Aufgaben:Exercise 4.4Z: Contour Lines of the "2D-PDF" - Revision history

Guenter at 15:43, 10 April 2022

2022-04-10T15:43:37Z

Guenter at 16:25, 31 March 2022

2022-03-31T16:25:55Z

Guenter at 13:33, 24 February 2022

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Guenter at 13:10, 24 February 2022

2022-02-24T13:10:54Z

Guenter at 17:07, 23 February 2022

2022-02-23T17:07:33Z

Guenter: Guenter moved page Aufgaben:Exercise 4.4Z: Contour Lines of the 2D PDF to Aufgaben:Exercise 4.4Z: Contour Lines of the "2D-PDF"

2022-01-24T09:07:28Z

Guenter moved page Exercise 4.4Z: Contour Lines of the 2D PDF to Exercise 4.4Z: Contour Lines of the "2D-PDF"

Noah at 21:37, 23 January 2022

2022-01-23T21:37:15Z

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Guenter at 10:58, 12 January 2022

2022-01-12T10:58:18Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.4Z: Höhenlinien der 2D-WDF to Aufgaben:Exercise 4.4Z: Contour Lines of the 2D PDF

2022-01-12T10:54:33Z

Guenter moved page Aufgabe 4.4Z: Höhenlinien der 2D-WDF to Exercise 4.4Z: Contour Lines of the 2D PDF

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

2021-03-23T12:43:53Z

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

@@ Line 12: / Line 12: @@
 * a contour line of this PDF for&nbsp; $f_{xy}(x,y) =0.2$,
 * the&nbsp; (dark blue)&nbsp; ellipse major axis&nbsp; $\rm (EA)$,&nbsp; and
-* the&nbsp; (red)&nbsp; correlation line&nbsp; $y=K\cdot x$.
+* the&nbsp; (red)&nbsp; "correlation line"&nbsp; or&nbsp; "regression line" $\rm (RL)$.

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID297__Sto_Z_4_4.png|right|frame|Gaussian 2D&ndash;PDF: Contour lines]]
+[[File:EN Sto Z 4 4.png|right|frame|Gaussian 2D&ndash;PDF: Contour lines]]
 Given a two-dimensional Gaussian random variable&nbsp; $(x, y)$&nbsp; with mean&nbsp; $(0, 0)$&nbsp; and the 2D&ndash;PDF
 :$$f_{xy}(x,\ y) = C\cdot{\rm e}^{-(x^{\rm 2} + y^{\rm 2} +\sqrt{\rm 2}\hspace{0.05cm}\cdot \hspace{0.05cm} x \hspace{0.05cm}\cdot \hspace{0.05cm} y)}.$$

