Aufgaben:Exercise 4.16Z: Multi-dimensional Data Reduction - Revision history

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	[[Category:Theory of Stochastic Signals: Exercises\|^4.7 N-dimensionale Zufallsgrößen^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.7 N-dimensionale Zufallsgrößen^]]
	[[de:~~Zusatzaufgaben~~:4.~~16_2D-_und_3D~~-Datenreduktion]]		[[de:Aufgaben:Aufgabe 4.16Z: Zwei- und dreidimensionale Datenreduktion]]

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 '''(3)'''&nbsp; The equation of determination of the eigenvalues of&nbsp; $\mathbf{K_z}$&nbsp; is:
-:$${\rm det} \left[ \begin{array}{ccc}
+:$${\rm det} \left[ \begin{array}{ccc}1-\lambda & 1/3 & 1/3\\1/3 & 1-\lambda & 1/3\\1/3 & 1/3 & 1-\lambda\end{array}\right] = 0 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}(1- \lambda) \left[(1- \lambda)^2 -\frac{1}{9} \right]- \frac{1}{3} \left[\frac{1}{3}(1- \lambda) -\frac{1}{9} \right] + \frac{1}{3} \left[\frac{1}{9} -\frac{1}{3}(1- \lambda)\right] = 0$$:$$\Rightarrow \hspace{0.3cm}(1- \lambda) (\lambda^2 -2\lambda+\frac{8}{9})- \frac{1}{9} (\frac{2}{3}- \lambda )+ \frac{1}{9} (\lambda - \frac{2}{3})= 0$$:$$\Rightarrow \hspace{0.3cm}\lambda^2 - 2\lambda + \frac{8}{9} -\lambda^3 + 2 \lambda^2 - \frac{8}{9}\lambda - \frac{4}{27} +\frac{2}{9}\lambda = 0 \hspace{0.3cm}\Rightarrow \hspace{0.3cm}\lambda^3 -  3 \lambda^2 +\frac{24}{9}\lambda - \frac{20}{27}  = 0.$$*This equation has already been given as a solution hint,&nbsp; as well as one of the solutions: &nbsp; $\lambda_1= 5/3$.*This gives the equation for the further eigenvalues&nbsp; $\lambda_2$&nbsp; and&nbsp; $\lambda_3$&nbsp; to:$$\frac{\lambda^3 - 3 \lambda^2 + {24}/{9}\lambda -{20}/{27}}{\lambda -{5}/{3}} = \lambda^2 -{4}/{3} \cdot \lambda + {4}/{9} =0.$$
 *This equation can be transformed as follows: &nbsp; $(\lambda - {2}/{3})^2 =0.$

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	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  From the condition   $\mathbf{K_y} - \lambda \cdot\mathbf{E} = 0$   follows:		'''(1)'''  From the condition   $\mathbf{K_y} - \lambda \cdot\mathbf{E} = 0$   follows:
	:$${\rm det}\left[ \begin{array}{cc}		:$${\rm det}\left[ \begin{array}{cc}1- \lambda & 1/3 \\1/3 & 1- \lambda\end{array} \right] = (1-\lambda)^2 -{1}/{9} = 0\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\lambda^2 -2\lambda+ {8}/{9}= 0\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\lambda_{1/2}= 1 \pm\sqrt{1-{8}/{9}}= 1 \pm {1}/{3}.$$
	1- \lambda & 1/3 \\
	1/3 & 1- \lambda
	\end{array} \right] = (1-\lambda)^2 -{1}/{9} = 0
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\lambda^2 -2\lambda+ {8}/{9}= 0
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\lambda_{1/2}= 1 \pm
	\sqrt{1-{8}/{9}}= 1 \pm {1}/{3}.$$

	*The eigenvalues of this  $2\times2$  matrix are thus  $\lambda_1 = 4/3\hspace{0.15cm}\underline{=1.333}$  and  $\lambda_2 = 2/3\hspace{0.15cm}\underline{=0.667}$.		*The eigenvalues of this  $2\times2$  matrix are thus  $\lambda_1 = 4/3\hspace{0.15cm}\underline{=1.333}$  and  $\lambda_2 = 2/3\hspace{0.15cm}\underline{=0.667}$.

