Chapter 12, Problem 37SE

To determine

Find the fractions that should be assigned to each value of x to minimize $V\left({\widehat{\beta }}_2\right)$.

Answer to Problem 37SE

The fractions of total number of observations that can be assigned at $x=-1,0\text{and }1$ to minimize $V\left({\widehat{\beta }}_2\right)$ are 0.25, 0.50, and 0.25, respectively.

Explanation of Solution

Calculation:

Consider that $k_1,k_2,k_3$ are the fractions of total number of observations that can be assigned at $x=-1,0\text{and }1$, respectively.

Thus, it can be aid that if total number of observation is n that is large enough, then there are $n{k}_1$ points at $x=-1$, there are $n{k}_2$ points at $x=0$, and there are $n{k}_3$ points at $x=1$.

Now, the Design matrix X can be written as,

$X={\left[\begin{array}{l}\begin{array}{ccc}1& -1& 1\\ 1& -1& 1\\ .& .& .\end{array}\\ \begin{array}{ccc}.& .& .\\ 1& -1& 1\\ 1& 0& 0\end{array}\\ \begin{array}{ccc}1& 0& 0\\ .& .& .\\ .& .& .\end{array}\\ \begin{array}{ccc}1& 0& 0\\ 1& 1& 1\\ 1& 1& 1\end{array}\\ \begin{array}{c}.\\ .\\ 1\end{array}\begin{array}{c}.\\ .\\ 1\end{array}\begin{array}{c}.\\ .\\ 1\end{array}\end{array}\right]}_{n\times n}$

Now,

$\begin{array}{c}{X}^{\prime }X=\left[\begin{array}{ccc}n& n\left(k_3-k_1\right)& n\left(k_1+k_3\right)\\ n\left(k_3-k_1\right)& n\left(k_1+k_3\right)& n\left(k_3-k_1\right)\\ n\left(k_1+k_3\right)& n\left(k_3-k_1\right)& n\left(k_1+k_3\right)\end{array}\right]\\ =n\left[\begin{array}{ccc}1& b& a\\ b& a& b\\ a& b& a\end{array}\right]\\ =nA\end{array}$, where $a=k_1+k_3,b=k_3-k_1$.

Now, it is needed to minimize $V\left({\widehat{\beta }}_2\right)={\sigma }^2{c}_{22}$, where $c_{33}$ is the $\left(3\times 3\right)$ element of ${\left(X^{\prime }X\right)}^{-1}$.

The determination of matrix A is obtained as,

$\begin{array}{c}A=\left[\begin{array}{ccc}1& b& a\\ b& a& b\\ a& b& a\end{array}\right]\\ \left|A\right|=1\left|\begin{array}{cc}a& b\\ b& a\end{array}\right|-b\left|\begin{array}{cc}b& b\\ a& a\end{array}\right|+a\left|\begin{array}{cc}b& a\\ a& b\end{array}\right|\\ =a^2-b^2-b\left(ab-ba\right)+a\left(b^2-a^2\right)\\ =a^2-b^2-0+a{b}^2-a^3\\ ={\left(k_1+k_3\right)}^2-{\left(k_3-k_1\right)}^2+\left(k_1+k_3\right){\left(k_3+k_1\right)}^2-{\left(k_1+k_3\right)}^3\\ =k_1^2+k_3^2+2k_1{k}_3-k_3^2-k_1^2+2k_1{k}_3+k_1^3+k_1{k}_3^2+2k_1^2{k}_3\\ +k_3{k}_1^2+k_3^3+2k_1{k}_3^2-k_1^3-3k_1^2{k}_3-3k_1{k}_3^2-k_3^3\\ =4k_1{k}_2{k}_3\end{array}$

