Commit cd362d13 authored by Maximilian Vitz's avatar Maximilian Vitz
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parent a4f26df8
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### Error Propagation
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1)
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$a = 6.1 \pm 0.1$ cm
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$b = 2.8 \pm 0.1$ cm
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With 100% correlation one ends up with $\rho = 1$.
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$\Rightarrow \text{cov}(a,b) = \rho \sigma_{a} \sigma_{b} = \sigma_{a} \sigma_{b}$
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$\sigma_{f}^{2} = \left( \frac{\delta f}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta f}{\delta b} \right)^{2} \sigma_{b}^{2} + 2 \left( \frac{\delta A}{\delta a} \right) \left( \frac{\delta f}{\delta b} \right) \text{cov}(a,b) = \left( \frac{\delta f}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta f}{\delta b} \right)^{2} \sigma_{b}^{2} + 2 \left( \frac{\delta f}{\delta a} \right) \left( \frac{\delta f}{\delta b} \right) \sigma_{a} \sigma_{b} = \left( \frac{\delta f}{\delta a} \sigma_{a} + \frac{\delta f}{\delta b} \sigma_{b} \right)^{2}$
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$f = A(a,b) = a \cdot b$
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$\Rightarrow \sigma_{A} = \frac{\delta A}{\delta a} \sigma_{a} + \frac{\delta A}{\delta b} \sigma_{b} = b \sigma_{a} + a \sigma_{b} = 0.89 \text{cm}^{2}$
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$f = C(a,b) = 2a + 2b$
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$\Rightarrow \sigma_{C} = \frac{\delta C}{\delta a} \sigma_{a} + \frac{\delta C}{\delta b} \sigma_{b} = 2 \sigma_{a} + 2 \sigma_{b} = 0.4 \text{cm}$
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2)
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With 0% correlation one ends up with $\rho = 0$.
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$\Rightarrow \text{cov}(a,b) = \rho \sigma_{a} \sigma_{b} = 0$
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$\sigma_{f}^{2} = \left( \frac{\delta f}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta f}{\delta b} \right)^{2} \sigma_{b}^{2} + 2 \left( \frac{\delta A}{\delta a} \right) \left( \frac{\delta f}{\delta b} \right) \text{cov}(a,b) = \left( \frac{\delta f}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta f}{\delta b} \right)^{2} \sigma_{b}^{2} $
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$f = A(a,b) = a \cdot b$
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$\Rightarrow \sigma_{A} = \sqrt{\left( \frac{\delta A}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta A}{\delta b} \right)^{2} \sigma_{b}^{2}} = \sqrt{b^{2} \sigma_{a}^{2} + a^{2} \sigma_{b}^{2}} = 0.67 \text{cm}^{2}$
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$f = C(a,b) = 2a + 2b$
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$\Rightarrow \sigma_{C} = \sqrt{\left( \frac{\delta C}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta C}{\delta b} \right)^{2} \sigma_{b}^{2}} = \sqrt{4 \sigma_{a}^{2} + 4 \sigma_{b}^{2}} = 0.28 \text{cm}$
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3)
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$\left(\begin{array}{cc}
\sigma_{A}^{2} & \text{cov}(A,C)\\
\text{cov}(A,C) & \sigma_{C}^{2}
\end{array}\right) = \left(\begin{array}{cc}
\frac{\delta A}{\delta a} & \frac{\delta A}{\delta b}\\
\frac{\delta C}{\delta a} & \frac{\delta C}{\delta b}
\end{array}\right) \cdot \left(\begin{array}{cc}
\sigma_{a}^{2} & \text{cov}(a,b)\\
\text{cov}(a,b) & \sigma_{b}^{2}
\end{array}\right) \cdot \left(\begin{array}{cc}
\frac{\delta A}{\delta a} & \frac{\delta C}{\delta a}\\
\frac{\delta A}{\delta b} & \frac{\delta C}{\delta b}
\end{array}\right) = \left(\begin{array}{cc}
b & a\\
2 & 2
\end{array}\right) \cdot \left(\begin{array}{cc}
\sigma_{a}^{2} & \text{cov}(a,b)\\
\text{cov}(a,b) & \sigma_{b}^{2}
\end{array}\right) \cdot \left(\begin{array}{cc}
b & 2\\
a & 2
\end{array}\right)$
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$\Rightarrow \left(\begin{array}{cc}
\sigma_{A}^{2} & \text{cov}(A,C)\\
\text{cov}(A,C) & \sigma_{C}^{2}
\end{array}\right) = \left(\begin{array}{cc}
\sigma_{a}^{2} b^{2} + \sigma_{b}^{2} a^{2} + 2 a b \text{cov} (a,b) & 2 (b \sigma_{2}^2 + a \sigma_{b}^2 + \text{cov}(a,b)(a+b))\\
2 (b \sigma_{2}^2 + a \sigma_{b}^2 + \text{cov}(a,b)(a+b)) & 4 \sigma_{a}^2 + 4 \sigma_{b}^2 + 8 \text{cov}(a,b)
\end{array}\right)$
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$\Rightarrow \text{cov}(A,C) = 2 (b \sigma_{2}^2 + a \sigma_{b}^2 + \text{cov}(a,b)(a+b))$
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$\rho_{A,C} = \frac{\text{cov}(A,C)}{\sigma_{A}\sigma_{C}}$
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With 100% correlation one ends up with $\rho_{a,b} = 1$, $\sigma_{a}= \sigma_{b}$.
