Aufgaben:Exercise 4.2: Triangle Area again - Revision history

Guenter at 15:43, 13 March 2023

2023-03-13T15:43:56Z

Guenter at 16:19, 31 March 2022

2022-03-31T16:19:02Z

Bene: Text replacement - "rms value" to "standard deviation"

2022-02-17T11:11:39Z

Text replacement - "rms value" to "standard deviation"

Guenter at 12:44, 7 February 2022

2022-02-07T12:44:19Z

Guenter at 12:32, 7 February 2022

2022-02-07T12:32:09Z

Noah at 18:18, 17 January 2022

2022-01-17T18:18:55Z

Noah at 19:15, 13 January 2022

2022-01-13T19:15:09Z

Guenter at 10:51, 12 January 2022

2022-01-12T10:51:10Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.2: Wieder Dreieckgebiet to Aufgaben:Exercise 4.2: Triangle Area again

2022-01-12T10:46:04Z

Guenter moved page Aufgabe 4.2: Wieder Dreieckgebiet to Exercise 4.2: Triangle Area again

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

2021-03-23T12:43:27Z

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

@@ Line 57: / Line 57: @@
-{What is the equation of the correlation line&nbsp; $y = K(x)$?&nbsp; At what point&nbsp; $y_0$&nbsp; does the straight line intersect the&nbsp; $y$&ndash;axis? <br>Show that the correlation line also passes through the point&nbsp; $(m_x, m_y)$.
+{What is the equation of the regression line&nbsp; $y = K(x)$?&nbsp; At what point&nbsp; $y_0$&nbsp; does the straight line intersect the&nbsp; $y$&ndash;axis? <br>Show that the correlation line also passes through the point&nbsp; $(m_x, m_y)$.
 |type="{}"}
 $y_0\ = \ $ { 1 3% }
@@ Line 80: / Line 80: @@
 '''(3)'''&nbsp; Since both random variables each have a nonzero mean,&nbsp; it follows for the covariance:
-[[File:EN_Sto_Z_4_2d.png|right|frame|Correlation line '''KORREKTUR''']]
+[[File:EN_Sto_Z_4_2d.png|right|frame|Regression line]]
 :$$\it \mu_{xy}=\it m_{xy}-m_{x}\cdot m_{y}=\frac{\rm 26}{\rm 3}-\frac{\rm 8}{\rm 3}\cdot\rm 3={2}/{3} \hspace{0.15cm}\underline{=0.667}.$$
@@ Line 86: / Line 86: @@
 :$$\rho_{xy}=\frac{\mu_{xy}}{\sigma_{x}\cdot\sigma_{y}}=\frac{{\rm 2}/{\rm 3}}{\sqrt{{\rm 8}/{\rm 9}}\cdot\sqrt{{\rm 2}/{\rm 3}}}=\sqrt{0.75}\hspace{0.15cm}\underline{=\rm 0.866}.$$
-'''(5)'''&nbsp; For the correlation line ("KG"&nbsp; in the graph),&nbsp; in general:
+'''(5)'''&nbsp; For the regression line ("RL"&nbsp; in the graph),&nbsp; in general:
 :$$ y-m_{y}=\rho_{xy}\cdot\frac{\sigma_{y}}{\sigma_ {x}}\cdot(x-m_{x}).$$
 *Using the numerical values calculated above,&nbsp; we obtain

@@ Line 80: / Line 80: @@
 '''(3)'''&nbsp; Since both random variables each have a nonzero mean,&nbsp; it follows for the covariance:
-[[File:P_ID223__Sto_A_4_2_d.png|right|frame|Correlation line]]
+[[File:EN_Sto_Z_4_2d.png|right|frame|Correlation line '''KORREKTUR''']]
 :$$\it \mu_{xy}=\it m_{xy}-m_{x}\cdot m_{y}=\frac{\rm 26}{\rm 3}-\frac{\rm 8}{\rm 3}\cdot\rm 3={2}/{3} \hspace{0.15cm}\underline{=0.667}.$$

@@ Line 83: / Line 83: @@
 :$$\it \mu_{xy}=\it m_{xy}-m_{x}\cdot m_{y}=\frac{\rm 26}{\rm 3}-\frac{\rm 8}{\rm 3}\cdot\rm 3={2}/{3} \hspace{0.15cm}\underline{=0.667}.$$
-'''(4)'''&nbsp; With the given rms values we obtain:
+'''(4)'''&nbsp; With the given standard deviations we obtain:
 :$$\rho_{xy}=\frac{\mu_{xy}}{\sigma_{x}\cdot\sigma_{y}}=\frac{{\rm 2}/{\rm 3}}{\sqrt{{\rm 8}/{\rm 9}}\cdot\sqrt{{\rm 2}/{\rm 3}}}=\sqrt{0.75}\hspace{0.15cm}\underline{=\rm 0.866}.$$

