Aufgaben:Exercise 4.7Z: Generation of a Joint PDF - Revision history

Guenter at 16:45, 25 February 2022

2022-02-25T16:45:53Z

Guenter at 16:39, 25 February 2022

2022-02-25T16:39:26Z

Guenter: Guenter moved page Aufgaben:Exercise 4.7Z: Generation of a joint PDF to Aufgaben:Exercise 4.7Z: Generation of a Joint PDF

2022-02-25T12:48:01Z

Guenter moved page Exercise 4.7Z: Generation of a joint PDF to Exercise 4.7Z: Generation of a Joint PDF

Guenter: Guenter moved page Aufgaben:Exercise 4.7Z: Generation of a two-dimensional PDF to Aufgaben:Exercise 4.7Z: Generation of a joint PDF

2022-02-25T12:40:00Z

Guenter moved page Exercise 4.7Z: Generation of a two-dimensional PDF to Exercise 4.7Z: Generation of a joint PDF

Guenter: Guenter moved page Aufgaben:Exercise 4.7Z: Generation of a "2D-PDF" to Aufgaben:Exercise 4.7Z: Generation of a two-dimensional PDF

2022-02-25T12:38:31Z

Guenter moved page Exercise 4.7Z: Generation of a "2D-PDF" to Exercise 4.7Z: Generation of a two-dimensional PDF

Noah at 16:11, 24 February 2022

2022-02-24T16:11:41Z

Noah at 16:11, 24 February 2022

2022-02-24T16:11:12Z

Guenter at 17:54, 24 January 2022

2022-01-24T17:54:27Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.7Z: Erzeugung einer 2D–WDF to Aufgaben:Exercise 4.7Z: Generation of a "2D-PDF"

2022-01-24T17:50:48Z

Guenter moved page Aufgabe 4.7Z: Erzeugung einer 2D–WDF to Exercise 4.7Z: Generation of a "2D-PDF"

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

2021-03-23T12:44:46Z

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

@@ Line 63: / Line 63: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Given the mean values given, it must hold:
+'''(1)'''&nbsp; Given the mean values,&nbsp; it must hold:
 :$$ C = m_x\hspace{0.15cm}\underline{ = 0},$$
 :$$ F = m_y\hspace{0.15cm}\underline{ = 1}.$$
@@ Line 71: / Line 71: @@
 :$$\sigma_x^2 = \sigma^2 \cdot ( A^2 + B^2)= {2}/{3} \cdot ( A^2 + B^2) .$$
-*Because&nbsp; $\sigma_x^2 = 4$&nbsp; it follows&nbsp; $A^2 + B^2= 6$.
+*Because of&nbsp; $\sigma_x^2 = 4$&nbsp; it follows&nbsp; $A^2 + B^2= 6$.
 *A triangular PDF means that&nbsp; $A = \pm B$&nbsp; must hold.
-*Thus, since negative coefficients have been excluded, we obtain:
+*Thus,&nbsp; since negative coefficients have been excluded,&nbsp; we obtain:
 :$$ A = B = \sqrt{3}\hspace{0.15cm}\underline{ = 1.732}.$$
 [[File:P_ID424__Sto_Z_4_7_d.png|right|frame|Rhombic joint PDF]]
-'''(3)'''&nbsp; With&nbsp; $ A = B = \sqrt{3}$&nbsp; corresponding to the last subtask, two equations of determination remain for&nbsp; $D$&nbsp; and&nbsp; $E$:
+'''(3)'''&nbsp; With&nbsp; $ A = B = \sqrt{3}$&nbsp; corresponding to the last subtask,&nbsp; two equations of determination remain for&nbsp; $D$&nbsp; and&nbsp; $E$:
 :$$\sigma_y^2 = \sigma^2 \cdot ( D^2 + E^2)= 10 \hspace{0.5cm} \Rightarrow \hspace{0.5cm} D^2 + E^2 = \frac {\sigma_y^2}{\sigma^2} = \frac {10}{2/3} \stackrel{!}{=}15,$$
 :$$\rho_{xy} = \frac{A \cdot D + B \cdot E}{\sqrt{(A^2 + B^2)(D^2 + E^2)}} = \frac{\sqrt{3} \cdot (D + E)}{\sqrt{6 \cdot (D^2 + E^2)}}  \stackrel{!}{=} \sqrt{0.9}.$$
 *From this it further follows:&nbsp; $D + E = \sqrt{1.8 \cdot ( D^2 + E^2)} = \sqrt{27} = 3 \cdot \sqrt{3}.$
-*The equation, in conjunction with&nbsp; $D^2 + E^2 = 15$&nbsp; and the constraint&nbsp; $(D>E)$&nbsp; leads to the result:
+*The equation,&nbsp; in conjunction with&nbsp; $D^2 + E^2 = 15$&nbsp; and the constraint&nbsp; $(D>E)$&nbsp; leads to the result:
 :$$ D= 2 \cdot \sqrt{3}\hspace{0.15cm}\underline{ = 3.464}, \hspace{0.5cm}E= \sqrt{3} \hspace{0.15cm}\underline{= 1.732}.$$
-'''(4)'''&nbsp; The random variables&nbsp; $x$&nbsp; and&nbsp; $y$&nbsp; respectively, take their maximum values when respectively&nbsp; $u= +1$ and&nbsp; $v= +1$&nbsp; holds:
+'''(4)'''&nbsp; The random variables&nbsp; $x$&nbsp; and&nbsp; $y$&nbsp; resp. take their maximum values when&nbsp; $u= +1$ and&nbsp; $v= +1$&nbsp; holds:
 :$$ x_\text{max}= A+B \hspace{0.15cm}\underline{ = +3.464}, \hspace{0.5cm} x_\text{min} = - A - B= -3.464.$$
 :$$ y_\text{max}= D+E+F \hspace{0.15cm}\underline{ = +6.196}, \hspace{0.5cm} y_\text{min} = -D-E+F= -4.196.$$

