Aufgaben:Exercise 4.15: PDF and Covariance Matrix - Revision history

Guenter at 11:20, 28 March 2022

2022-03-28T11:20:38Z

Guenter at 11:09, 28 March 2022

2022-03-28T11:09:58Z

Guenter at 10:54, 28 March 2022

2022-03-28T10:54:36Z

Guenter: Guenter moved page Aufgaben:Exercise 4.15: PDF and Correlation Matrix to Aufgaben:Exercise 4.15: PDF and Covariance Matrix

2022-03-28T10:50:37Z

Guenter moved page Exercise 4.15: PDF and Correlation Matrix to Exercise 4.15: PDF and Covariance Matrix

Noah at 11:04, 15 March 2022

2022-03-15T11:04:13Z

Noah at 11:03, 15 March 2022

2022-03-15T11:03:27Z

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Guenter: Guenter moved page Aufgaben:Aufgabe 4.15: WDF und Korrelationsmatrix to Aufgaben:Exercise 4.15: PDF and Correlation Matrix

2022-02-28T09:57:00Z

Guenter moved page Aufgabe 4.15: WDF und Korrelationsmatrix to Exercise 4.15: PDF and Correlation Matrix

Javier: Text replacement - "„" to """

2021-05-28T14:40:19Z

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Javier: Text replacement - "”" to """

2021-05-28T14:32:10Z

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Javier: Text replacement - "Category:Aufgaben zu Stochastische Signaltheorie" to "Category:Theory of Stochastic Signals: Exercises"

2021-03-23T12:44:25Z

Text replacement - "Category:Aufgaben zu Stochastische Signaltheorie" to "Category:Theory of Stochastic Signals: Exercises"

@@ Line 86: / Line 86: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Only <u>the proposed solution 2</u> is correct:
+'''(1)'''&nbsp; Only&nbsp; <u>the proposed solution 2</u>&nbsp; is correct:
-*On the basis of the covariance matrix&nbsp; $\mathbf{K}_{\mathbf{x}}$&nbsp; it is not possible to make any statement about whether the underlying random variable&nbsp; $\mathbf{x}$&nbsp; is zero mean or mean-invariant, since any mean value&nbsp; $\mathbf{m}$&nbsp; is factored out.
+*On the basis of the covariance matrix&nbsp; $\mathbf{K}_{\mathbf{x}}$&nbsp; it is not possible to make any statement about whether the underlying random variable&nbsp; $\mathbf{x}$&nbsp; is zero mean or mean-invariant,&nbsp; since any mean value&nbsp; $\mathbf{m}$&nbsp; is factored out.
-*To make statements about the mean, the correlation matrix&nbsp; $\mathbf{R}_{\mathbf{x}}$&nbsp; would have to be known.
+*To make statements about the mean,&nbsp; the correlation matrix&nbsp; $\mathbf{R}_{\mathbf{x}}$&nbsp; would have to be known.
 *From&nbsp; $K_{22} = \sigma_2^2 = 0$&nbsp; it follows necessarily that all other elements in the second row&nbsp; $(K_{21}, K_{23})$&nbsp; and the second column&nbsp; $(K_{12}, K_{32})$&nbsp; are also zero.
-*On the other hand, the third statement is false: &nbsp; The elements are symmetric about the main diagonal, so that always&nbsp; $K_{31} = K_{13}$&nbsp; must hold.
+*On the other hand,&nbsp; the third statement is false: &nbsp; The elements are symmetric about the main diagonal,&nbsp; so that always&nbsp; $K_{31} = K_{13}$&nbsp; must hold.
@@ Line 96: / Line 96: @@
 [[File:P_ID2915__Sto_A_4_15a.png|right|frame|Complete covariance matrix]]
 '''(2)'''&nbsp; From&nbsp; $K_{11} = 1$&nbsp; and&nbsp; $K_{33} = 0.25$&nbsp; follow directly&nbsp; $\sigma_1 = 1$&nbsp; and&nbsp; $\sigma_3 = 0.5$.
-*Taken together with the correlation coefficient&nbsp; $\rho_{13} = 0.8$&nbsp; (see specification sheet), we thus obtain:
+*Taken together with the correlation coefficient&nbsp; $\rho_{13} = 0.8$&nbsp; (see specification sheet),&nbsp; we thus obtain:
 :$$K_{13} = K_{31} = \sigma_1 \cdot \sigma_2 \cdot \rho_{13}\hspace{0.15cm}\underline{= 0.4}.$$
@@ Line 121: / Line 121: @@
 '''(5)'''&nbsp; A comparison of&nbsp; $\mathbf{K_y}$&nbsp; and&nbsp; $\mathbf{K_x}$&nbsp; with constraint&nbsp; $K_{22} = 0$&nbsp; shows that&nbsp; $\mathbf{x}$&nbsp; and&nbsp; $\mathbf{y}$&nbsp; are identical random variables if one sets&nbsp; $y_1 = x_1$&nbsp; and&nbsp; $y_2 = x_3$&nbsp; .
 *Thus, for the PDF parameters:
-:$$\sigma_1 =1, \hspace{0.3cm} \sigma_2 =0.5, \hspace{0.3cm} \rho =.
+:$$\sigma_1 =1, \hspace{0.3cm} \sigma_2 =0.5, \hspace{0.3cm} \rho =0.8.$$
 *The prefactor according to the general PDF definition is thus:
@@ Line 150: / Line 149: @@
 y_2\right).$$
-*By comparing coefficients, we get:
+*By comparing coefficients,&nbsp; we get:
 :$$\gamma_1 = \frac{25}{18} \approx 1.389; \hspace{0.3cm} \gamma_2 =
 \frac{100}{18} \approx 5.556; \hspace{0.3cm} \gamma_{12} = -
 \frac{80}{18} \approx -4,444.$$
-*According to the conventional procedure, the same numerical values result:
+*According to the conventional procedure,&nbsp; the same numerical values result:
 :$$\gamma_1 =\frac{\rm 1}{\rm 2\cdot \sigma_1^2 \cdot ({\rm
 -\rho^2})}=

