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Chapter 6: Q16E (page 263)

Suppose the true average growth\({\rm{\mu }}\)of one type of plant during a l-year period is identical to that of a second type, but the variance of growth for the first type is\({{\rm{\sigma }}^{\rm{2}}}\), whereas for the second type the variance is\({\rm{4}}{{\rm{\sigma }}^{\rm{2}}}{\rm{. Let }}{{\rm{X}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{m}}}\)be\({\rm{m}}\)independent growth observations on the first type (so\({\rm{E}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ = \mu ,V}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ = \sigma\hat 2}}\)$ ), and let\({{\rm{Y}}_{\rm{1}}}{\rm{, \ldots ,}}{{\rm{Y}}_{\rm{n}}}\)be\({\rm{n}}\)independent growth observations on the second type\(\left( {{\rm{E}}\left( {{{\rm{Y}}_{\rm{i}}}} \right){\rm{ = \mu ,V}}\left( {{{\rm{Y}}_{\rm{j}}}} \right){\rm{ = 4}}{{\rm{\sigma }}^{\rm{2}}}} \right)\)

a. Show that the estimator\({\rm{\hat \mu = \delta \bar X + (1 - \delta )\bar Y}}\)is unbiased for\({\rm{\mu }}\)(for\({\rm{0 < \delta < 1}}\), the estimator is a weighted average of the two individual sample means).

b. For fixed\({\rm{m}}\)and\({\rm{n}}\), compute\({\rm{V(\hat \mu ),}}\)and then find the value of\({\rm{\delta }}\)that minimizes\({\rm{V(\hat \mu )}}\). (Hint: Differentiate\({\rm{V(\hat \mu )}}\)with respect to\({\rm{\delta }}{\rm{.)}}\)

Short Answer

Expert verified

a) \({\rm{\hat \mu is an unbiased estimator for \mu }}{\rm{. }}\)

b) The value \({\rm{\delta = }}\frac{{{\rm{4m}}}}{{{\rm{n + 4m}}}}\) minimized\({\rm{V(\hat \mu )}}\).

Step by step solution

01

Introduction

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.

02

Proofing estimator unbiased

a)

Consider the given information,

\(\begin{aligned}{\rm{E}}\left( {{{\rm{X}}_{\rm{i}}}} \right) &= \mu V \left( {{{\rm{X}}_{\rm{i}}}} \right)\\& = {{\rm{\sigma }}^{\rm{2}}}{\rm{E}}\left( {{{\rm{Y}}_{\rm{i}}}} \right) & = \mu V \left( {{{\rm{Y}}_{\rm{i}}}} \right)\\ & = {{\rm{\sigma }}^{\rm{2}}}{\rm{\hat \mu }}\\ &= \delta \bar X + (1 - \delta )\bar Y\end{aligned}\)

The mean of the sampling distribution of the sample mean is the population mean:

\(\begin{array}{c}{\rm{E(\bar X) = E}}\left( {{{\rm{X}}_{\rm{i}}}} \right){\rm{ = \mu }}\\{\rm{E(\bar Y) = E}}\left( {{{\rm{Y}}_{\rm{i}}}} \right){\rm{ = \mu }}\end{array}\)

Determine the expected value of \({\rm{\hat \mu }}\) :

\(\begin{aligned} E(\hat \mu ) &= E(\delta \bar X + (1 - \delta )\bar Y) \\ &= \delta E(\bar X) + (1 - \delta )E(\bar Y)\\ &= \delta \mu + (1 - \delta )\mu \\ & = \delta \mu + \mu - \delta \mu \\ &= \mu \end{aligned}\)

Since, the expected value of \({\rm{\hat \mu }}\) is \({\rm{\mu ,\hat \mu }}\)is called an unbiased estimator for\({\rm{\mu }}\).