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID423__Sto_Z_4_7.png|right|frame|Requirements for the generation of a <br>2D random variable]]
+[[File:P_ID423__Sto_Z_4_7.png|right|frame|Requirements for the generation of a <br>two-dimensional random variable]]
-Given statistically independent quantities&nbsp; $u$&nbsp; and&nbsp; $v$, both of which are uniformly distributed between&nbsp; $-1$&nbsp; and&nbsp; $+1$&nbsp; and thus each have variance&nbsp; $\sigma^2 = 2/3$&nbsp;, generate a 2D random variable&nbsp; $(x, y)$&nbsp; where for the components:
+Given statistically independent quantities&nbsp; $u$&nbsp; and&nbsp; $v$,
 :$$x = A \cdot u + B \cdot v + C,$$
 :$$y= D \cdot u + E \cdot v + F.$$
-The 2D&ndash;random variable&nbsp; $(x, y)$&nbsp; to be generated should have the following statistical properties:
+The two-dimensional random variable&nbsp; $(x,\hspace{0.08cm} y)$&nbsp; to be generated should have the following statistical properties:
 * Let the variances be&nbsp; $\sigma_x^2 = 4$&nbsp; and&nbsp; $\sigma_y^2 = 10$.
 * Let the random variable&nbsp; $x$&nbsp; be mean-free&nbsp; $(m_x =0)$.
-* For the mean of&nbsp; $y$&nbsp; let&nbsp; $m_y = 1$ hold.
+* For the mean of&nbsp; $y$&nbsp; let&nbsp; $m_y = 1$&nbsp; hold.
 * The correlation coefficient between&nbsp; $x$&nbsp; and&nbsp; $y$&nbsp; is&nbsp; $\rho_{xy} = \sqrt{0.9} = 0.949.$
 * The random variable&nbsp; $x$&nbsp; possess a triangular PDF $f_x(x)$&nbsp; corresponding to the above graph.
@@ Line 19: / Line 24: @@
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Linear_Combinations_of_Random_Variables|Linear Combinations of Random Variables]].
-*In particular, reference is made to the page&nbsp; [Theory_of_Stochastic_Signals/Linear_Combinations_of_Random_Variables#Generation_of_correlated_random_variables|Generation of correlated random variables]].
+*In particular,&nbsp; reference is made to the page&nbsp; [[Theory_of_Stochastic_Signals/Linear_Combinations_of_Random_Variables#Generation_of_correlated_random_variables|Generation of correlated random variables]].
-*To avoid ambiguity, it is specified that all coefficients&nbsp; $A$, ... , $F$&nbsp; should be non-negative.
+*To avoid ambiguity,&nbsp; it is specified that all coefficients&nbsp; $A$, ... , $F$&nbsp; should be non-negative.
@@ Line 44: / Line 46: @@
-{Determine the coefficients&nbsp; $D$&nbsp; and&nbsp; $E$, where&nbsp; $D > E$&nbsp; should hold.
+{Determine the coefficients&nbsp; $D$&nbsp; and&nbsp; $E$,&nbsp; where&nbsp; $D > E$&nbsp; should hold.
 |type="{}"}
 $D \ = \ $ { 3.464 3% }
@@ Line 50: / Line 52: @@
-{Specify the maximum values for&nbsp; $x$&nbsp; and&nbsp; $y$&nbsp;.
+{Specify the maximum values for&nbsp; $x$&nbsp; and&nbsp; $y$.
 |type="{}"}
 $x_\text{max}\ = \ $ { 3.464 3% }