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	'''(2)'''  Without considering correlations,  there are  $N_2 = \left({8}/{ \Delta_x}\right)^2= 256^2 = 65536$  different pairs of values.		'''(2)'''  Without considering correlations,  there are  $N_2 = \left({8}/{ \Delta_x}\right)^2= 256^2 = 65536$  different pairs of values.
	*Taking into account the correlations and the fact that the two components created by coordinate rotation  $\eta_1$  and  $\eta_2$  are each in the range  $-4\cdot \sigma_1$  to  $+4\cdot \sigma_1$  $($resp. from  $-4\cdot \sigma_2$  to  $+4\cdot \sigma_2)$ ) are to be quantized,  one obtains		*Taking into account the correlations and the fact that the two components created by coordinate rotation  $\eta_1$  and  $\eta_2$  are each in the range  $-4\cdot \sigma_1$  to  $+4\cdot \sigma_1$  $($resp. from  $-4\cdot \sigma_2$  to  $+4\cdot \sigma_2)$ ) are to be quantized,  one obtains
	:$$N_2\hspace{0.01cm}' = \frac{8 \hspace{0.05cm}\sigma_1}{\it \Delta_x}\cdot\frac{8		:$$N_2\hspace{0.01cm}' = \frac{8 \hspace{0.05cm}\sigma_1}{\it \Delta_x}\cdot\frac{8\hspace{0.05cm}\sigma_2}{\it \Delta_y}= N_2 \cdot \sigma_1 \cdot\sigma_2 .$$
	\hspace{0.05cm}\sigma_2}{\it \Delta_y}= N_2 \cdot \sigma_1 \cdot
	\sigma_2 .$$

	*The quotient is thus with  $\sigma_1^2 = \lambda_1$  and  $\sigma_2^2 = \lambda_2$:		*The quotient is thus with  $\sigma_1^2 = \lambda_1$  and  $\sigma_2^2 = \lambda_2$:
	:$${N_2\hspace{0.01cm}'}/{N_2} = \sigma_1 \cdot \sigma_2 = \sqrt{{4}/{3}}		:$${N_2\hspace{0.01cm}'}/{N_2} = \sigma_1 \cdot \sigma_2 = \sqrt{{4}/{3}}\cdot \sqrt{{2}/{3}} = \frac{2 \cdot \sqrt{2}}{3} \hspace{0.15cm}\underline{ \approx 0.943}.$$
	\cdot \sqrt{{2}/{3}} = \frac{2 \cdot \sqrt{2}}{3} \hspace{0.15cm}\underline{ \approx 0.943}.$$


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	\frac{1}{9} \right] + \frac{1}{3} \left[\frac{1}{9} -		\frac{1}{9} \right] + \frac{1}{3} \left[\frac{1}{9} -
	\frac{1}{3}(1- \lambda)		\frac{1}{3}(1- \lambda)
	\right] = 0$$		\right] = 0$$:$$\Rightarrow \hspace{0.3cm}(1- \lambda) (\lambda^2 -2\lambda+
	:$$\Rightarrow \hspace{0.3cm}(1- \lambda) (\lambda^2 -2\lambda+
	\frac{8}{9})- \frac{1}{9} (\frac{2}{3}- \lambda )+ \frac{1}{9} (		\frac{8}{9})- \frac{1}{9} (\frac{2}{3}- \lambda )+ \frac{1}{9} (
	\lambda - \frac{2}{3})= 0$$		\lambda - \frac{2}{3})= 0$$:$$\Rightarrow \hspace{0.3cm}\lambda^2 - 2\lambda + \frac{8}{9} -
	:$$\Rightarrow \hspace{0.3cm}\lambda^2 - 2\lambda + \frac{8}{9} -
	\lambda^3 + 2 \lambda^2 - \frac{8}{9}\lambda - \frac{4}{27} +		\lambda^3 + 2 \lambda^2 - \frac{8}{9}\lambda - \frac{4}{27} +
	\frac{2}{9}\lambda = 0 \hspace{0.3cm}		\frac{2}{9}\lambda = 0 \hspace{0.3cm}
	\Rightarrow \hspace{0.3cm}\lambda^3 - 3 \lambda^2 +		\Rightarrow \hspace{0.3cm}\lambda^3 - 3 \lambda^2 +
	\frac{24}{9}\lambda - \frac{20}{27} = 0.$$		\frac{24}{9}\lambda - \frac{20}{27} = 0.$$This equation has already been given as a solution hint,  as well as one of the solutions:   $\lambda_1= 5/3$.This gives the equation for the further eigenvalues  $\lambda_2$  and  $\lambda_3$  to:$$\frac{\lambda^3 - 3 \lambda^2 + {24}/{9}\lambda -