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$\rho_{A,C} = \frac{\text{cov}(A,C)}{\sigma_{A}\sigma_{C}} = \frac{4(a+b)}{4(a+b)} = 1$
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With 0% correlation one ends up with $\rho_{a,b} = 0$, $\sigma_{a}= \sigma_{b}$.
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$\rho_{A,C} = \frac{\text{cov}(A,C)}{\sigma_{A}\sigma_{C}} = \frac{2(a+b)}{\sqrt{8(a^{2}+b^{2})}} = 0.938$
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### Triangulation
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1)
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$l = 101.2$ m, $\sigma_{l} = 0.1$ m
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$\alpha = 87.4^{\circ}$, $\sigma_{\alpha} = 0.1^{\circ}$
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$\beta = 87.8^{\circ}$, $\sigma_{\beta} = 0.1^{\circ}$
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$d = \frac{l\sin{(\alpha)} \sin{(\beta)}}{\sin{(\alpha +\beta)}}$
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$\sigma_{d}^{2} = \sigma_{\alpha}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\cos{(\alpha)} \sin{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2} + \sigma_{\beta}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\sin{(\alpha)} \cos{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2} + \sigma_{l}^{2} \left( \frac{\sin{(\alpha)} \sin{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2}$
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$\Rightarrow d =1207.36$ m, $\sigma_{d} = 35.63$ m
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2)
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Assuming $\sigma_{l} = 0$.
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$\Rightarrow \sigma_{d}^{2} = \sigma_{\alpha}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\cos{(\alpha)} \sin{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2} + \sigma_{\beta}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\sin{(\alpha)} \cos{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2}$
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$\Rightarrow \sigma_{d} = 35.61$ m
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The error on the distance is dominated by the error on the angles.
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3)
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$\alpha = \beta$, $\sigma_{\alpha} = \sigma_{\beta} = \sigma$
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$\sigma_{d}^{2} = \sigma_{\alpha}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\cos{(\alpha)} \sin{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2} + \sigma_{\beta}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\sin{(\alpha)} \cos{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2}$
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$\Rightarrow \sigma_{d}^{2} = 2 \sigma^{2} l^{2} \left( \frac{-\sin{(\alpha)} \sin{(\alpha)} \cos{(2\alpha)}}{\sin^{2}{(2\alpha)}} + \frac{\cos{(\alpha)} \sin{(\alpha)}}{\sin{(2 \alpha)}} \right)^{2} = 2 \sigma^{2} l^{2} \left( \frac{-\sin^{2}{(\alpha)} \cos{(2\alpha)} + \cos{(\alpha)} \sin{(\alpha)}\sin{(2\alpha)}}{\sin^{2}{(2\alpha)}} \right)^{2}$