@@ Line 61: / Line 61: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Even without specifying&nbsp; $\sigma_x = \sigma_y = 1$&nbsp; one could see that the standard deviations&nbsp; $\sigma_x$&nbsp; and&nbsp; $\sigma_y$&nbsp; are equal,
+'''(1)'''&nbsp; Even without specifying&nbsp; $\sigma_x = \sigma_y = 1$&nbsp; one could see that the standard deviations&nbsp; $\sigma_x$&nbsp; and&nbsp; $\sigma_y$&nbsp; are equal,&nbsp; <br>since in the exponent of $f_{xy}(x, y)$&nbsp; the coefficients at&nbsp; $x^2$&nbsp; and&nbsp; $y^2$&nbsp; are equal.
-*since in the exponent of $f_{xy}(x, y)$&nbsp; the coefficients at&nbsp; $x^2$&nbsp; and&nbsp; $y^2$&nbsp; are equal.
+*By comparing coefficients,&nbsp; we thus obtain:
 :$$\frac{- 2 \rho_{xy}}{\sigma_x\cdot\sigma_y} = \sqrt{2}\hspace{0.3cm}\Rightarrow\hspace{0.3cm}
 \rho_{xy}=\frac{-1}{\sqrt{2}} \hspace{0.15cm}\underline{\approx -0.707}.$$
-'''(2)'''&nbsp; Using the numerical values calculated in point&nbsp; '''(1)'''&nbsp; we also obtain:
+'''(2)'''&nbsp; Using the numerical values calculated in point&nbsp; '''(1)''',&nbsp; we obtain:
 :$$C=\frac{\rm 1}{\rm 2\it\pi\cdot\sigma_x\cdot\sigma_y\cdot\sqrt{\rm 1 - \rho_{xy}^{\rm 2}}}
 =\frac{\rm 1}{\rm 2\pi\cdot\rm 1\cdot 1\cdot\sqrt{0.5}}=\frac{\rm 1}{\sqrt{\rm 2}\cdot \pi}\hspace{0.15cm}\underline{\approx \rm 0.225}.$$
@@ Line 74: / Line 73: @@
 '''(3)'''&nbsp; The general equation is:
-:$$\alpha = {\rm 1}/{\rm 2}\cdot \rm arctan \ (\rm 2 \cdot\it \rho_{xy}\cdot \frac{\sigma_x\cdot\sigma_y}{\sigma_x^{\rm 2} - \sigma_y^{\rm 2}}{\rm ).}.$$
+:$$\alpha = {\rm 1}/{\rm 2}\cdot \rm arctan \ (\rm 2 \cdot\it \rho_{xy}\cdot \frac{\sigma_x\cdot\sigma_y}{\sigma_x^{\rm 2} - \sigma_y^{\rm 2}}{\rm ).}$$
-*Applies&nbsp; $\sigma_x = \sigma_y$&nbsp; and&nbsp; $\rho_{xy} \ne 0$,&nbsp; then the angle is always&nbsp; $\alpha = \pm 45^\circ$, where the sign is equal to the sign of&nbsp; $\rho_{xy}$&nbsp; .
+*Applies&nbsp; $\sigma_x = \sigma_y$&nbsp; and&nbsp; $\rho_{xy} \ne 0$,&nbsp; then the angle is always&nbsp; $\alpha = \pm 45^\circ$,&nbsp; where the sign is equal to the sign of&nbsp; $\rho_{xy}$.
 *In the present case&nbsp; $\alpha\hspace{0.15cm}\underline{ = -45^\circ}$ holds.
@@ Line 83: / Line 82: @@
 '''(4)'''&nbsp; For the plotted contour line holds:
 :$$f_{xy}(x, y)=\frac{1}{\sqrt{2}\cdot \pi}\cdot {\rm e}^{(x^{2} + y^{2} + \sqrt{2}\hspace{0.05cm}\cdot \hspace{0.05cm} x \hspace{0.05cm}\cdot \hspace{0.05cm}y)}=0.2\hspace{0.3cm}
-\rightarrow \hspace{0.3cm}{\rm e}^{-(x^{2} + y^{2} + \sqrt{2}\hspace{0.05cm} \cdot \hspace{0.05cm} x \hspace{0.05cm} \cdot \hspace{0.05cm}y)} = 0.8885
+\Rightarrow \hspace{0.3cm}{\rm e}^{-(x^{2} + y^{2} + \sqrt{2}\hspace{0.05cm} \cdot \hspace{0.05cm} x \hspace{0.05cm} \cdot \hspace{0.05cm}y)} = 0.8885
 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} x^{\rm 2} + y^{\rm 2} + \sqrt{\rm 2}\cdot\hspace{0.05cm} x \hspace{0.05cm} \cdot \hspace{0.05cm}y = -{\rm ln(0.8885)} \approx\rm 0.118.$$
-*The angle of the ellipse major axis is&nbsp; $\alpha = -45^\circ$.&nbsp; Therefore&nbsp; $y_0 = - x_0$&nbsp; must hold. It further follows:
+*The angle of the ellipse major axis is&nbsp; $\alpha = -45^\circ$.&nbsp; Therefore&nbsp; $y_0 = - x_0$&nbsp; must hold.&nbsp; It further follows:
 :$$x_{\rm 0}^{\rm 2} + (-x_{\rm 0})^{\rm 2} + \sqrt{\rm 2}\cdot x_{\rm 0}(-x_{\rm 0}) = 0.118$$
 :$$\Rightarrow \hspace{0.3cm}(\rm 2 - \sqrt{\rm 2})\cdot \it x_{\rm 0}^{\rm 2} = {\rm 0.118}
@@ Line 97: / Line 96: @@
-'''(5)'''&nbsp; Correct are <u>the proposed solutions 2 and 3</u>:
+'''(5)'''&nbsp; Correct are&nbsp; <u>the proposed solutions 2 and 3</u>:
 *With&nbsp; $\sigma_x = \sigma_y$&nbsp; and the result of the subtask&nbsp; '''(1)'''&nbsp; holds for the angle of the correlation line:
 :$$\theta_{y\hspace{0.05cm}\rightarrow \hspace{0.05cm}x} = \arctan (\rho_{\it xy})=\arctan(-{\rm 1}/{\sqrt{\rm 2}})\approx -\rm 35.3^{\circ}.$$
@@ Line 103: / Line 102: @@
-The following is the <u>proof of the correctness of the last statement</u>:
+The following is the&nbsp; <u>proof of the correctness of the last statement</u>:
-*Solving the elliptic equation&nbsp; $($with&nbsp; $z = 0. 118)$,&nbsp; so&nbsp; $x^{\rm 2}+ y^{\rm 2} +\sqrt{\rm 2}\cdot \it x\cdot \it y - \it z = \rm 0$,&nbsp; to&nbsp; $y$&nbsp; we get after solving a quadratic equation:
+*Solving the elliptic equation&nbsp; $($with&nbsp; $z = 0. 118)$,&nbsp; so&nbsp;
 :$$y_{\rm 1, \ 2}={\sqrt{\rm 2}}/ {\rm 2} \cdot x\pm\sqrt{{x^{\rm 2}}/{\rm 2}-x^{\rm 2}+{\it z}}
-\hspace{0.5cm}\rightarrow \hspace{0.5cm} y_{\rm 1, \ 2}={\it x}/{\sqrt{\rm 2}}\pm \sqrt{z-{x^{\rm 2}}/{\rm 2}}.$$
+\hspace{0.5cm}\Rightarrow \hspace{0.5cm} y_{\rm 1, \ 2}={\it x}/{\sqrt{\rm 2}}\pm \sqrt{z-{x^{\rm 2}}/{\rm 2}}.$$
-*The vertical tangent results for the case that the two solutions&nbsp; $y_{\rm 1, \rm 2}$&nbsp; are identical.&nbsp; That is: &nbsp; The root expression must result in zero.
+*The vertical tangent results for the case that the two solutions&nbsp; $y_{\rm 1,\ \rm 2}$&nbsp; are identical.&nbsp; That is: &nbsp; The root expression must result in zero.
 *The solution for positive&nbsp; $x$&nbsp; is then: &nbsp; $x_{\rm T}=\sqrt{\rm 2\cdot \it z}=\rm \rm 0.485.$
-*Inserted into the ellipse equation one obtains f&uuml;r the&nbsp; $y$&ndash;value of the tangent point:
+*Inserted into the ellipse equation one obtains for the&nbsp; $y$&ndash;value of the tangent point:
 :$$x_{\rm T}^{\rm 2} + y_{\rm T}^{\rm 2} + \sqrt{2} \cdot x_{\rm T} \cdot y_{\rm T} - z = 0
 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} 2 z + y_{\rm T}^{\rm 2} + 2\sqrt{ z}\cdot y_{\rm T} - z = 0$$

