Chapter 9, Problem 103SE

a.

To determine

Find the MLE of θ.

Answer to Problem 103SE

TheMLE of θ is, $\widehat{\theta }=\frac{1}{n}{\displaystyle \sum _{i=1}^n{Y}_i^2}$.

Explanation of Solution

The density function of Rayleigh’s distribution is given as follows:

$f\left(y\right)=\{\begin{array}{l}\left(\frac{2y}{\theta }\right)e^{-\frac{y^2}{\theta }}\text{ ; }y>0,\\ 0\text{ ; Otherwise}\text{.}\end{array}$

From Exercise 9.82, the MLE of θ for Weibull’s distribution is given as $\widehat{\theta }={\displaystyle \sum _{i=1}^n\frac{Y_i^r}{n}}$.

It is known that the Rayleigh’s distribution is a special case of Weibull’s distribution with $r=2$.

Therefore, by substituting $r=2$ in the MLE of θ for Weibull’s distribution, one can get the MLE of θ for Rayleigh’sdistribution.

Hence, the MLE of θ for Rayleigh’sdistribution is obtained as follows:

$\begin{array}{c}\widehat{\theta }={\displaystyle \sum _{i=1}^n\frac{Y_i^r}{n}}\\ =\frac{1}{n}{\displaystyle \sum _{i=1}^n{Y}_i^2}\left[\because r=2\right]\end{array}$

Thus, the MLE of θ is, $\widehat{\theta }=\frac{1}{n}{\displaystyle \sum _{i=1}^n{Y}_i^2}$.

b.

To determine

Find the asymptotic variance of the MLEof θ.

Answer to Problem 103SE

The asymptotic variance of the MLE is $\frac{1}{E\left[-\frac{{\partial }^{2}l}{\partial {\theta }^2}\right]}=\frac{{\theta }^2}{n}$ .

Explanation of Solution

Under the regulatory conditions, the formula for asymptotic variance is given as follows:

$V\left(\widehat{p}\right)=\frac{1}{E\left[-\frac{{\partial }^{2}l}{\partial p^2}\right]}$

The log-likelihood function of θ is obtained as follows:

$\begin{array}{c}L\left(\theta \right)={\displaystyle \prod _{i=1}^n\left(\frac{2y}{\theta }\right)\exp \left(-\frac{y^2}{\theta }\right)}\\ =2^n{\theta }^{-n}{\displaystyle \prod _{i=1}^n{Y}_i}\exp \left(-\frac{{\displaystyle \sum _{i=1}^n{Y}_i^2}}{\theta }\right)\\ l=n\log \left(2\right)-n\log \left(\theta \right)+\log \left({\displaystyle \sum _{i=1}^n{Y}_i}\right)-\frac{1}{\theta }\left({\displaystyle \sum _{i=1}^n{Y}_i^2}\right)\end{array}$

By obtaining the first- and second-order partial derivations of the log-likelihood function with respect to θ, the asymptotic variance is found as follows:

$\begin{array}{c}\frac{\partial l}{\partial \theta }=0-\frac{n}{\theta }+0+\frac{1}{{\theta }^2}\left({\displaystyle \sum _{i=1}^n{Y}_i^2}\right)\\ \frac{{\partial }^{2}l}{\partial {\theta }^2}=\frac{n}{{\theta }^2}-\frac{1}{{\theta }^3}\left(2{\displaystyle \sum _{i=1}^n{Y}_i^2}\right)\\ E\left[-\frac{{\partial }^{2}l}{\partial {\theta }^2}\right]=-\frac{n}{{\theta }^2}+\frac{1}{{\theta }^3}\left(2{\displaystyle \sum _{i=1}^{n}E\left(Y_i^2\right)}\right)\cdots \cdots \left(1\right)\end{array}$

Consider,

$\begin{array}{c}E\left(Y^2\right)={\displaystyle {\int }_0^{\infty }{y}^2\left(\frac{2y}{\theta }\right)e^{-\frac{y^2}{\theta }}dy}\\ \left[Let\frac{y^2}{\theta }=t\Rightarrow 2ydy=\theta dt\right]\\ E\left(Y^2\right)=\theta {\displaystyle {\int }_0^{\infty }t{e}^{-t}dt}\\ E\left(Y^2\right)=\theta {\displaystyle {\int }_0^{\infty }t{e}^{-t}dt}\left[\because {\displaystyle {\int }_0^{\infty }{x}^{n-1}{e}^{-x}dx}=\Gamma \left(n\right)\right]\\ =\theta \left(\Gamma \left(2\right)\right)\\ =\theta \cdots \cdots \left(2\right)\end{array}$

By substituting Equation $\left(2\right)$ in Equation $\left(1\right)$,

$\begin{array}{c}E\left[-\frac{{\partial }^{2}l}{\partial {\theta }^2}\right]=-\frac{n}{{\theta }^2}+\frac{1}{{\theta }^3}\left(2n\theta \right)\\ =\frac{n}{{\theta }^2}\\ \frac{1}{E\left[-\frac{{\partial }^{2}l}{\partial {\theta }^2}\right]}=\frac{{\theta }^2}{n}\end{array}$

Therefore, the asymptotic variance of the MLE is $\frac{1}{E\left[-\frac{{\partial }^{2}l}{\partial {\theta }^2}\right]}=\frac{{\theta }^2}{n}$.