Q34E The mean squared error of an est... [FREE SOLUTION]

91影视

Probability And Statistics For Engineering And Sciences

Jay L. Devore

$Math Studyset 91影视 Explanations$ Math

9th Edition 路 714 pages

ISBN: 9781305251809

Chapter 6: Q34E (page 275)

The mean squared error of an estimator ${\rm{\hat \theta }}$ is ${\rm{MSE(\hat \theta ) = E(\hat \theta - \hat \theta }}{{\rm{)}}^{\rm{2}}}$. If ${\rm{\hat \theta }}$ is unbiased, then ${\rm{MSE(\hat \theta ) = V(\hat \theta )}}$, but in general ${\rm{MSE(\hat \theta ) = V(\hat \theta ) + (bias}}{{\rm{)}}^{\rm{2}}}$ . Consider the estimator ${{\rm{\hat \sigma }}^{\rm{2}}}{\rm{ = K}}{{\rm{S}}^{\rm{2}}}$, where ${{\rm{S}}^{\rm{2}}}{\rm{ = }}$ sample variance. What value of K minimizes the mean squared error of this estimator when the population distribution is normal? (Hint: It can be shown that ${\rm{E}}\left( {{{\left( {{{\rm{S}}^{\rm{2}}}} \right)}^{\rm{2}}}} \right){\rm{ = (n + 1)}}{{\rm{\sigma }}^{\rm{4}}}{\rm{/(n - 1)}}$ In general, it is difficult to find ${\rm{\hat \theta }}$ to minimize ${\rm{MSE(\hat \theta )}}$, which is why we look only at unbiased estimators and minimize ${\rm{V(\hat \theta )}}$.)

Short Answer

Expert verified

The value of ${\rm{K = }}\frac{{{\rm{n - 1}}}}{{{\rm{n - 3}}}}$.

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

To begin, calculate the mean squared error of${{\rm{\hat \sigma }}^{\rm{2}}}{\rm{ = K}}{{\rm{S}}^{\rm{2}}}$. Take note of this:

${\rm{MSE}}\left( {{{{\rm{\hat \sigma }}}^{\rm{2}}}} \right){\rm{ = V}}\left( {\widehat {{{\rm{\sigma }}^{\rm{2}}}}} \right){\rm{ + Bias}}\left( {{{{\rm{\hat \sigma }}}^{\rm{2}}}} \right)$

where there is a bias.

$\begin{aligned}{\rm{Bias}}\left( {{{{\rm{\hat \sigma }}}^{\rm{2}}}} \right) &= E \left( {{{{\rm{\hat \sigma }}}^{\rm{2}}}} \right){\rm{ - }}{{\rm{\sigma }}^{\rm{2}}}\\&= E \left( {{\rm{K}}{{\rm{S}}^{\rm{2}}}} \right){\rm{ - }}{{\rm{\sigma }}^{\rm{2}}}\\ &= K {{\rm{\sigma }}^{\rm{2}}}{\rm{ - }}{{\rm{\sigma }}^{\rm{2}}}\\ & = {{\rm{\sigma }}^{\rm{2}}}{\rm{(K - 1)}}\end{aligned}$

We used the notion that ${{\rm{S}}^{\rm{2}}}$ is an unbiased ${{\rm{\sigma }}^{\rm{2}}}$ estimator and the definition of bias in this example. The variance can be computed as follows:

$\begin{aligned}{\rm{V}}\left( {\widehat {{{\rm{\sigma }}^{\rm{2}}}}} \right) &= V \left( {{\rm{K}}{{\rm{S}}^{\rm{2}}}} \right)\\ &= {{\rm{K}}^{\rm{2}}}{\rm{V}}\left( {{{\rm{S}}^{\rm{2}}}} \right)\\ &= {{\rm{K}}^{\rm{2}}}\left( {{\rm{E}}{{\left( {{{\rm{S}}^{\rm{2}}}} \right)}^{\rm{2}}}{\rm{ - }}{{\left( {{\rm{E}}\left( {{{\rm{S}}^{\rm{2}}}} \right)} \right)}^{\rm{2}}}} \right)\\ &= {{\rm{K}}^{\rm{2}}}\left( {\frac{{{\rm{n + 1}}}}{{{\rm{n - 1}}}}{{\rm{\sigma }}^{\rm{4}}}{\rm{ - }}{{\left( {{{\rm{\sigma }}^{\rm{2}}}} \right)}^{\rm{2}}}} \right)\\ & = {{\rm{K}}^{\rm{2}}}{{\rm{\sigma }}^{\rm{4}}}\left( {\frac{{{\rm{(n + 1)}}}}{{{\rm{n - 1}}}}{\rm{ - 1}}} \right)\end{aligned}$