@@ Line 67: / Line 67: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Correct is the <u>middle proposition</u>:
+'''(1)'''&nbsp; Correct is the&nbsp; <u>middle proposition</u>:
 *Both&nbsp; $y_1(x)$&nbsp; and&nbsp; $y_2(x)$&nbsp; intersect the&nbsp; $y$-axis at&nbsp; $y= 1$.
-*The lower boundary line has slope&nbsp; $0.5$, the upper has slope&nbsp; $1$.
+*The lower boundary line has slope&nbsp; $0.5$,&nbsp; the upper has slope&nbsp; $1$.
@@ Line 78: / Line 78: @@
 *This leads to the integral or final result:
 :$$m_{xy}={1}/{8}\int_{\rm 0}^{\rm 4}(\frac{3}{4}\cdot x^{3}{\rm +} x^2\,{\rm d}x = \rm \frac{1}{8} \cdot (\frac{3}{16}\cdot 4^4+\rm \frac{4^3}{3})=\frac{26}{3}\hspace{0.15cm}\underline{ \approx 8.667}.$$
-'''(3)'''&nbsp; Since both random variables each have a nonzero mean, it follows f&uuml;r the covariance:
-[[File:P_ID223__Sto_A_4_2_d.png|right|frame|correlation line]]
+'''(3)'''&nbsp; Since both random variables each have a nonzero mean,&nbsp; it follows for the covariance:
 :$$\it \mu_{xy}=\it m_{xy}-m_{x}\cdot m_{y}=\frac{\rm 26}{\rm 3}-\frac{\rm 8}{\rm 3}\cdot\rm 3={2}/{3} \hspace{0.15cm}\underline{=0.667}.$$
+'''(4)'''&nbsp; With the given rms values we obtain:
 :$$\rho_{xy}=\frac{\mu_{xy}}{\sigma_{x}\cdot\sigma_{y}}=\frac{{\rm 2}/{\rm 3}}{\sqrt{{\rm 8}/{\rm 9}}\cdot\sqrt{{\rm 2}/{\rm 3}}}=\sqrt{0.75}\hspace{0.15cm}\underline{=\rm 0.866}.$$
-'''(5)'''&nbsp; For the correlation line (KG), in general:
+'''(5)'''&nbsp; For the correlation line ("KG"&nbsp; in the graph),&nbsp; in general:
 :$$ y-m_{y}=\rho_{xy}\cdot\frac{\sigma_{y}}{\sigma_ {x}}\cdot(x-m_{x}).$$
+*Using the numerical values calculated above,&nbsp; we obtain
 :$$y={\rm 3}/{\rm 4}\cdot x +\rm 1.$$
+*The correlation line intersects the&nbsp; $y$-axis at&nbsp; $\underline{y=1}$&nbsp; and also passes through the point&nbsp; $(4, 4)$.&nbsp;
-The correlation line intersects the&nbsp; $y$-axis at $\underline{y=1}$ and also passes through the point&nbsp; $(4, 4)$.&nbsp; Any other result would also be impossible to interpret considering the definition area:
+*Any other result would be impossible to interpret considering the definition area:
-*If one sets&nbsp; $m_x = 8/3$&nbsp;, one obtains&nbsp; $y = m_y = 3$.
+::If one sets&nbsp; $m_x = 8/3$,&nbsp; one obtains&nbsp; $y = m_y = 3$.
-*This means: &nbsp; The calculated correlation line actually passes through the point&nbsp; $(m_x, m_y)$, as the theory says.
+*This means: &nbsp; The calculated correlation line actually passes through the point&nbsp; $(m_x, m_y)$,&nbsp; as the theory says.
 {{ML-Fuß}}