@@ Line 4: / Line 4: @@
 [[File:P_ID297__Sto_Z_4_4.png|right|frame|Gaussian 2D&ndash;PDF: Contour lines]]
-Given a two-dimensional Gaussian random variable&nbsp; $(x,\ y)$&nbsp; with mean&nbsp; $(0,\ 0)$&nbsp; and the 2D PDF
+Given a two-dimensional Gaussian random variable&nbsp; $(x, y)$&nbsp; with mean&nbsp; $(0, 0)$&nbsp; and the 2D&ndash;PDF
 :$$f_{xy}(x,\ y) = C\cdot{\rm e}^{-(x^{\rm 2} + y^{\rm 2} +\sqrt{\rm 2}\hspace{0.05cm}\cdot \hspace{0.05cm} x \hspace{0.05cm}\cdot \hspace{0.05cm} y)}.$$
-It is further known that the two standard deviations are&nbsp; $\sigma_x=\sigma_y=1.$&nbsp;
+It is further known that the standard deviations are&nbsp; $\sigma_x=\sigma_y=1.$&nbsp;
 Entered in the sketch are:
-* a height line of this PDF for&nbsp; $f_{xy}(x,y) =0.2$,
+* a contour line of this PDF for&nbsp; $f_{xy}(x,y) =0.2$,
-* the (dark blue) ellipse major axis&nbsp; $\rm (EA)$, and
+* the&nbsp; (dark blue)&nbsp; ellipse major axis&nbsp; $\rm (EA)$,&nbsp; and
-* the (red) correlation line&nbsp; $y=K(x)$.
+* the&nbsp; (red)&nbsp; correlation line&nbsp; $y=K\cdot x$.
@@ Line 22: / Line 22: @@
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Two-Dimensional_Gaussian_Random_Variables|Two-dimensional Gaussian Random Variables]].
+*More information on this topic is provided in the&nbsp; (German language)&nbsp;learning video&nbsp; [[Gaußsche_2D-Zufallsgrößen_(Lernvideo)|"Gaußsche 2D-Zufallsgrößen"]]:
 ::Part 1: &nbsp; Gaussian random variables without statistical bindings,
 ::Part 2: &nbsp; Gaussian random variables with statistical bindings.
@@ Line 33: / Line 30: @@
 <quiz display=simple>
-{How large is the correlation coefficient $\rho_{xy}$?
+{What is the correlation coefficient&nbsp; $\rho_{xy}$?
 |type="{}"}
 $\rho_{xy} \ = \ $ { -0.727--0.687 }
-{What is the maximum value&nbsp; $C = f_{xy}(0, 0)$&nbsp; of the PDF?
+{What is the maximum PDF value&nbsp; $C = f_{xy}(0, 0)$?
 |type="{}"}
 $C \ = \ $ { 0.225 3% }
@@ Line 48: / Line 45: @@
-{At what values&nbsp; $x_0$&nbsp; and&nbsp; $y_0$&nbsp; respectively, does the hemline&nbsp; $f_{xy}(x,y) = 0.2$&nbsp; intersect the ellipse major axis?&nbsp; What is the relationship between&nbsp; $x_0$&nbsp; and&nbsp; $y_0$?
+{At what values&nbsp; $x_0$&nbsp; and&nbsp; $y_0$&nbsp; does the contour line&nbsp; $f_{xy}(x,y) = 0.2$&nbsp; intersect the ellipse major axis?&nbsp; What is the relationship between&nbsp; $x_0$&nbsp; and&nbsp; $y_0$?
 |type="{}"}
 $x_0/y_0 \ = \ $ { -1.03--0.97 }
-{Which statements are true regarding the correlation line&nbsp; $K(x)$&nbsp;?
+{Which statements are true regarding the correlation line&nbsp; $y=K\cdot x$&nbsp;?
 |type="[]"}
 - The correlation line is steeper than the ellipse major axis.
-+ The angle of&nbsp; $K(x)$&nbsp; with respect to&nbsp; $x$&ndash;axis is about&nbsp; $-35^\circ$.
++ The angle ofthe correlation line with respect to the&nbsp; $x$&ndash;axis is about&nbsp; $-35^\circ$.
-+ The correlation line intersects all hemlines where a vertical tangent can be applied to the ellipse.
++ The correlation line intersects all contour line where a vertical tangent can be applied to the ellipse.

