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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

01

Define exponential function

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

02

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:

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

03

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.

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Most popular questions from this chapter

The accompanying data on flexural strength (MPa) for concrete beams of a certain type was introduced in Example 1.2.

\(\begin{array}{*{20}{r}}{{\rm{5}}{\rm{.9}}}&{{\rm{7}}{\rm{.2}}}&{{\rm{7}}{\rm{.3}}}&{{\rm{6}}{\rm{.3}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{6}}{\rm{.8}}}&{{\rm{7}}{\rm{.0}}}\\{{\rm{7}}{\rm{.6}}}&{{\rm{6}}{\rm{.8}}}&{{\rm{6}}{\rm{.5}}}&{{\rm{7}}{\rm{.0}}}&{{\rm{6}}{\rm{.3}}}&{{\rm{7}}{\rm{.9}}}&{{\rm{9}}{\rm{.0}}}\\{{\rm{3}}{\rm{.2}}}&{{\rm{8}}{\rm{.7}}}&{{\rm{7}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{7}}{\rm{.7}}}&{{\rm{9}}{\rm{.7}}}\\{{\rm{7}}{\rm{.3}}}&{{\rm{7}}{\rm{.7}}}&{{\rm{11}}{\rm{.6}}}&{{\rm{11}}{\rm{.3}}}&{{\rm{11}}{\rm{.8}}}&{{\rm{10}}{\rm{.7}}}&{}\end{array}\)

Calculate a point estimate of the mean value of strength for the conceptual population of all beams manufactured in this fashion, and state which estimator you used\({\rm{(Hint:\Sigma }}{{\rm{x}}_{\rm{i}}}{\rm{ = 219}}{\rm{.8}}{\rm{.)}}\)

b. Calculate a point estimate of the strength value that separates the weakest 50% of all such beams from the strongest 50 %, and state which estimator you used.

c. Calculate and interpret a point estimate of the population standard deviation\({\rm{\sigma }}\). Which estimator did you use?\({\rm{(Hint:}}\left. {{\rm{\Sigma x}}_{\rm{i}}^{\rm{2}}{\rm{ = 1860}}{\rm{.94}}{\rm{.}}} \right)\)

d. Calculate a point estimate of the proportion of all such beams whose flexural strength exceeds\({\rm{10MPa}}\). (Hint: Think of an observation as a "success" if it exceeds 10.)

e. Calculate a point estimate of the population coefficient of variation\({\rm{\sigma /\mu }}\), and state which estimator you used.

Urinary angiotensinogen (AGT) level is one quantitative indicator of kidney function. The article 鈥淯rinary Angiotensinogen as a Potential Biomarker of Chronic Kidney Diseases鈥� (J. of the Amer. Society of Hypertension, \({\rm{2008: 349 - 354}}\)) describes a study in which urinary AGT level \({\rm{(\mu g)}}\) was determined for a sample of adults with chronic kidney disease. Here is representative data (consistent with summary quantities and descriptions in the cited article):

An appropriate probability plot supports the use of the lognormal distribution (see Section \({\rm{4}}{\rm{.5}}\)) as a reasonable model for urinary AGT level (this is what the investigators did).

a. Estimate the parameters of the distribution. (Hint: Rem ember that \({\rm{X}}\) has a lognormal distribution with parameters \({\rm{\mu }}\) and \({{\rm{\sigma }}^{\rm{2}}}\) if \({\rm{ln(X)}}\) is normally distributed with mean \({\rm{\mu }}\) and variance \({{\rm{\sigma }}^{\rm{2}}}\).)

b. Use the estimates of part (a) to calculate an estimate of the expected value of AGT level. (Hint: What is \({\rm{E(X)}}\)?)

Of \({{\rm{n}}_{\rm{1}}}\)randomly selected male smokers, \({{\rm{X}}_{\rm{1}}}\) smoked filter cigarettes, whereas of \({{\rm{n}}_{\rm{2}}}\) randomly selected female smokers, \({{\rm{X}}_{\rm{2}}}\) smoked filter cigarettes. Let \({{\rm{p}}_{\rm{1}}}\) and \({{\rm{p}}_{\rm{2}}}\) denote the probabilities that a randomly selected male and female, respectively, smoke filter cigarettes.

a. Show that \({\rm{(}}{{\rm{X}}_{\rm{1}}}{\rm{/}}{{\rm{n}}_{\rm{1}}}{\rm{) - (}}{{\rm{X}}_{\rm{2}}}{\rm{/}}{{\rm{n}}_{\rm{2}}}{\rm{)}}\) is an unbiased estimator for \({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\). (Hint: \({\rm{E(}}{{\rm{X}}_{\rm{i}}}{\rm{) = }}{{\rm{n}}_{\rm{i}}}{{\rm{p}}_{\rm{i}}}\) for \({\rm{i = 1,2}}\).)

b. What is the standard error of the estimator in part (a)?

c. How would you use the observed values \({{\rm{x}}_{\rm{1}}}\) and \({{\rm{x}}_{\rm{2}}}\) to estimate the standard error of your estimator?

d. If \({{\rm{n}}_{\rm{1}}}{\rm{ = }}{{\rm{n}}_{\rm{2}}}{\rm{ = 200, }}{{\rm{x}}_{\rm{1}}}{\rm{ = 127}}\), and \({{\rm{x}}_{\rm{2}}}{\rm{ = 176}}\), use the estimator of part (a) to obtain an estimate of \({{\rm{p}}_{\rm{1}}}{\rm{ - }}{{\rm{p}}_{\rm{2}}}\).

e. Use the result of part (c) and the data of part (d) to estimate the standard error of the estimator.

\({{\rm{X}}_{\rm{1}}}{\rm{,}}.....{\rm{,}}{{\rm{X}}_{\rm{n}}}\)be a random sample from a gamma distribution with parameters \({\rm{\alpha }}\) and \({\rm{\beta }}\). a. Derive the equations whose solutions yield the maximum likelihood estimators of \({\rm{\alpha }}\) and \({\rm{\beta }}\). Do you think they can be solved explicitly? b. Show that the mle of \({\rm{\mu = \alpha \beta }}\) is \(\widehat {\rm{\mu }}{\rm{ = }}\overline {\rm{X}} \).

Consider randomly selecting \({\rm{n}}\) segments of pipe and determining the corrosion loss (mm) in the wall thickness for each one. Denote these corrosion losses by \({{\rm{Y}}_{\rm{1}}}{\rm{,}}.....{\rm{,}}{{\rm{Y}}_{\rm{n}}}\). The article 鈥淎 Probabilistic Model for a Gas Explosion Due to Leakages in the Grey Cast Iron Gas Mains鈥� (Reliability Engr. and System Safety (\({\rm{(2013:270 - 279)}}\)) proposes a linear corrosion model: \({{\rm{Y}}_{\rm{i}}}{\rm{ = }}{{\rm{t}}_{\rm{i}}}{\rm{R}}\), where \({{\rm{t}}_{\rm{i}}}\) is the age of the pipe and \({\rm{R}}\), the corrosion rate, is exponentially distributed with parameter \({\rm{\lambda }}\). Obtain the maximum likelihood estimator of the exponential parameter (the resulting mle appears in the cited article). (Hint: If \({\rm{c > 0}}\) and \({\rm{X}}\) has an exponential distribution, so does \({\rm{cX}}\).)
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