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Use: $\hspace{0.5cm}$ $\cos(2 \alpha) = \cos^{2}{(\alpha)} - \sin^{2}{(\alpha)}$ $\hspace{0.5cm}$ and $\hspace{0.5cm}$ $\sin(2 \alpha) = 2\sin{(\alpha)}\cos{(\alpha)}$
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$\Rightarrow \sigma_{d}^{2} = 2 \sigma^{2} l^{2} \left( \frac{-\sin^{2}{(\alpha)} \cos^{2}{(\alpha)} + \sin^{2}{(\alpha)} \sin^{2}{(\alpha)} + 2 \cos^{2}{(\alpha)} \sin^{2}{(\alpha)}}{\sin^{2}{(2\alpha)}} \right)^{2} = 2 \sigma^{2} l^{2} \left( \frac{ \sin^{2}{(\alpha)} \cdot [ - \cos^{2}{(\alpha)} + \sin^{2}{(\alpha)} + 2 \cos^{2}{(\alpha)} ]}{\sin^{2}{(2\alpha)}} \right)^{2}= 2 \sigma^{2} l^{2} \left( \frac{ \sin^{2}{(\alpha)} \cdot [ \cos^{2}{(\alpha)} + \sin^{2}{(\alpha)]} }{\sin^{2}{(2\alpha)}} \right)^{2}$
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Use: $\hspace{0.5cm}$ $\sin(2 \alpha) = 2\sin{(\alpha)}\cos{(\alpha)}$
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$\Rightarrow \sigma_{d}^{2} = 2 \sigma^{2} l^{2} \left( \frac{ \sin^{2}{(\alpha)}}{\sin^{2}{(2\alpha)}} \right)^{2} = 2 \sigma^{2} l^{2} \left( \frac{ \sin^{2}{(\alpha)}}{4 \sin^{2}{(\alpha)}\cos^{2}{(\alpha)}} \right)^{2} = \frac{ \sigma^{2} l^{2}}{8 \cos^{4}{(\alpha)}}$
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$\Rightarrow \sigma_{d} \approx \frac{ \sigma l}{\sqrt{8} } \frac{(l/2)^{2} + d}{(l/2)^{2}} \approx \frac{ \sigma l}{\sqrt{8} } \frac{4 d^{2}}{l^{2}} = \sqrt{2} \sigma \frac{d^{2}}{l}$
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``` python
```
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### Error Propagation
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1)
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$a = 6.1 \pm 0.1$ cm
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$b = 2.8 \pm 0.1$ cm
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With 100% correlation one ends up with $\rho = 1$.
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$\Rightarrow \text{cov}(a,b) = \rho \sigma_{a} \sigma_{b} = \sigma_{a} \sigma_{b}$
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$\sigma_{f}^{2} = \left( \frac{\delta f}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta f}{\delta b} \right)^{2} \sigma_{b}^{2} + 2 \left( \frac{\delta A}{\delta a} \right) \left( \frac{\delta f}{\delta b} \right) \text{cov}(a,b) = \left( \frac{\delta f}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta f}{\delta b} \right)^{2} \sigma_{b}^{2} + 2 \left( \frac{\delta f}{\delta a} \right) \left( \frac{\delta f}{\delta b} \right) \sigma_{a} \sigma_{b} = \left( \frac{\delta f}{\delta a} \sigma_{a} + \frac{\delta f}{\delta b} \sigma_{b} \right)^{2}$
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$f = A(a,b) = a \cdot b$
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$\Rightarrow \sigma_{A} = \frac{\delta A}{\delta a} \sigma_{a} + \frac{\delta A}{\delta b} \sigma_{b} = b \sigma_{a} + a \sigma_{b} = 0.89 \text{cm}^{2}$
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$f = C(a,b) = 2a + 2b$
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$\Rightarrow \sigma_{C} = \frac{\delta C}{\delta a} \sigma_{a} + \frac{\delta C}{\delta b} \sigma_{b} = 2 \sigma_{a} + 2 \sigma_{b} = 0.4 \text{cm}$
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2)
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With 0% correlation one ends up with $\rho = 0$.
