Question

Each of \({\rm{n}}\) specimens is to be weighed twice on the same scale. Let \({{\rm{X}}_{\rm{i}}}\) and \({{\rm{Y}}_{\rm{i}}}\) denote the two observed weights for the \({\rm{ith}}\) specimen. Suppose\({{\rm{X}}_{\rm{i}}}\) and \({{\rm{Y}}_{\rm{i}}}\) are independent of one another, each normally distributed with mean value \({{\rm{\mu }}_{\rm{i}}}\) (the true weight of specimen \({\rm{i}}\)) and variance \({{\rm{\sigma }}^{\rm{2}}}\) . a. Show that the maximum likelihood estimator of \({{\rm{\sigma }}^{\rm{2}}}\) is \({\widehat {\rm{\sigma }}^{\rm{2}}}{\rm{ = \Sigma }}{\left( {{{\rm{X}}_{\rm{i}}}{\rm{ - }}{{\rm{Y}}_{\rm{i}}}} \right)^{\rm{2}}}{\rm{/(4n)}}\). (Hint: If \({\rm{\bar z = }}\left( {{{\rm{z}}_{\rm{1}}}{\rm{ + }}{{\rm{z}}_{\rm{2}}}} \right){\rm{/2}}\), then \({\rm{\Sigma }}{\left( {{{\rm{z}}_{\rm{i}}}{\rm{ - \bar z}}} \right)^{\rm{2}}}{\rm{ = }}{\left( {{{\rm{z}}_{\rm{1}}}{\rm{ - }}{{\rm{z}}_{\rm{2}}}} \right)^{\rm{2}}}{\rm{/2}}\)) b. Is the mle \({{\rm{\hat \sigma }}^{\rm{2}}}\) an unbiased estimator of \({{\rm{\sigma }}^{\rm{2}}}\)? Find an unbiased estimator of \({{\rm{\sigma }}^{\rm{2}}}\). (Hint: For any \({\rm{rv Z,E}}\left( {{{\rm{Z}}^{\rm{2}}}} \right){\rm{ = V(Z) + (E(Z)}}{{\rm{)}}^{\rm{2}}}\). Apply this to \({\rm{Z = }}{{\rm{X}}_{\rm{i}}}{\rm{ - }}{{\rm{Y}}_{\rm{i}}}\).)

Short answer

a.It is shown that \({{\rm{\hat \sigma }}^{\rm{2}}}{\rm{ = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{X}}_{\rm{i}}}{\rm{ - }} {{\rm{Y}}_{\rm{i}}}} \right)}^{\rm{2}}}} }}{{{\rm{4n}}}}\).

b.It isn't objective. \({\rm{\hat \sigma }}_{\rm{1}}^{\rm{2}}{\rm{ = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{X}}_{\rm{i}}}{\rm{ - }} {{\rm{Y}}_{\rm{i}}}} \right)}^{\rm{2}}}} }}{{{\rm{2n}}}}{\rm{ }}\)

Step-by-step solution

Define exponential function

A function that increases or decays at a rate proportional to its present value is called an exponential function.

Explanation

1. The pdf of random variables\({{\rm{X}}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,n}}\),

\({{\rm{f}}_{{{\rm{X}}_{\rm{i}}}}}{\rm{(x) = }}\frac{{\rm{1}}}{{\sqrt {{\rm{2\pi }}{{\rm{\sigma }}^{\rm{2}}}} }}{\rm{exp}}\left\{ {{\rm{ - }}\frac{{\left( {{\rm{x - }}{{\rm{\mu }}_{\rm{i}}}} \right)}}{{{\rm{2}}{{\rm{\sigma }}^{\rm{2}}}}}} \right\}{\rm{,x}} \in {\rm{R}}\)

And the pdf of random variables\({{\rm{Y}}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,n}}\)

\({{\rm{f}}_{{{\rm{Y}}_{\rm{i}}}}}{\rm{(x) = }}\frac{{\rm{1}}}{{\sqrt {{\rm{2\pi }}{{\rm{\sigma }}^{\rm{2}}}} }}{\rm{exp}}\left\{ {{\rm{ - }}\frac{{\left( {{{\rm{y}}_{\rm{i}}}{\rm{ - }}{{\rm{\mu }}_{\rm{i}}}} \right)}}{{{\rm{2}}{{\rm{\sigma }}^{\rm{2}}}}}} \right\}{\rm{,x}} \in {\rm{R}}\)

because they are all evenly distributed.

