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Exercise 5.5.1

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Write down the two equations as $Ax = b$ , we have: $[\begin{matrix} 1 \\ 1 \end{matrix}] [\begin{matrix} x \end{matrix}] = [\begin{matrix} b_{1} \\ b_{2} \end{matrix}]$ , the covariance matrix for the measurements is $V = [\begin{matrix} σ_{1}^{2} & 0 \\ 0 & σ_{2}^{2} \end{matrix}]$ .

Multiply $A \hat{x} = b$ by $V^{- \frac{1}{2}}$ on both sides, we have $A^{T} V^{- 1} A \hat{x} = A^{T} V^{- 1} b$ , i.e.

\begin{array}{l} [\begin{matrix} 1 & 1 \end{matrix}] [\begin{matrix} \frac{1}{σ_{1}^{2}} & 0 \\ 0 & \frac{1}{σ_{2}^{2}} \end{matrix}] [\begin{matrix} 1 \\ 1 \end{matrix}] \hat{x} & = [\begin{matrix} 1 & 1 \end{matrix}] [\begin{matrix} \frac{1}{σ_{1}^{2}} & 0 \\ 0 & \frac{1}{σ_{2}^{2}} \end{matrix}] [\begin{matrix} b_{1} \\ b_{2} \end{matrix}] \\ (\frac{1}{σ_{1}^{2}} + \frac{1}{σ_{2}^{2}}) \hat{x} & = \frac{b_{1}}{σ_{1}^{2}} + \frac{b_{2}}{σ_{2}^{2}} \\ \hat{x} & = \frac{\frac{b_{1}}{σ_{1}^{2}} + \frac{b_{2}}{σ_{2}^{2}}}{\frac{1}{σ_{1}^{2}} + \frac{1}{σ_{2}^{2}}} \end{array}

Then we can compute the variance of $\hat{x}$ , i.e. $W = {(A^{T} V^{- 1} A)}^{- 1} = {(\frac{1}{σ_{1}^{2}} + \frac{1}{σ_{2}^{2}})}^{- 1}$

niuers

2020-03-20 00:00

Exercise 5.5.1

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