Question

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’s 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

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

Define equations

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

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.