@@ Line 75: / Line 75: @@
-[[File:P_ID424__Sto_Z_4_7_d.png|right|frame|Rhombic two dimensional PDF]]
+[[File:P_ID424__Sto_Z_4_7_d.png|right|frame|Rhombic joint PDF]]
 '''(3)'''&nbsp; With&nbsp; $ A = B = \sqrt{3}$&nbsp; corresponding to the last subtask, two equations of determination remain for&nbsp; $D$&nbsp; and&nbsp; $E$:
 :$$\sigma_y^2 = \sigma^2 \cdot ( D^2 + E^2)= 10 \hspace{0.5cm} \Rightarrow \hspace{0.5cm} D^2 + E^2 = \frac {\sigma_y^2}{\sigma^2} = \frac {10}{2/3} \stackrel{!}{=}15,$$

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Stochastische Signaltheorie/Linearkombinationen von Zufallsgrößen
+{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Linear_Combinations_of_Random_Variables
 }}
-[[File:P_ID423__Sto_Z_4_7.png|right|frame|Vorgaben zur Erzeugung einer <br>2D-Zufallsgröße]]
+[[File:P_ID423__Sto_Z_4_7.png|right|frame|Requirements for the generation of a <br>2D random variable]]
-Ausgehend von statistisch unabhängigen Größen&nbsp; $u$&nbsp; und&nbsp; $v$, die beide zwischen&nbsp; $-1$&nbsp; und&nbsp; $+1$&nbsp; gleichverteilt sind und somit jeweils die Varianz&nbsp; $\sigma^2 = 2/3$&nbsp; besitzen, soll eine 2D-Zufallsgröße&nbsp; $(x, y)$&nbsp; generiert werden, wobei für die Komponenten gilt:
+Given statistically independent quantities&nbsp; $u$&nbsp; and&nbsp; $v$, both of which are uniformly distributed between&nbsp; $-1$&nbsp; and&nbsp; $+1$&nbsp; and thus each have variance&nbsp; $\sigma^2 = 2/3$&nbsp;, generate a 2D random variable&nbsp; $(x, y)$&nbsp; where for the components:
-:$$x = A \cdot u + B \cdot  v + C,$$
+:$$x = A \cdot u + B \cdot v + C,$$
-:$$y= D \cdot u + E \cdot  v + F.$$
+:$$y= D \cdot u + E \cdot v + F.$$
-Die zu erzeugende 2D&ndash;Zufallsgröße&nbsp; $(x, y)$&nbsp; soll die folgenden statistischen Eigenschaften aufweisen:
+The 2D&ndash;random variable&nbsp; $(x, y)$&nbsp; to be generated should have the following statistical properties:
-* Die Varianzen seien&nbsp; $\sigma_x^2 = 4$&nbsp; und&nbsp; $\sigma_y^2 = 10$.
+* Let the variances be&nbsp; $\sigma_x^2 = 4$&nbsp; and&nbsp; $\sigma_y^2 = 10$.
-* Die Zufallsgröße&nbsp; $x$&nbsp; sei mittelwertfrei&nbsp; $(m_x =0)$.
+* Let the random variable&nbsp; $x$&nbsp; be mean-free&nbsp; $(m_x =0)$.
-* Für den Mittelwert von&nbsp; $y$&nbsp; gelte&nbsp; $m_y = 1$.
+* For the mean of&nbsp; $y$&nbsp; let&nbsp; $m_y = 1$ hold.
-* Der Korrelationskoeffizient zwischen&nbsp; $x$&nbsp; und&nbsp; $y$&nbsp; betrage&nbsp; $\rho_{xy} = \sqrt{0.9} = 0.949.$
+* The correlation coefficient between&nbsp; $x$&nbsp; and&nbsp; $y$&nbsp; is&nbsp; $\rho_{xy} = \sqrt{0.9} = 0.949.$
-* Die Zufallsgröße&nbsp; $x$&nbsp; besitze eine dreieckförmige WDF  $f_x(x)$&nbsp; entsprechend der oberen Grafik.
+* The random variable&nbsp; $x$&nbsp; possess a triangular PDF $f_x(x)$&nbsp; corresponding to the above graph.
-* Die Zufallsgröße&nbsp; $y$&nbsp; besitze eine trapezförmige WDF  $f_y(y)$&nbsp; entsprechend der unteren Grafik.