Now, finding the value of\({\rm{\delta }}\),

\(\begin{aligned}{\rm{E}}\left( {{{\rm{X}}_{\rm{i}}}} \right) &= \mu V\left( {{{\rm{X}}_{\rm{i}}}} \right)\\ &= {{\rm{\sigma }}^{\rm{2}}}{\rm{E}}\left( {{{\rm{Y}}_{\rm{i}}}} \right)\\ &= \mu V \left( {{{\rm{Y}}_{\rm{i}}}} \right)\\ & = {{\rm{\sigma }}^{\rm{2}}}{\rm{\hat \mu = \delta \bar X + (1 - \delta )\bar Y}}\end{aligned}\)

The mean of the sampling distribution of the sample mean is the population mean:

\(\begin{aligned}E(\bar X) &= E\left( {{{\rm{X}}_{\rm{i}}}} \right)\\ &= \mu E(\bar Y) = E \left( {{{\rm{Y}}_{\rm{i}}}} \right)\\ &= \mu \end{aligned}\)

The population variance divided by the sample size equals the variance of the sampling distribution of the sample mean:

\(\begin{array}{c}{\rm{V(\bar X) = }}\frac{{{\rm{V}}\left( {{{\rm{X}}_{\rm{i}}}} \right)}}{{\rm{m}}}\\{\rm{ = }}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{m}}}{\rm{V(\bar Y)}}\\{\rm{ = }}\frac{{{\rm{V}}\left( {{{\rm{Y}}_{\rm{i}}}} \right)}}{{\rm{n}}}\\{\rm{ = }}\frac{{{\rm{4}}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}\end{array}\)

Determine the variance of \({\rm{\hat \mu }}\)(using the property \({\rm{V(aX + bY) = }}{{\rm{a}}^{\rm{2}}}{\rm{V(x) + }}{{\rm{b}}^{\rm{2}}}{\rm{V(Y)}}\)for the variance, when \({\rm{X}}\)and \({\rm{Y}}\)are independent):

03

Calculation

(b)

Consider the given information,

\(\begin{array}{l}{\rm{V(\hat \mu ) = V(\delta \bar X + (1 - \delta )\bar Y)}}\\{\rm{ = }}{{\rm{\delta }}^{\rm{2}}}{\rm{V(\bar X) + (1 - \delta }}{{\rm{)}}^{\rm{2}}}{\rm{V(\bar Y)}}\\{\rm{ = }}{{\rm{\delta }}^{\rm{2}}}\frac{{{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{m}}}{\rm{ + (1 - \delta }}{{\rm{)}}^{\rm{2}}}\frac{{{\rm{4}}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}\\{\rm{ = }}\frac{{{{\rm{\delta }}^{\rm{2}}}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{m}}}{\rm{ + }}\frac{{{\rm{4}}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}{\rm{ - }}\frac{{{\rm{8\delta }}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}{\rm{ + }}\frac{{{\rm{4}}{{\rm{\delta }}^{\rm{2}}}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}\\{\rm{ = }}\frac{{{{\rm{\delta }}^{\rm{2}}}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{m}}}{\rm{ + }}\frac{{{\rm{4}}{{\rm{\delta }}^{\rm{2}}}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}{\rm{ - }}\frac{{{\rm{8\delta }}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}{\rm{ + }}\frac{{{\rm{4}}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}\end{array}\)

Differentiate with respect to\({\rm{\delta }}\):

\(\begin{aligned}\frac{{\rm{d}}}{{{\rm{d\delta }}}}V(\hat \mu ) &= \frac{{\rm{d}}}{{{\rm{d\delta }}}}\left( {\frac{{{{\rm{\delta }}^{\rm{2}}}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{m}}}{\rm{ + }}\frac{{{\rm{4}}{{\rm{\delta }}^{\rm{2}}}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}{\rm{ - }}\frac{{{\rm{8\delta }}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}{\rm{ + }}\frac{{{\rm{4}}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}} \right)\\ &= \frac{{{\rm{2\delta }}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{m}}}{\rm{ + }}\frac{{{\rm{8\delta }}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}{\rm{ - }}\frac{{{\rm{8}}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}\end{aligned}\)

The minimum is the value for \({\rm{\delta }}\)for which the expression becomes zero:

\(\frac{{{\rm{2\delta }}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{m}}}{\rm{ + }}\frac{{{\rm{8\delta }}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}{\rm{ - }}\frac{{{\rm{8}}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}{\rm{ = 0}}\)

Add \(\frac{{{\rm{8}}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}\) to each side of the equation:

\(\frac{{{\rm{2n\delta }}{{\rm{\sigma }}^{\rm{2}}}{\rm{ + 8m\delta }}{{\rm{\sigma }}^{\rm{2}}}}}{{{\rm{mn}}}}{\rm{ = }}\frac{{{\rm{8}}{{\rm{\sigma }}^{\rm{2}}}}}{{\rm{n}}}\)

