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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

At time \({\rm{t = 0}}\), there is one individual alive in a certain population. A pure birth process then unfolds as follows. The time until the first birth is exponentially distributed with parameter \({\rm{\lambda }}\). After the first birth, there are two individuals alive. The time until the first gives birth again is exponential with parameter \({\rm{\lambda }}\), and similarly for the second individual. Therefore, the time until the next birth is the minimum of two exponential (\({\rm{\lambda }}\)) variables, which is exponential with parameter \({\rm{2\lambda }}\). Similarly, once the second birth has occurred, there are three individuals alive, so the time until the next birth is an exponential \({\rm{rv}}\) with parameter \({\rm{3\lambda }}\), and so on (the memoryless property of the exponential distribution is being used here). Suppose the process is observed until the sixth birth has occurred and the successive birth times are \({\rm{25}}{\rm{.2,41}}{\rm{.7,51}}{\rm{.2,55}}{\rm{.5,59}}{\rm{.5,61}}{\rm{.8}}\) (from which you should calculate the times between successive births). Derive the mle of l. (Hint: The likelihood is a product of exponential terms.)

When the sample standard deviation S is based on a random sample from a normal population distribution, it can be shown that \({\rm{E(S) = }}\sqrt {{\rm{2/(n - 1)}}} {\rm{\Gamma (n/2)\sigma /\Gamma ((n - 1)/2)}}\)

Use this to obtain an unbiased estimator for \({\rm{\sigma }}\) of the form \({\rm{cS}}\). What is \({\rm{c}}\) when \({\rm{n = 20}}\)?

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.

At time \({\rm{t = 0, 20}}\) identical components are tested. The lifetime distribution of each is exponential with parameter \({\rm{\lambda }}\). The experimenter then leaves the test facility unmonitored. On his return \({\rm{24}}\) hours later, the experimenter immediately terminates the test after noticing that \({\rm{y = 15}}\) of the \({\rm{20}}\) components are still in operation (so \({\rm{5}}\) have failed). Derive the mle of \({\rm{\lambda }}\). (Hint: Let \({\rm{Y = }}\) the number that survive \({\rm{24}}\) hours. Then \({\rm{Y}} \sim {\rm{Bin(n,p)}}\). What is the mle of \({\rm{p}}\)? Now notice that \({\rm{p = P(}}{{\rm{X}}_{\rm{i}}} \ge {\rm{24)}}\), where \({{\rm{X}}_{\rm{i}}}\) is exponentially distributed. This relates \({\rm{\lambda }}\) to \({\rm{p}}\), so the former can be estimated once the latter has been.)

\({{\rm{X}}_{\rm{1}}}{\rm{,}}.....{\rm{,}}{{\rm{X}}_{\rm{n}}}\)be a random sample from a gamma distribution with parameters \({\rm{\alpha }}\) and \({\rm{\beta }}\). a. Derive the equations whose solutions yield the maximum likelihood estimators of \({\rm{\alpha }}\) and \({\rm{\beta }}\). Do you think they can be solved explicitly? b. Show that the mle of \({\rm{\mu = \alpha \beta }}\) is \(\widehat {\rm{\mu }}{\rm{ = }}\overline {\rm{X}} \).
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