Aufgaben:Exercise 4.1: Triangular (x, y) Area - Revision history

Guenter at 11:05, 7 February 2022

2022-02-07T11:05:37Z

Guenter at 09:44, 7 February 2022

2022-02-07T09:44:42Z

Noah at 14:31, 17 January 2022

2022-01-17T14:31:40Z

Noah at 14:31, 17 January 2022

2022-01-17T14:31:21Z

Noah at 14:29, 17 January 2022

2022-01-17T14:29:15Z

Show changes

Noah at 20:01, 15 January 2022

2022-01-15T20:01:18Z

Noah at 19:28, 13 January 2022

2022-01-13T19:28:06Z

Guenter at 10:50, 12 January 2022

2022-01-12T10:50:06Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.1: Dreieckiges (x, y)-Gebiet to Aufgaben:Exercise 4.1: Triangular (x, y) Area

2022-01-12T10:43:34Z

Guenter moved page Aufgabe 4.1: Dreieckiges (x, y)-Gebiet to Exercise 4.1: Triangular (x, y) Area

Javier: Text replacement - "”" to """

2021-05-28T14:39:26Z

Text replacement - "”" to """

@@ Line 64: / Line 64: @@
 {{ML-Kopf}}
 '''(1)'''&nbsp; The volume under the two dimensional PDF is by definition equal to&nbsp; $1$:
-[[File:P_ID219__Sto_A_4_1_b.png|right|frame|Triangular 2D PDF]]
+[[File:P_ID219__Sto_A_4_1_b.png|right|frame|Triangular 2D&ndash;PDF]]
 :$$\int_{-\infty}^{+\infty}\int_{-\infty}^{+\infty}f_{xy}(x,y)\, {\rm d}x\, {\rm d}y=1.$$
 *The triangle area is&nbsp; $D = 0.5 \cdot 2 \cdot 4 = 4$.
-*Since in this definition area the PDF is constantly equal to&nbsp; $A$&nbsp;, we get.
+*Since in this definition area the PDF is constantly equal to&nbsp; $A$,&nbsp; we get.
 :$$A= 1/D\hspace{0.15cm}\underline{= 0.25}.$$
@@ Line 74: / Line 74: @@
 '''(2)'''&nbsp; For the solution we start from the adjacent sketch.
 *The area&nbsp; $x>y$&nbsp; lies to the right of the angle bisector&nbsp; $x=y$&nbsp; and is marked in green.
 *This green triangular area is&nbsp; $D_{\rm (2)} = 0.5 \cdot 1 \cdot 2 = 1 $, i.e. exactly one quarter of the total area&nbsp; $D$&nbsp; of the definition area.
 *From this follows&nbsp; ${\rm Pr}(x > y)\hspace{0.15cm}\underline{= 0.25}$.
@@ Line 81: / Line 80: @@
 '''(3)'''&nbsp; For the wanted marginal PDF holds in this case:
-[[File:P_ID220__Sto_A_4_1_c.png|right|frame|marginal PDF with respect to&nbsp; $x$]]
+[[File:P_ID220__Sto_A_4_1_c.png|right|frame|Marginal PDF with respect to&nbsp; $x$]]
 :$$f_x(x)=\int_{-\infty}^{+\infty}f_{xy}(x,y)\, {\rm d}y=A\cdot B_y (x).$$
@@ Line 90: / Line 89: @@
 :$$\rm Pr(\it x\ge \rm 2) = \rm 1-\rm Pr(\it x < \rm 2) = \rm 1-\frac{1}{2}\cdot2\cdot 0.25\hspace{0.15cm}\underline{ =0.75}. $$
-*The same result is obtained using 2D PDF:&nbsp; To the right of&nbsp; $x = 2$&nbsp; lies&nbsp; $3/4$&nbsp; of the total definition area.
+*The same result is obtained using 2D&ndash;PDF:&nbsp; To the right of&nbsp; $x = 2$&nbsp; lies&nbsp; $3/4$&nbsp; of the total definition area.
 <br clear=all>
-[[File:P_ID221__Sto_A_4_1_d.png|right|frame|marginal PDF with respect to&nbsp; $y$]]
+[[File:P_ID221__Sto_A_4_1_d.png|right|frame|Marginal PDF with respect to&nbsp; $y$]]
-'''(4)'''&nbsp; Analogous to the sample solution to the subtask&nbsp; '''(3)'''&nbsp; holds:
+'''(4)'''&nbsp; Analogous to the solution of the subtask&nbsp; '''(3)'''&nbsp; holds:
 :$$f_y(y)=\int_{-\infty}^{+\infty}f_{xy}(x,y)\, {\rm d}x=A\cdot B_x (y).$$
-*The spread of PDF area in&nbsp; $x$&ndash;direction is zero for&nbsp; $y \le 1$&nbsp; and&nbsp; $y \ge 5$&nbsp; respectively.
+*The width of PDF area in&nbsp; $x$&ndash;direction is zero for&nbsp; $y \le 1$&nbsp; and&nbsp; $y \ge 5$&nbsp; respectively.
 *The maximum is at&nbsp; $y=3$&nbsp; and gives&nbsp; $B_x(y=3) = 2$.
-*In between, the increase&nbsp; and decrease of&nbsp; $B_x(y)$&nbsp; is linear, yielding a triangular-shaped PDF.
+*In between,&nbsp; the increase&nbsp; and decrease of&nbsp; $B_x(y)$&nbsp; is linear,&nbsp; yielding a triangular-shaped PDF.
-*The probability that&nbsp; $y \is 3$&nbsp; corresponds to the green shaded area in the adjacent sketch.
+*The probability that&nbsp; $y=3$&nbsp; corresponds to the green shaded area in the adjacent sketch.
-* Because of the symmetry, one obtains:
+* Because of the symmetry,&nbsp; one obtains:
 :$${\rm Pr}(y ≥ 3)\hspace{0.15cm}\underline{ =0.5}. $$
-The same result is obtained using 2D PDF: &nbsp; Above the horizontal&nbsp; $y= 3$&nbsp; lies half of the total definition area.
+The same result is obtained using 2D&ndash;PDF: &nbsp; Above the horizontal&nbsp; $y= 3$&nbsp; lies half of the total definition area.
 [[File:P_ID222__Sto_A_4_1_e.png|right|frame|On subtask '''(5)''']]
-'''(5)'''&nbsp; If&nbsp; $y \ge 3$&nbsp; is&nbsp; $($red highlighted triangle&nbsp; $D)$,&nbsp; is always also true&nbsp; $x \ge 2$&nbsp; $($green outlined trapezoid&nbsp; $T)$.
+'''(5)'''&nbsp; If&nbsp; $y \ge 3$&nbsp; $($red highlighted triangle&nbsp; $D)$,&nbsp; it is always true&nbsp; $x \ge 2$&nbsp; $($green outlined trapezoid&nbsp; $T)$.
-*This means:&nbsp; In this example&nbsp; $D$&nbsp; is a subset of&nbsp; $T$, and it holds:
+*This means:&nbsp; In this example&nbsp; $D$&nbsp; is a subset of&nbsp; $T$,&nbsp; and it holds:
 :$${\rm Pr}[(x ≥ 2) ∩ (y ≥ 3)] = {\rm Pr}(y ≥ 3) \hspace{0.15cm}\underline{= 0.50}.$$
-'''(6)'''&nbsp; According to the solution to the subtask&nbsp; '''(5)'''&nbsp; it follows from&nbsp; $y \ge 3$&nbsp; with certainty also&nbsp; $x \ge 2$.
+'''(6)'''&nbsp; According to the solution to the subtask&nbsp; '''(5)''',&nbsp; it follows from&nbsp; $y \ge 3$&nbsp; with certainty also&nbsp; $x \ge 2$.
 *So the conditional probability we are looking for is:
 :$${\rm Pr}[x ≥ 2\hspace{0.05cm} | \hspace{0.05cm} y ≥ 3]\hspace{0.15cm}\underline{= 1}.$$
@@ Line 120: / Line 119: @@
-'''(7)'''&nbsp; This subtask can be solved using Bayes' theorem and the results from&nbsp; '''(2)'''&nbsp; and&nbsp; '''(5)'''&nbsp; :
+'''(7)'''&nbsp; This subtask can be solved using Bayes' theorem and the results from&nbsp; '''(2)'''&nbsp; and&nbsp; '''(5)''':
 :$$\rm Pr(\it y \ge \rm 3\hspace{0.1cm}|\hspace{0.1cm} \it x \ge \rm 2) = \frac{ \rm Pr((\it x \ge \rm 2)\cap(\it y \ge \rm 3))} {\rm Pr(\it x \ge \rm 2)}=2/3\hspace{0.15cm}\underline{=0.667}.$$
-* Or expressed differently: &nbsp; The area&nbsp; $D$&nbsp; of the triangle with red background makes&nbsp; $2/3$&nbsp; of the area of the trapezoid with green border&nbsp; $T$&nbsp;.