The random variables\({{\rm{X}}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,n}}\)are unrelated to each other.

Allow joint pdf or pmf for random variables\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\).

\({\rm{f}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}{\rm{;}}{{\rm{\theta }}_{\rm{1}}}{\rm{,}}{{\rm{\theta }}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{\theta }}_{\rm{m}}}} \right){\rm{, n,m}} \in {\rm{N}}\)

where \({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,m}}\) are unknown parameters. The likelihood function is defined as a function of parameters \({{\rm{\theta }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,m}}\) where function f is a function of parameter. The maximum likelihood estimates (mle's), or values \(\widehat {{{\rm{\theta }}_{\rm{i}}}}\) for which the likelihood function is maximised, are the maximum likelihood estimates,

\({\rm{f}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}{\rm{;}}{{{\rm{\hat \theta }}}_{\rm{1}}}{\rm{,}}{{{\rm{\hat \theta }}}_{\rm{2}}}{\rm{, \ldots ,}}{{{\rm{\hat \theta }}}_{\rm{m}}}} \right) \ge {\rm{f}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{x}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{x}}_{\rm{n}}}{\rm{;}}{{\rm{\theta }}_{\rm{1}}}{\rm{,}}{{\rm{\theta }}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{\theta }}_{\rm{m}}}} \right)\)

As,\({\rm{i = 1,2, \ldots ,m}}\)for every\({{\rm{\theta }}_{\rm{i}}}\). Maximum likelihood estimators are derived by replacing\({{\rm{X}}_{\rm{i}}}\)with\({{\rm{x}}_{\rm{i}}}\).

