91Ó°ÊÓ

A random variable X takes five values\(x = 1,2,3,4,5\)with probabilities,

\(\begin{aligned}{}f\left( {1\left| \theta \right.} \right) &= {\theta ^3},\,\,\,\,f\left( {2\left| \theta \right.} \right) &= {\theta ^2}\left( {1 - \theta } \right),\\f\left( {3\left| \theta \right.} \right) &= 2\theta \left( {1 - \theta } \right),\,\,\,f\left( {4\left| \theta \right.} \right) &= \theta {\left( {1 - \theta } \right)^2}\\f\left( {5\left| \theta \right.} \right) &= {\left( {1 - \theta } \right)^3}\end{aligned}\)

An estimator\({\delta _c}\left( X \right)\)the form,

\(\begin{aligned}{}{\delta _c}\left( 1 \right) &= 1,\,\,{\delta _c}\left( 2 \right) &= 2 - 2c,\,\,{\delta _c}\left( 3 \right) &= c,\\{\delta _c}\left( 4 \right) &= 1 - 2c,\,\,{\delta _c}\left( 5 \right) &= 0.\end{aligned}\)

\(\begin{aligned}{}f\left( {1\left| \theta \right.} \right) + f\left( {2\left| \theta \right.} \right) &= {\theta ^3} + {\theta ^2}\left( {1 - \theta } \right)\\ &= {\theta ^3} + {\theta ^2} - {\theta ^3}\\ &= {\theta ^2}\end{aligned}\)

\(\begin{aligned}{}f\left( {4\left| \theta \right.} \right) + f\left( {5\left| \theta \right.} \right) &= \theta {\left( {1 - \theta } \right)^2} + {\left( {1 - \theta } \right)^3}\\ &= {\left( {1 - \theta } \right)^2}\left( {\theta + 1 - \theta } \right)\\ &= {\left( {1 - \theta } \right)^2}\end{aligned}\)

\(f\left( {3\left| \theta \right.} \right) = 2\theta \left( {1 - \theta } \right)\)

The total of the five probabilities on the left sides of these equations equals the sum of the probabilities on the right sides, which is

\(\begin{aligned}{}{\theta ^2} + {\left( {1 - \theta } \right)^2} + 2\theta \left( {1 - \theta } \right) &= {\theta ^2} + 1 - 2\theta + {\theta ^2} + 2\theta - 2{\theta ^2}\\ &= 1\end{aligned}\)

Hence, the sum of given probabilities is 1 for every value of \(\theta \)

\(\begin{aligned}{}{E_\theta }\left( {{\delta _c}\left( X \right)} \right) &= \sum\limits_{x = 1}^5 {{\delta _c}\left( x \right)f\left( {x\left| \theta \right.} \right)} \\ &= \left( {1 \times {\theta ^3}} \right) + \left( {\left( {2 - 2c} \right) \times {\theta ^2} \times \left( {1 - \theta } \right)} \right) + \left( {c \times 2\theta \left( {1 - \theta } \right)} \right) + \left( {\left( {1 - 2c} \right) \times \theta \times {{\left( {1 - \theta } \right)}^2}} \right) + 0\\ &= \theta \end{aligned}\)

The sum of the coefficients of\({\theta ^3}\)is 0, the sum of the coefficients of\({\theta ^2}\)is 0, the sum of coefficients of\(\theta \)is 1, and the constant term is 0.

Hence,\(E\left[ {{\delta _c}\left( X \right)} \right] = \theta \)

Hence, for each constant c, \({\delta _c}\left( X \right)\) is an unbiased estimator of \(\theta \)

For every value of c,

\(\begin{aligned}{}Va{r_{{\theta _0}}}\left( {{\delta _c}} \right) &= {E_{{\theta _0}}}\left( {\delta _c^2} \right) - {\left( {{E_{{\theta _0}}}\left( {{\delta _c}} \right)} \right)^2}\\ &= {E_{{\theta _0}}}\left( {\delta _c^2} \right) - {\theta ^2}\end{aligned}\)

Hence, the value of c for which\(Va{r_{{\theta _0}}}\left( {{\delta _c}} \right)\)is a minimum will be the value of c for which\({E_{{\theta _0}}}\left( {\delta _c^2} \right)\)is a minimum

Now,

\(\begin{aligned}{}{E_{{\theta _0}}}\left( {\delta _c^2} \right) &= \left( {{1^2} \times \theta _0^3} \right) + {\left( {2 - 2c} \right)^2}\left( {1 - {\theta _0}} \right) + {c^2}2{\theta _0}\left( {1 - {\theta _0}} \right) + {\left( {1 - 2c} \right)^2}{\theta _0}{\left( {1 - {\theta _0}} \right)^2} + 0\\ = 2{c^2}\left[ {2\theta _0^2\left( {1 - {\theta _0}} \right) + {\theta _0}\left( {1 - {\theta _0}} \right) + 2{\theta _0}{{\left( {1 - {\theta _0}} \right)}^2}} \right] - 4c\left[ {2\theta _0^2\left( {1 - {\theta _0}} \right) + {\theta _0}{{\left( {1 - {\theta _0}} \right)}^2}} \right] + \,\,terms\,not\,\,c\end{aligned}\)

After further simplification of the coefficients of\({c^2}\,\,and\,\,c\), we obtain the relation

\({E_{{\theta _0}}}\left( {\delta _c^2} \right) = 6{\theta _0}\left( {1 - {\theta _0}} \right){c^2} + 4{\theta _0}\left( {1 - \theta _0^2} \right)c + \,terms\,\,not\,\,involving\,\,c\)

By differentiating concerning c and setting the derivative equal to 0, it is found that the value of c for which\({E_{{\theta _0}}}\left( {\delta _c^2} \right)\)is minimum is\(c = \frac{{\left( {1 + {\theta _0}} \right)}}{3}\)

Therefore, the value of c is \(\frac{{\left( {1 + {\theta _0}} \right)}}{3}\)