+* Or expressed differently: &nbsp; The area&nbsp; $D$&nbsp; of the triangle with red background makes&nbsp; $2/3$&nbsp; of the area of the trapezoid with green border&nbsp; $T$.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID172__Sto_A_4_1.png|right|frame|Triangular 2D area]]
+[[File:P_ID172__Sto_A_4_1.png|right|frame|Triangular two-dimensional area]]
-A 2D random variable is defined by the adjacent sketch:
+A two-dimensional random variable is defined by the adjacent sketch:
-*F&nbsp; $(x, \ y)$&nbsp; only values within the triangular-shaped region defined by the three vertices&nbsp; $(0,\ 1)$, $(4,\ 3)$, and $(4,\ 5)$&nbsp; can occur.
+*For&nbsp; $(x, \ y)$&nbsp; only values within the triangular-shaped region defined by the three vertices&nbsp; $(0,\ 1)$,$&nbsp; $(4,\ 3)$,$&nbsp; and&nbsp; $(4,\ 5)$&nbsp; can occur.
-*Within the triangle, all the random variables&nbsp; $(x,\ y)$&nbsp; are equally probable.
+*Within the triangle,&nbsp; all the random variables&nbsp; $(x,\ y)$&nbsp; are equally probable.
-*For the 2D&ndash;WDF, in this domain:
+*For the 2D&ndash;PDF,&nbsp; in this domain:
 :$$f_{xy}(x,y) = A .$$
-In addition, the straight line&nbsp; $x = y$ &nbsp; &#8658; &nbsp; "angle bisector" &nbsp; is drawn in the sketch above &nbsp; &#8658; &nbsp;see subtask&nbsp; '''(2)'''.
+In addition,&nbsp; the straight line&nbsp; $x = y$&nbsp; &#8658; &nbsp; "angle bisector"&nbsp; is drawn in the graphic&nbsp; &#8658; &nbsp;see subtask&nbsp; '''(2)'''.
@@ Line 17: / Line 17: @@
+Hint:
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Two-Dimensional_Random_Variables|Two-Dimensional Random Variables]].
@@ Line 32: / Line 30: @@
-{Calculate the probability that&nbsp; $x$&nbsp; is greater than&nbsp; $y$&nbsp; .
+{Calculate the probability that&nbsp; $x$&nbsp; is greater than&nbsp; $y$.
 |type="{}"}
 ${\rm Pr}(x > y) \ = \ $ { 0.25 3% }
-{Determine the marginal PDF&nbsp; $f_x(x)$.&nbsp; Calculate the probability that&nbsp; $x$&nbsp; is greater than or equal to&nbsp; $2$&nbsp; . <br>Check the value against the 2D PDF.
+{Determine the marginal PDF&nbsp; $f_x(x)$.&nbsp; Calculate the probability that&nbsp; $x$&nbsp; is greater than or equal to&nbsp; $2$. <br>Check the value with the 2D&ndash;PDF.
 |type="{}"}
 ${\rm Pr}(x ≥ 2)\ = \ $ { 0.75 3% }
-{Determine the marginal PDF&nbsp; $f_y(y)$.&nbsp; Calculate the probability that&nbsp; $y$&nbsp; is greater than or equal to&nbsp; $3$&nbsp; . <br>Check the value against the 2D PDF.
+{Determine the marginal PDF&nbsp; $f_y(y)$.&nbsp; Calculate the probability that&nbsp; $y$&nbsp; is greater than or equal to&nbsp; $3$. <br>Check the value with the 2D&ndash;PDF.
 |type="{}"}
 ${\rm Pr}(y ≥ 3)\ = \ $ { 0.5 3% }
-{What is the probability that the random variable is $x$&nbsp; greater than or equal to $2$&nbsp; and at the same time the random size is $y$&nbsp; greater than or equal to $3$&nbsp?
+{What is the probability that the random variable&nbsp; $x$&nbsp; is greater than or equal to&nbsp; $2$&nbsp; and at the same time the random variable&nbsp; $y$&nbsp; is greater than or equal to $3$?
 |type="{}"}
 ${\rm Pr}\big[(x ≥ 2) ∩ (y ≥ 3)\big]\ = \ $ { 0.5 3% }

