误差理论与平差作业：第二周周一

教材 3-1

相关观测值向量 $\underset{t\times 1}{\mathit{X}}$ 的协方差阵是如何定义的？试说明 $\underset{t\times t}{{\mathit{D}}_{XX}}$ 中各元素的含义。当向量 $\underset{t\times 1}{\mathit{X}}$ 的各个分量两两独立时，其协方差阵有什么特点？

解：定义 $\underset{t\times 1}{\mathit{X}}$ 的方差 - 协方差阵：

$$\begin{array}{rl}\underset{t\times t}{{\mathit{D}}_{XX}}& =E\{[\mathit{X}-E(\mathit{X})][\mathit{X}-E(\mathit{X}){]}^{\mathrm{T}}\}\\ & =\left[\begin{array}{cccc}{\sigma }_{X_1}^2& {\sigma }_{X_1{X}_2}& \cdots & {\sigma }_{X_1{X}_t}\\ {\sigma }_{X_2{X}_1}& {\sigma }_{X_2}^2& \cdots & {\sigma }_{X_2{X}_t}\\ ⋮& ⋮& \ddots & ⋮\\ {\sigma }_{X_t{X}_1}& {\sigma }_{X_t{X}_2}& \cdots & {\sigma }_{X_t}^2\end{array}\right]\end{array}$$

其各元素的含义：

• 主对角元是方差：$i$ 行 $i$ 列是随机变量 $X_i$ 的方差
• 非主对角元是协方差：$i$ 行 $j$ 列（$i\ne j$）是随机变量 $X_i$ 和 $X_j$ 的协方差

当向量 $\underset{t\times 1}{\mathit{X}}$ 的各个分量两两独立时，所有协方差为零，即非主对角元均为零，协方差阵变为对角阵：$\underset{t\times t}{{\mathit{D}}_{XX}}=\mathrm{diag}[{\sigma }_{X_1}^2,\cdots ,{\sigma }_{X_t}^2]$

教材 3-5

设有函数关系式

$$\begin{array}{rl}\mathit{Y}& =\mathit{A}\mathit{X}\\ \mathit{Z}& =\mathit{B}\mathit{Y}\\ \mathit{Y}& =\mathit{C}\mathit{Y}+\mathit{G}\mathit{Z}\\ \mathit{W}& =\mathit{F}\mathit{Z}+\mathit{H}\mathit{R}\end{array}$$

式中，$\mathit{A},\mathit{B},\mathit{C},\mathit{G},\mathit{F},\mathit{H}$ 均为常数阵。已知 ${\mathit{D}}_{XX}$，求：${\mathit{D}}_{RR},{\mathit{D}}_{RZ},{\mathit{D}}_{RW}$。

解：

$$\begin{array}{rl}\mathit{R}& =\mathit{C}\mathit{Y}+\mathit{G}\mathit{Z}\\ & =\mathit{C}\mathit{A}\mathit{X}+\mathit{G}\mathit{B}\mathit{A}\mathit{X}\\ & =(\mathit{C}+\mathit{G}\mathit{B})\mathit{A}\mathit{X}\\ \Rightarrow {\mathit{D}}_{RR}& =(\mathit{C}+\mathit{G}\mathit{B})\mathit{A}\mathit{X}[(\mathit{C}+\mathit{G}\mathit{B})\mathit{A}{]}^{\mathrm{T}}\\ & =(\mathit{C}+\mathit{G}\mathit{B})\mathit{A}\mathit{X}{\mathit{A}}^{\mathrm{T}}(\mathit{C}+\mathit{G}\mathit{B}{)}^{\mathrm{T}}\end{array}$$$$\begin{array}{rl}\mathit{R}& =(\mathit{C}+\mathit{G}\mathit{B})\mathit{A}\mathit{X}\\ \mathit{Z}& =\mathit{B}\mathit{A}\mathit{X}\\ \Rightarrow {\mathit{D}}_{RZ}& =(\mathit{C}+\mathit{G}\mathit{B})\mathit{A}\mathit{X}(\mathit{B}\mathit{A}{)}^{\mathrm{T}}\\ & =(\mathit{C}+\mathit{G}\mathit{B})\mathit{A}\mathit{X}{\mathit{A}}^{\mathrm{T}}{\mathit{B}}^{\mathrm{T}}\end{array}$$$$\begin{array}{rl}\mathit{W}& =\mathit{F}\mathit{Z}+\mathit{H}\mathit{R}\\ & =\mathit{F}\mathit{B}\mathit{A}\mathit{X}+\mathit{H}(\mathit{C}+\mathit{G}\mathit{B})\mathit{A}\mathit{X}\\ & =(\mathit{F}\mathit{B}+\mathit{H}\mathit{C}+\mathit{H}\mathit{G}\mathit{B})\mathit{A}\mathit{X}\\ \Rightarrow {\mathit{D}}_{RW}& =(\mathit{C}+\mathit{G}\mathit{B})\mathit{A}\mathit{X}{\mathit{A}}^{\mathrm{T}}(\mathit{F}\mathit{B}+\mathit{H}\mathit{C}+\mathit{H}\mathit{G}\mathit{B}{)}^{\mathrm{T}}\end{array}$$