	*This equation has already been given as a solution hint,  as well as one of the solutions:   $\lambda_1= 5/3$.
	*This gives the equation for the further eigenvalues  $\lambda_2$  and  $\lambda_3$  to
	:$$\frac{\lambda^3 - 3 \lambda^2 + {24}/{9}\lambda -
	{20}/{27}}{\lambda -{5}/{3}} = \lambda^2 -		{20}/{27}}{\lambda -{5}/{3}} = \lambda^2 -
	{4}/{3} \cdot \lambda + {4}/{9} =0.$$		{4}/{3} \cdot \lambda + {4}/{9} =0.$$
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	'''(4)'''  Analogous to the procedure in the subtask  '''(2)'''  results here:		'''(4)'''  Analogous to the procedure in the subtask  '''(2)'''  results here:
	:$${N_3\hspace{0.01cm}'}/{N_3} = \sqrt{\lambda_1 \cdot \lambda_2\cdot		:$${N_3\hspace{0.01cm}'}/{N_3} = \sqrt{\lambda_1 \cdot \lambda_2\cdot\lambda_3} = \sqrt{\frac{5}{3} \cdot \frac{2}{3}\cdot \frac{2}{3}}= \sqrt{\frac{20}{27}} \hspace{0.15cm}\underline{ \approx 0.861}.$$
	\lambda_3} = \sqrt{\frac{5}{3} \cdot \frac{2}{3}\cdot \frac{2}{3}}
	= \sqrt{\frac{20}{27}} \hspace{0.15cm}\underline{ \approx 0.861}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

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	[[Category:Theory of Stochastic Signals: Exercises\|^4.7 N-dimensionale Zufallsgrößen^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.7 N-dimensionale Zufallsgrößen^]]
			[[de:Zusatzaufgaben:4.16_2D-_und_3D-Datenreduktion]]

Guenter at 14:58, 29 March 2022

2022-03-29T14:58:47Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  From the condition  $\mathbf{K_y} - \lambda \cdot\mathbf{E} = 0$  follows:		'''(1)'''  From the condition   $\mathbf{K_y} - \lambda \cdot\mathbf{E} = 0$   follows:
	:$${\rm det}\left[ \begin{array}{cc}		:$${\rm det}\left[ \begin{array}{cc}
	1- \lambda & 1/3 \		1- \lambda & 1/3 \\
	1/3 & 1- \lambda		1/3 & 1- \lambda
	\end{array} \right] = (1-\lambda)^2 -{1}/{9} = 0		\end{array} \right] = (1-\lambda)^2 -{1}/{9} = 0
	\hspace{0.3cm}\~~rightarrow~~ \hspace{0.3cm}\lambda^2 -2\lambda+ {8}/{9}= 0		\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\lambda^2 -2\lambda+ {8}/{9}= 0
	\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\lambda_{1/2}= 1 \pm		\hspace{0.3cm}\Rightarrow \hspace{0.3cm}\lambda_{1/2}= 1 \pm
	\sqrt{1-{8}/{9}}= 1 \pm {1}/{3}.$$		\sqrt{1-{8}/{9}}= 1 \pm {1}/{3}.$$

	*The eigenvalues of this  $2\times2$ matrix are thus  $\lambda_1 = 4/3\hspace{0.15cm}\underline{=1.333}$  and  $\lambda_2 = 2/3\hspace{0.15cm}\underline{=0.667}$.		*The eigenvalues of this  $2\times2$  matrix are thus  $\lambda_1 = 4/3\hspace{0.15cm}\underline{=1.333}$  and  $\lambda_2 = 2/3\hspace{0.15cm}\underline{=0.667}$.



