Aufgaben:Exercise 4.15: PDF and Covariance Matrix

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Two correlation matrices

We consider here the three-dimensional random variable  $\mathbf{x}$, whose commonly represented covariance matrix  $\mathbf{K}_{\mathbf{x}}$  is given in the graph.  The random variable has the following properties:

• The three components are Gaussian distributed and it holds for the elements of the covariance matrix:

$$K_{ij} = \sigma_i \cdot \sigma_j \cdot \rho_{ij}.$$

• Let the elements on the main diagonal be known:

$$ K_{11} =1, \ K_{22} =0, \ K_{33} =0.25.$$

• The correlation coefficient between the coefficients  $x_1$  and  $x_3$  is  $\rho_{13} = 0.8$.

In the second part of the exercise, consider the random variable  $\mathbf{y}$  with the two components  $y_1$  and  $y_2$  whose covariance matrix  $\mathbf{K}_{\mathbf{y}}$  is determined by the given numerical values  $(1, \ 0.4, \ 0.25)$  .

The probability density function of a zero mean Gaussian two-dimensional random variable  $\mathbf{y}$  is as specified on page  Relationship between covariance matrix and PDF  with  $N = 2$:

$$\mathbf{f_y}(\mathbf{y}) = \frac{1}{{2 \pi \cdot \sqrt{|\mathbf{K_y}|}}}\cdot {\rm e}^{-{1}/{2} \hspace{0.05cm}\cdot\hspace{0.05cm} \mathbf{y} ^{\rm T}\hspace{0.05cm}\cdot\hspace{0.05cm}\mathbf{K_y}^{-1} \hspace{0.05cm}\cdot\hspace{0.05cm} \mathbf{y} }= C \cdot {\rm e}^{-\gamma_1 \hspace{0.05cm}\cdot\hspace{0.05cm} y_1^2 \hspace{0.1cm}+\hspace{0.1cm} \gamma_2 \hspace{0.05cm}\cdot\hspace{0.05cm} y_2^2 \hspace{0.1cm}+\hspace{0.1cm}\gamma_{12} \hspace{0.05cm}\cdot\hspace{0.05cm} y_1 \hspace{0.05cm}\cdot\hspace{0.05cm} y_2 }.$$

• In the subtasks  (5)  and  (6)  the prefactor  $C$  and the further PDF coefficients  $\gamma_1$,  $\gamma_2$  and  $\gamma_{12}$  are to be calculated according to this vector representation.
• In contrast, the corresponding equation in  conventional approach according to the chapter  Two-dimensional Gaussian Random Variables  would be:

$$f_{y_1,\hspace{0.1cm}y_2}(y_1,y_2)=\frac{\rm 1}{\rm 2\pi \sigma_1 \sigma_2 \sqrt{\rm 1-\rho^2}}\cdot\exp\Bigg[-\frac{\rm 1}{\rm 2 (1-\rho^{\rm 2})}\cdot(\frac { y_1^{\rm 2}}{\sigma_1^{\rm 2}}+\frac { y_2^{\rm 2}}{\sigma_2^{\rm 2}}-\rm 2\rho \frac{{\it y}_1{\it y}_2}{\sigma_1 \cdot \sigma_2}) \rm \Bigg].$$

Hints:

• The exercise belongs to the chapter  Generalization to N-Dimensional Random Variables.
• Some basics on the application of vectors and matrices can be found on the pages  Determinant of a Matrix  and  Inverse of a Matrix .
• Reference is also made to the chapter  Two-Dimensional Gaussian Random Variables.

Questions

1

Which of the following statements are true?

	The random variable  $\mathbf{x}$  is zero mean with certainty.
	The matrix elements  $K_{12}$,  $K_{21}$,  $K_{23}$  and  $K_{32}$  are zero.
	It holds that $K_{31} = -K_{13}$.

2

Calculate the matrix element of the last row and first column.

$K_\text{31} \ = \ $

3

Calculate the determinant  $|\mathbf{K}_{\mathbf{y}}|$.

$|\mathbf{K}_{\mathbf{y}}| \ = \ $

4

Calculate the inverse matrix  $\mathbf{I}_{\mathbf{y}} = \mathbf{K}_{\mathbf{y}}^{-1}$  with matrix elements. $I_{ij}$ :

$I_\text{11} \ = \ $

$I_\text{12} \ = \ $

$I_\text{21} \ = \ $

$I_\text{22} \ = \ $

5

Calculate the prefactor  $C$  of the two-dimensional probability density function.  Compare the result with the formula given in the theory section.

$C\ = \ $

6

Determine the coefficients in the argument of the exponential function.  Compare the result with the 2D–PDF–equation.