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID297__Sto_Z_4_4.png|right|frame|Gaussian 2D PDF: &nbsp; contour lines]]
+[[File:P_ID297__Sto_Z_4_4.png|right|frame|Gaussian 2D&ndash;PDF: Contour lines]]
-Given a two-dimensional Gaussian random variable&nbsp; $(x, y)$&nbsp; with mean&nbsp; $(0, 0)$&nbsp; and the 2D PDF.
+Given a two-dimensional Gaussian random variable&nbsp; $(x,\ y)$&nbsp; with mean&nbsp; $(0,\ 0)$&nbsp; and the 2D PDF
-:$$f_{xy}(x, y) = C\cdot{\rm e}^{-(x^{\rm 2} + y^{\rm 2} +\sqrt{\rm 2}\hspace{0.05cm}\cdot \hspace{0.05cm} x \hspace{0.05cm}\cdot \hspace{0.05cm} y)}.$$
+:$$f_{xy}(x,\ y) = C\cdot{\rm e}^{-(x^{\rm 2} + y^{\rm 2} +\sqrt{\rm 2}\hspace{0.05cm}\cdot \hspace{0.05cm} x \hspace{0.05cm}\cdot \hspace{0.05cm} y)}.$$
-It is further known that the two standard deviations&nbsp; $\sigma_x$&nbsp; and&nbsp; $\sigma_y$&nbsp; are respectively equal&nbsp; $1$&nbsp;.
+It is further known that the two standard deviations are&nbsp; $\sigma_x=\sigma_y=1.$&nbsp;
 Entered in the sketch are:

← Older revision	Revision as of 09:07, 24 January 2022
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← Older revision		Revision as of 10:58, 12 January 2022
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−	[[Category:Theory of Stochastic Signals: Exercises\|^4.2 ~~Gaußsche~~ 2D~~-Zufallsgrößen~~^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^4.2 Gaussian 2D Random Variables^]]

← Older revision	Revision as of 10:54, 12 January 2022
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← Older revision		Revision as of 12:43, 23 March 2021
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−	[[Category:~~Aufgaben zu Stochastische Signaltheorie~~\|^4.2 Gaußsche 2D-Zufallsgrößen^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^4.2 Gaußsche 2D-Zufallsgrößen^]]