Aufgaben:Exercise 4.8: Diamond-shaped Joint PDF - Revision history

Guenter at 13:07, 27 February 2022

2022-02-27T13:07:28Z

Guenter at 12:45, 27 February 2022

2022-02-27T12:45:33Z

Noah at 19:16, 25 February 2022

2022-02-25T19:16:08Z

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Guenter at 17:09, 25 February 2022

2022-02-25T17:09:28Z

Guenter: Guenter moved page Aufgaben:Exercise 4.8: Diamond shaped Joint PDF to Aufgaben:Exercise 4.8: Diamond-shaped Joint PDF

2022-02-25T12:49:58Z

Guenter moved page Exercise 4.8: Diamond shaped Joint PDF to Exercise 4.8: Diamond-shaped Joint PDF

Guenter: Guenter moved page Aufgaben:Exercise 4.8: Diamond shaped joint PDF to Aufgaben:Exercise 4.8: Diamond shaped Joint PDF

2022-02-25T12:48:31Z

Guenter moved page Exercise 4.8: Diamond shaped joint PDF to Exercise 4.8: Diamond shaped Joint PDF

Guenter: Guenter moved page Aufgaben:Exercise 4.8: Diamond shaped "2D-PDF" to Aufgaben:Exercise 4.8: Diamond shaped joint PDF

2022-02-25T12:40:37Z

Guenter moved page Exercise 4.8: Diamond shaped "2D-PDF" to Exercise 4.8: Diamond shaped joint PDF

Noah at 16:19, 24 February 2022

2022-02-24T16:19:42Z

Guenter at 17:55, 24 January 2022

2022-01-24T17:55:11Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.8: Rautenförmige 2D-WDF to Aufgaben:Exercise 4.8: Diamond shaped "2D-PDF"

2022-01-24T17:52:27Z

Guenter moved page Aufgabe 4.8: Rautenförmige 2D-WDF to Exercise 4.8: Diamond shaped "2D-PDF"

