Chapter 9.6, Problem 69E

To determine

Provide an estimator of $\theta$ by the method of moments.

Prove that the estimator is consistent.

Explain whether the estimator is a function of the sufficient statistic $-{\displaystyle \sum _{i=1}^n\ln \left(Y_i\right)}$, that can be obtained from the factorization criterion.

Also provide the implication.

Answer to Problem 69E

The method of moments estimator of $\theta$ is $\frac{2\overline{Y}-1}{1-\overline{Y}}$.

Explanation of Solution

Calculation:

The expectation of the given random variable Y is obtained as follows:

$\begin{array}{c}E\left(Y\right)={\displaystyle \underset{0}{\overset{1}{\int }}y\left(\theta +1\right)y^{\theta }dy}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}\left(\theta +1\right)y^{\theta +1}dy}\\ =\left(\theta +1\right){\left[\frac{y^{\theta +2}}{\theta +2}\right]}_0^1\\ =\frac{\theta +1}{\theta +2}\end{array}$

Consider that the method of moments estimator of $\theta$ is $\widehat{\theta }$.

Now, to obtained the method of moments estimator of $\theta$, it is needed to equate the expectation with the sample mean. That is,

$\begin{array}{c}\frac{\widehat{\theta }+1}{\widehat{\theta }+2}=\overline{Y}\\ \widehat{\theta }+1=\overline{Y}\widehat{\theta }+2\overline{Y}\\ \widehat{\theta }\left(1-\overline{Y}\right)=2\overline{Y}-1\\ \widehat{\theta }=\frac{2\overline{Y}-1}{1-\overline{Y}}\end{array}$

Hence, the method of moments estimator of $\theta$ is $\frac{2\overline{Y}-1}{1-\overline{Y}}$.

Now consider,

$\begin{array}{c}E\left(\overline{Y}\right)=E\left(\frac{\theta +1}{\theta +2}\right)\\ =\frac{\theta +1}{\theta +2}\end{array}$

Thus, $\overline{Y}$ is an unbiased estimator of $\frac{\theta +1}{\theta +2}.$

Now,

$\begin{array}{c}E\left(Y^2\right)={\displaystyle \underset{0}{\overset{1}{\int }}{y}^2\left(\theta +1\right)y^{\theta }dy}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}\left(\theta +1\right)y^{\theta +2}dy}\\ =\left(\theta +1\right){\left[\frac{y^{\theta +3}}{\theta +3}\right]}_0^1\\ =\frac{\theta +1}{\theta +3}\end{array}$

The variance of Y is given as follows:

$\begin{array}{c}V\left(Y\right)=E\left(Y^2\right)-{\left[E\left(Y\right)\right]}^2\\ =\frac{\theta +1}{\theta +3}-\frac{{\left(\theta +1\right)}^2}{{\left(\theta +2\right)}^2}\\ =\frac{\left(\theta +1\right){\left(\theta +2\right)}^2-\left(\theta +3\right){\left(\theta +1\right)}^2}{{\left(\theta +2\right)}^2\left(\theta +3\right)}\\ =\frac{\left(\theta +1\right)\left({\theta }^2+2\theta +4\right)-\left(\theta +3\right)\left({\theta }^2+2\theta +1\right)}{{\left(\theta +2\right)}^2\left(\theta +3\right)}\\ =\frac{{\theta }^3+2{\theta }^2+4\theta +{\theta }^2+2\theta +4-{\theta }^3-2{\theta }^2-\theta -3{\theta }^2-6\theta -3}{{\left(\theta +2\right)}^2\left(\theta +3\right)}\\ =\frac{1-2{\theta }^2-\theta }{{\left(\theta +2\right)}^2\left(\theta +3\right)}\end{array}$

The variance of $\overline{Y}$ is given as follows:

$\begin{array}{c}V\left(\overline{Y}\right)=V\left(\frac{1}{n}{\displaystyle \sum _{i=1}^{n}Y}\right)\\ =\frac{1}{n^2}V\left({\displaystyle \sum _{i=1}^{n}Y}\right)\\ =\frac{1}{n^2}{\displaystyle \sum _{i=1}^{n}V\left(Y\right)}\left(\because Y_i\text{'}\text{s are independent}\right)\\ \text{=}\frac{1}{n^2}\left[\frac{1-2{\theta }^2-\theta }{{\left(\theta +2\right)}^2\left(\theta +3\right)}+\mathrm{...}+\frac{1-2{\theta }^2-\theta }{{\left(\theta +2\right)}^2\left(\theta +3\right)}\left(n\text{ times}\right)\right]\\ =\frac{n}{n^2}\left(\frac{1-2{\theta }^2-\theta }{{\left(\theta +2\right)}^2\left(\theta +3\right)}\right)\\ =\frac{1-2{\theta }^2-\theta }{n{\left(\theta +2\right)}^2\left(\theta +3\right)}\\ \underset{n\to \infty }{\lim }V\left(\overline{Y}\right)=\underset{n\to \infty }{\lim }\frac{1-2{\theta }^2-\theta }{n{\left(\theta +2\right)}^2\left(\theta +3\right)}\\ =0\end{array}$

Thus, the estimator $\overline{Y}$ is a consistent estimator of $\frac{\theta +1}{\theta +2}.$

Now, using Law of large numbers it can be aid that $\overline{Y}$ converges in probability to $\frac{\theta +1}{\theta +2}.$

That is,

$\begin{array}{c}\widehat{\theta }=\frac{2\overline{Y}-1}{1-\overline{Y}}\\ \widehat{\theta }\to \frac{2\left(\frac{\theta +1}{\theta +2}\right)-1}{1-\left(\frac{\theta +1}{\theta +2}\right)}\\ \widehat{\theta }\to \theta \end{array}$

Hence, $\widehat{\theta }$ converges to $\theta$.

The likelihood function of α can be written as follows:

$\begin{array}{c}L\left(\alpha \right)={\left(\theta +1\right)}^n{\left({\displaystyle \prod _{i=1}^n{y}_i}\right)}^{\theta }\\ =h\left(y\right)g\left({\displaystyle \prod _{i=1}^n{y}_i},\theta \right)\end{array}$

Where $h\left(y\right)=1$ and $g\left({\displaystyle \prod _{i=1}^n{y}_i},\theta \right)={\left(\theta +1\right)}^n{\left({\displaystyle \prod _{i=1}^n{y}_i}\right)}^{\theta }$.

By Theorem 9.4 (Factorization theorem), it can be said that, ${\displaystyle \prod _{i=1}^n{Y}_i}$ is sufficient for $\theta$.

Hence, the estimator is estimator is not a function of the sufficient statistic $-{\displaystyle \sum _{i=1}^n\ln \left(Y_i\right)}$. Moreover, it is the function of ${\displaystyle \prod _{i=1}^n{Y}_i}$.

Therefore, it implies that it is not a minimum variance unbiased estimator.