Because of the independence, the likelihood function becomes,

\(\begin{aligned}{\rm{f(x,y;}}\mu {\rm{,}}\sigma ) &= \frac{{\rm{1}}}{{\sqrt {{\rm{2p}}{\sigma ^{\rm{2}}}} }}{\rm{exp}}\left\{ {{\rm{ - }}\frac{{\left( {{{\rm{x}}_{\rm{1}}}{\rm{ - }}{\mu _{\rm{1}}}} \right)}}{{{\rm{2}}{\sigma ^{\rm{2}}}}}} \right\}{\rm{ \times }}\frac{{\rm{1}}}{{\sqrt {{\rm{2}}\pi {\sigma ^{\rm{2}}}} }}{\rm{exp}}\left\{ {{\rm{ - }}\frac{{\left( {{{\rm{x}}_{\rm{2}}}{\rm{ - }}{\mu _{\rm{2}}}} \right)}}{{{\rm{2}}{\sigma ^{\rm{2}}}}}} \right\}\\{\rm{ \times \ldots \times }}\frac{{\rm{1}}}{{\sqrt {{\rm{2p}}{\sigma ^{\rm{2}}}} }}{\rm{exp}}\left\{ {{\rm{ - }}\frac{{\left( {{{\rm{x}}_{\rm{n}}}{\rm{ - }}{\mu _{\rm{n}}}} \right)}}{{{\rm{2}}{\sigma ^{\rm{2}}}}}} \right\}{\rm{ \times }}\frac{{\rm{1}}}{{\sqrt {{\rm{2}}\pi {\sigma ^{\rm{2}}}} }}{\rm{exp}}\left\{ {{\rm{ - }}\frac{{\left( {{{\rm{y}}_{\rm{1}}}{\rm{ - }}{\mu _{\rm{1}}}} \right)}}{{{\rm{2}}{\sigma ^{\rm{2}}}}}} \right\}\\ \times \frac{{\rm{1}}}{{\sqrt {{\rm{2}}\pi {\sigma ^{\rm{2}}}} }}{\rm{exp}}\left\{ {{\rm{ - }}\frac{{\left( {{{\rm{y}}_{\rm{2}}}{\rm{ - }}{\mu _{\rm{2}}}} \right)}}{{{\rm{2}}{\sigma ^{\rm{2}}}}}} \right\}{\rm{ \times \ldots \times }}\frac{{\rm{1}}}{{\sqrt {{\rm{2}}\pi {\sigma ^{\rm{2}}}} }}{\rm{exp}}\left\{ {{\rm{ - }}\frac{{\left( {{{\rm{y}}_{\rm{n}}}{\rm{ - }}{\mu _{\rm{n}}}} \right)}}{{{\rm{2}}{\sigma ^{\rm{2}}}}}} \right\}\\& = \prod\limits_{{\rm{i = 1}}}^{\rm{n}} {\left( {\frac{{\rm{1}}}{{\sqrt {{\rm{2}}\pi {\sigma ^{\rm{2}}}} }}{\rm{exp}}\left\{ {{\rm{ - }}\frac{{\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}{\mu _{\rm{i}}}} \right)}}{{{\rm{2}}{\sigma ^{\rm{2}}}}}} \right\}{\rm{ \times }}\frac{{\rm{1}}}{{\sqrt {{\rm{2}}\pi {\sigma ^{\rm{2}}}} }}{\rm{exp}}\left\{ {{\rm{ - }}\frac{{\left( {{{\rm{y}}_{\rm{i}}}{\rm{ - }}{\mu _{\rm{i}}}} \right)}}{{{\rm{2}}{\sigma ^{\rm{2}}}}}} \right\}} \right)} \\ &= \frac{{\rm{1}}}{{{{\left( {\sqrt {{\rm{2}}\pi {\sigma ^{\rm{2}}}} } \right)}^{{\rm{2n}}}}}}{\rm{ \times exp}}\left\{ {{\rm{ - }}\frac{{\rm{1}}}{{{\rm{2}}{\sigma ^{\rm{2}}}}}{\rm{ \times }}\left( {\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}{\mu _{\rm{i}}}} \right)}^{\rm{2}}}} {\rm{ + }}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{y}}_{\rm{i}}}{\rm{ - }}{\mu _{\rm{i}}}} \right)}^{\rm{2}}}} } \right)} \right\}\\&= \frac{{\rm{1}}}{{{{\left( {{\rm{2}}\pi {\sigma ^{\rm{2}}}} \right)}^{\rm{n}}}}}{\rm{ \times exp}}\left\{ {{\rm{ - }}\frac{{\rm{1}}}{{{\rm{2}}{\sigma ^{\rm{2}}}}}{\rm{ \times }}\left( {\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}{\mu _{\rm{i}}}} \right)}^{\rm{2}}}} {\rm{ + }}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{y}}_{\rm{i}}}{\rm{ - }}{\mu _{\rm{i}}}} \right)}^{\rm{2}}}} } \right)} \right\}\end{aligned}\)

Evaluating the maximum likelihood function

Look at the log likelihood function to locate the maximum.

\(\begin{array}{c}{\rm{lnf(x,y;\mu ,\sigma ) = ln}}\left( {\frac{{\rm{1}}}{{{{\left( {{\rm{2\pi }}{{\rm{\sigma }}^{\rm{2}}}} \right)}^{\rm{n}}}}}{\rm{ \times exp}}\left\{ {{\rm{ - }}\frac{{\rm{1}}}{{{\rm{2}}{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ \times }}\left( {\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}{{\rm{\mu }}_{\rm{i}}}} \right)}^{\rm{2}}}} {\rm{ + }}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{y}}_{\rm{i}}}{\rm{ - }}{{\rm{\mu }}_{\rm{i}}}} \right)}^{\rm{2}}}} } \right)} \right\}} \right)\\{\rm{ = - ln}}\left( {{\rm{2\pi }}{{\rm{\sigma }}^{\rm{2}}}} \right){\rm{ - }}\frac{{\rm{1}}}{{{\rm{2}}{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ \times }}\left( {\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}{{\rm{\mu }}_{\rm{i}}}} \right)}^{\rm{2}}}} {\rm{ + }}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{y}}_{\rm{i}}}{\rm{ - }}{{\rm{\mu }}_{\rm{i}}}} \right)}^{\rm{2}}}} } \right)\end{array}\)

