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Question: Suppose that \({{\bf{X}}_{\bf{1}}}{\bf{, }}{\bf{. }}{\bf{. }}{\bf{. , }}{{\bf{X}}_{\bf{n}}}\)form a random sample ofsize n from the uniform distribution on the interval \(\left( {{\bf{0,\theta }}} \right)\),where the value of \({\bf{\theta }}\) is unknown. Show that the sequence of M.L.E.’s of \({\bf{\theta }}\) is a consistent sequence.

Short Answer

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Step by step solution

01

Define the pdf of Uniform Distribution 

The pdf of a uniform distribution is obtained by using the formula:\(\frac{1}{{b - a}};a \le x \le b\).

Therefore, the PDF of X is expressed as,

\({f_x} = \left\{ \begin{array}{l}\frac{1}{{\theta - 0}} = \frac{1}{\theta }\;\;\;\;\;\;\;\;\;\;0 \le x \le \theta \\0;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;otherwise\end{array} \right.\)

02

Define the order statistics and MLE

Here,\({X_1} \ldots {X_n}\)is a random sample from uniform distribution\(U\left( {0,\theta } \right)\).

\(\begin{array}{l}{X_{\left( 1 \right)}} = \min \left\{ {{X_1} \ldots {X_n}} \right\}\\{X_{\left( n \right)}} = \max \left\{ {{X_1} \ldots {X_n}} \right\}\end{array}\)

Note that the sequence of the random variables would be as follows,

\(0 \le {X_{\left( 1 \right)}} \le \ldots \le {X_{\left( n \right)}} \le \theta \)

Defining the likelihood function:

\(\begin{array}{c}L\left( \theta \right) = \prod\limits_{i = 1}^n {f\left( {{x_i}} \right)} \\ = \frac{1}{{{\theta ^n}}},0 \le x \le \theta \end{array}\)

The log-likelihood is:

\(\begin{array}{c}\ln L\left( \theta \right) = \ln \left( {\frac{1}{{{\theta ^n}}}} \right)\\ = \ln {\left( {{\theta ^n}} \right)^{ - 1}}\\ = - n\ln \left( \theta \right)\end{array}\)

Setting its derivative with respect to parameter

\(\begin{array}{c}\frac{\partial }{{\partial \theta }}\ln L\left( \theta \right) = - \frac{\partial }{{\partial \theta }}n\ln \left( \theta \right)\\ = - \frac{n}{\theta }\end{array}\)

By equating the equation to 0, the parameter vanishes, hence we cannot use this approach.

The log likelihood function is a decreasing function. Also,\(0 \le {X_{\left( 1 \right)}} \le \ldots \le {X_{\left( n \right)}} \le \theta \).

The log likelihood function\( - \frac{n}{\theta }\)is maximised at\(\theta = {X_{\left( n \right)}}\)since as value of denominator increases the fraction value decreases. Due to the negative sign, the value becomes maximum.

Therefore, the M.L.E. is

\(\widehat \theta = {X_{\left( n \right)}}\), which lies in the parameter space.

Therefore, M.L.E. is\(\widehat {\bf{\theta }}{\bf{ = }}{{\bf{X}}_{\left( {\bf{n}} \right)}}\)

03

To show MLE is a consistent estimator

\(\begin{array}{c}P\left( {\left| {{X_{\left( n \right)}} - \theta } \right| > \varepsilon } \right) = P\left( {{X_{\left( n \right)}} \ge \theta + \varepsilon } \right) + P\left( {{X_{\left( n \right)}} \le \theta - \varepsilon } \right)\\ = \left\{ \begin{array}{l}{\left( {\frac{{\theta - \varepsilon }}{\theta }} \right)^n}\,\,if\,\varepsilon < \theta \,\\0\,\,if\,\,\varepsilon \ge \theta \end{array} \right.\\ = 0\,\,as\,\,n \to \infty \end{array}\)

Therefore, it is a consistent estimator.

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