91影视

An estimator is a rule for computing an estimate of a given quantity based on observable data: the rule (estimator), the quantity of interest (estimate), and the output (estimate) are all distinct.

(a)

Let random variables \({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)have same distribution with pmf or pdf\({\rm{f}}\left( {{\rm{x;}}{{\rm{\theta }}_{\rm{1}}}{\rm{,}}{{\rm{\theta }}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{\theta }}_{\rm{m}}}} \right){\rm{,m\hat I N}}\), where the parameters \({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,m}}\) are unknown. The moment estimators

\(\widehat {{{\rm{\theta }}_{\rm{1}}}}{\rm{,}}\widehat {{{\rm{\theta }}_{\rm{2}}}}{\rm{, \ldots ,}}\widehat {{{\rm{\theta }}_{\rm{m}}}}\)can be obtaining by equating sample moment to the corresponding population moments and solving for unknown parameters\({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,m}}\).

There is only one unknown parameter\({\rm{\theta }}\), therefore, by solving equation

\({\rm{\bar X = E(X)}}\)for\({\rm{\theta }}\), the moment estimator \({\rm{\hat \theta }}\) will be obtained. Remember that \({\rm{\bar X}}\)is the sample moment of first order, and \({\rm{E(X)}}\)is the population moment of first order.

The following is true for the expected value

\(\begin{aligned}E(X) &= \int_{\rm{0}}^{\rm{1}} {\rm{x}} {\rm{(\theta + 1)}}{{\rm{x}}^{\rm{\theta }}}{\rm{dx}}\\&= \left. {{\rm{(\theta + 1) \times }}\frac{{{{\rm{x}}^{\rm{\theta }}}{\rm{ + 2}}}}{{{\rm{\theta + 2}}}}} \right|_{\rm{0}}^{\rm{1}}\\&= \frac{{{\rm{\theta + 1}}}}{{{\rm{\theta + 2}}}}{\rm{.}}\end{aligned}\)

The solution of equation\({\rm{\bar X = E(X)}}\)is:

\(\begin{aligned} \bar X &= \frac{{{\rm{\hat \theta + 1}}}}{{{\rm{\hat \theta + 2}}}}\\ \bar X &= \frac{{{\rm{\hat \theta + 1 + (1 - 1)}}}}{{{\rm{\hat \theta + 2}}}}\\ \bar X &= \frac{{{\rm{\hat \theta + 2}}}}{{{\rm{\hat \theta + 2}}}}{\rm{ - }}\frac{{\rm{1}}}{{{\rm{\hat \theta + 2}}}}\\\ \bar X - 1 &= - \frac{{\rm{1}}}{{{\rm{\hat \theta + 2}}}}{\rm{\hat \theta + 2}}\\ &= \frac{{\rm{1}}}{{{\rm{1 - \bar X}}}}\end{aligned}\)

Which, yields the moment estimator \({\rm{\hat \theta }}\)

\({\rm{\hat \theta = }}\frac{{\rm{1}}}{{{\rm{1 - \bar X}}}}{\rm{ - 2}}\)

The Sample Mean \({\rm{\bar x}}\)of observations \({{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}\) is given by:

\({\rm{\bar x = }}\frac{{{{\rm{x}}_{\rm{1}}}{\rm{ + }}{{\rm{x}}_{\rm{2}}}{\rm{ + \ldots + }}{{\rm{x}}_{\rm{n}}}}}{{\rm{n}}}{\rm{ = }}\frac{{\rm{1}}}{{\rm{n}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\rm{x}}_{\rm{i}}}} \)

Using this, the sample moment of first order is

\({\rm{\bar x = }}\frac{{\rm{1}}}{{{\rm{10}}}}{\rm{(0}}{\rm{.92 + 0}}{\rm{.79 + \ldots + 0}}{\rm{.88) = 0}}{\rm{.8}}\)

Therefore, the estimate \({\rm{\hat \theta }}\)is\(\frac{{\rm{1}}}{{{\rm{1 - 0}}{\rm{.8}}}}{\rm{ - 2 = 3}}\).

(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

\({\rm{\hat \theta = - }}\frac{{{\rm{10}}}}{{{\rm{ - 2}}{\rm{.4295}}}}{\rm{ - 1 = 3}}{\rm{.12}}\)