@@ Line 94: / Line 94: @@
 *The same result is obtained using 2D PDF:&nbsp; To the right of&nbsp; $x = 2$&nbsp; lies&nbsp; $3/4$&nbsp; of the total definition area.
 <br clear=all>
-[[File:P_ID221__Sto_A_4_1_d.png|right|frame|marginal PDF with respect&nbsp; $y$]]
+[[File:P_ID221__Sto_A_4_1_d.png|right|frame|marginal PDF with respect to&nbsp; $y$]]
 '''(4)'''&nbsp; Analogous to the sample solution to the subtask&nbsp; '''(3)'''&nbsp; holds:
 :$$f_y(y)=\int_{-\infty}^{+\infty}f_{xy}(x,y)\, {\rm d}x=A\cdot B_x (y).$$

@@ Line 94: / Line 94: @@
 *The same result is obtained using 2D PDF:&nbsp; To the right of&nbsp; $x = 2$&nbsp; lies&nbsp; $3/4$&nbsp; of the total definition area.
 <br clear=all>
-[[File:P_ID221__Sto_A_4_1_e.png|right|frame|marginal PDF with respect&nbsp; $y$]]
+[[File:P_ID221__Sto_A_4_1_d.png|right|frame|marginal PDF with respect&nbsp; $y$]]
 '''(4)'''&nbsp; Analogous to the sample solution to the subtask&nbsp; '''(3)'''&nbsp; holds:
 :$$f_y(y)=\int_{-\infty}^{+\infty}f_{xy}(x,y)\, {\rm d}x=A\cdot B_x (y).$$

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID172__Sto_A_4_1.png|right|frame|Dreieckigförmiges 2D-Gebiet]]
+[[File:P_ID172__Sto_A_4_1.png|right|frame|Triangular 2D area]]
-Eine 2D-Zufallsgr&ouml;&szlig;e ist durch die nebenstehende Skizze definiert:
+A 2D random variable is defined by the adjacent sketch:
-*F&uuml;r&nbsp; $(x, \ y)$&nbsp; k&ouml;nnen nur Werte innerhalb des durch die drei Eckpunkte&nbsp; $(0,\ 1)$, $(4,\ 3)$ und $(4,\ 5)$&nbsp;  festgelegten dreieckf&ouml;rmigen Gebietes auftreten.
+*F&nbsp; $(x, \ y)$&nbsp; only values within the triangular-shaped region defined by the three vertices&nbsp; $(0,\ 1)$, $(4,\ 3)$, and $(4,\ 5)$&nbsp; can occur.
-*Innerhalb des Dreiecks sind alle Zufallsgr&ouml;&szlig;en&nbsp; $(x, \ y)$&nbsp; gleichwahrscheinlich.
+*Within the triangle, all the random variables&nbsp; $(x,\ y)$&nbsp; are equally probable.
-*Für die 2D&ndash;WDF gilt in diesem Bereich:
+*For the 2D&ndash;WDF, in this domain:
 :$$f_{xy}(x,y) = A .$$
-Zusätzlich ist die Gerade&nbsp; $x = y$ &nbsp; &#8658; &nbsp; "Winkelhalbierende" &nbsp; in obiger Skizze eingezeichnet &nbsp; &#8658; &nbsp;siehe Teilaufgabe&nbsp; '''(2)'''.
+In addition, the straight line&nbsp; $x = y$ &nbsp; &#8658; &nbsp; "angle bisector" &nbsp; is drawn in the sketch above &nbsp; &#8658; &nbsp;see subtask&nbsp; '''(2)'''.
@@ Line 19: / Line 19: @@
-''Hinweis:''
+Hints:
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Zweidimensionale_Zufallsgrößen|Zweidimensionale Zufallsgrößen]].
+*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Two-Dimensional_Random_Variables|Two-Dimensional Random Variables]].
-===Fragebogen===
+===Questions===
 <quiz display=simple>

← Older revision		Revision as of 19:28, 13 January 2022
Line 1:		Line 1:

−	{{quiz-Header\|Buchseite=~~Stochastische Signaltheorie~~/~~Zweidimensionale Zufallsgrößen~~	+	{{quiz-Header\|Buchseite=Theory_of_Stochastic_Signals/Two-Dimensional_Random_Variables
	}}		}}

← Older revision		Revision as of 10:50, 12 January 2022
Line 132:		Line 132:


−	[[Category:Theory of Stochastic Signals: Exercises\|^4.1 ~~Zweidimensionale Zufallsgrößen~~^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^4.1 Two-Dimensional Random Variables^]]

← Older revision		Revision as of 14:39, 28 May 2021
Line 10:		Line 10:
	:$$f_{xy}(x,y) = A .$$		:$$f_{xy}(x,y) = A .$$

−	Zusätzlich ist die Gerade  $x = y$   ⇒   "Winkelhalbierende~~”~~   in obiger Skizze eingezeichnet   ⇒  siehe Teilaufgabe  '''(2)'''.	+	Zusätzlich ist die Gerade  $x = y$   ⇒   "Winkelhalbierende"   in obiger Skizze eingezeichnet   ⇒  siehe Teilaufgabe  '''(2)'''.