@@ Line 68: / Line 68: @@
 *The area of the triangle $(1,0)\ (1,4)\ (-1,3)$&nbsp; gives&nbsp; $0.5 &middot; 4 &middot; 2 = 4$.
 *The total area is double: &nbsp; $F = 8$.
-*Since the PDF&ndash;volume is always&nbsp; $1$&nbsp;, then&nbsp; $H= 1/F\hspace{0.15cm}\underline{ = 0.125}$.
+*Since the PDF volume is always&nbsp; $1$&nbsp;, then&nbsp; $H= 1/F\hspace{0.15cm}\underline{ = 0.125}$.
 '''(2)'''&nbsp; The minimum value of&nbsp; $x$&nbsp; is obtained for&nbsp; $\underline{ u=0}$&nbsp; and&nbsp; $\underline{ v=1}$.
-*From the above equations, the results&nbsp; $x= -1$&nbsp; and&nbsp; $y= +3$ follow.
+*From the above equations,&nbsp; the results&nbsp; $x= -1$&nbsp; and&nbsp; $y= +3$&nbsp; follow.
 '''(3)'''&nbsp; The equation given in the theory section is valid in general, i.e., for any PDF of the two statistically independent variables&nbsp; $u$&nbsp; and&nbsp; $v$,
-*as long as they have equal standard deviations&nbsp; $(\sigma_u = \sigma_v)$.
 *With&nbsp; $A = 2$,&nbsp; $B = -2$,&nbsp; $D = 1$&nbsp; and&nbsp; $E = 3$&nbsp; we obtain:
 :$$\rho_{xy } =  \frac {\it A \cdot D + B \cdot E}{\sqrt{(\it A^{\rm 2}+\it B^{\rm 2})(\it D^{\rm 2}+\it E^{\rm 2})}} =\frac {2 \cdot 1 -2 \cdot 3}{\sqrt{(4 +4)(1+9)}} = \frac {-4}{\sqrt{80}} = \frac {-1}{\sqrt{5}}\hspace{0.15cm}\underline{ = -0.447}. $$
 '''(4)'''&nbsp; The regression line is generally:
-:$$y=K(x)=\frac{\sigma_y}{\sigma_x}\cdot\rho_{xy}\cdot(x-m_x)+m_y.$$
+:$$y=\frac{\sigma_y}{\sigma_x}\cdot\rho_{xy}\cdot(x-m_x)+m_y.$$
-*From the linear means&nbsp; $m_u = m_v = 0.5$&nbsp; and the equations given in the problem statement, we obtain&nbsp; $m_x = 1$&nbsp; and&nbsp; $m_y = 2$.
+*From the linear means&nbsp; $m_u = m_v = 0.5$&nbsp; and the equations given in the problem statement,&nbsp; we obtain&nbsp; $m_x = 1$&nbsp; and&nbsp; $m_y = 2$.
 *The variances of&nbsp; $u$&nbsp; and&nbsp; $v$&nbsp; are respectively&nbsp; $\sigma_u^2 = \sigma_v^2 =1/12$.&nbsp; It follows:
@@ Line 88: / Line 92: @@
 :$$\sigma_y^2 = \sigma_u^2 + 9\cdot \sigma_v^2 = 5/6.$$
-*Substituting these values into the equation of the regression line, we get:
+*Substituting these values into the equation of the regression line,&nbsp; we get:
-:$$y=K(x)=\frac{\sqrt{5/6}}{\sqrt{2/3}}\cdot (\frac{-1}{\sqrt{5}})\cdot(x-1)+2= - x/{2} + 2.5.$$
+:$$y=\frac{\sqrt{5/6}}{\sqrt{2/3}}\cdot (\frac{-1}{\sqrt{5}})\cdot(x-1)+2= - x/{2} + 2.5.$$
-*From this follows the value&nbsp; $y_0=K(x=0)\hspace{0.15cm}\underline{ = 2.5}$
+*From this follows with&nbsp; $x=0$&nbsp; the value&nbsp; $y_0=\hspace{0.15cm}\underline{ = 2.5}$
-'''(5)'''&nbsp; With the auxiliary quantities &nbsp; $q= 2u$, &nbsp; $r= -2v$ &nbsp; and &nbsp; $s= x-1$&nbsp; the relation holds: &nbsp; $s= q+r$.
+'''(5)'''&nbsp; With the auxiliary quantities &nbsp; $q= 2u$, &nbsp; $r= -2v$ &nbsp; and &nbsp; $s= x-1$:&nbsp;  $s= q+r$.
 [[File:P_ID414__Sto_A_4_8_e.png|right|frame|Triangular PDF $f_x(x)$]]
 :$$f_s(s) = f_q(q) \star f_r(r).$$
 *The addition&nbsp; $x = s+1$&nbsp; leads to a shift of the triangular&ndash;PDF by&nbsp; $1$&nbsp; to the right.
-*For the sought probability&nbsp; (highlighted in green in the following image)&nbsp; therefore holds: &nbsp; ${\rm Pr}(x < 0)\hspace{0.15cm}\underline{ = 0.125}$.
+*For the sought probability&nbsp; (highlighted in green in the graphic)&nbsp; therefore holds: &nbsp;
-<br clear=all>
+:$${\rm Pr}(x < 0)\hspace{0.15cm}\underline{ = 0.125}.$$
 [[File: P_ID415__Sto_A_4_8_f.png|right|frame|Trapezoidal PDF $f_y(y)$]]
-'''(6)'''&nbsp; Analogous to the sample solution for the subtask&nbsp; '''(5)'''&nbsp; holds with&nbsp; $t = 3v$:
+'''(6)'''&nbsp; Analogous to the solution for the subtask&nbsp; '''(5)'''&nbsp; holds with&nbsp; $t = 3v$:
 :$$f_y(y) = f_u(u) \star f_t(t).$$
-*The convolution between two rectangles of different widths results in a trapezoid.
+*The convolution between two rectangles of different widths results in a trapezoid.&nbsp; For the probability we are looking for,&nbsp; we get:
-*For the probability we are looking for, we get&nbsp; ${\rm Pr}(y>3) =1/6\hspace{0.15cm}\underline{ \approx 0.167}$.
+:$${\rm Pr}(y>3) =1/6\hspace{0.15cm}\underline{ \approx 0.167}.$$
 *This probability is highlighted in green in the right sketch.

