91Ó°ÊÓ

It is given that X is a random variable which denotes defective items in large lots. It follows beta distribution with parameters \(\alpha \,\,and\,\,\beta \).

\(f\left( x \right) = \frac{{{x^{\alpha - 1}}{{\left( {1 - x} \right)}^{\beta - 1}}}}{{\beta \left( {\alpha ,\beta } \right)}},0 < x < 1\)

a. In the method of moments method,

\(\begin{align}{\mu _j}\left( \theta \right) &= E\left( {X_i^j} \right) \ldots \left( 1 \right)\\{m_j} &= \frac{1}{n}\sum\limits_{i = 1}^n {X_i^j} \ldots \left( 2 \right)\\Equating\,\,\left( 1 \right)\,\,and\,\,\left( 2 \right)\\{\mu _j} &= {m_j}\end{align}\)

By properties of beta distribution,

\(\begin{align}{m_1} &= E\left( X \right) &= \frac{\alpha }{{\alpha + \beta }}\\{m_2} &= E\left( {{X^2}} \right) &= \frac{{\alpha \left( {\alpha + 1} \right)}}{{\left( {\alpha + \beta } \right)\left( {\alpha + \beta + 1} \right)}}\end{align}\)

Now, therefore, for equating the first-order moment and second order moment with their method of moments estimator,

\(\begin{align}{m_1} &= \frac{\alpha }{{\alpha + \beta }} \ldots \left( 3 \right)\\{m_2} &= \frac{{\alpha \left( {\alpha + 1} \right)}}{{\left( {\alpha + \beta } \right)\left( {\alpha + \beta + 1} \right)}} \ldots \left( 4 \right)\end{align}\)

By solving (3),

\(\begin{align}{m_1} &= \frac{\alpha }{{\alpha + \beta }}\\\alpha &= {m_1}\left( {\alpha + \beta } \right)\\\beta &= \frac{{\alpha \left( {1 - {m_1}} \right)}}{{{m_1}}} \ldots \left( 5 \right)\end{align}\)

Substituting in (4),

\(\begin{align}{m_2} &= \frac{{\alpha \left( {\alpha + 1} \right)}}{{\left( {\alpha + \frac{{\alpha \left( {1 - {m_1}} \right)}}{{{m_1}}}} \right)\left( {\alpha + \frac{{\alpha \left( {1 - {m_1}} \right)}}{{{m_1}}} + 1} \right)}}\\ &= \frac{{\alpha \left( {\alpha + 1} \right)}}{{\left( {\frac{{\alpha {m_1} + \alpha \left( {1 - {m_1}} \right)}}{{{m_1}}}} \right)\left( {\frac{{\alpha {m_1} + \alpha \left( {1 - {m_1}} \right) + {m_1}}}{{{m_1}}}} \right)}}\\ &= \frac{{\alpha \left( {\alpha + 1} \right)}}{{\left( {\frac{\alpha }{{{m_1}}}} \right)\left( {\frac{{\alpha + {m_1}}}{{{m_1}}}} \right)}}\\ &= \frac{{{m_1}^2\left( {\alpha + 1} \right)}}{{\left( {\alpha + {m_1}} \right)}}\end{align}\)

\(\begin{align}{m_2}\left( {\alpha + {m_1}} \right) &= {m_1}^2\left( {\alpha + 1} \right)\\\alpha &= \frac{{{m_1}\left( {{m_1} - {m_2}} \right)}}{{{m_2} - {m_1}^2}}\end{align}\)

Replace the value in (5)

\(\begin{align}\beta &= \frac{{\left( {\frac{{{m_1}\left( {{m_1} - {m_2}} \right)}}{{{m_2} - {m_1}^2}}} \right)\left( {1 - {m_1}} \right)}}{{{m_1}}}\\ &= \frac{{\left( {1 - {m_1}} \right)\left( {{m_1} - {m_2}} \right)}}{{{m_2} - {m_1}^2}}\end{align}\)

b.

Defining the likelihood function:

\(\begin{align}L\left( \theta \right) &= \prod\limits_{i = 1}^n {f\left( {{x_i}} \right)} \\ &= \prod\limits_{i = 1}^n {\frac{{{x^{\alpha - 1}}{{\left( {1 - x} \right)}^{\beta - 1}}}}{{\beta \left( {\alpha ,\beta } \right)}}} \\ &= \frac{{\prod\limits_{i = 1}^n {{x^{\alpha - 1}}} \prod\limits_{i = 1}^n {{{\left( {1 - x} \right)}^{\beta - 1}}} }}{{\beta {{\left( {\alpha ,\beta } \right)}^n}}}\end{align}\)

The log-likelihood is:

\(\ln L\left( \theta \right) = \left( {\alpha - 1} \right)\sum {\log {x_i} \times } \left( {\beta - 1} \right)\sum {\log \left( {1 - {x_i}} \right)} - n\log \beta \left( {\alpha ,\beta } \right)\)

Setting its derivative with respect to parameter:

\(\begin{align}\frac{\partial }{{\partial \alpha }}\ln L\left( \theta \right) &= \frac{\partial }{{\partial \alpha }}\left( {\alpha - 1} \right)\sum {\log {x_i} \times } \left( {\beta - 1} \right)\sum {\log \left( {1 - {x_i}} \right)} - n\log \beta \left( {\alpha ,\beta } \right)\\ &= \sum {\log {x_i}} + 0 + 0\\ &= \sum {\log {x_i}} \end{align}\)

\(\begin{align}\frac{\partial }{{\partial \beta }}\ln L\left( \theta \right) &= \frac{\partial }{{\partial \beta }}\left( {\alpha - 1} \right)\sum {\log {x_i} \times } \left( {\beta - 1} \right)\sum {\log \left( {1 - {x_i}} \right)} - n\log \beta \left( {\alpha ,\beta } \right)\\ &= 0 + \sum {\log \left( {1 - {x_i}} \right)} + 0\\ &= \sum {\log \left( {1 - {x_i}} \right)} \end{align}\)

Equating both of the equations to 0

\(\begin{align}\sum {\log {x_i}} &= 0\\\sum {\log \left( {1 - {x_i}} \right)} &= 0\end{align}\)

Applying antilog on both sides we get

\(\prod {{X_i}} \,\,and\,\,\prod {\left( {1 - {X_i}} \right)} \)

The log likelihood function is a decreasing function and it is maximised at \(\begin{align}\alpha &= \prod {{X_i}} \\\beta &= \prod {\left( {1 - {X_i}} \right)} \end{align}\)

Therefore, the M.L.E. is

\(\begin{align}\alpha &= \prod {{X_i}} \\\beta &= \prod {\left( {1 - {X_i}} \right)} \end{align}\)

Therefore, Method of moments estimator is\(\alpha = \frac{{{m_1}\left( {{m_1} - {m_2}} \right)}}{{{m_2} - {m_1}^2}}\)\(\beta = \frac{{\left( {1 - {m_1}} \right)\left( {{m_1} - {m_2}} \right)}}{{{m_2} - {m_1}^2}}\)and M.L.E. is\(\alpha = \prod {{X_i}} ,\beta = \prod {\left( {1 - {X_i}} \right)} \)

They are not equal.