Multiply each side of the equation by\({\rm{mn}}\):

\({\rm{2n\delta }}{{\rm{\sigma }}^{\rm{2}}}{\rm{ + 8m\delta }}{{\rm{\sigma }}^{\rm{2}}}{\rm{ = 8}}{{\rm{\sigma }}^{\rm{2}}}{\rm{m}}\)

Factor out\({\rm{\delta }}\):

\(\left( {{\rm{2n}}{{\rm{\sigma }}^{\rm{2}}}{\rm{ + 8m}}{{\rm{\sigma }}^{\rm{2}}}} \right){\rm{\delta = 8}}{{\rm{\sigma }}^{\rm{2}}}{\rm{m}}\)

Divide each side of the equation by\({\rm{2n}}{{\rm{\sigma }}^{\rm{2}}}{\rm{ + 8m}}{{\rm{\sigma }}^{\rm{2}}}\):

\({\rm{\delta = }}\frac{{{\rm{8}}{{\rm{\sigma }}^{\rm{2}}}{\rm{m}}}}{{{\rm{2n}}{{\rm{\sigma }}^{\rm{2}}}{\rm{ + 8m}}{{\rm{\sigma }}^{\rm{2}}}}}\)

Divide the numerator and denominator by\({\rm{2}}{{\rm{\sigma }}^{\rm{2}}}\):

\({\rm{\delta = }}\frac{{{\rm{4m}}}}{{{\rm{n + 4m}}}}\)

Thus, the value \({\rm{\delta = }}\frac{{{\rm{4m}}}}{{{\rm{n + 4m}}}}\) minimized\({\rm{V(\hat \mu )}}\).

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Most popular questions from this chapter

Let\({{\rm{X}}_{\rm{1}}}{\rm{,}}{{\rm{X}}_{\rm{2}}}{\rm{, \ldots ,}}{{\rm{X}}_{\rm{n}}}\)represent a random sample from a Rayleigh distribution with pdf

\({\rm{f(x,\theta ) = }}\frac{{\rm{x}}}{{\rm{\theta }}}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/(2\theta )}}}}\quad {\rm{x > 0}}\)a. It can be shown that\({\rm{E}}\left( {{{\rm{X}}^{\rm{2}}}} \right){\rm{ = 2\theta }}\). Use this fact to construct an unbiased estimator of\({\rm{\theta }}\)based on\({\rm{\Sigma X}}_{\rm{i}}^{\rm{2}}\)(and use rules of expected value to show that it is unbiased).

b. Estimate\({\rm{\theta }}\)from the following\({\rm{n = 10}}\)observations on vibratory stress of a turbine blade under specified conditions:

\(\begin{array}{*{20}{l}}{{\rm{16}}{\rm{.88}}}&{{\rm{10}}{\rm{.23}}}&{{\rm{4}}{\rm{.59}}}&{{\rm{6}}{\rm{.66}}}&{{\rm{13}}{\rm{.68}}}\\{{\rm{14}}{\rm{.23}}}&{{\rm{19}}{\rm{.87}}}&{{\rm{9}}{\rm{.40}}}&{{\rm{6}}{\rm{.51}}}&{{\rm{10}}{\rm{.95}}}\end{array}\)

a. A random sample of 10 houses in a particular area, each of which is heated with natural gas, is selected and the amount of gas (therms) used during the month of January is determined for each house. The resulting observations are 103, 156, 118, 89, 125, 147, 122, 109, 138, 99. Let m denote the average gas usage during January by all houses in this area. Compute a point estimate of m. b. Suppose there are 10,000 houses in this area that use natural gas for heating. Let t denote the total amount of gas used by all of these houses during January. Estimate t using the data of part (a). What estimator did you use in computing your estimate? c. Use the data in part (a) to estimate p, the proportion of all houses that used at least 100 therms. d. Give a point estimate of the population median usage (the middle value in the population of all houses) based on the sample of part (a). What estimator did you use?

