Chapter 4, Problem 200SE

a.

To determine

If $a$ is any positive or negative value such that $\alpha +a>0$, then prove that $E\left(Y^a\right)=\frac{\Gamma \left(\alpha +a\right)\Gamma \left(\alpha +\beta \right)}{\Gamma \left(\alpha \right)\Gamma \left(\alpha +\beta +a\right)}$.

Explanation of Solution

 It is given that Y has a beta distribution with parameters $\alpha \text{ and }\beta$.

The probability density function of Y is given as follows:

$f\left(y\right)=\{\begin{array}{l}\frac{y^{\alpha -1}{\left(1-y\right)}^{{}^{\beta -1}}}{B\left(\alpha ,\beta \right)}\text{, 0}<y<1,\\ \text{ 0 ,elsewhere}\text{.}\end{array}$

The expected value of $y^a$ is as follows:

$\begin{array}{c}E\left(Y^a\right)={\displaystyle \underset{0}{\overset{1}{\int }}{y}^{a}f\left(y\right)}dy\\ ={\displaystyle \underset{0}{\overset{1}{\int }}{y}^a\frac{y^{\alpha -1}{\left(1-y\right)}^{\beta -1}}{B\left(\alpha ,\beta \right)}dy}\\ =\frac{\Gamma \left(\alpha +\beta \right)}{\Gamma \left(\alpha \right)\Gamma \left(\beta \right)}{\displaystyle \underset{0}{\overset{1}{\int }}{y}^{\alpha +a-1}{\left(1-y\right)}^{\beta -1}}dy\\ =\frac{\Gamma \left(\alpha +\beta \right)}{\Gamma \left(\alpha \right)\Gamma \left(\beta \right)}\times B\left(\alpha +a,\beta \right)\left\{\because B\left(\alpha +a,\beta \right)={\displaystyle \underset{0}{\overset{1}{\int }}{y}^{\alpha +a-1}{\left(1-y\right)}^{\beta -1}}dy\right\}\\ =\frac{\Gamma \left(\alpha +\beta \right)}{\Gamma \left(\alpha \right)\Gamma \left(\beta \right)}\times \frac{\Gamma \left(\alpha +a\right)\Gamma \left(\beta \right)}{\Gamma \left(\alpha +\beta +a\right)}\\ =\frac{\Gamma \left(\alpha +\beta \right)\Gamma \left(\alpha +a\right)}{\Gamma \left(\alpha \right)\Gamma \left(\alpha +\beta +a\right)}\end{array}$

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

b.

To determine

Explain the reason for Part (a), which requires that $\alpha +a>0$.

Explanation of Solution

The answer in Part (a) has to be satisfied that $\alpha +a>0$ because the gamma function is not defined for non-positive values and thus equality will not hold if $\alpha +a\le 0$.

c.

To determine

Show that, with $a=1$, the result in Part (a) yields $E\left(Y\right)=\frac{\alpha }{\alpha +\beta }$.

Answer to Problem 200SE

It is proved that $E\left(Y\right)=\frac{\alpha }{\alpha +\beta }$.

Explanation of Solution

The answer in Part (a) is as follows:

$E\left(Y^a\right)=\frac{\Gamma \left(\alpha +a\right)\Gamma \left(\alpha +\beta \right)}{\Gamma \left(\alpha \right)\Gamma \left(\alpha +\beta +a\right)}$

Substitute $a=1$ in the above expression, then one can get the following result:

$\begin{array}{c}E\left(Y^1\right)=\frac{\Gamma \left(\alpha +1\right)\Gamma \left(\alpha +\beta \right)}{\Gamma \left(\alpha \right)\Gamma \left(\alpha +\beta +1\right)}\\ =\frac{\alpha \Gamma \left(\alpha \right)\Gamma \left(\alpha +\beta \right)}{\Gamma \left(\alpha \right)\left(\alpha +\beta \right)\Gamma \left(\alpha +\beta \right)}\left\{\because \Gamma \left(n+1\right)=n\Gamma \left(n\right)\forall n>0\right\}\\ =\frac{\alpha }{\alpha +\beta }\end{array}$

Therefore, for $a=1$, the result in Part (a) yields $E\left(Y\right)=\frac{\alpha }{\alpha +\beta }$.

d.

To determine

Find an expression for $E\left(\sqrt{Y}\right)$ using the results in Part (a).

Find the expression that one needs to assume about $\alpha$.

