.94,}}{{\rm{x}}_{{\rm{10}}}}{\rm{ = }}{\rm{.77}}\).

a. Use the method of moments to obtain an estimator of\({\rm{\theta }}\), and then compute the estimate for this data.

b. Obtain the maximum likelihood estimator of\({\rm{\theta }}\), and then compute the estimate for the given data.
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