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID669__Sto_A_4_15.png |right|frame|Two correlation matrices]]
+[[File:P_ID669__Sto_A_4_15.png |right|frame|Two covariance matrices]]
 We consider here the three-dimensional random variable &nbsp; $\mathbf{x}$,&nbsp; whose commonly represented covariance matrix &nbsp; $\mathbf{K}_{\mathbf{x}}$ &nbsp; is given in the upper graph.&nbsp; The random variable has the following properties:
@@ Line 15: / Line 15: @@
 In the second part of the exercise, consider the random variable&nbsp; $\mathbf{y}$&nbsp; with the two components&nbsp; $y_1$&nbsp; and&nbsp; $y_2$&nbsp; whose covariance matrix&nbsp; $\mathbf{K}_{\mathbf{y}}$&nbsp; is determined by the given numerical values &nbsp;$(1, \ 0.4, \ 0.25)$&nbsp; .
-The probability density function of a zero mean Gaussian two-dimensional random variable&nbsp; $\mathbf{y}$&nbsp; is as specified on page&nbsp; [[Theory_of_Stochastic_Signals/Generalization_to_N-Dimensional_Random_Variables#Relationship_between_covariance_matrix_and_PDF|Relationship between covariance matrix and PDF]]&nbsp; with&nbsp; $N = 2$:
+The probability density function&nbsp; $\rm (PDF)$&nbsp; of a zero mean Gaussian two-dimensional random variable&nbsp; $\mathbf{y}$&nbsp; is as specified on page&nbsp; [[Theory_of_Stochastic_Signals/Generalization_to_N-Dimensional_Random_Variables#Relationship_between_covariance_matrix_and_PDF|"Relationship between covariance matrix and PDF"]]&nbsp; with&nbsp; $N = 2$:
 :$$\mathbf{f_y}(\mathbf{y})   =  \frac{1}{{2 \pi \cdot
 \sqrt{|\mathbf{K_y}|}}}\cdot {\rm e}^{-{1}/{2} \hspace{0.05cm}\cdot\hspace{0.05cm} \mathbf{y} ^{\rm T}\hspace{0.05cm}\cdot\hspace{0.05cm}\mathbf{K_y}^{-1} \hspace{0.05cm}\cdot\hspace{0.05cm} \mathbf{y}  }=   C \cdot  {\rm e}^{-\gamma_1 \hspace{0.05cm}\cdot\hspace{0.05cm} y_1^2 \hspace{0.1cm}+\hspace{0.1cm} \gamma_2 \hspace{0.05cm}\cdot\hspace{0.05cm} y_2^2 \hspace{0.1cm}+\hspace{0.1cm}\gamma_{12} \hspace{0.05cm}\cdot\hspace{0.05cm} y_1 \hspace{0.05cm}\cdot\hspace{0.05cm} y_2 }.$$
 *In the subtasks&nbsp; '''(5)'''&nbsp; and&nbsp; '''(6)'''&nbsp; the prefactor&nbsp; $C$&nbsp; and the further PDF coefficients&nbsp; $\gamma_1$,&nbsp; $\gamma_2$&nbsp; and&nbsp; $\gamma_{12}$&nbsp; are to be calculated according to this vector representation.
-*In contrast, the corresponding equation in &nbsp;conventional approach according to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Two-Dimensional_Gaussian_Random_Variables#Probability_density_function_and_cumulative_distribution_function|Two-dimensional Gaussian Random Variables]]&nbsp; would be:
+*In contrast,&nbsp; the corresponding equation in &nbsp;conventional approach according to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Two-Dimensional_Gaussian_Random_Variables#Probability_density_function_and_cumulative_distribution_function|"Two-dimensional Gaussian Random Variables"]]&nbsp; would be:
 :$$f_{y_1,\hspace{0.1cm}y_2}(y_1,y_2)=\frac{\rm 1}{\rm 2\pi \sigma_1
 \sigma_2 \sqrt{\rm 1-\rho^2}}\cdot\exp\Bigg[-\frac{\rm 1}{\rm 2
@@ Line 33: / Line 33: @@
 Hints:
 *The exercise belongs to the chapter &nbsp;[[Theory_of_Stochastic_Signals/Generalization_to_N-Dimensional_Random_Variables|Generalization to N-Dimensional Random Variables]].
-*Some basics on the application of vectors and matrices can be found on the pages &nbsp;[[Theory_of_Stochastic_Signals/Generalization_to_N-Dimensional_Random_Variables#Basics_of_matrix_operations:_Determinant_of_a_matrix|Determinant of a Matrix]]&nbsp; and &nbsp;[[Theory_of_Stochastic_Signals/Generalization_to_N-Dimensional_Random_Variables#Basics_of_matrix_operations:_Inverse_of_a_matrix|Inverse of a Matrix]]&nbsp;.
+*Some basics on the application of vectors and matrices can be found in the sections &nbsp;[[Theory_of_Stochastic_Signals/Generalization_to_N-Dimensional_Random_Variables#Basics_of_matrix_operations:_Determinant_of_a_matrix|"Determinant of a Matrix"]]&nbsp; and &nbsp;[[Theory_of_Stochastic_Signals/Generalization_to_N-Dimensional_Random_Variables#Basics_of_matrix_operations:_Inverse_of_a_matrix|Inverse of a Matrix]]&nbsp;.
-*Reference is also made to the chapter &nbsp;[[Theory_of_Stochastic_Signals/Two-Dimensional_Gaussian_Random_Variables|Two-Dimensional Gaussian Random Variables]].
+*Reference is also made to the chapter &nbsp;[[Theory_of_Stochastic_Signals/Two-Dimensional_Gaussian_Random_Variables|"Two-Dimensional Gaussian Random Variables"]].
@@ Line 74: / Line 74: @@
-{Determine the coefficients in the argument of the exponential function.&nbsp; Compare the result with the 2D&ndash;PDF&ndash;equation.
+{Determine the coefficients in the argument of the exponential function.&nbsp; Compare the result with the two-dimensional PDF equation.
 |type="{}"}
 $\gamma_1 \ = \ $ { 1.389 3% }