教材 3-6

已知独立观测值 $L_1,L_2$ 的中误差分别为 ${\sigma }_1,{\sigma }_2$，求下列函数的中误差：

1. $x=L_1-2L_2$
2. $y={\displaystyle \frac{1}{2}}{L}_1^2+L_1{L}_2$
3. $z={\displaystyle \frac{\sin L_1}{\sin (L_1+L_2)}}$

解：由于 $L_1,L_2$ 相互独立，协方差为零，因此观测值向量 $\mathit{L}=[L_1,L_2{]}^{\mathrm{T}}$ 的自协方差阵为

$${\mathit{D}}_{LL}=\left[\begin{array}{cc}{\sigma }_1^2& 0\\ 0& {\sigma }_2^2\end{array}\right]$$

1. $x=L_1-2L_2$

   $$\begin{array}{rl}x& =[1,-2]\mathit{L}\\ \Rightarrow {\mathit{D}}_{xx}& =[1,-2]{\mathit{D}}_{LL}[1,-2{]}^{\mathrm{T}}\\ & =[1,-2]\left[\begin{array}{cc}{\sigma }_1^2& 0\\ 0& {\sigma }_2^2\end{array}\right]\left[\begin{array}{c}1\\ -2\end{array}\right]\\ & ={\sigma }_1^2+4{\sigma }_2^2\\ \Rightarrow {\sigma }_x& =\sqrt{{\sigma }_1^2+4{\sigma }_2^2}\end{array}$$

2. $y={\displaystyle \frac{1}{2}}{L}_1^2+L_1{L}_2$

   $$\begin{array}{rl}\mathrm{d}y& =L_1\mathrm{d}{L}_1+L_2\mathrm{d}{L}_1+L_1\mathrm{d}{L}_2\\ & =[L_1+L_2,L_1]\mathrm{d}\mathit{L}\\ \Rightarrow {\mathit{D}}_{yy}& =[L_1+L_2,L_1]{\mathit{D}}_{LL}[L_1+L_2,L_1{]}^{\mathrm{T}}\\ & =[L_1+L_2,L_1]\left[\begin{array}{cc}{\sigma }_1^2& 0\\ 0& {\sigma }_2^2\end{array}\right]\left[\begin{array}{c}{L}_1+L_2\\ L_1\end{array}\right]\\ & =(L_1+L_2{)}^2{\sigma }_1^2+L_1^2{\sigma }_2^2\\ \Rightarrow {\sigma }_y& =\sqrt{(L_1+L_2{)}^2{\sigma }_1^2+L_1^2{\sigma }_2^2}\end{array}$$