	'''(2)'''  Without considering correlations, there are  $N_2 = \left({8}/{ \Delta_x}\right)^2= 256^2 = 65536$  different pairs of values.
	*Taking into account the correlations and the fact that the two components created by coordinate rotation  $\eta_1$  and  $\eta_2$  are each in the range  $-4\sigma_1$  to  $+4\sigma_1$  $($~~bzw~~. from  $-4\sigma_2$  to  $+4\sigma_2)$ ) are to be quantized, one obtains		'''(2)'''  Without considering correlations,  there are  $N_2 = \left({8}/{ \Delta_x}\right)^2= 256^2 = 65536$  different pairs of values.
			*Taking into account the correlations and the fact that the two components created by coordinate rotation  $\eta_1$  and  $\eta_2$  are each in the range  $-4\cdot \sigma_1$  to  $+4\cdot \sigma_1$  $($resp. from  $-4\cdot \sigma_2$  to  $+4\cdot \sigma_2)$ ) are to be quantized,  one obtains
	:$$N_2\hspace{0.01cm}' = \frac{8 \hspace{0.05cm}\sigma_1}{\it \Delta_x}\cdot\frac{8		:$$N_2\hspace{0.01cm}' = \frac{8 \hspace{0.05cm}\sigma_1}{\it \Delta_x}\cdot\frac{8
	\hspace{0.05cm}\sigma_2}{\it \Delta_y}= N_2 \cdot \sigma_1 \cdot		\hspace{0.05cm}\sigma_2}{\it \Delta_y}= N_2 \cdot \sigma_1 \cdot
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	\frac{24}{9}\lambda - \frac{20}{27} = 0.$$		\frac{24}{9}\lambda - \frac{20}{27} = 0.$$

	*This equation has already been given as a solution hint, as well as one of the solutions:   $\lambda_1= 5/3$.		*This equation has already been given as a solution hint,  as well as one of the solutions:   $\lambda_1= 5/3$.
	*This gives the equation ~~of determination~~ for the further eigenvalues  $\lambda_2$  and  $\lambda_3$  to.		*This gives the equation for the further eigenvalues  $\lambda_2$  and  $\lambda_3$  to
	:$$\frac{\lambda^3 - 3 \lambda^2 + {24}/{9}\lambda -		:$$\frac{\lambda^3 - 3 \lambda^2 + {24}/{9}\lambda -
	{20}/{27}}{\lambda -{5}/{3}} = \lambda^2 -		{20}/{27}}{\lambda -{5}/{3}} = \lambda^2 -
	{4}/{3} \cdot \lambda + {4}/{9} =0.$$		{4}/{3} \cdot \lambda + {4}/{9} =0.$$

	*This equation ~~of determination~~ can be transformed as follows:   $(\lambda - {2}/{3})^2 =0.$		*This equation can be transformed as follows:   $(\lambda - {2}/{3})^2 =0.$

	*The other eigenvalues besides  $\lambda_1= 5/3$  are thus equal and result in   $\lambda_2 = \lambda_3 =2/3\hspace{0.15cm}\underline{=0.667}$.		*The other eigenvalues besides  $\lambda_1= 5/3$  are thus equal and result in   $\lambda_2 = \lambda_3 =2/3\hspace{0.15cm}\underline{=0.667}$.