+* The random variable&nbsp; $y$&nbsp; has a trapezoidal PDF $f_y(y)$&nbsp; according to the lower graph.
@@ Line 22: / Line 22: @@
+''Hints:''
-''Hinweise:''
+*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Linear_Combinations_of_Random_Variables|Linear Combinations of Random Variables]].
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Linearkombinationen_von_Zufallsgrößen|Linearkombinationen von Zufallsgrößen]].
+*In particular, reference is made to the page&nbsp; [Theory_of_Stochastic_Signals/Linear_Combinations_of_Random_Variables#Generation_of_correlated_random_variables|Generation of correlated random variables]].
-*Insbesondere wird Bezug genommen auf die Seite&nbsp; [[Theory_of_Stochastic_Signals/Linearkombinationen_von_Zufallsgrößen#Erzeugung_korrelierter_Zufallsgr.C3.B6.C3.9Fen|Erzeugung korrelierter Zufallsgrößen]].
+*To avoid ambiguity, it is specified that all coefficients&nbsp; $A$, ... , $F$&nbsp; should be non-negative.
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Bestimmen Sie die Koeffizienten&nbsp; $C$&nbsp; und&nbsp; $F$.
+{Determine the coefficients&nbsp; $C$&nbsp; and&nbsp; $F$.
 |type="{}"}
 $C \ = \ $ { 0. }
@@ Line 39: / Line 38: @@
-{Bestimmen Sie die Koeffizienten&nbsp; $A$&nbsp; und&nbsp; $B$.
+{Determine the coefficients&nbsp; $A$&nbsp; and&nbsp; $B$.
 |type="{}"}
 $A \ = \ $ { 1.732 3% }
@@ Line 45: / Line 44: @@
-{Bestimmen Sie die Koeffizienten&nbsp; $D$&nbsp; und&nbsp; $E$, wobei&nbsp; $D > E$&nbsp; gelten soll.
+{Determine the coefficients&nbsp; $D$&nbsp; and&nbsp; $E$, where&nbsp; $D > E$&nbsp; should hold.
 |type="{}"}
 $D \ = \ $ { 3.464 3% }
@@ Line 51: / Line 50: @@
-{Geben Sie die Maximalwerte f&uuml;r&nbsp; $x$&nbsp; und&nbsp; $y$&nbsp; an.
+{Specify the maximum values for&nbsp; $x$&nbsp; and&nbsp; $y$&nbsp;.
 |type="{}"}
 $x_\text{max}\ = \ $ { 3.464 3% }
 $y_\text{max}\ = \ $ { 6.196 3% }
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Aufgrund der angegebenen Mittelwerte muss gelten:
+'''(1)'''&nbsp; Given the mean values given, it must hold:
 :$$ C = m_x\hspace{0.15cm}\underline{ = 0},$$
 :$$ F = m_y\hspace{0.15cm}\underline{ = 1}.$$
-'''(2)'''&nbsp; Unter Ber&uuml;cksichtigung von&nbsp; $\sigma^2 = 2/3$&nbsp; gilt:
+'''(2)'''&nbsp; Taking into account&nbsp; $\sigma^2 = 2/3$&nbsp; holds:
-:$$\sigma_x^2 =  \sigma^2 \cdot ( A^2 + B^2)= {2}/{3} \cdot ( A^2 + B^2) .$$
+:$$\sigma_x^2 = \sigma^2 \cdot ( A^2 + B^2)= {2}/{3} \cdot ( A^2 + B^2) .$$
-*Wegen&nbsp; $\sigma_x^2 = 4$&nbsp; folgt&nbsp; $A^2 + B^2= 6$.
+*Because&nbsp; $\sigma_x^2 = 4$&nbsp; it follows&nbsp; $A^2 + B^2= 6$.
-*Eine dreieckf&ouml;rmige WDF bedeutet, dass&nbsp; $A = \pm B$&nbsp; gelten muss.
+*A triangular PDF means that&nbsp; $A = \pm B$&nbsp; must hold.
-*Somit erh&auml;lt man, da  negative Koeffizienten  ausgeschlossen wurden:
+*Thus, since negative coefficients have been excluded, we obtain:
 :$$ A = B = \sqrt{3}\hspace{0.15cm}\underline{ = 1.732}.$$
