Chapter 9.3, Problem 31E

To determine

Prove that $\overline{Y}$ converges in probability to some constant and obtain the constant.

Explanation of Solution

It is given that $Y_1,Y_2,\mathrm{...},Y_n$ denote independent variables. The probability density function of each random variable is $\Gamma \left(\alpha ,\beta \right)$. PDF is given as follows:

$f\left(y\right)=\{\begin{array}{l}\frac{y^{\alpha -1}{e}^{\frac{-y}{\beta }}}{{\beta }^{\alpha }\Gamma \left(a\right)}\text{ ; }0\le y,\\ 0\text{ ; Otherwise}\text{.}\end{array}$

To prove that $\overline{Y}$ converges in probability, it is necessary to show that $\overline{Y}$ is an unbiased estimator.

Hence,

$\begin{array}{c}E\left(Y\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}y\frac{y^{\alpha -1}{e}^{\frac{-y}{\beta }}}{{\beta }^{\alpha }\Gamma \left(a\right)}}dy\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{y^{\alpha }{e}^{\frac{-y}{\beta }}}{{\beta }^{\alpha }\Gamma \left(a\right)}}dy\left[Put\frac{y}{\beta }=t\Rightarrow dy=\beta dt\right]\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{{\beta }^{\alpha }{t}^{\alpha }{e}^{-t}}{{\beta }^{\alpha }\Gamma \left(a\right)}}\beta dt\\ =\frac{\beta }{\Gamma \left(a\right)}{\displaystyle \underset{0}{\overset{\infty }{\int }}{t}^{\alpha }{e}^{-t}}dt\left[\because \Gamma \left(z\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{t}^{z-1}{e}^{-t}}dt\right]\\ =\beta \frac{\Gamma \left(a+1\right)}{\Gamma \left(a\right)}\left[\because \Gamma \left(z+1\right)=z\Gamma \left(z\right)\right]\\ =\beta \frac{a\Gamma \left(a\right)}{\Gamma \left(a\right)}\\ =\alpha \beta \end{array}$

Again,

$\begin{array}{c}E\left(\overline{Y}\right)=E\left(\frac{1}{n}{\displaystyle \sum _{i=1}^n{Y}_i}\right)\\ =\frac{1}{n}\left(E{\displaystyle \sum _{i=1}^n{Y}_i}\right)\\ \text{=}\frac{1}{n}\left({\displaystyle \sum _{i=1}^{n}E\left(Y_i\right)}\right)\left(\because Y_i\text{'}\text{s are identically distributed}\right)\\ \text{=}\frac{1}{n}\left(\alpha \beta +\alpha \beta +\cdots +\alpha \beta \left(n\text{ terms}\right)\right)\\ =\frac{n}{n}\left(\alpha \beta \right)\\ =\alpha \beta \end{array}$

Thus, $\overline{Y}$ is an unbiased estimator of $\alpha \beta$.

The claim is that $\overline{Y}$ converges in probability to $\alpha \beta$. It implies that $\overline{Y}$ is a consistent estimator of $\alpha \beta$.

Hence,

$\begin{array}{c}E\left(Y\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{y}^2\frac{y^{\alpha -1}{e}^{\frac{-y}{\beta }}}{{\beta }^{\alpha }\Gamma \left(a\right)}}dy\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{y^{\alpha +1}{e}^{\frac{-y}{\beta }}}{{\beta }^{\alpha }\Gamma \left(a\right)}}dy\left[Put\frac{y}{\beta }=t\Rightarrow dy=\beta dt\right]\\ ={\displaystyle \underset{0}{\overset{\infty }{\int }}\frac{{\beta }^{\alpha +1}{t}^{\alpha +1}{e}^{-t}}{{\beta }^{\alpha }\Gamma \left(a\right)}}\beta dt\\ =\frac{{\beta }^2}{\Gamma \left(a\right)}{\displaystyle \underset{0}{\overset{\infty }{\int }}{t}^{\alpha +1}{e}^{-t}}dt\left[\because \Gamma \left(z\right)={\displaystyle \underset{0}{\overset{\infty }{\int }}{t}^{z-1}{e}^{-t}}dt\right]\\ ={\beta }^2\frac{\Gamma \left(a+2\right)}{\Gamma \left(a\right)}\left[\because \Gamma \left(z+1\right)=z\Gamma \left(z\right)\right]\\ ={\beta }^2\frac{a\left(a+1\right)\Gamma \left(a\right)}{\Gamma \left(a\right)}\\ =\alpha \left(a+1\right){\beta }^2\end{array}$

The variance is calculated as follows:

$\begin{array}{c}V\left(Y\right)=E\left(Y^2\right)-{\left[E\left(Y\right)\right]}^2\\ =\alpha \left(a+1\right){\beta }^2-{\left(\alpha \beta \right)}^2\\ =\left(a^2+a\right){\beta }^2-{\left(\alpha \beta \right)}^2\\ ={\left(\alpha \beta \right)}^2+\alpha {\beta }^2-{\left(\alpha \beta \right)}^2\\ =\alpha {\beta }^2\end{array}$

Again,

$\begin{array}{c}V\left(\overline{Y}\right)=V\left(\frac{1}{n}{\displaystyle \sum _{i=1}^n{Y}_i}\right)\\ =\frac{1}{n^2}V\left({\displaystyle \sum _{i=1}^n{Y}_i}\right)\\ =\frac{1}{n^2}{\displaystyle \sum _{i=1}^{n}V\left(Y_i\right)}\left(\because Y_i\text{'}\text{s are independent}\right)\\ \text{=}\frac{1}{n^2}\left(\alpha {\beta }^2+\mathrm{...}+\alpha {\beta }^2\left(n\text{ terms}\right)\right)\\ \left(\because Y_i\text{'}\text{s are identically distributed}\right)\\ =\frac{n}{n^2}\alpha {\beta }^2\\ =\frac{\alpha {\beta }^2}{n}\\ \underset{n\to \infty }{\lim }V\left(\overline{Y}\right)=\underset{n\to \infty }{\lim }\left(\frac{\alpha {\beta }^2}{n}\right)\\ =0\left(asn\to \infty ,\frac{1}{n}\to 0\right)\end{array}$

Therefore, $\overline{Y}$ is a consistent estimator of $\alpha \beta$.

Hence, $\overline{Y}$ converges in probability to $\alpha \beta$.