(b)
Let random variables \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)have joint pdf or pmb,
\({\rm{f}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}{\rm{;}}{{\rm{\theta }}_{\rm{1}}}{\rm{,}}{{\rm{\theta }}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{\theta }}_{\rm{m}}}} \right){\rm{,}}\quad {\rm{n,m\hat I N}}\)
Where, the parameters \({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,m}}\)are unknown. When function \({\rm{f}}\)is a function of parameters\({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,m}}\), it is called the
likelihood function
Values \({{\rm{\hat \theta }}_{\rm{i}}}\)that maximize the likelihood function are the maximum likelihood estimates (mle's), or equally values \({{\rm{\hat \theta }}_{\rm{i}}}\)for which
for every\({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,m}}\). By substituting\({{\rm{X}}_{\rm{i}}}{\rm{ with }}{{\rm{x}}_{\rm{i}}}\), the
maximum likelihood estimators
are obtained.
The pdf is given in the exercise. The likelihood function (assuming independence) becomes
\(\begin{aligned}{\rm{f}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}{\rm{;\theta }}} \right)\\&= (\theta + 1)x_{\rm{1}}^{\rm{\theta }}{\rm{ \times (\theta + 1)x}}_{\rm{2}}^{\rm{\theta }}{\rm{ \times \ldots \times (\theta + 1)x}}_{\rm{n}}^{\rm{\theta }}\\&= (\theta + 1 {{\rm{)}}^{\rm{2}}}{\rm{ \times }}{\left( {{{\rm{x}}_{\rm{1}}}{\rm{ \times }}{{\rm{x}}_{\rm{2}}}{\rm{ \times \ldots \times }}{{\rm{x}}_{\rm{n}}}} \right)^{\rm{\theta }}}\end{aligned}\)
In order to find maximum, look at the log likelihood function
\(\begin{aligned}{\rm{lnf}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}{\rm{;\theta }}} \right)\\&= ln \left( {{{{\rm{(\theta + 1)}}}^{\rm{n}}}{\rm{ \times }}{{\left( {{{\rm{x}}_{\rm{1}}}{\rm{ \times }}{{\rm{x}}_{\rm{2}}}{\rm{ \times \ldots \times }}{{\rm{x}}_{\rm{n}}}} \right)}^{\rm{\theta }}}} \right)\\ &= n \times ln(\theta + 1) + \theta \times \sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{ln}}} {{\rm{x}}_{\rm{i}}}{\rm{.}}\end{aligned}\)
By taking derivative of log likelihood function in respect to \({\rm{\theta }}\)and equating it to \({\rm{0}}\) the maximum likelihood estimator is obtained. Therefore, the derivative is
\(\begin{aligned}\frac{{\rm{d}}}{{{\rm{d\theta }}}}{\rm{f}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}{\rm{;\theta }}} \right)\\& = \frac{{\rm{d}}}{{{\rm{d\theta }}}}\left( {{\rm{2 \times ln(\theta + 1) + \theta \times }}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{ln}}} {{\rm{x}}_{\rm{i}}}} \right)\\{\rm{ = n \times }}\frac{{\rm{1}}}{{{\rm{\theta + 1}}}}{\rm{ + }}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{ln}}} {{\rm{x}}_{\rm{i}}}{\rm{.}}\end{aligned}\)
Therefore, the maximum likelihood estimator is obtained by solving equation
\({\rm{n \times }}\frac{{\rm{1}}}{{{\rm{\hat \theta + 1}}}}{\rm{ + }}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{ln}}} {{\rm{x}}_{\rm{i}}}{\rm{ = 0}}\)
For\({\rm{\hat \theta }}\). Obviously, the solution is
\({\rm{\hat \theta = - }}\frac{{\rm{n}}}{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{ln}}} {{\rm{X}}_{\rm{i}}}}}{\rm{ - 1}}{\rm{.}}\)
which is the maximum likelihood estimator.
By taking \({\rm{ln}}{{\rm{x}}_{\rm{i}}}\)for every\({\rm{i = 1,2, \ldots ,10}}\), and summing the values, the maximum likelihood estimate is obtained as
\({\rm{\hat \theta = - }}\frac{{{\rm{10}}}}{{{\rm{ - 2}}{\rm{.4295}}}}{\rm{ - 1 = 3}}{\rm{.12}}\)