Guenter at 14:42, 29 March 2022

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Guenter: Guenter moved page Aufgaben:Exercise 4.16Z: Multidimensional Data Reduction to Aufgaben:Exercise 4.16Z: Multi-dimensional Data Reduction

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Guenter moved page Aufgaben:Exercise 4.16Z: Multidimensional Data Reduction to Aufgaben:Exercise 4.16Z: Multi-dimensional Data Reduction

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Noah at 11:40, 15 March 2022

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Guenter: Guenter moved page Aufgaben:Exercise 4.16Z: Two-dimensional and Three-dimensional Data Reduction to Aufgaben:Exercise 4.16Z: Multidimensional Data Reduction

2022-02-28T10:03:53Z

Guenter moved page Aufgaben:Exercise 4.16Z: Two-dimensional and Three-dimensional Data Reduction to Aufgaben:Exercise 4.16Z: Multidimensional Data Reduction

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	}}		}}

	[[File:P_ID678__Sto_Z_4_16.png\|right\|frame\|~~Correlation matrices <br>~~$\mathbf{K_y}$  and  $\mathbf{K_z}$ ]]		[[File:P_ID678__Sto_Z_4_16.png\|right\|frame\|Matrices  $\mathbf{K_y}$  and  $\mathbf{K_z}$ ]]
	We consider Gaussian zero mean random variables  $\mathbf{x}$,  $\mathbf{y}$  and  $\mathbf{z}$  with dimensions  $N= 1$,  $N= 2$  and  $N= 3$:		We consider Gaussian zero mean random variables   $\mathbf{x}$,  $\mathbf{y}$   and   $\mathbf{z}$   with dimensions  $N= 1$,  $N= 2$  and  $N= 3$:
	* The one-dimensional random variable  $\mathbf{x}$  is characterized by the variance  $\sigma^2 = 1$  and the standard deviation   $\sigma = 1$  respectively. <br>Because of the dimension  $N= 1$  $\mathbf{x} = x$ ~~holds~~.		* The one-dimensional random variable  $\mathbf{x}$  is characterized by the variance  $\sigma^2 = 1$  and the standard deviation   $\sigma = 1$  respectively. <br>Because of the dimension  $N= 1$   ⇒   $\mathbf{x} = x$.

	* The correlation coefficient between the components  $y_1$  and  $y_2$  of the 2D random variable  $\mathbf{y}$  is  $\rho = 1/3$  $($see matrix  $\mathbf{K_y})$. <br>$y_1$ and $y_2$ also have the standard deviation $\sigma = 1$.		* The correlation coefficient between the components  $y_1$  and  $y_2$  of the 2D random variable  $\mathbf{y}$  is  $\rho = 1/3$  $($see matrix  $\mathbf{K_y})$. <br>$y_1$ and $y_2$ also have the standard deviation $\sigma = 1$.
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	If one quantizes the random variable  $\mathbf{x}$  in the range between  $-4$  and  $+4$  with interval width  $\Delta_x = 1/32$, there are altogether  $N_1 = 256$  different quantization values, for whose transmission thus  $n_1 = 8\ \rm {bit}$  would be needed.		If one quantizes the random variable  $\mathbf{x}$  in the range between  $-4$  and  $+4$  with interval width  $\Delta_x = 1/32$,  there are altogether  $N_1 = 256$  different quantization values,  for whose transmission thus  $n_1 = 8\ \rm {bit}$  would be needed.

	Similarly, the random variable  $\mathbf{y}$  results in a total of  $N_2 = 256^2 = 65536$  different quantized value pairs, if the correlation between  $y_1$  and  $y_2$  is not taken into account.		Similarly,  the random variable  $\mathbf{y}$  results in a total of  $N_2 = 256^2 = 65536$  different quantized value pairs,  if the correlation between  $y_1$  and  $y_2$  is not taken into account.

	Exploiting this correlation– for example, by coordinate transformation from the original system  $(y_1, y_2)$  to the new system  $(\eta_1, \eta_2)$  – results in a smaller number  $N_2\hspace{0.01cm}'$  of quantized value pairs.		Exploiting this correlation  – for example, by coordinate transformation from the original system  $(y_1, y_2)$  to the new system  $(\eta_1, \eta_2)$  –  results in a smaller number  $N_2\hspace{0.01cm}'$  of quantized value pairs.

	*Here, it is to be considered that each component is to be quantized according to its respective ~~dispersion~~  $(\sigma_1$  resp.  $\sigma_2)$  in the range of  $-4$  to  $+4$  and the quantization intervals should be the same in both directions:   $\Delta_x = \Delta_y =1/32$.		*Here, it is to be considered that each component is to be quantized according to its respective standard deviation  $(\sigma_1$  resp.  $\sigma_2)$  in the range of  $-4$  to  $+4$  and the quantization intervals should be the same in both directions:   $\Delta_x = \Delta_y =1/32$.