@@ Line 30: / Line 30: @@
 <quiz display=simple>
-{What is the height $H$&nbsp; of the joint PDF within the parallelogram?
+{What is the height&nbsp; $H$&nbsp; of the joint PDF within the parallelogram?
 |type="{}"}
 $H \ = \ $ { 0.125 3% }
-{What values of&nbsp; $u$&nbsp; and&nbsp; $v$&nbsp; underlie the vertex&nbsp; $(-1, 3)$&nbsp;?
+{What values of&nbsp; $u$&nbsp; and&nbsp; $v$&nbsp; underlie the corner point&nbsp; $(-1, 3)$?
 |type="{}"}
 $u \ = \ $ { 0. }
@@ Line 46: / Line 46: @@
-{What is the regression line&nbsp; $y=K(x)$?&nbsp; At what point&nbsp; $y_0$&nbsp; does it intersect the&nbsp; $y$ axis?
+{What is the regression line&nbsp; $\rm (RL)$?&nbsp; At what point&nbsp; $y_0$&nbsp; does it intersect the&nbsp; $y$&ndash;axis?
 |type="{}"}
 $y_0\ = \ $ { 2.5 3% }
@@ Line 56: / Line 56: @@
-{Calculate the marginal probability density function $f_y(y)$.&nbsp; What is the probability that the random variable&ouml;&aerospace;e&nbsp; $y >3$&nbsp;?
+{Calculate the marginal probability density function $f_y(y)$.&nbsp; What is the probability that the random variablee&nbsp; $y >3$?
 |type="{}"}
 ${\rm Pr}(y > 3)\ = \ $ { 0.167 3% }

@@ Line 4: / Line 4: @@
 [[File:P_ID412__Sto_A_4_8.png|right|frame|Diamond shaped joint PDF]]
-We consider a 2D&ndash;random variable&nbsp; $(x, y)$ whose components arise respectively as linear combinations of two random variables&nbsp; $u$&nbsp; and&nbsp; $v$&nbsp; :
+We consider a two-dimensional random variable&nbsp; $(x,\hspace{0.08cm} y)$&nbsp; whose components arise as linear combinations of two random variables&nbsp; $u$&nbsp; and&nbsp; $v$:
 :$$x=2u-2v+1,$$
 :$$y=u+3v.$$
-Further, note:
+Further,&nbsp; note:
-*The two statistically independent random variables&nbsp; $u$&nbsp; and&nbsp; $v$&nbsp; are each equally distributed between&nbsp; $0$&nbsp; and&nbsp; $1$.
+*The two statistically independent random variables&nbsp; $u$&nbsp; and&nbsp; $v$&nbsp; are each uniformly distributed between&nbsp; $0$&nbsp; and&nbsp; $1$.
-*In the figure you can see the 2D&ndash;PDF.&nbsp; Within the parallelogram drawn in blue holds:
+*In the figure you can see the joint PDF.&nbsp; Within the parallelogram drawn in blue holds:
-:$$f_{xy}(x, y) = H = {\rm const.}$$
+:$$f_{xy}(x,\hspace{0.08cm} y) = H = {\rm const.}$$
-*Outside the parallelogram no values are possible: $f_{xy}(x, y) = 0$.
+*Outside the parallelogram no values are possible:&nbsp; $f_{xy}(x,\hspace{0.08cm} y) = 0$.
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Linear_Combinations_of_Random_Variables|Linear_Combinations of Random Variables]].
-*Reference is also made to the page&nbsp; [[/Theory_of_Stochastic_Signals/Two-Dimensional_Random_Variables#Regression_line|regression line]].
+*Reference is also made to the page&nbsp; [[Theory_of_Stochastic_Signals/Two-Dimensional_Random_Variables#Regression_line|Regression line]].
-*We also refer here to the interactive applet&nbsp; [Applets:Korrelationskoeffizient_%26_Regressionsgerade|Correlation coefficient and regression line]]
+*We also refer here to the interactive applet&nbsp; [[Applets:Korrelationskoeffizient_%26_Regressionsgerade|Correlation coefficient and regression line]]
-*Assume - if possible - the given equations&nbsp; Use the information of the above sketch mainly only to check your results.
+*Assume - if possible - the given equations.&nbsp; Use the information of the above sketch mainly only to check your results.