3. $z={\displaystyle \frac{\sin L_1}{\sin (L_1+L_2)}}$

   $$\begin{array}{rl}\mathrm{d}z& =\frac{\sin L_2}{{\sin }^2(L_1+L_2)}\mathrm{d}{L}_1-\frac{\sin L_1\cos (L_1+L_2)}{{\sin }^2(L_1+L_2)}\mathrm{d}{L}_2\\ \Rightarrow {\mathit{D}}_{ZZ}& =[\frac{\sin L_2}{{\sin }^2(L_1+L_2)},-\frac{\sin L_1\cos (L_1+L_2)}{{\sin }^2(L_1+L_2)}]{\mathit{D}}_{LL}{[\frac{\sin L_2}{{\sin }^2(L_1+L_2)},-\frac{\sin L_1\cos (L_1+L_2)}{{\sin }^2(L_1+L_2)}]}^{\mathrm{T}}\\ & =[\frac{\sin L_2}{{\sin }^2(L_1+L_2)},-\frac{\sin L_1\cos (L_1+L_2)}{{\sin }^2(L_1+L_2)}]\left[\begin{array}{cc}{\sigma }_1^2& 0\\ 0& {\sigma }_2^2\end{array}\right]\left[\begin{array}{c}\frac{\sin L_2}{{\sin }^2(L_1+L_2)}\\ -\frac{\sin L_1\cos (L_1+L_2)}{{\sin }^2(L_1+L_2)}\end{array}\right]\\ & =\frac{{\sigma }_1^2{\sin }^2{L}_2}{{\sin }^4(L_1+L_2)}+\frac{{\sigma }_2^2{\sin }^2{L}_1{\cos }^2(L_1+L_2)}{{\sin }^4(L_1+L_2)}\\ \Rightarrow {\sigma }_z& =\sqrt{\frac{{\sigma }_1^2{\sin }^2{L}_2}{{\sin }^4(L_1+L_2)}+\frac{{\sigma }_2^2{\sin }^2{L}_1{\cos }^2(L_1+L_2)}{{\sin }^4(L_1+L_2)}}\end{array}$$

教材 3-7

有函数

$$\begin{array}{rl}{\psi }_1& =3L_1-2L_3\\ {\psi }_2& =L_1{L}_2+L_3^2\end{array}$$

已知观测值的协因数阵为

$${\mathit{Q}}_{LL}=\left[\begin{array}{ccc}6& -2& 1\\ -2& 4& 0\\ 1& 0& 3\end{array}\right]$$

单位权方差为 ${\sigma }_0^2=0.2$。当 $L_1=6,L_2=8,L_3=10$ 时，试求 ${\psi }_1,{\psi }_2$ 的方差 ${\sigma }_{{\psi }_1}^2,{\sigma }_{{\psi }_2}^2$ 以及 ${\psi }_1$ 对 ${\psi }_2$ 的协方差 ${\sigma }_{{\psi }_1{\psi }_2}$。

解：

$$\begin{array}{rl}\mathrm{d}{\psi }_1& =3\mathrm{d}{L}_1-2\mathrm{d}{L}_3\\ \mathrm{d}{\psi }_2& =L_2\mathrm{d}{L}_1+L_1\mathrm{d}{L}_2+2L_3\mathrm{d}{L}_3\end{array}\Rightarrow \mathrm{d}\mathit{\Psi }=\mathrm{d}\left[\begin{array}{c}{\psi }_1\\ {\psi }_2\end{array}\right]=\left[\begin{array}{ccc}3& 0& -2\\ 1& 1& 2\end{array}\right]\mathrm{d}\mathit{L}$$

不妨设 $\mathit{K}=\left[\begin{array}{ccc}3& 0& -2\\ 1& 1& 2\end{array}\right]$，则 $\mathrm{d}\mathit{\Psi }=\mathit{K}\mathrm{d}\mathit{L}$，故有

$$\begin{array}{rl}{\mathit{D}}_{\Psi \Psi }& =\mathit{K}{\mathit{D}}_{LL}{\mathit{K}}^{\mathrm{T}}\\ & ={\sigma }_0^2\mathit{K}{\mathit{Q}}_{LL}{\mathit{K}}^{\mathrm{T}}\\ & =\frac{1}{5}\left[\begin{array}{ccc}3& 0& -2\\ 1& 1& 2\end{array}\right]\left[\begin{array}{ccc}6& -2& 1\\ -2& 4& 0\\ 1& 0& 3\end{array}\right]\left[\begin{array}{cc}3& 1\\ 0& 1\\ -2& 2\end{array}\right]\\ & =\left[\begin{array}{cc}10.8& 0.8\\ 0.8& 4.4\end{array}\right]\end{array}$$

故有

$$\{\begin{array}{rl}& {\sigma }_{{\psi }_1}^2=10.8\\ & {\sigma }_{{\psi }_2}^2=4.4\\ & {\sigma }_{{\psi }_1{\psi }_2}=0.8\end{array}$$