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$\Rightarrow \text{cov}(a,b) = \rho \sigma_{a} \sigma_{b} = 0$
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$\sigma_{f}^{2} = \left( \frac{\delta f}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta f}{\delta b} \right)^{2} \sigma_{b}^{2} + 2 \left( \frac{\delta A}{\delta a} \right) \left( \frac{\delta f}{\delta b} \right) \text{cov}(a,b) = \left( \frac{\delta f}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta f}{\delta b} \right)^{2} \sigma_{b}^{2} $
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$f = A(a,b) = a \cdot b$
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$\Rightarrow \sigma_{A} = \sqrt{\left( \frac{\delta A}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta A}{\delta b} \right)^{2} \sigma_{b}^{2}} = \sqrt{b^{2} \sigma_{a}^{2} + a^{2} \sigma_{b}^{2}} = 0.67 \text{cm}^{2}$
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$f = C(a,b) = 2a + 2b$
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$\Rightarrow \sigma_{C} = \sqrt{\left( \frac{\delta C}{\delta a} \right)^{2} \sigma_{a}^{2} + \left( \frac{\delta C}{\delta b} \right)^{2} \sigma_{b}^{2}} = \sqrt{4 \sigma_{a}^{2} + 4 \sigma_{b}^{2}} = 0.28 \text{cm}$
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3)
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$\left(\begin{array}{cc}
\sigma_{A}^{2} & \text{cov}(A,C)\\
\text{cov}(A,C) & \sigma_{C}^{2}
\end{array}\right) = \left(\begin{array}{cc}
\frac{\delta A}{\delta a} & \frac{\delta A}{\delta b}\\
\frac{\delta C}{\delta a} & \frac{\delta C}{\delta b}
\end{array}\right) \cdot \left(\begin{array}{cc}
\sigma_{a}^{2} & \text{cov}(a,b)\\
\text{cov}(a,b) & \sigma_{b}^{2}
\end{array}\right) \cdot \left(\begin{array}{cc}
\frac{\delta A}{\delta a} & \frac{\delta C}{\delta a}\\
\frac{\delta A}{\delta b} & \frac{\delta C}{\delta b}
\end{array}\right) = \left(\begin{array}{cc}
b & a\\
2 & 2
\end{array}\right) \cdot \left(\begin{array}{cc}
\sigma_{a}^{2} & \text{cov}(a,b)\\
\text{cov}(a,b) & \sigma_{b}^{2}
\end{array}\right) \cdot \left(\begin{array}{cc}
b & 2\\
a & 2
\end{array}\right)$
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$\Rightarrow \left(\begin{array}{cc}
\sigma_{A}^{2} & \text{cov}(A,C)\\
\text{cov}(A,C) & \sigma_{C}^{2}
\end{array}\right) = \left(\begin{array}{cc}
\sigma_{a}^{2} b^{2} + \sigma_{b}^{2} a^{2} + 2 a b \text{cov} (a,b) & 2 (b \sigma_{2}^2 + a \sigma_{b}^2 + \text{cov}(a,b)(a+b))\\
2 (b \sigma_{2}^2 + a \sigma_{b}^2 + \text{cov}(a,b)(a+b)) & 4 \sigma_{a}^2 + 4 \sigma_{b}^2 + 8 \text{cov}(a,b)
\end{array}\right)$
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$\Rightarrow \text{cov}(A,C) = 2 (b \sigma_{2}^2 + a \sigma_{b}^2 + \text{cov}(a,b)(a+b))$
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$\rho_{A,C} = \frac{\text{cov}(A,C)}{\sigma_{A}\sigma_{C}}$
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With 100% correlation one ends up with $\rho_{a,b} = 1$, $\sigma_{a}= \sigma_{b}$.
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$\rho_{A,C} = \frac{\text{cov}(A,C)}{\sigma_{A}\sigma_{C}} = \frac{4(a+b)}{4(a+b)} = 1$
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With 0% correlation one ends up with $\rho_{a,b} = 0$, $\sigma_{a}= \sigma_{b}$.
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$\rho_{A,C} = \frac{\text{cov}(A,C)}{\sigma_{A}\sigma_{C}} = \frac{2(a+b)}{\sqrt{8(a^{2}+b^{2})}} = 0.938$
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### Triangulation
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1)
%% Cell type:markdown id: tags:
$l = 101.2$ m, $\sigma_{l} = 0.1$ m
%% Cell type:markdown id: tags:
$\alpha = 87.4^{\circ}$, $\sigma_{\alpha} = 0.1^{\circ}$
%% Cell type:markdown id: tags:
$\beta = 87.8^{\circ}$, $\sigma_{\beta} = 0.1^{\circ}$
%% Cell type:markdown id: tags:
$d = \frac{l\sin{(\alpha)} \sin{(\beta)}}{\sin{(\alpha +\beta)}}$
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$\sigma_{d}^{2} = \sigma_{\alpha}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\cos{(\alpha)} \sin{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2} + \sigma_{\beta}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\sin{(\alpha)} \cos{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2} + \sigma_{l}^{2} \left( \frac{\sin{(\alpha)} \sin{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2}$
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$\Rightarrow d =1207.36$ m, $\sigma_{d} = 35.63$ m
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2)
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Assuming $\sigma_{l} = 0$.
