Chapter 9.7, Problem 85E

Let Y₁, Y₂,…, Y_n denote a random sample from the density function given by

$f(y|\alpha ,\theta )=\{\begin{array}{ll}\left(\frac{1}{\Gamma (\alpha ){\theta }^{\alpha }}\right)y^{\alpha -1}{e}^{-y/\theta },\hfill & y>0,\hfill \\ 0,\hfill & \text{elsewhere}.\hfill \end{array}$

where α > 0 is known.

1. a Find the MLE $\hat{\theta }$ of θ.
2. b Find the expected value and variance of $\hat{\theta }$.
3. c Show that $\hat{\theta }$ is consistent for θ.
4. d What is the best (minimal) sufficient statistic for θ in this problem?
5. e Suppose that n = 5 and α = 2. Use the minimal sufficient statistic to construct a 90% confidence interval for θ. [Hint: Transform to a χ² distribution.]

To determine

Provide the MLE $\widehat{\theta }$ for $\theta$.

Answer to Problem 85E

The MLE $\widehat{\theta }$ for $\theta$ is $\underset{\_}{\frac{\overline{Y}}{a}}.$

Explanation of Solution

Calculation:

The likelihood function of $\theta$ is obtained as follows:

$\begin{array}{c}L\left(\theta \right)=L\left(Y_1,Y_2,\mathrm{...},Y_n|\theta \right)\\ ={\displaystyle \prod _{i=1}^n\frac{1}{\Gamma \alpha {\theta }^{\alpha }}{Y}_i^{\alpha -1}{e}^{-\frac{Y_i}{\theta }}}\\ =e^{-\frac{{\displaystyle \sum _{i=1}^n{Y}_i}}{\theta }}\frac{{\displaystyle \prod _{i=1}^n{Y}_i^{\alpha -1}}}{{\left(\Gamma \alpha \right)}^n{\theta }^{n\alpha }}\end{array}$

The log likelihood function is obtained as follows:

$\begin{array}{c}l\left(\theta \right)=\log \left(e^{-\frac{{\displaystyle \sum _{i=1}^n{Y}_i}}{\theta }}\frac{{\displaystyle \prod _{i=1}^n{Y}_i^{\alpha -1}}}{{\left(\Gamma \alpha \right)}^n{\theta }^{n\alpha }}\right)\\ =-n\log \left(\Gamma \alpha \right)-n\alpha \log \left(\theta \right)+\left(\alpha -1\right){\displaystyle \sum _{i=1}^n\log Y_i}-\frac{{\displaystyle \sum _{i=1}^n{Y}_i}}{\theta }\end{array}$

Obtain the first order derivative of the log likelihood function and equate it with 0.

Hence,

$\begin{array}{c}\frac{d\left(l\left(\theta \right)\right)}{d\theta }=0\\ \frac{d\left(-n\log \left(\Gamma \alpha \right)-n\alpha \log \left(\theta \right)+\left(\alpha -1\right){\displaystyle \sum _{i=1}^n\log Y_i}-\frac{{\displaystyle \sum _{i=1}^n{Y}_i}}{\theta }\right)}{d\theta }=0\\ -\frac{n\alpha }{\theta }+\frac{{\displaystyle \sum _{i=1}^n{Y}_i}}{{\theta }^2}=0\\ \widehat{\theta }={\displaystyle \sum _{i=1}^n\frac{Y_i}{n\alpha }}\\ =\frac{\overline{Y}}{\alpha }\end{array}$

Thus, the MLE $\widehat{\theta }$ for $\theta$ is $\underset{\_}{\frac{\overline{Y}}{a}}.$

To determine

Compute the expected value and variance of $\widehat{\theta }$.

Answer to Problem 85E

The expected value and variance of $\widehat{\theta }$ are $\underset{\_}{\theta }\text{and }\underset{\_}{\frac{{\theta }^2}{\alpha n}}$, respectively.

Explanation of Solution

Calculation:

It is known that if a random variable Y follows Gamma distribution with parameters $\alpha$ and $\beta$, then the expected value and the variance of Y are $\alpha \beta \text{and }\frac{\alpha {\beta }^2}{n}$, respectively.

From the Part (a), the expression of $\widehat{\theta }$ is obtained as $\widehat{\theta }=\frac{\overline{Y}}{\alpha }$.

