Question

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.

Short answer

(a) It is shown that \({{\rm{f}}_{\rm{Y}}}{\rm{(y) = F}}_{\rm{Y}}'{\rm{(y) = }}\frac{{{\rm{n}}{{\rm{y}}^{{\rm{n - 1}}}}}}{{{{\rm{\theta }}^{\rm{n}}}}}{\rm{, 0}} \le {\rm{y}} \le {\rm{\theta }}\).

(b) It is shown that \({\rm{E(}}\widetilde {\rm{Y}}{\rm{) = \theta }}\).

Step-by-step solution

Define uniform distribution

A uniform distribution is a probability distribution in which every event inside a certain interval has the same probability of occurring. It's a graphical representation of a set of data in the form of a graph or a list.

Explanation

(a) The cdf of random variable Y can be calculated in the following manner.

\(\begin{array}{l}{{\rm{F}}_{\rm{Y}}}{\rm{(y) = P(Y}} \le {\rm{y)}}\\{\rm{ = P}}\left( {{\rm{max}}\left( {{{\rm{X}}_{\rm{i}}}} \right) \le {\rm{y}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{P}}\left( {{{\rm{X}}_{\rm{1}}} \le {\rm{y,}}{{\rm{X}}_{\rm{2}}} \le {\rm{y, \ldots ,}}{{\rm{X}}_{\rm{n}}} \le {\rm{y}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(2)}}} {\rm{P}}\left( {{{\rm{X}}_{\rm{1}}} \le {\rm{y}}} \right){\rm{ \times P}}\left( {{{\rm{X}}_{\rm{2}}} \le {\rm{y}}} \right){\rm{ \times \ldots \times P}}\left( {{{\rm{X}}_{\rm{n}}} \le {\rm{y}}} \right)\\\mathop {\rm{ = }}\limits^{{\rm{(3)}}} {\left( {\frac{{\rm{y}}}{{\rm{\theta }}}} \right)^{\rm{n}}}{\rm{,0}} \le {\rm{y}} \le {\rm{\theta ,}}\end{array}\)

(1): if maximum is less than\({\rm{y}}\), then all\({{\rm{X}}_{\rm{i}}}\)are less.

(2): as a result of one's independence,

(3): a uniform distribution cdf on\(\left( {{\rm{0,\theta }}} \right)\).

It is simple to obtain pdf as a derivative of cdf if you have cdf,

\({{\rm{f}}_{\rm{Y}}}{\rm{(y) = F}}_{\rm{Y}}'{\rm{(y) = }}\frac{{{\rm{n}}{{\rm{y}}^{{\rm{n - 1}}}}}}{{{{\rm{\theta }}^{\rm{n}}}}}{\rm{, 0}} \le {\rm{y}} \le {\rm{\theta }}\)

Otherwise, it is zero.

Explanation

(b) The estimator is unbiased if \({\rm{E(Y) = \theta }}\), however

\(\begin{array}{c}{\rm{E(Y) = }}\int_{\rm{0}}^{\rm{\theta }} {\rm{y}} {\rm{ \times }}\frac{{{\rm{n}}{{\rm{y}}^{{\rm{n - 1}}}}}}{{{{\rm{\theta }}^{\rm{n}}}}}{\rm{dy}}\\{\rm{ = }}\left. {\frac{{\rm{n}}}{{{{\rm{\theta }}^{\rm{n}}}}}\frac{{{{\rm{y}}^{{\rm{n + 1}}}}}}{{{\rm{n + 1}}}}} \right|_{\rm{0}}^{\rm{\theta }}\\{\rm{ = }}\frac{{\rm{n}}}{{{\rm{n + 1}}}}{\rm{\theta }} \ne {\rm{\theta }}\end{array}\)

This indicates that the estimator isn't completely objective. Estimator,

however,

\({\rm{\tilde Y = }}\frac{{{\rm{n + 1}}}}{{\rm{n}}}\)

is objective because

\(\begin{array}{c}{\rm{E(\tilde Y) = E}}\left( {\frac{{{\rm{n + 1}}}}{{\rm{n}}}} \right)\\{\rm{ = }}\frac{{{\rm{n + 1}}}}{{\rm{n}}}{\rm{E(Y)}}\\{\rm{ = }}\frac{{{\rm{n + 1}}}}{{\rm{n}}}\frac{{\rm{n}}}{{{\rm{n + 1}}}}{\rm{\theta }}\\{\rm{ = \theta }}\end{array}\)

This indicates that the estimator \({\rm{\tilde Y}}\) is unbiased.