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$\Rightarrow \sigma_{d}^{2} = \sigma_{\alpha}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\cos{(\alpha)} \sin{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2} + \sigma_{\beta}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\sin{(\alpha)} \cos{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2}$
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$\Rightarrow \sigma_{d} = 36.61$ m
$\Rightarrow \sigma_{d} = 35.61$ m
%% Cell type:markdown id: tags:
The error on the distance is dominated by the error on the angles.
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3)
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$\alpha = \beta$, $\sigma_{\alpha} = \sigma_{\beta} = \sigma$
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$\sigma_{d}^{2} = \sigma_{\alpha}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\cos{(\alpha)} \sin{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2} + \sigma_{\beta}^{2} \left( - \frac{l\sin{(\alpha)} \sin{(\beta)} \cos{(\alpha +\beta)}}{\sin^{2}{(\alpha +\beta)}} + \frac{l\sin{(\alpha)} \cos{(\beta)}}{\sin{(\alpha +\beta)}} \right)^{2}$
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$\Rightarrow \sigma_{d}^{2} = 2 \sigma^{2} l^{2} \left( \frac{-\sin{(\alpha)} \sin{(\alpha)} \cos{(2\alpha)}}{\sin^{2}{(2\alpha)}} + \frac{\cos{(\alpha)} \sin{(\alpha)}}{\sin{(2 \alpha)}} \right)^{2} = 2 \sigma^{2} l^{2} \left( \frac{-\sin^{2}{(\alpha)} \cos{(2\alpha)} + \cos{(\alpha)} \sin{(\alpha)}\sin{(2\alpha)}}{\sin^{2}{(2\alpha)}} \right)^{2}$
%% Cell type:markdown id: tags:
Use: $\hspace{0.5cm}$ $\cos(2 \alpha) = \cos^{2}{(\alpha)} - \sin^{2}{(\alpha)}$ $\hspace{0.5cm}$ and $\hspace{0.5cm}$ $\sin(2 \alpha) = 2\sin{(\alpha)}\cos{(\alpha)}$
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$\Rightarrow \sigma_{d}^{2} = 2 \sigma^{2} l^{2} \left( \frac{-\sin^{2}{(\alpha)} \cos^{2}{(\alpha)} + \sin^{2}{(\alpha)} \sin^{2}{(\alpha)} + 2 \cos^{2}{(\alpha)} \sin^{2}{(\alpha)}}{\sin^{2}{(2\alpha)}} \right)^{2} = 2 \sigma^{2} l^{2} \left( \frac{ \sin^{2}{(\alpha)} \cdot [ - \cos^{2}{(\alpha)} + \sin^{2}{(\alpha)} + 2 \cos^{2}{(\alpha)} ]}{\sin^{2}{(2\alpha)}} \right)^{2}= 2 \sigma^{2} l^{2} \left( \frac{ \sin^{2}{(\alpha)} \cdot [ \cos^{2}{(\alpha)} + \sin^{2}{(\alpha)]} }{\sin^{2}{(2\alpha)}} \right)^{2}$
%% Cell type:markdown id: tags:
Use: $\hspace{0.5cm}$ $\sin(2 \alpha) = 2\sin{(\alpha)}\cos{(\alpha)}$
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$\Rightarrow \sigma_{d}^{2} = 2 \sigma^{2} l^{2} \left( \frac{ \sin^{2}{(\alpha)}}{\sin^{2}{(2\alpha)}} \right)^{2} = 2 \sigma^{2} l^{2} \left( \frac{ \sin^{2}{(\alpha)}}{4 \sin^{2}{(\alpha)}\cos^{2}{(\alpha)}} \right)^{2} = \frac{ \sigma^{2} l^{2}}{8 \cos^{4}{(\alpha)}}$
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$\Rightarrow \sigma_{d} \approx \frac{ \sigma l}{\sqrt{8} } \frac{(l/2)^{2} + d}{(l/2)^{2}} \approx \frac{ \sigma l}{\sqrt{8} } \frac{4 d^{2}}{l^{2}} = \sqrt{2} \sigma \frac{d^{2}}{l}$
%% Cell type:code id: tags:
``` python
```
......