The maximum likelihood estimator is generated by calculating the derivative of the log likelihood function in respect of\({{\rm{\mu }}_{\rm{i}}}{\rm{,i = 1,2, \ldots ,n}}\)and equating it with\({\rm{0}}\)the maximum likelihood estimator\({{\rm{\hat \mu }}_{\rm{i}}}\). As a result, the derivatives are,

\(\begin{aligned}\frac{{\rm{d}}}{{{\rm{d}}{{\rm{\mu }}_{\rm{i}}}}}f(x,y;\mu ,\sigma ) &= \frac{{\rm{d}}}{{{\rm{d}}{{\rm{\mu }}_{\rm{i}}}}}\left( {{\rm{ - ln}}\left( {{\rm{2\pi }}{{\rm{\sigma }}^{\rm{2}}}} \right){\rm{ - }}\frac{{\rm{1}}}{{{\rm{2}}{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ \times }}\left( {\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}{{\rm{\mu }}_{\rm{i}}}} \right)}^{\rm{2}}}} {\rm{ + }}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{y}}_{\rm{i}}}{\rm{ - }}{{\rm{\mu }}_{\rm{i}}}} \right)}^{\rm{2}}}} } \right)} \right)\\&= 0 - \frac{{\rm{1}}}{{{\rm{2}}{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ \times }}\left( {{\rm{2 \times }}\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}{{\rm{\mu }}_{\rm{i}}}} \right){\rm{ \times ( - 1) + 2 \times }}\left( {{{\rm{y}}_{\rm{i}}}{\rm{ - }}{{\rm{\mu }}_{\rm{i}}}} \right){\rm{ \times ( - 1)}}} \right)\\&= \frac{{\rm{1}}}{{{{\rm{\sigma }}^{\rm{2}}}}}\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}{{\rm{\mu }}_{\rm{i}}}{\rm{ + }}{{\rm{y}}_{\rm{i}}}{\rm{ - }}{{\rm{\mu }}_{\rm{i}}}} \right)\\ & = \frac{{\rm{1}}}{{{{\rm{\sigma }}^{\rm{2}}}}}\left( {{{\rm{x}}_{\rm{i}}}{\rm{ + }}{{\rm{y}}_{\rm{i}}}{\rm{ - 2}}{{\rm{\mu }}_{\rm{i}}}} \right)\end{aligned}\)

As a result, solving equation gives the maximum likelihood estimator\({{\rm{\hat \mu }}_{\rm{i}}}\).

\(\begin{array}{c}\frac{{\rm{1}}}{{{{\rm{\sigma }}^{\rm{2}}}}}\left( {{{\rm{x}}_{\rm{i}}}{\rm{ + }}{{\rm{y}}_{\rm{i}}}{\rm{ - 2}}{{{\rm{\hat \mu }}}_{\rm{i}}}} \right){\rm{ = 0}}\\{\rm{2}}{{{\rm{\hat \mu }}}_{\rm{i}}}{\rm{ = }}{{\rm{x}}_{\rm{i}}}{\rm{ + }}{{\rm{y}}_{\rm{y}}}\end{array}\)

For\({{\rm{\hat \mu }}_{\rm{i}}}\). The maximum likelihood estimator of parameter\({{\rm{\mu }}_{{\rm{i,}}}}{\rm{i = 1,2, \ldots ,n}}\)is,

\({{\rm{\hat \mu }}_{\rm{i}}}{\rm{ = }}\frac{{{{\rm{X}}_{\rm{i}}}{\rm{ + }}{{\rm{Y}}_{\rm{i}}}}}{{\rm{2}}}\)

To get the greatest likelihood \({{\rm{\sigma }}^{\rm{2}}}\) estimator, first substitute \({{\rm{\hat \mu }}_{\rm{i}}}\) in the log likelihood function, then take the derivative in respect of \({{\rm{\sigma }}^{\rm{2}}}\), equal it with zero, and solve the equation to get the result.

Evaluating the value

The log likelihood function is transformed by replacing\({{\rm{\hat \mu }}_{\rm{i}}}\).

