Question

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

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

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.

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):

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\({\rm{V(\hat \mu )}}\).