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### Likelihood Estimation
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1)
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$\mathcal{L}= \exp{\left(-\frac{[\sin{(\theta)}-s]^{2}}{2\sigma^{2}}\right)} \exp{\left(-\frac{[\cos{(\theta)}-c]^{2}}{2\sigma^{2}}\right)}$
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$\mathcal{l} = \log{(\mathcal{L})} = -\left(\frac{[\sin{(\theta)}-s]^{2}}{2\sigma^{2}} + \frac{[\cos{(\theta)}-c]^{2}}{2\sigma^{2}}\right)$
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$\frac{\delta \mathcal{l}}{\delta \theta} = - \frac{1}{\sigma^{2}} [\cos{(\theta)}(\sin{(\theta)}-s) - \sin{(\theta)}(\cos{(\theta)}-c) ] = \frac{\cos{(\theta)}s - \sin{(\theta)}c}{\sigma^{2}} \overset{\mathrm{!}}{=} 0$
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$\Rightarrow \hat{\theta} = \text{atan}\left( \frac{s}{c} \right)$
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2)
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$\frac{\delta^{2} \mathcal{l}}{\delta \theta^{2}} = - \frac{\sin{(\theta)}s + \cos{(\theta)}c}{\sigma^{2}}$
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$\left\langle \frac{\delta^{2} \mathcal{l}}{\delta \theta^{2}} \right\rangle = - \frac{1}{\sigma^{2}}$
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### Maximum Likelihood Method
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1)
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$\mathcal{L} = \left(\begin{array}{c}
N \\ k
\end{array}\right) \cdot p^{k} \cdot (1-p)^{N-k}$
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$\mathcal{l} = \log{(\mathcal{L})} = \sum_{i} \left[ \text{log} \left(\begin{array}{c}
N \\ k
\end{array}\right) + k_{i} \text{log}(p) + (N-k_{i}) \text{log}(1-p) \right]$
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$\frac{\delta \mathcal{l}}{\delta p} = \sum_{i} \left( \frac{k_{i}}{p} - (N-k_{i}) \frac{1}{1-p} \right) \overset{\mathrm{!}}{=} 0$
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$(1-p) \sum_{i} k_{i} - nNp + \sum_{i} k_{i} p = 0$
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$\Rightarrow \hat{p} = \frac{\sum_{i} k_{i}}{nN}$
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2)
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$\frac{\delta^{2} \mathcal{l}}{\delta p^{2}} = \sum_{i=1}^{n} \left( \frac{k_{i}}{p^{2}} - (N-k_{i}) \frac{1}{(1-p)^{2}} \right)$
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$\left\langle \frac{\delta^{2} \mathcal{l}}{\delta p^{2}} \right\rangle = n \left( -\frac{pN}{p^{2}} - \frac{N-pN}{(1-p)^{2}} \right) = nN \left( - \frac{1}{p} - \frac{1}{1-p} \right) = -nN \frac{1}{p(1-p)} $
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### Likelihood and Exponential
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1)
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$\mathcal{L}= \frac{1}{\tau} \exp{\left(-\frac{t}{\tau} \right)}$
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$\mathcal{l} = \ln{(\mathcal{L})} = \sum_{i} \ln \left[ \frac{1}{\tau} \exp{\left(-\frac{t_{i}}{\tau} \right)} \right] = \sum_{i} \left[ - \frac{t_{i}}{\tau} - \ln(\tau) \right]$
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$\frac{\delta \mathcal{l}}{\delta \tau} = \sum_{i} \left[ \frac{t_{i}}{\tau^{2}} - \frac{1}{\tau} \right] \overset{\mathrm{!}}{=} 0$
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$\Rightarrow \hat{\tau} = \sum_{i} \frac{t_{i}}{N}$
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$\frac{\delta^{2} \mathcal{l}}{\delta \tau^{2}} = \sum_{i} \left[ - \frac{2t_{i}}{\tau^{3}} + \frac{1}{\tau^{2}} \right]$
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$\left\langle \frac{\delta^{2} \mathcal{l}}{\delta \tau^{2}} \right\rangle = - \frac{2N}{\tau^{2}} + \frac{N}{\tau^{2}} = -\frac{N}{\tau^{2}}$
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2)
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Show that $\int_{0}^{T} f(t;\tau) dt = 1$:
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$\int_{0}^{T} f(t;\tau) dt = \frac{1}{\tau} \frac{1}{1-\exp{\left(-\frac{T}{\tau}\right)}} \left[ - \tau \exp{\left(-\frac{t}{\tau}\right)} \right]_{0}^{T} = 1$
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``` python
```
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``` python
```
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### Likelihood Estimation
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1)
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$\mathcal{L}= \exp{\left(-\frac{[\sin{(\theta)}-s]^{2}}{2\sigma^{2}}\right)} \exp{\left(-\frac{[\cos{(\theta)}-c]^{2}}{2\sigma^{2}}\right)}$