Thus, the value of $E\left(\widehat{\theta }\right)$ is obtained as follows:

$\begin{array}{c}E\left(\widehat{\theta }\right)=E\left(\frac{\overline{Y}}{\alpha }\right)\\ =\frac{E\left(\overline{Y}\right)}{\alpha }\\ =\frac{\alpha \theta }{\alpha }\\ =\theta \end{array}$

Similarly, the value of $V\left(\widehat{\theta }\right)$ is obtained as follows:

$\begin{array}{c}V\left(\widehat{\theta }\right)=V\left(\frac{\overline{Y}}{\alpha }\right)\\ =\frac{V\left(\overline{Y}\right)}{{\alpha }^2}\\ =\frac{\alpha {\theta }^2}{{\alpha }^{2}n}\\ =\frac{{\theta }^2}{\alpha n}\end{array}$

Thus, the expected value and variance of $\widehat{\theta }$ are $\underset{\_}{\theta }\text{and }\underset{\_}{\frac{{\theta }^2}{\alpha n}}$, respectively.

To determine

Prove that $\widehat{\theta }$ is an consistent estimator of $\theta$.

Explanation of Solution

Calculation:

From Part (b), it is obtained that the expected value and variance of $\widehat{\theta }$ are $\theta \text{and }\frac{{\theta }^2}{\alpha n}$, respectively.

Hence, for any positive number $\epsilon$ it is obtained that,

$\begin{array}{c}\underset{n\to \infty }{\lim }P\left(\left|\widehat{\theta }-\theta \right|>\epsilon \right)=P\left(\left|\overline{Y}-\theta \right|>\epsilon \right)\\ \le \underset{n\to \infty }{\lim }\frac{V\left(\overline{Y}\right)}{{\epsilon }^2}\\ =\underset{n\to \infty }{\lim }\frac{{\theta }^2}{n\alpha {\epsilon }^2}\\ \to 0\end{array}$

According to Definition 9.2, it can be said that $\widehat{\theta }$ is an consistent estimator of $\theta$.

To determine

Find the best (minimal) sufficient statistic for $\theta$ in this problem.

Answer to Problem 85E

The best (minimal) sufficient statistic for $\theta$ in this problem is $\underset{\_}{{\displaystyle \sum _{i=1}^n{Y}_i}}$.

Explanation of Solution

Calculation:

From Part (a), the likelihood function of $\theta$ is obtained as follows:

$\begin{array}{c}L\left(\theta \right)=e^{-\frac{{\displaystyle \sum _{i=1}^n{Y}_i}}{\theta }}\frac{{\displaystyle \prod _{i=1}^n{Y}_i^{\alpha -1}}}{{\left(\Gamma \alpha \right)}^n{\theta }^{n\alpha }}\\ =g\left(u,\theta \right)h\left(Y_1,Y_2,\mathrm{...},Y_n\right)\end{array}$

Here, $g\left(u,\theta \right)=e^{-\frac{{\displaystyle \sum _{i=1}^n{Y}_i}}{\theta }}\frac{1}{{\left(\Gamma \alpha \right)}^n{\theta }^{n\alpha }}$,  $h\left(Y_1,Y_2,\mathrm{...},Y_n\right)={\displaystyle \prod _{i=1}^n{Y}_i^{\alpha -1}}$ and $u={\displaystyle \sum _{i=1}^n{Y}_i}$.

Thus, according to Theorem 9.4 (factorization criterion), it can be said that the sufficient statistic for $\theta$ is ${\displaystyle \sum _{i=1}^n{Y}_i}$.

Consider another sample of random variables, such that, $X_1,X_2,\mathrm{...},X_n$, that follows Gamma distribution with parameters $\alpha \text{and }\theta$.

Hence, the ration of the likelihood function for this two samples are:

 $\begin{array}{c}\frac{L\left(Y_1,Y_2,\mathrm{...},Y_n|\theta \right)}{L\left(X_1,X_2,\mathrm{...},X_n|\theta \right)}=\frac{e^{-\frac{{\displaystyle \sum _{i=1}^n{Y}_i}}{\theta }}\frac{{\displaystyle \prod _{i=1}^n{Y}_i^{\alpha -1}}}{{\left(\Gamma \alpha \right)}^n{\theta }^{n\alpha }}}{e^{-\frac{{\displaystyle \sum _{i=1}^n{X}_i}}{\theta }}\frac{{\displaystyle \prod _{i=1}^n{X}_i^{\alpha -1}}}{{\left(\Gamma \alpha \right)}^n{\theta }^{n\alpha }}}\\ =\frac{{\displaystyle \prod _{i=1}^n{Y}_i^{\alpha -1}}}{{\displaystyle \prod _{i=1}^n{X}_i^{\alpha -1}}}{e}^{-\left(\frac{{\displaystyle \sum _{i=1}^n{Y}_i}}{\theta }-\frac{{\displaystyle \sum _{i=1}^n{X}_i}}{\theta }\right)}\end{array}$

The above ratio will be independent of θ if and only if the term in the power of exponential term is equal to 0.

