Exercise 3.1.11

With $F=1$, divide both sides of those three equations by $\sigma ,s,\sigma$, we have

$\left[\begin{array}{cc}\hfill \frac{1}{\sigma }\hfill & \hfill 0\hfill \\ \hfill -\frac{1}{s}\hfill & \hfill \frac{1}{s}\hfill \\ \hfill 0\hfill & \hfill \frac{1}{\sigma }\hfill \end{array}\right]\left[\begin{array}{c}\hfill x_0\hfill \\ \hfill x_1\hfill \end{array}\right]=\left[\begin{array}{c}\hfill \frac{b_0}{\sigma }\hfill \\ \hfill 0\hfill \\ \hfill \frac{b_1}{\sigma }\hfill \end{array}\right]$

$A^T=\left[\begin{array}{ccc}\hfill \frac{1}{\sigma }\hfill & \hfill -\frac{1}{s}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \frac{1}{s}\hfill & \hfill \frac{1}{\sigma }\hfill \end{array}\right]$,

so $A^{T}A=\left[\begin{array}{cc}\hfill \frac{1}{{\sigma }^2}+\frac{1}{s^2}\hfill & \hfill -\frac{1}{s^2}\hfill \\ \hfill -\frac{1}{s^2}\hfill & \hfill \frac{1}{{\sigma }^2}+\frac{1}{s^2}\hfill \end{array}\right]$

${(A^{T}A)}^{-1}=\frac{{\sigma }^2}{\frac{1}{{\sigma }^2}+\frac{2}{s^2}}\left[\begin{array}{cc}\hfill \frac{1}{{\sigma }^2}+\frac{1}{s^2}\hfill & \hfill \frac{1}{s^2}\hfill \\ \hfill \frac{1}{s^2}\hfill & \hfill \frac{1}{{\sigma }^2}+\frac{1}{s^2}\hfill \end{array}\right]$

So $x = (A^TA){-1}A^Tb = ${\sigma }^2$$\frac{1}{{\sigma }^2}+\frac{2}{s^2}$

$\left[\begin{array}{cc}\hfill \frac{1}{{\sigma }^2}+\frac{1}{s^2}\hfill & \hfill \frac{1}{s^2}\hfill \\ \hfill \frac{1}{s^2}\hfill & \hfill \frac{1}{{\sigma }^2}+\frac{1}{s^2}\hfill \end{array}\right]$$\left[\begin{array}{ccc}\hfill \frac{1}{\sigma }\hfill & \hfill -\frac{1}{s}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \frac{1}{s}\hfill & \hfill \frac{1}{\sigma }\hfill \end{array}\right]$$\left[\begin{array}{c}\hfill \frac{b_0}{\sigma }\hfill \\ \hfill 0\hfill \\ \hfill \frac{b_1}{\sigma }\hfill \end{array}\right]$

= 1$\frac{1}{{\sigma }^2}+\frac{2}{s^2}$

$\left[\begin{array}{c}\hfill b_0(\frac{1}{{\sigma }^2}+\frac{1}{s^2})+b_1\frac{1}{s^2}\hfill \\ \hfill b_0\frac{1}{s^2}+b_1(\frac{1}{{\sigma }^2}+\frac{1}{s^2})\hfill \end{array}\right]$

$

So it says ${\widehat{x}}_0$ has more weight to $b_0$, while ${\widehat{x}}_1$ has more weight to $b_1$.