-[[File:P_ID424__Sto_Z_4_7_d.png|right|frame|Rautenförmige 2D-WDF]]
+[[File:P_ID424__Sto_Z_4_7_d.png|right|frame|Rhombic two dimensional PDF]]
-'''(3)'''&nbsp; Mit&nbsp; $ A = B = \sqrt{3}$&nbsp; entsprechend der letzten Teilaufgabe verbleiben zwei Bestimmungsgleichungen f&uuml;r&nbsp; $D$&nbsp; und&nbsp; $E$:
+'''(3)'''&nbsp; With&nbsp; $ A = B = \sqrt{3}$&nbsp; corresponding to the last subtask, two equations of determination remain for&nbsp; $D$&nbsp; and&nbsp; $E$:
-:$$\sigma_y^2 =  \sigma^2 \cdot ( D^2 + E^2)= 10 \hspace{0.5cm} \Rightarrow \hspace{0.5cm} D^2 + E^2 = \frac {\sigma_y^2}{\sigma^2}  = \frac {10}{2/3} \stackrel{!}{=}15,$$
+:$$\sigma_y^2 = \sigma^2 \cdot ( D^2 + E^2)= 10 \hspace{0.5cm} \Rightarrow \hspace{0.5cm} D^2 + E^2 = \frac {\sigma_y^2}{\sigma^2} = \frac {10}{2/3} \stackrel{!}{=}15,$$
-:$$\rho_{xy} = \frac{A \cdot D + B \cdot E}{\sqrt{(A^2 + B^2)(D^2 + E^2)}} = \frac{\sqrt{3} \cdot (D +  E)}{\sqrt{6 \cdot (D^2 + E^2)}}  \stackrel{!}{=} \sqrt{0.9}.$$
+:$$\rho_{xy} = \frac{A \cdot D + B \cdot E}{\sqrt{(A^2 + B^2)(D^2 + E^2)}} = \frac{\sqrt{3} \cdot (D + E)}{\sqrt{6 \cdot (D^2 + E^2)}}  \stackrel{!}{=} \sqrt{0.9}.$$
-*Daraus folgt weiter:&nbsp; $D + E = \sqrt{1.8 \cdot ( D^2 + E^2)} = \sqrt{27} = 3 \cdot \sqrt{3}.$
+*From this it further follows:&nbsp; $D + E = \sqrt{1.8 \cdot ( D^2 + E^2)} = \sqrt{27} = 3 \cdot \sqrt{3}.$
-*Die Gleichung führt in Verbindung mit&nbsp; $D^2 + E^2 = 15$&nbsp; und der Nebenbedingung&nbsp; $(D>E)$&nbsp;  zum Ergebnis:
+*The equation, in conjunction with&nbsp; $D^2 + E^2 = 15$&nbsp; and the constraint&nbsp; $(D>E)$&nbsp; leads to the result:
 :$$ D= 2 \cdot \sqrt{3}\hspace{0.15cm}\underline{ = 3.464}, \hspace{0.5cm}E= \sqrt{3} \hspace{0.15cm}\underline{= 1.732}.$$
-'''(4)'''&nbsp; Die Zufallsgr&ouml;&szlig;e&nbsp; $x$&nbsp; bzw.&nbsp; $y$&nbsp; nehmen ihre maximalen Werte  an, wenn jeweils&nbsp; $u= +1$ und&nbsp; $v= +1$&nbsp; gilt:
+'''(4)'''&nbsp; The random variables&nbsp; $x$&nbsp; and&nbsp; $y$&nbsp; respectively, take their maximum values when respectively&nbsp; $u= +1$ and&nbsp; $v= +1$&nbsp; holds:
-:$$ x_\text{max}= A+B \hspace{0.15cm}\underline{ = +3.464}, \hspace{0.5cm} x_\text{min} = -  A - B= -3.464.$$
+:$$ x_\text{max}= A+B \hspace{0.15cm}\underline{ = +3.464}, \hspace{0.5cm} x_\text{min} = - A - B= -3.464.$$
 :$$ y_\text{max}= D+E+F \hspace{0.15cm}\underline{ = +6.196}, \hspace{0.5cm} y_\text{min} = -D-E+F= -4.196.$$

← Older revision		Revision as of 17:54, 24 January 2022
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−	[[Category:Theory of Stochastic Signals: Exercises\|^4.3 ~~Linearkombinationen~~^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^4.3 Linear Combinations^]]

← Older revision		Revision as of 12:44, 23 March 2021
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−	[[Category:~~Aufgaben zu Stochastische Signaltheorie~~\|^4.3 Linearkombinationen^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^4.3 Linearkombinationen^]]