Question

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:

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

Short answer

a) The estimator is unbiased estimator\(\frac{{\sum {{\rm{X}}_{\rm{i}}^{\rm{2}}} }}{{{\rm{2n}}}}\).

b) The estimated value is\({\rm{\hat \theta = 74}}{\rm{.505}}\).

Step-by-step solution

Definition

An estimator is a rule for computing an estimate of a given quantity based on observable data: the rule (estimator), the quantity of interest (estimate), and the output (estimate) are all distinct.

Proofing estimator unbiased

a)

This part is practically solved-because of,

\({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ = 2\theta }}\)

It is easy to notice that the unbiased estimator of parameter \(\theta \)is

\(\frac{{\sum {{\rm{X}}_{\rm{i}}^{\rm{2}}} }}{{{\rm{2n}}}}\)

The proof of this claim follows

\(\begin{aligned}E(\hat \theta ) &= E\left( {\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{X}}_{\rm{i}}^{\rm{2}}} }}{{{\rm{2n}}}}} \right)\\ &= \frac{{\rm{1}}}{{{\rm{2n}}}}{\rm{E}}\left( {\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{X}}_{\rm{i}}^{\rm{2}}} } \right)\\ & = \frac{{\rm{1}}}{{{\rm{2n}}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{E}} \left( {{\rm{X}}_{\rm{i}}^{\rm{2}}} \right)\\&= \frac{{\rm{1}}}{{{\rm{2n}}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{2}} {\rm{\theta }}\\ &= \frac{{\rm{1}}}{{\rm{n}}}{\rm{ \times n \times \theta }}\\ &= \theta \end{aligned}\)

Which, proofs that the estimator is unbiased for\({\rm{\theta }}\).

Finding the value of \({\rm{\delta }}\)

b)

Given \({\rm{n = 10}}\)observations, the sum of squared data is:

\(\begin{aligned}\sum\limits_{{\rm{i = 1}}}^{{\rm{10}}} {{\rm{x}}_{\rm{i}}^{\rm{2}}} &= 16{\rm{.8}}{{\rm{8}}^{\rm{2}}}{\rm{ + 10}}{\rm{.2}}{{\rm{3}}^{\rm{2}}}{\rm{ + \ldots + 10}}{\rm{.9}}{{\rm{5}}^{\rm{2}}}\\ &= 1490{\rm{.1058}}\end{aligned}\)

Therefore, the estimate is\({\rm{\hat \theta = }}\frac{{\rm{1}}}{{{\rm{20}}}}{\rm{ \times 1490}}{\rm{.1058 = 74}}{\rm{.505}}\).