@@ 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_ID412__Sto_A_4_8.png|right|frame|Rautenförmige 2D-WDF]]
+[[File:P_ID412__Sto_A_4_8.png|right|frame|Diamond shaped joint PDF]]
-Wir betrachten eine 2D&ndash;Zufallsgr&ouml;&szlig;e&nbsp; $(x, y)$, deren Komponenten sich jeweils als Linearkombinationen zweier Zufallsgr&ouml;&szlig;en&nbsp; $u$&nbsp; und&nbsp; $v$&nbsp; ergeben:
+We consider a 2D&ndash;random variable&nbsp; $(x, y)$ whose components arise respectively as linear combinations of two random variables&nbsp; $u$&nbsp; and&nbsp; $v$&nbsp; :
 :$$x=2u-2v+1,$$
 :$$y=u+3v.$$
-Weiter ist zu beachten:
+Further, note:
-*Die zwei statistisch unabh&auml;ngigen Zufallsgr&ouml;&szlig;en&nbsp; $u$&nbsp; und&nbsp; $v$&nbsp; sind jeweils gleichverteilt zwischen&nbsp; $0$&nbsp; und&nbsp; $1$.
+*The two statistically independent random variables&nbsp; $u$&nbsp; and&nbsp; $v$&nbsp; are each equally distributed between&nbsp; $0$&nbsp; and&nbsp; $1$.
-*In der Abbildung sehen Sie die 2D&ndash;WDF.&nbsp; Innerhalb des blau eingezeichneten Parallelogramms gilt:
+*In the figure you can see the 2D&ndash;PDF.&nbsp; Within the parallelogram drawn in blue holds:
 :$$f_{xy}(x, y) = H = {\rm const.}$$
-*Au&szlig;erhalb des Parallelogramms sind keine Werte m&ouml;glich: &nbsp; $f_{xy}(x, y) = 0$.
+*Outside the parallelogram no values are possible: $f_{xy}(x, y) = 0$.
@@ Line 20: / Line 20: @@
-''Hinweise:''
+''Hints:''
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Linearkombinationen_von_Zufallsgrößen|Linearkombinationen von Zufallsgrößen]].
+*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Linear_Combinations_of_Random_Variables|Linear_Combinations of Random Variables]].
-*Bezug genommen wird auch auf die Seite&nbsp; [[Theory_of_Stochastic_Signals/Zweidimensionale_Zufallsgrößen#Korrelationsgerade|Korrelationsgerade]].
+*Reference is also made to the page&nbsp; [[/Theory_of_Stochastic_Signals/Two-Dimensional_Random_Variables#Regression_line|regression line]].
-*Wir verweisen hier auch auf das interaktive Applet&nbsp; [[Applets:Korrelationskoeffizient_%26_Regressionsgerade|Korrelationskoeffizient und Regressionsgerade]].
+*We also refer here to the interactive applet&nbsp; [Applets:Korrelationskoeffizient_%26_Regressionsgerade|Correlation coefficient and regression line]]
-*Gehen Sie - wenn m&ouml;glich - von den angegebenen Gleichungen aus.&nbsp; Nutzen Sie die Informationen der obigen Skizze vorwiegend nur zur Kontrolle Ihrer Ergebnisse.
+*Assume - if possible - the given equations&nbsp; Use the information of the above sketch mainly only to check your results.
@@ Line 30: / Line 30: @@
-===Fragebogen===
+===Questions===
 <quiz display=simple>

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