\(\begin{array}{c}{\rm{lnf(x,y;\hat \mu ,\sigma ) = - ln}}\left( {{\rm{2\pi }}{{\rm{\sigma }}^{\rm{2}}}} \right){\rm{ - }}\frac{{\rm{1}}}{{{\rm{2}}{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ \times }}\left( {\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}\frac{{{{\rm{x}}_{\rm{i}}}{\rm{ + }}{{\rm{y}}_{\rm{i}}}}}{{\rm{2}}}} \right)}^{\rm{2}}}} {\rm{ + }}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{y}}_{\rm{i}}}{\rm{ - }}\frac{{{{\rm{x}}_{\rm{i}}}{\rm{ + }}{{\rm{y}}_{\rm{i}}}}}{{\rm{2}}}} \right)}^{\rm{2}}}} } \right)\\\mathop {\rm{ = }}\limits^{{\rm{(1)}}} {\rm{ - ln}}\left( {{\rm{2\pi }}{{\rm{\sigma }}^{\rm{2}}}} \right){\rm{ - }}\frac{{\rm{1}}}{{{\rm{2}}{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ \times }}\left( {\frac{{\rm{1}}}{{\rm{2}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}{{\rm{y}}_{\rm{i}}}} \right)}^{\rm{2}}}} } \right)\end{array}\)

1. : It can also be proven from the hint.

The maximum likelihood estimator\({\rm{\hat \sigma }}\)is generated by taking the derivative of the log likelihood function in relation to\({{\rm{\sigma }}^{\rm{2}}}\)and equating it with\({\rm{0}}\).

As a result, the derivative,

\(\begin{array}{c}\frac{{\rm{d}}}{{{\rm{d}}{{\rm{\sigma }}^{\rm{2}}}}}{\rm{f(x,y;\hat \mu ,\sigma ) = }}\frac{{\rm{d}}}{{{\rm{d}}{{\rm{\sigma }}^{\rm{2}}}}}\left( {{\rm{ - ln}}\left( {{\rm{2\pi }}{{\rm{\sigma }}^{\rm{2}}}} \right){\rm{ - }}\frac{{\rm{1}}}{{{\rm{2}}{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ \times }}\left( {\frac{{\rm{1}}}{{\rm{2}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}{{\rm{y}}_{\rm{i}}}} \right)}^{\rm{2}}}} } \right)} \right)\\{\rm{ = - n}}\frac{{\rm{1}}}{{{{\rm{\sigma }}^{\rm{2}}}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{2}}{{\rm{\sigma }}^{\rm{4}}}}}\left( {\frac{{\rm{1}}}{{\rm{2}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}{{\rm{y}}_{\rm{i}}}} \right)}^{\rm{2}}}} } \right)\end{array}\)

As a result, solving equation yields the maximum likelihood estimator\({\rm{\hat \sigma }}\).

\(\begin{aligned}{\rm{ - n}}\frac{{\rm{1}}}{{{{{\rm{\hat \sigma }}}^{\rm{2}}}}}{\rm{ + }}\frac{{\rm{1}}}{{{\rm{2}}{{{\rm{\hat \sigma }}}^{\rm{4}}}}}\left( {\frac{{\rm{1}}}{{\rm{2}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}{{\rm{y}}_{\rm{i}}}} \right)}^{\rm{2}}}} } \right) &= 0 \\\frac{{\rm{1}}}{{{\rm{4}}{{{\rm{\hat \sigma }}}^{\rm{4}}}}}\left( {\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}{{\rm{y}}_{\rm{i}}}} \right)}^{\rm{2}}}} } \right) & = \frac{{\rm{n}}}{{{{{\rm{\hat \sigma }}}^{\rm{2}}}}}\\{\rm{4n}}{{{\rm{\hat \sigma }}}^{\rm{2}}} & = \sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{x}}_{\rm{i}}}{\rm{ - }}{{\rm{y}}_{\rm{i}}}} \right)}^{\rm{2}}}} \end{aligned}\)

For\({\rm{\hat \sigma }}\). The maximum likelihood estimator of parameter\({\rm{\hat \sigma }}\)is,