Now, the inverse of A is obtained as,

$\begin{array}{c}{A}^{-1}=\frac{1}{n\left|A\right|}\left[\begin{array}{ccc}{\left(-1\right)}^{1+1}\left|\begin{array}{cc}a& b\\ b& a\end{array}\right|& {\left(-1\right)}^{1+2}\left|\begin{array}{cc}b& b\\ a& a\end{array}\right|& {\left(-1\right)}^{1+3}\left|\begin{array}{cc}b& a\\ a& b\end{array}\right|\\ {\left(-1\right)}^{2+1}\left|\begin{array}{cc}b& a\\ b& a\end{array}\right|& {\left(-1\right)}^{2+2}\left|\begin{array}{cc}b& b\\ a& a\end{array}\right|& {\left(-1\right)}^{2+3}\left|\begin{array}{cc}1& b\\ a& b\end{array}\right|\\ {\left(-1\right)}^{3+1}\left|\begin{array}{cc}b& a\\ a& b\end{array}\right|& {\left(-1\right)}^{3+2}\left|\begin{array}{cc}1& b\\ a& b\end{array}\right|& {\left(-1\right)}^{3+3}\left|\begin{array}{cc}1& b\\ b& a\end{array}\right|\end{array}\right]\\ =\frac{1}{4n{k}_1{k}_2{k}_3}\left[\begin{array}{ccc}{a}^2-b^2& 0& b^2-a^2\\ 0& a-a^2& ab-b\\ b^2-a^2& ab-b& a-b^2\end{array}\right]\end{array}$

Now, the $c_{22}$ element of $A^{-1}$ matrix is $\frac{a-b^2}{4n{k}_1{k}_2{k}_3}$.

Hence, the $V\left({\widehat{\beta }}_2\right)$ can be written as,

$\begin{array}{c}V\left({\widehat{\beta }}_2\right)={\sigma }^2\frac{a-b^2}{4n{k}_1{k}_2{k}_3}\\ =\frac{{\sigma }^2}{n}\left[\frac{k_1+k_3-{\left(k_3-k_1\right)}^2}{4k_1{k}_2{k}_3}\right]\\ =\frac{{\sigma }^2}{n}\left[\frac{k_1+k_3-\left[{\left(k_3+k_1\right)}^2-4k_1{k}_3\right]}{4k_1{k}_2{k}_3}\right]\\ =\frac{{\sigma }^2}{n}\left[\frac{\left(k_1+k_3\right)\left[1-k_1-k_3\right]}{4k_1{k}_2{k}_3}-\frac{4k_1{k}_3}{4k_1{k}_2{k}_3}\right]\\ =\frac{{\sigma }^2}{n}\left[\frac{\left(k_1+k_3\right)}{4k_1{k}_3}-\frac{1}{k_2}\right]\\ =\frac{{\sigma }^2}{n}\left[\frac{\left(k_1+k_3\right)}{4k_1{k}_3}-\frac{1}{1-k_1-k_3}\right]\left[\text{as }{k}_1+k_2+k_3=1\right]\end{array}$

Now, it is needed to partial differentiate $V\left({\widehat{\beta }}_2\right)$ with respect to $k_1\text{and }{k}_3$ and equate to .

That is,

$\begin{array}{c}\frac{\partial \left(V\left({\widehat{\beta }}_2\right)\right)}{\partial k_1}=0\\ \frac{\partial \left(\frac{{\sigma }^2}{n}\left[\frac{\left(k_1+k_3\right)}{4k_1{k}_3}-\frac{1}{1-k_1-k_3}\right]\right)}{\partial k_1}=0\\ 4k_1^2={\left(1-k_1-k_3\right)}^2\mathrm{............}\left(1\right)\end{array}$,

And

$\begin{array}{c}\frac{\partial \left(V\left({\widehat{\beta }}_2\right)\right)}{\partial k_3}=0\\ \frac{\partial \left(\frac{{\sigma }^2}{n}\left[\frac{\left(k_1+k_3\right)}{4k_1{k}_3}-\frac{1}{1-k_1-k_3}\right]\right)}{\partial k_3}=0\\ 4k_3^2={\left(1-k_1-k_3\right)}^2\mathrm{..........}\left(2\right)\end{array}$.

As $k_1,k_2,k_3$ are positive constants and by symmetry of $k_1=k_3$, the equation (1) can be written as,

$\begin{array}{c}4k_1^2={\left(1-2k_1\right)}^2\\ k_1=0.25\end{array}$

Similarly, $k_3=0.25$.

Thus,

$\begin{array}{c}1-k_1-k_3=1-0.25-0.25\\ =0.50\end{array}$.

Thus, the fractions of total number of observations that can be assigned at $x=-1,0\text{and }1$ to minimize $V\left({\widehat{\beta }}_2\right)$ are 0.25, 0.50, and 0.25, respectively.