Evaluating the value

The derivative of the MSE is required to obtain the value of K that minimises the MSE. The estimator${\rm{\hat \theta }}$mean squared error is,

$\begin{array}{c}{\rm{MSE}}\left( {{{{\rm{\hat \sigma }}}^{\rm{2}}}} \right){\rm{ = }}{{\rm{K}}^{\rm{2}}}{{\rm{\sigma }}^{\rm{4}}}\left( {\frac{{{\rm{(n + 1)}}}}{{{\rm{n - 1}}}}{\rm{ - 1}}} \right){\rm{ - }}{\left( {{{\rm{\sigma }}^{\rm{2}}}{\rm{(K - 1)}}} \right)^{\rm{2}}}\\{\rm{ = }}{{\rm{K}}^{\rm{2}}}{{\rm{\sigma }}^{\rm{4}}}\left( {\frac{{{\rm{(n + 1)}}}}{{{\rm{n - 1}}}}{\rm{ - 1}}} \right){\rm{ - }}{{\rm{\sigma }}^{\rm{4}}}{{\rm{(K - 1)}}^{\rm{2}}}\end{array}$

In terms of K, its derivative is,

$\begin{array}{c}\frac{{\rm{d}}}{{{\rm{dK}}}}{\rm{MSE}}\left( {{{{\rm{\hat \sigma }}}^{\rm{2}}}} \right){\rm{ = }}\frac{{\rm{d}}}{{{\rm{dK}}}}\left( {{{\rm{K}}^{\rm{2}}}{{\rm{\sigma }}^{\rm{4}}}\left( {\frac{{{\rm{(n + 1)}}}}{{{\rm{n - 1}}}}{\rm{ - 1}}} \right){\rm{ - }}{{\rm{\sigma }}^{\rm{4}}}{{{\rm{(K - 1)}}}^{\rm{2}}}} \right)\\{\rm{ = 2K}}{{\rm{\sigma }}^{\rm{4}}}\left( {\frac{{{\rm{(n + 1)}}}}{{{\rm{n - 1}}}}{\rm{ - 1}}} \right){\rm{ - 2}}{{\rm{\sigma }}^{\rm{4}}}{\rm{(K - 1)}}\end{array}$

and theminimum comes from,

${\rm{2K}}{{\rm{\sigma }}^{\rm{4}}}\left( {\frac{{{\rm{(n + 1)}}}}{{{\rm{n - 1}}}}{\rm{ - 1}}} \right){\rm{ - 2}}{{\rm{\sigma }}^{\rm{4}}}{\rm{(K - 1) = 0}}$

or, similarly,

$\begin{aligned}{\rm{2}}{{\rm{\sigma }}^{\rm{4}}}{\rm{K}}\left( {\frac{{{\rm{(n + 1)}}}}{{{\rm{n - 1}}}}{\rm{ - 1}}} \right) &= 2 {{\rm{\sigma }}^{\rm{4}}}{\rm{(K - 1)}}\\{\rm{K}}\left( {\frac{{{\rm{(n + 1)}}}}{{{\rm{n - 1}}}}{\rm{ - 1}}} \right)&= K - 1 \\{\rm{K}}\left( {\frac{{{\rm{(n + 1)}}}}{{{\rm{n - 1}}}}{\rm{ - 1 - 1}}} \right) &= - 1 \\{\rm{K}}\left( {\frac{{{\rm{(n + 1)}}}}{{{\rm{n - 1}}}}{\rm{ - }}\frac{{{\rm{2n - 2}}}}{{{\rm{n - 1}}}}} \right) &= - 1\\{\rm{K}}\left( {\frac{{{\rm{ - n + 3}}}}{{{\rm{n - 1}}}}} \right) &= - 1 \end{aligned}$

As a result, the K value that minimises MSE is,

${\rm{K = }}\frac{{{\rm{n - 1}}}}{{{\rm{n - 3}}}}$.

The unbiased estimator (which was derived earlier) is obtained for ${\rm{K = 1}}$ and the maximum likelihood estimator is produced for ${\rm{K = }}\frac{{{\rm{n - 1}}}}{{\rm{n}}}$. As a result, this one is distinct from both, and it is neither unbiased nor mle.