\({{\rm{\hat \sigma }}^{\rm{2}}}{\rm{ = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{X}}_{\rm{i}}}{\rm{ - }}{{\rm{Y}}_{\rm{i}}}} \right)}^{\rm{2}}}} }}{{{\rm{4n}}}}\)

Explanation

b.If \({\rm{E}}\left( {{{{\rm{\hat \sigma }}}^{\rm{2}}}} \right){\rm{ = }}{{\rm{\sigma }}^{\rm{2}}}\), the mle \({{\rm{\hat \sigma }}^{\rm{2}}}\) will be unbiased. But, since

\(\begin{aligned}{\rm{E}}\left( {{{{\rm{\hat \sigma }}}^{\rm{2}}}} \right)&= E \left( {\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{X}}_{\rm{i}}}{\rm{ - }}{{\rm{Y}}_{\rm{i}}}} \right)}^{\rm{2}}}} }}{{{\rm{4n}}}}} \right)\\& = \frac{{\rm{1}}}{{{\rm{4n}}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{E}} {\left( {{{\rm{X}}_{\rm{i}}}{\rm{ - }}{{\rm{Y}}_{\rm{i}}}} \right)^{\rm{2}}}\\& = \frac{{\rm{1}}}{{{\rm{4n}}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left( {{\rm{V}}\left( {{{\rm{X}}_{\rm{i}}}{\rm{ - }}{{\rm{Y}}_{\rm{i}}}} \right){\rm{ + }}{{\left( {{\rm{E}}\left( {{{\rm{X}}_{\rm{i}}}{\rm{ - }}{{\rm{Y}}_{\rm{i}}}} \right)} \right)}^{\rm{2}}}} \right)} \\ & = \frac{{\rm{1}}}{{{\rm{4n}}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left( {{\rm{V}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ + ( - 1}}{{\rm{)}}^{\rm{2}}}{\rm{V}}\left( {{{\rm{Y}}_{\rm{i}}}} \right){\rm{ + }}{{\left( {{\rm{E}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ - E}}\left( {{{\rm{Y}}_{\rm{i}}}} \right)} \right)}^{\rm{2}}}} \right)} \\& = \frac{{\rm{1}}}{{{\rm{4n}}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\left( {{\rm{2}}{{\rm{\sigma }}^{\rm{2}}}{\rm{ + }}{{\left( {{{\rm{\mu }}_{\rm{i}}}{\rm{ - }}{{\rm{\mu }}_{\rm{i}}}} \right)}^{\rm{2}}}} \right)} \\&= \frac{{\rm{1}}}{{{\rm{4n}}}}\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {\rm{2}} {{\rm{\sigma }}^{\rm{2}}} &= \frac{{\rm{1}}}{{{\rm{2n}}}}{\rm{ \times n}}{{\rm{\sigma }}^{\rm{2}}}\\ &= \frac{{\rm{1}}}{{\rm{2}}}{{\rm{\sigma }}^{\rm{2}}} \ne {{\rm{\sigma }}^{\rm{2}}}\end{aligned}\)

The maximum likelihood estimator isn't completely objective. Nonetheless, the estimator,

\({\rm{\hat \sigma }}_{\rm{1}}^{\rm{2}}{\rm{ = }}\frac{{\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{{\left( {{{\rm{X}}_{\rm{i}}}{\rm{ - }}{{\rm{Y}}_{\rm{i}}}} \right)}^{\rm{2}}}} }}{{{\rm{2n}}}}{\rm{ }}\)

will be genuine This is a result of

\(\begin{array}{c}{\rm{E}}\left( {{\rm{\hat \sigma }}_{\rm{1}}^{\rm{2}}} \right){\rm{ = E}}\left( {{\rm{2 \times }}{{{\rm{\hat \sigma }}}^{\rm{2}}}} \right)\\{\rm{ = 2 \times }}\frac{{\rm{1}}}{{\rm{2}}}{{\rm{\sigma }}^{\rm{2}}}\\{\rm{ = }}{{\rm{\sigma }}^{\rm{2}}}\end{array}\)