Exercise 5.5.6

Divide both sides by $\sigma ,s,\sigma$, we have $\left[\begin{array}{cc}\hfill \frac{1}{\sigma }\hfill & \hfill 0\hfill \\ \hfill -\frac{1}{s}\hfill & \hfill \frac{1}{s}\hfill \\ \hfill 0\hfill & \hfill \frac{1}{\sigma }\hfill \end{array}\right]\left[\begin{array}{c}\hfill x_0\hfill \\ \hfill x_1\hfill \end{array}\right]=\left[\begin{array}{c}\hfill \frac{b_0}{\sigma }\hfill \\ \hfill 0\hfill \\ \hfill \frac{b_1}{\sigma }\hfill \end{array}\right]$, then we have matrix $A=\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]$, $V^{-\frac{1}{2}}=\left[\begin{array}{ccc}\hfill \frac{1}{\sigma }\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \frac{1}{s}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{1}{\sigma }\hfill \end{array}\right]$

And we need solve $\widehat{x}$ from this equation $A^T{V}^{-1}A\widehat{x}=A^T{V}^{-1}b$, that is

$\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill \frac{1}{{\sigma }^2}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \frac{1}{s^2}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{1}{{\sigma }^2}\hfill \end{array}\right]\left[\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill -1\hfill & \hfill 1\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right]\widehat{x}=\left[\begin{array}{ccc}\hfill 1\hfill & \hfill -1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill & \hfill 1\hfill \end{array}\right]\left[\begin{array}{ccc}\hfill \frac{1}{{\sigma }^2}\hfill & \hfill 0\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill \frac{1}{s^2}\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 0\hfill & \hfill \frac{1}{{\sigma }^2}\hfill \end{array}\right]\left[\begin{array}{c}\hfill b_0\hfill \\ \hfill 0\hfill \\ \hfill b_1\hfill \end{array}\right]$, i.e. 

$

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

x^ =

$\left[\begin{array}{c}\hfill \frac{b_0}{{\sigma }^2}\hfill \\ \hfill \frac{b_1}{{\sigma }^2}\hfill \end{array}\right]$

$

So we have $\widehat{x}=\frac{{\sigma }^2}{\frac{1}{{\sigma }^2}+\frac{2}{s^2}}\left[\begin{array}{cc}\hfill \frac{1}{{\sigma }^2}+\frac{1}{s^2}\hfill & \hfill \frac{1}{s^2}\hfill \\ \hfill \frac{1}{s^2}\hfill & \hfill \frac{1}{{\sigma }^2}+\frac{1}{s^2}\hfill \end{array}\right]\left[\begin{array}{c}\hfill \frac{b_0}{{\sigma }^2}\hfill \\ \hfill \frac{b_1}{{\sigma }^2}\hfill \end{array}\right]=\frac{1}{\frac{1}{{\sigma }^2}+\frac{2}{s^2}}\left[\begin{array}{c}\hfill b_0(\frac{1}{{\sigma }^2}+\frac{1}{s^2})+b_1\frac{1}{s^2}\hfill \\ \hfill b_0\frac{1}{s^2}+b_1(\frac{1}{{\sigma }^2}+\frac{1}{s^2})\hfill \end{array}\right]$

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

This is exactly the same as problem III.1.11