Chapter 5.11, Problem 142E

a.

To determine

Prove that $E\left(Y\right)=\frac{n\alpha }{\alpha +\beta }$.

Explanation of Solution

Calculation:

Binomial probability distribution:

A discrete random variable Y is said to follow a Binomial distribution with mean $np$ and variance $np\left(1-p\right)$, if the probability mass function of Y is as follows:

$p\left(y\right)=\left(\begin{array}{l}n\\ y\end{array}\right)p^y{\left(1-p\right)}^{n-y}$, $y=0,1,\mathrm{...},n$ where n is the number of trials and p is the probability of success.

The mean and variance of Y are $np\text{and }np\left(1-p\right)$, respectively.

Beta distribution:

A continuous random variable Y is said to follow a Beta distribution with parameters $\alpha$ and $\beta$, if the probability density function of Y is as follows:

$f\left(y\right)=\frac{x^{\alpha -1}{\left(1-x\right)}^{\beta -1}}{\frac{\overline{)\alpha }\overline{)\beta }}{\overline{)\alpha +\beta }}},0<x<1$.

The mean and variance of Y are $\frac{a}{\alpha +\beta }\text{and }\frac{\alpha \beta }{{\left(\alpha +\beta \right)}^2\left(\alpha +\beta +1\right)}$, respectively.

The conditional expectation of any real valued function $g\left(Y_1\right)$ given that $Y_2=y_2$ is defined as follows:

$E\left[g\left(Y_1\right)|Y_2=y_2\right]={\displaystyle \sum _{y_1}g\left(y_1\right)f\left(y_1|y_2\right)}$, where $Y_1$ only follows discrete distribution.

It is given that the conditional distribution of Y given p follows Binomial distribution with parameters n and p where p  follows Beta distribution with parameters $\alpha$ and $\beta$.

Thus, the expected value of the conditional distribution of $Y_2$ given $Y_1=y_1$ is as follows:

$\begin{array}{c}E\left(Y|p\right)={\displaystyle \sum _{y=0}^{n}y\left(\begin{array}{l}n\\ y\end{array}\right)p^y{\left(1-p\right)}^{n-y}}\\ ={\displaystyle \sum _{y=0}^{n}y\frac{n!}{\left(n-y\right)!y!}{p}^y{\left(1-p\right)}^{n-y}}\\ ={\displaystyle \sum _{y=1}^n\frac{n!}{\left(n-y\right)!\left(y-1\right)!}{p}^y{\left(1-p\right)}^{n-y}}\\ =np{\displaystyle \sum _{y-1=0}^{n-1}\frac{\left(n-1\right)!}{\left(n-y-2\right)!\left(y-1\right)!}{p}^{y-1}{\left(1-p\right)}^{n-y+1}}\\ =np{\left(p+1-p\right)}^{m-1}\\ =np\left(1\right)\\ =np\end{array}$.

For two random variables $Y_1\text{and }{Y}_2$, it is known that $E\left(Y_1\right)=E\left[E\left(Y_1|Y_2\right)\right]$ and $V\left(Y_1\right)=E\left[V\left(Y_1|Y_2\right)\right]+V\left[E\left(Y_1|Y_2\right)\right]$.

Hence,

$\begin{array}{c}E\left(Y\right)=E\left[E\left(Y|p\right)\right]\\ =E\left[np\right]\\ =nE\left(p\right)\\ =n{\displaystyle \underset{0}{\overset{1}{\int }}p\frac{p^{\alpha -1}{\left(1-p\right)}^{\beta -1}}{\frac{\overline{)\alpha }\overline{)\beta }}{\overline{)\alpha +\beta }}}dp}\\ =n\frac{\overline{)\alpha +\beta }}{\overline{)\alpha }\overline{)\beta }}{\displaystyle \underset{0}{\overset{1}{\int }}{p}^{\alpha +1-1}{\left(1-p\right)}^{\beta -1}dp}\\ =n\frac{\overline{)\alpha +\beta }}{\overline{)\alpha }\overline{)\beta }}\frac{\overline{)\alpha +1}\overline{)\beta }}{\overline{)\alpha +1+\beta }}\\ =n\frac{\left(\alpha +\beta -1\right)!}{\left(\alpha -1\right)!\left(\beta -1\right)!}\frac{\alpha !\left(\beta -1\right)!}{\left(\alpha +\beta \right)!}\\ =n\frac{\alpha }{\alpha +\beta }\end{array}$

Therefore, it is proved that $E\left(Y\right)=\frac{n\alpha }{\alpha +\beta }$.

b.

To determine

Prove that $V\left(Y\right)=\frac{n\alpha \beta \left(\alpha +\beta +n\right)}{{\left(\alpha +\beta \right)}^2\left(\alpha +\beta +1\right)}$.

Explanation of Solution

Calculation:

For two random variables $Y_1\text{and }{Y}_2$, it is known that $V\left(Y_1\right)=E\left[V\left(Y_1|Y_2\right)\right]+V\left[E\left(Y_1|Y_2\right)\right]$.

From Part (a), it is obtained that $E\left(Y|p\right)=np$, $V\left(Y|p\right)=np\left(1-p\right)$, $E\left(p\right)=\frac{\alpha }{\alpha +\beta }$ and $V\left(p\right)=\frac{\alpha \beta }{{\left(\alpha +\beta \right)}^2\left(\alpha +\beta +1\right)}$.

Hence,

$\begin{array}{c}V\left(Y\right)=E\left[V\left(Y|p\right)\right]+V\left[E\left(Y|p\right)\right]\\ =E\left[np\left(1-p\right)\right]+V\left[np\right]\\ =nE\left(p\right)-nE\left(p^2\right)+n^{2}V\left(p\right)\\ =n\frac{\alpha }{\alpha +\beta }-n\left[\frac{\alpha \beta }{{\left(\alpha +\beta \right)}^2\left(\alpha +\beta +1\right)}-\frac{{\alpha }^2}{{\left(\alpha +\beta \right)}^2}\right]+n^2\frac{\alpha \beta }{{\left(\alpha +\beta \right)}^2\left(\alpha +\beta +1\right)}\\ =\frac{n\alpha \left(\alpha +\beta \right)\left(\alpha +\beta +1\right)-n\alpha \beta -{\alpha }^2\left(\alpha +\beta +1\right)+n^2\alpha \beta }{{\left(\alpha +\beta \right)}^2\left(\alpha +\beta +1\right)}\\ =\frac{n\alpha \beta \left(\alpha +\beta +n\right)}{{\left(\alpha +\beta \right)}^2\left(\alpha +\beta +1\right)}\end{array}$

Therefore, it is proved that $V\left(Y\right)=\frac{n\alpha \beta \left(\alpha +\beta +n\right)}{{\left(\alpha +\beta \right)}^2\left(\alpha +\beta +1\right)}$.