That is,

$\frac{{\displaystyle \sum _{i=1}^n{Y}_i}}{\theta }-\frac{{\displaystyle \sum _{i=1}^n{X}_i}}{\theta }=0\text{ or }{\displaystyle \sum _{i=1}^n{Y}_i}={\displaystyle \sum _{i=1}^n{X}_i}$.

By the method of Lehmann-Scheffe, $g\left(Y_1,Y_2,\cdots ,Y_n\right)={\displaystyle \sum _{i=1}^n{Y}_i}$.

Therefore, the best (minimal) sufficient statistic for $\theta$ in this problem is $\underset{\_}{{\displaystyle \sum _{i=1}^n{Y}_i}}$.

To determine

Construct a 90% confidence interval of $\theta$ for $n=5\text{and }\alpha =2$.

Answer to Problem 85E

The 90% confidence interval of $\theta$ for $n=5\text{and }\alpha =2$ is $\left(\frac{{\displaystyle \sum _{i=1}^5{Y}_i}}{15.7052},\frac{{\displaystyle \sum _{i=1}^5{Y}_i}}{5.4254}\right)$.

Explanation of Solution

Calculation:

From Part (d), it is obtained that the minimal sufficient statistic for $\theta$ is ${\displaystyle \sum _{i=1}^n{Y}_i}$.

According to question, $Y_i~\Gamma \left(2,\theta \right)$.

As, $Y_i\text{'}\text{s}$ are independent and identically distributed it can be said that,

$\begin{array}{c}{\displaystyle \sum _{i=1}^5{Y}_i}~\Gamma \left(2\times 5,\theta \right)\\ \frac{2{\displaystyle \sum _{i=1}^5{Y}_i}}{\theta }~\Gamma \left(10,\theta \right)\\ \frac{2{\displaystyle \sum _{i=1}^5{Y}_i}}{\theta }~{\chi }_{20}^2\end{array}$

For 90% confidence interval the level of significate $\alpha$ is $1-0.90=0.10$. It implies that there will be 5% area beyond both sides of the confidence limits.

From Table 6 Percentage Points of the ${\chi }^2$ Distribution the chi square critical value corresponding to level of significance of 0.05 and degrees of freedom 20 is obtained as ${\chi }_{0.05,20}^2=31.4104$ and the chi square critical value corresponding to level of significance of 0.95 and degrees of freedom 20 is obtained as ${\chi }_{0.95,20}^2=10.8508$.

The required 90% confidence interval is obtained as follows:

$\begin{array}{c}P\left({\chi }_{1-\frac{\alpha }{2},20}^2\le \frac{2{\displaystyle \sum _{i=1}^5{Y}_i}}{\theta }\le {\chi }_{\frac{\alpha }{2},20}^2\right)=0.90\\ P\left(\frac{1}{{\chi }_{0.95,20}^2}\ge \frac{\theta }{2{\displaystyle \sum _{i=1}^5{Y}_i}}\ge \frac{1}{{\chi }_{0.05,20}^2}\right)=0.9\\ P\left(\frac{2{\displaystyle \sum _{i=1}^5{Y}_i}}{{\chi }_{0.05,20}^2}\le \theta \le \frac{2{\displaystyle \sum _{i=1}^5{Y}_i}}{{\chi }_{0.95,20}^2}\right)=0.9\\ P\left(\frac{2{\displaystyle \sum _{i=1}^5{Y}_i}}{31.4104}\le \theta \le \frac{2{\displaystyle \sum _{i=1}^5{Y}_i}}{10.8508}\right)=0.9\\ P\left(\frac{{\displaystyle \sum _{i=1}^5{Y}_i}}{15.7052}\le \theta \le \frac{{\displaystyle \sum _{i=1}^5{Y}_i}}{5.4254}\right)=0.9\end{array}$

Thus, the 90% confidence interval of $\theta$ for $n=5\text{and }\alpha =2$ is $\left(\frac{{\displaystyle \sum _{i=1}^5{Y}_i}}{15.7052},\frac{{\displaystyle \sum _{i=1}^5{Y}_i}}{5.4254}\right)$.