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Chapter 6: Q31E (page 274)

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 \({\rm{\bar X}}\).)

Short Answer

Expert verified

It is proved.

\({\rm{P(|\bar X - \mu |}} \ge \in {\rm{)}} \le \frac{{{\rm{V(\bar X)}}}}{ \in }{\rm{ = }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{ \in {\rm{ \times }}\sqrt {\rm{n}} }} \to {\rm{0}}\)

Step by step solution

01

Define equations

A mathematical language that asserts that two algebraic expressions must be equal in nature is known as an equation.

02

Explanation

In this instance,

\({\rm{\hat \theta = \bar X}}\)

and the value of the parameter is\({\rm{\theta = \mu }}\). The question is the following true,

\({\rm{P(|\bar X - \mu |}} \ge \in {\rm{)}} \to {\rm{0}}\)

we know that\({\rm{E(\bar X) = \mu }}\)and\({\rm{V(\bar X) = }}{{\rm{\sigma }}^{\rm{2}}}{\rm{/}}\sqrt {\rm{n}} \). As a result of the Chebyshev's inequality,

\({\rm{P(|\bar X - \mu |}} \ge \in {\rm{)}} \le \frac{{{\rm{V(\bar X)}}}}{ \in }{\rm{ = }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{ \in {\rm{ \times }}\sqrt {\rm{n}} }} \to {\rm{0}}\)

when \({\rm{n}} \to \infty \), which brings the proof to a close.

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Most popular questions from this chapter

Let\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)represent a random sample from a Rayleigh distribution with pdf

\({\rm{f(x,\theta ) = }}\frac{{\rm{x}}}{{\rm{\theta }}}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/(2\theta )}}}}\quad {\rm{x > 0}}\)a. It can be shown that\({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ = 2\theta }}\). Use this fact to construct an unbiased estimator of\({\rm{\theta }}\)based on\({\rm{\Sigma X}}_{\rm{i}}^{\rm{2}}\)(and use rules of expected value to show that it is unbiased).

b. Estimate\({\rm{\theta }}\)from the following\({\rm{n = 10}}\)observations on vibratory stress of a turbine blade under specified conditions:

\(\begin{array}{*{20}{l}}{{\rm{16}}{\rm{.88}}}&{{\rm{10}}{\rm{.23}}}&{{\rm{4}}{\rm{.59}}}&{{\rm{6}}{\rm{.66}}}&{{\rm{13}}{\rm{.68}}}\\{{\rm{14}}{\rm{.23}}}&{{\rm{19}}{\rm{.87}}}&{{\rm{9}}{\rm{.40}}}&{{\rm{6}}{\rm{.51}}}&{{\rm{10}}{\rm{.95}}}\end{array}\)

In a random sample of 80 components of a certain type, 12 are found to be defective.

a. Give a point estimate of the proportion of all such components that are not defective.

b. A system is to be constructed by randomly selecting two of these components and connecting them in series, as shown here.

The series connection implies that the system will function if and only if neither component is defective (i.e., both components work properly). Estimate the proportion of all such systems that work properly. (Hint: If p denotes the probability that a component works properly, how can P (system works) be expressed in terms of p ?)

a. Let \({{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\) be a random sample from a uniform distribution on \({\rm{(0,\theta )}}\). Then the mle of \({\rm{\theta }}\) is \({\rm{\hat \theta = Y = max}}\left( {{{\rm{X}}_{\rm{i}}}} \right)\). Use the fact that \({\rm{Y}} \le {\rm{y}}\) if each \({{\rm{X}}_{\rm{i}}} \le {\rm{y}}\) to derive the cdf of Y. Then show that the pdf of \({\rm{Y = max}}\left( {{{\rm{X}}_{\rm{i}}}} \right)\) is \({{\rm{f}}_{\rm{Y}}}{\rm{(y) = }}\left\{ {\begin{array}{*{20}{c}}{\frac{{{\rm{n}}{{\rm{y}}^{{\rm{n - 1}}}}}}{{{{\rm{\theta }}^{\rm{n}}}}}}&{{\rm{0}} \le {\rm{y}} \le {\rm{\theta }}}\\{\rm{0}}&{{\rm{ otherwise }}}\end{array}} \right.\)

b. Use the result of part (a) to show that the mle is biased but that \({\rm{(n + 1)}}\)\({\rm{max}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{/n}}\) is unbiased.

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.

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 \({{\rm{\sigma }}^{\rm{2}}}\)
See all solutions

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