$\gamma_1 \ = \ $

$\gamma_2 \ = \ $

$\gamma_{12}\ = \ $

Solution

Solution

(1)  Only the proposed solution 2 is correct:

• On the basis of the covariance matrix  $\mathbf{K}_{\mathbf{x}}$  it is not possible to make any statement about whether the underlying random variable  $\mathbf{x}$  is zero mean or mean-invariant, since any mean value  $\mathbf{m}$  is factored out.
• To make statements about the mean, the correlation matrix  $\mathbf{R}_{\mathbf{x}}$  would have to be known.
• From  $K_{22} = \sigma_2^2 = 0$  it follows necessarily that all other elements in the second row  $(K_{21}, K_{23})$  and the second column  $(K_{12}, K_{32})$  are also zero.
• On the other hand, the third statement is false:   The elements are symmetric about the main diagonal, so that always  $K_{31} = K_{13}$  must hold.

Complete covariance matrix

(2)  From  $K_{11} = 1$  and  $K_{33} = 0.25$  follow directly  $\sigma_1 = 1$  and  $\sigma_3 = 0.5$.

• Taken together with the correlation coefficient  $\rho_{13} = 0.8$  (see specification sheet), we thus obtain:

$$K_{13} = K_{31} = \sigma_1 \cdot \sigma_2 \cdot \rho_{13}\hspace{0.15cm}\underline{= 0.4}.$$

(3)  The determinant of the matrix  $\mathbf{K_y}$  is:

$$|{\mathbf{K_y}}| = 1 \cdot 0.25 - 0.4 \cdot 0.4 \hspace{0.15cm}\underline{= 0.09}.$$

(4)  According to the statements on the pages "Determinant of a Matrix" and "Inverse of a Matrix" holds:

$${\mathbf{I_y}} = {\mathbf{K_y}}^{-1} = \frac{1}{|{\mathbf{K_y}}|}\cdot \left[ \begin{array}{cc} 0.25 & -0.4 \\ -0.4 & 1 \end{array} \right].$$

• With  $|\mathbf{K_y}|= 0.09$  therefore holds further:

$$I_{11} = {25}/{9}\hspace{0.15cm}\underline{ = 2.777};\hspace{0.3cm} I_{12} = I_{21} = -40/9 \hspace{0.15cm}\underline{ = -4.447};\hspace{0.3cm}I_{22} = {100}/{9} \hspace{0.15cm}\underline{= 11.111}.$$

(5)  A comparison of  $\mathbf{K_y}$  and  $\mathbf{K_x}$  with constraint  $K_{22} = 0$  shows that  $\mathbf{x}$  and  $\mathbf{y}$  are identical random variables if one sets  $y_1 = x_1$  and  $y_2 = x_3$  .

• Thus, for the PDF parameters:

$$\sigma_1 =1, \hspace{0.3cm} \sigma_2 =0.5, \hspace{0.3cm} \rho =. 0.8.$$

• The prefactor according to the general PDF definition is thus:

$$C =\frac{\rm 1}{\rm 2\pi \cdot \sigma_1 \cdot \sigma_2 \cdot \sqrt{\rm 1-\rho^2}}= \frac{\rm 1}{\rm 2\pi \cdot 1 \cdot 0.5 \cdot 0.6}= \frac{1}{0.6 \cdot \pi} \hspace{0.15cm}\underline{\approx 0.531}.$$

• With the determinant calculated in subtask  (3)  we get the same result:

$$C =\frac{\rm 1}{\rm 2\pi \sqrt{|{\mathbf{K_y}}|}}= \frac{\rm 1}{\rm 2\pi \sqrt{0.09}} = \frac{1}{0.6 \cdot \pi}.$$

(6)  The inverse matrix computed in subtask  (4)  can also be written as follows:

$${\mathbf{I_y}} = \frac{5}{9}\cdot \left[ \begin{array}{cc} 5 & -8 \\ -8 & 20 \end{array} \right].$$

• So the argument  $A$  of the exponential function is:

$$A = \frac{5}{18}\cdot{\mathbf{y}}^{\rm T}\cdot \left[ \begin{array}{cc} 5 & -8 \\ -8 & 20 \end{array} \right]\cdot{\mathbf{y}} =\frac{5}{18}\left( 5 \cdot y_1^2 + 20 \cdot y_2^2 -16 \cdot y_1 \cdot y_2\right).$$

• By comparing coefficients, 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:

$$\gamma_1 =\frac{\rm 1}{\rm 2\cdot \sigma_1^2 \cdot ({\rm 1-\rho^2})}= \frac{\rm 1}{\rm 2 \cdot 1 \cdot 0.36} \hspace{0.15cm}\underline{ \approx 1.389},$$

$$\gamma_2 =\frac{\rm 1}{\rm 2 \cdot\sigma_2^2 \cdot ({\rm 1-\rho^2})}= \frac{\rm 1}{\rm 2 \cdot 0.25 \cdot 0.36} = 4 \cdot \gamma_1 \hspace{0.15cm}\underline{\approx 5.556},$$

$$\gamma_{12} =-\frac{\rho}{ \sigma_1 \cdot \sigma_2 \cdot ({\rm 1-\rho^2})}= -\frac{\rm 0.8}{\rm 1 \cdot 0.5 \cdot 0.36} \hspace{0.15cm}\underline{ \approx -4.444}.$$