@@ Line 4: / Line 4: @@
 [[File:P_ID669__Sto_A_4_15.png |right|frame|Two correlation matrices]]
-We consider here the three-dimensional random variable&nbsp; $\mathbf{x}$, whose commonly represented covariance matrix&nbsp; $\mathbf{K}_{\mathbf{x}}$&nbsp; is given in the graph.&nbsp; The random variable has the following properties:
+We consider here the three-dimensional random variable &nbsp; $\mathbf{x}$,&nbsp; whose commonly represented covariance matrix &nbsp; $\mathbf{K}_{\mathbf{x}}$ &nbsp; is given in the upper graph.&nbsp; The random variable has the following properties:
 * The three components are Gaussian distributed and it holds for the elements of the covariance matrix:

@@ Line 87: / Line 87: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; Only <u>the proposed solution 2</u> is correct:
-*On the basis of the covariance matrix&nbsp; $\mathbf{K}_{\mathbf{x}}$&nbsp; it is not possible to make any statement about whether the underlying random variable&nbsp; $\mathbf{x}$&nbsp; is mean-free or mean-invariant, since any mean value&nbsp; $\mathbf{m}$&nbsp; is factored out.
+*On the basis of the covariance matrix&nbsp; $\mathbf{K}_{\mathbf{x}}$&nbsp; it is not possible to make any statement about whether the underlying random variable&nbsp; $\mathbf{x}$&nbsp; is zero mean or mean-invariant, since any mean value&nbsp; $\mathbf{m}$&nbsp; is factored out.
 *To make statements about the mean, the correlation matrix&nbsp; $\mathbf{R}_{\mathbf{x}}$&nbsp; would have to be known.
 *From&nbsp; $K_{22} = \sigma_2^2 = 0$&nbsp; it follows necessarily that all other elements in the second row&nbsp; $(K_{21}, K_{23})$&nbsp; and the second column&nbsp; $(K_{12}, K_{32})$&nbsp; are also zero.

@@ Line 105: / Line 105: @@
-'''(4)'''&nbsp; Entsprechend den Angaben auf den Seiten &bdquo;Determinante einer Matrix"  und &bdquo;Inverse einer Matrix" gilt:
+'''(4)'''&nbsp; Entsprechend den Angaben auf den Seiten "Determinante einer Matrix"  und "Inverse einer Matrix" gilt:
 :$${\mathbf{I_y}} = {\mathbf{K_y}}^{-1} =
 \frac{1}{|{\mathbf{K_y}}|}\cdot \left[

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−	[[Category:~~Aufgaben zu Stochastische Signaltheorie~~\|^4.7 N-dimensionale Zufallsgrößen^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^4.7 N-dimensionale Zufallsgrößen^]]