	*We denote the quotient  $N_2\hspace{0.01cm}'/N_2$  as the data reduction factor with respect to the 2D random variable  $\mathbf{y}$.		*We denote the quotient  $N_2\hspace{0.01cm}'/N_2$  as the data reduction factor with respect to the two-dimensional random variable  $\mathbf{y}$.
	*In analogous definition  $N_3'/N_3$  is the corresponding reduction factor of the 3D random variable  $\mathbf{z}$  for  $\Delta_x = \Delta_y =\Delta_z =1/32.$		*In analogous definition  $N_3'/N_3$  is the corresponding reduction factor of the three-dimensional random variable  $\mathbf{z}$  for  $\Delta_x = \Delta_y =\Delta_z =1/32.$
	*Note that in both cases the smallest possible value of this quotient would be favorable.		*Note that in both cases the smallest possible value of this quotient would be favorable.






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	Hints:		Hints:
	*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Generalization_to_N-Dimensional_Random_Variables\|Generalization to N-Dimensional Random Variables]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Generalization_to_N-Dimensional_Random_Variables\|Generalization to N-Dimensional Random Variables]].
	*In particular, reference is made to the ~~page~~  [[Theory_of_Stochastic_Signals/Generalization_to_N-Dimensional_Random_Variables#Eigenvalues_and_eigenvectors\|Eigenvalues and Eigenvectors]] .		*In particular,  reference is made to the section  [[Theory_of_Stochastic_Signals/Generalization_to_N-Dimensional_Random_Variables#Eigenvalues_and_eigenvectors\|Eigenvalues and Eigenvectors]] .
	*Some basics on the application of vectors and matrices can be found on the pages  [[Theory_of_Stochastic_Signals/Generalization_to_N-Dimensional_Random_Variables#Basics_of_matrix_operations:_Determinant_of_a_matrix\|Determinant of a Matrix]]  ~~and  ~~[[Theory_of_Stochastic_Signals/Generalization_to_N-Dimensional_Random_Variables#Basics_of_matrix_operations:_Inverse_of_a_matrix\|Inverse of a Matrix]] .		*Some basics on the application of vectors and matrices can be found on the pages
			**[[Theory_of_Stochastic_Signals/Generalization_to_N-Dimensional_Random_Variables#Basics_of_matrix_operations:_Determinant_of_a_matrix\|Determinant of a Matrix]],
	*The equation of determination of the eigenvalues of  $\mathbf{K_z}$  is:   $\lambda^3 - 3 \lambda^2 + {24}/{9}\lambda - {20}/{27} = 0.$		**[[Theory_of_Stochastic_Signals/Generalization_to_N-Dimensional_Random_Variables#Basics_of_matrix_operations:_Inverse_of_a_matrix\|Inverse of a Matrix]] .
			*The equation for determination the eigenvalues of  $\mathbf{K_z}$  is:   $\lambda^3 - 3 \lambda^2 + {24}/{9}\lambda - {20}/{27} = 0.$
	*One of the three solutions of this equation is  $\lambda_1 = 5/3$.		*One of the three solutions of this equation is  $\lambda_1 = 5/3$.

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	{What is the data reduction factor for the 2D random variable  $\mathbf{y}$?		{What is the data reduction factor for the two-dimensional random variable  $\mathbf{y}$?
	\|type="{}"}		\|type="{}"}
	$N_2\hspace{0.01cm}'/N_2 \ = $ { 0.943 3% }		$N_2\hspace{0.01cm}'/N_2 \ = $ { 0.943 3% }
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	{What is the data reduction factor for the 3D random variable  $\mathbf{z}$?		{What is the data reduction factor for the three-dimensional random variable  $\mathbf{z}$?
	\|type="{}"}		\|type="{}"}
	$N_3\hspace{0.01cm}'/N_3 \ = $ { 0.861 3% }		$N_3\hspace{0.01cm}'/N_3 \ = $ { 0.861 3% }