Each of 150 newly manufactured items is examined and the number of scratches per item is recorded (the items are supposed to be free of scratches), yielding the following data:

Assume that X has a Poisson distribution with parameter \({\bf{\mu }}.\)and that X represents the number of scratches on a randomly picked item.

a. Calculate the estimate for the data using an unbiased \({\bf{\mu }}.\)estimator. (Hint: for X Poisson, \({\rm{E(X) = \mu }}\) ,therefore \({\rm{E(\bar X) = ?)}}\)

c. What is your estimator's standard deviation (standard error)? Calculate the standard error estimate. (Hint: \({\rm{\sigma }}_{\rm{X}}^{\rm{2}}{\rm{ = \mu }}\), \({\rm{X}}\))

The article from which the data in Exercise 1 was extracted also gave the accompanying strength observations for cylinders:

\(\begin{array}{l}\begin{array}{*{20}{r}}{{\rm{6}}{\rm{.1}}}&{{\rm{5}}{\rm{.8}}}&{{\rm{7}}{\rm{.8}}}&{{\rm{7}}{\rm{.1}}}&{{\rm{7}}{\rm{.2}}}&{{\rm{9}}{\rm{.2}}}&{{\rm{6}}{\rm{.6}}}&{{\rm{8}}{\rm{.3}}}&{{\rm{7}}{\rm{.0}}}&{{\rm{8}}{\rm{.3}}}\\{{\rm{7}}{\rm{.8}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{8}}{\rm{.5}}}&{{\rm{8}}{\rm{.9}}}&{{\rm{9}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{14}}{\rm{.1}}}&{{\rm{12}}{\rm{.6}}}&{{\rm{11}}{\rm{.2}}}\end{array}\\\begin{array}{*{20}{l}}{{\rm{7}}{\rm{.8}}}&{{\rm{8}}{\rm{.1}}}&{{\rm{7}}{\rm{.4}}}&{{\rm{8}}{\rm{.5}}}&{{\rm{8}}{\rm{.9}}}&{{\rm{9}}{\rm{.8}}}&{{\rm{9}}{\rm{.7}}}&{{\rm{14}}{\rm{.1}}}&{{\rm{12}}{\rm{.6}}}&{{\rm{11}}{\rm{.2}}}\end{array}\end{array}\)

Prior to obtaining data, denote the beam strengths by X1, 鈥� ,Xm and the cylinder strengths by Y1, . . . , Yn. Suppose that the Xi 鈥檚 constitute a random sample from a distribution with mean m1 and standard deviation s1 and that the Yi 鈥檚 form a random sample (independent of the Xi 鈥檚) from another distribution with mean m2 and standard deviation\({{\rm{\sigma }}_{\rm{2}}}\).

a. Use rules of expected value to show that \({\rm{\bar X - \bar Y}}\)is an unbiased estimator of \({{\rm{\mu }}_{\rm{1}}}{\rm{ - }}{{\rm{\mu }}_{\rm{2}}}\). Calculate the estimate for the given data.

b. Use rules of variance from Chapter 5 to obtain an expression for the variance and standard deviation (standard error) of the estimator in part (a), and then compute the estimated standard error.

c. Calculate a point estimate of the ratio \({{\rm{\sigma }}_{\rm{1}}}{\rm{/}}{{\rm{\sigma }}_{\rm{2}}}\)of the two standard deviations.

d. Suppose a single beam and a single cylinder are randomly selected. Calculate a point estimate of the variance of the difference \({\rm{X - Y}}\) between beam strength and cylinder strength.

An estimator \({\rm{\hat \theta }}\) is said to be consistent if for any \( \in {\rm{ > 0}}\), \({\rm{P(|\hat \theta - \theta |}} \ge \in {\rm{)}} \to {\rm{0}}\) as \({\rm{n}} \to \infty \). That is, \({\rm{\hat \theta }}\) is consistent if, as the sample size gets larger, it is less and less likely that \({\rm{\hat \theta }}\) will be further than \( \in \) from the true value of \({\rm{\theta }}\). Show that \({\rm{\bar X}}\) is a consistent estimator of \({\rm{\mu }}\) when \({{\rm{\sigma }}^{\rm{2}}}{\rm{ < }}\infty \) , by using Chebyshev鈥檚 inequality from Exercise \({\rm{44}}\) of Chapter \({\rm{3}}\). (Hint: The inequality can be rewritten in the form \({\rm{P}}\left( {\left| {{\rm{Y - }}{{\rm{\mu }}_{\rm{Y}}}} \right| \ge \in } \right) \le {\rm{\sigma }}_{\rm{Y}}^{\rm{2}}{\rm{/}} \in \). Now identify \({\rm{Y}}\) with \({\rm{\bar X}}\).)
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