Explanation of Solution

The answer in Part (a) is as follows:

$E\left(Y^a\right)=\frac{\Gamma \left(\alpha +a\right)\Gamma \left(\alpha +\beta \right)}{\Gamma \left(\alpha \right)\Gamma \left(\alpha +\beta +a\right)}$

Substitute $a=\frac{1}{2}$ in the above expression, then one can get the expression for $E\left(\sqrt{Y}\right)$.

$\begin{array}{c}E\left(\sqrt{y}\right)=E\left(y^{\frac{1}{2}}\right)\\ =\frac{\Gamma \left(\alpha +\frac{1}{2}\right)\Gamma \left(\alpha +\beta \right)}{\Gamma \left(\alpha \right)\Gamma \left(\alpha +\beta +\frac{1}{2}\right)}\left\{\because \Gamma \left(\alpha \right) \text{value is defined only if }\alpha >-\frac{1}{2}\right\}\end{array}$

Therefore, the value of $\alpha >-\frac{1}{2}$ should be assumed.

e.

To determine

Use the results in Part (a) to find an expression for $E\left(\frac{1}{Y}\right)$, $E\left(\frac{1}{\sqrt{Y}}\right)$, and $E\left(\frac{1}{Y^2}\right)$.

Identify the value of $\alpha$ that needs to be assumed in each case.

Explanation of Solution

The result in Part (a) is as follows:

$E\left(Y^a\right)=\frac{\Gamma \left(\alpha +a\right)\Gamma \left(\alpha +\beta \right)}{\Gamma \left(\alpha \right)\Gamma \left(\alpha +\beta +a\right)}$

Case1:

Substitute $a=-1$ in the above expression to get the expression for $E\left(\frac{1}{Y}\right)$.

$\begin{array}{c}E\left(\frac{1}{Y}\right)==\frac{\Gamma \left(\alpha -1\right)\Gamma \left(\alpha +\beta \right)}{\Gamma \left(\alpha \right)\Gamma \left(\alpha +\beta -1\right)}\\ =\frac{\Gamma \left(\alpha -1\right)\left(\alpha +\beta -1\right)\Gamma \left(\alpha +\beta -1\right)}{\left(\alpha -1\right)\Gamma \left(\alpha -1\right)\Gamma \left(\alpha +\beta -1\right)}\\ =\frac{\alpha +\beta -1}{\alpha -1}\left\{\because \Gamma \left(\alpha -1\right) \text{value is defined only if}\alpha >1\right\}\end{array}$

Thus, in this case one needs to assume that the value of $\alpha >1$.

Case 2:

Substitute $a=-\frac{1}{2}$ in the above expression to get the expression for $E\left(\frac{1}{\sqrt{Y}}\right)$.

$E\left(\frac{1}{\sqrt{Y}}\right)=\frac{\Gamma \left(\alpha -\frac{1}{2}\right)\Gamma \left(\alpha +\beta \right)}{\Gamma \left(\alpha \right)\Gamma \left(\alpha +\beta -\frac{1}{2}\right)}\left\{\because \Gamma \left(\alpha -\frac{1}{2}\right) \text{value is defined only if}\alpha >\frac{1}{2}\right\}$

Therefore, the value of $\alpha >\frac{1}{2}$ should be assumed.

Case3:

Substitute $a=-2$ in the above expression to get the expression for $E\left(\frac{1}{Y^2}\right)$.

$\begin{array}{c}E\left(\frac{1}{Y^2}\right)=\frac{\Gamma \left(\alpha -2\right)\Gamma \left(\alpha +\beta \right)}{\Gamma \left(\alpha \right)\Gamma \left(\alpha +\beta -2\right)}\\ =\frac{\left(\alpha +\beta -1\right)\left(\alpha +\beta -2\right)\Gamma \left(\alpha +\beta -2\right)\Gamma \left(\alpha -2\right)}{\left(\alpha -1\right)\left(\alpha -2\right)\Gamma \left(\alpha -2\right)\Gamma \left(\alpha +\beta -2\right)}\\ \text{= }\frac{\left(\alpha +\beta -1\right)\left(\alpha +\beta -2\right)}{\left(\alpha -1\right)\left(\alpha -2\right)}\left\{\because \left(\alpha -2\right)>0 \text{value is defined only if}\alpha >2\right\}\end{array}$

Thus, value of $\alpha >2$ should be assumed in this case.