@@ Line 4: / Line 4: @@
 [[File:P_ID226__Sto_A_4_2.png|right|frame|Triangular 2D area and the two marginal probability densities]]
-We consider the same random variable&nbsp; $(x, \ y)$&nbsp; as in the&nbsp; [[Aufgaben:Exercise_4.2:_Triangle_Area_again|Exercise 4.1]]:
+We consider the same random variable&nbsp; $(x, \ y)$&nbsp; as in&nbsp; [[Aufgaben:Exercise_4.1:_Triangular_(x,_y)_Area|Exercise 4.1]]:
-*In a domain defined by vertices&nbsp; $(0,\ 1)$, $(4,\ 3)$, and $(4,\ 5)$&nbsp; let the 2D&ndash;PDF&nbsp; $f_{xy} (x, y) = 0.25$.
+*In a domain defined by vertices&nbsp; $(0,\ 1)$,&nbsp; $(4,\ 3)$,&nbsp; and $(4,\ 5)$&nbsp; let the 2D&ndash;PDF&nbsp; $f_{xy} (x, y) = 0.25$.
 *There are no values outside this definition area&nbsp; $D$&nbsp; marked in red in the graph.
-Furthermore, the two marginal probability densities with respect to the quantities&nbsp; $x$&nbsp; and&nbsp; $y$&nbsp; are drawn in the graph, which have already been determined in Exercise 4.1.
+Furthermore,&nbsp; the two marginal probability densities with respect to the quantities&nbsp; $x$&nbsp; and&nbsp; $y$&nbsp; are drawn in the graph,&nbsp; which have already been determined in Exercise 4.1.
-From this, the equations of chapter&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|Expected Values and Moments]]&nbsp; can be used to determine the characteristics of the two random variables:
+From this,&nbsp; the equations of chapter&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|Expected Values and Moments]]&nbsp; can be used to determine the characteristics of the two random variables:
 :$$m_x=8/3 ,\hspace{0.5cm} \sigma_x=\sqrt{8/9},$$
 :$$ m_y= 3,\hspace{0.95cm} \sigma_y = \sqrt{\rm 2/3}.$$
-Due to the fact that the domain of definition&nbsp; $D$&nbsp; is bounded by two straight lines&nbsp; $y_1(x)$&nbsp; and&nbsp; $y_2(x)$&nbsp;, the first order joint moment can be calculated here as follows.
+Due to the fact that the definition domain&nbsp; $D$&nbsp; is bounded by two straight lines&nbsp; $y_1(x)$&nbsp; and&nbsp; $y_2(x)$&nbsp;, the first order joint moment can be calculated here as follows.
 :$$m_{xy}={\rm E}\big[x\cdot y\big]=\int_{x_{1}}^{x_{2}}x\cdot \int_{y_{1}(x)}^{y_{2}(x)}y \cdot f_{xy}(x,y) \, \,{\rm d}y\, {\rm d}x.$$
@@ Line 35: / Line 35: @@
 <quiz display=simple>
-{What are the limit lines of the inner integral for&nbsp; $m_{xy}$&ndash;calculation?
+{What are the limit lines of the inner integral for the&nbsp; $m_{xy}$ calculation?
 |type="()"}
 - $y_1(x) = x+1, $ &nbsp; &nbsp; $y_2(x) = 2x+1.$
@@ Line 47: / Line 47: @@
-{What is the value of the covariance&nbsp; $\mu_{xy}$ ?
+{What is the covariance&nbsp; $\mu_{xy}$ ?
 |type="{}"}
 $\mu_{xy}\ = \ $ { 0.667 3% }
-{What is the value of the correlation coefficient $\rho_{xy}$?
+{What is the correlation coefficient $\rho_{xy}$?
 |type="{}"}
 $\rho_{xy}\ = \ $ { 0.866 3% }
-{What is the equation of the correlation line&nbsp; $y = K(x)$&nbsp;? At what point&nbsp; $y_0$&nbsp; does the straight line intersect the&nbsp; $y$ axis? <br>Show that the correlation line also passes through the point&nbsp; $(m_x, m_y)$&nbsp;.
+{What is the equation of the correlation line&nbsp; $y = K(x)$?&nbsp; At what point&nbsp; $y_0$&nbsp; does the straight line intersect the&nbsp; $y$&ndash;axis? <br>Show that the correlation line also passes through the point&nbsp; $(m_x, m_y)$.
 |type="{}"}
 $y_0\ = \ $ { 1 3% }

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID226__Sto_A_4_2.png|right|frame|Dreieckiges 2D-Gebiet und die beiden Randwahrscheinlichkeitsdichten]]
+[[File:P_ID226__Sto_A_4_2.png|right|frame|Triangular 2D area and the two marginal probability densities]]
-Wir betrachten die gleiche Zufallsgr&ouml;&szlig;e&nbsp; $(x, \ y)$&nbsp; wie in der&nbsp; [[Aufgaben:4.1_Dreieckiges_(x,_y)-Gebiet|Aufgabe 4.1]]:
+We consider the same random variable&nbsp; $(x, \ y)$&nbsp; as in the&nbsp; [[Aufgaben:Exercise_4.2:_Triangle_Area_again|Exercise 4.1]]:
-*In einem durch die Eckpunkte&nbsp; $(0,\ 1)$, $(4,\ 3)$ und $(4,\ 5)$&nbsp; definierten Gebiet&nbsp; $D$&nbsp; sei die 2D&ndash;WDF&nbsp; $f_{xy} (x, y) = 0.25$.
+*In a domain defined by vertices&nbsp; $(0,\ 1)$, $(4,\ 3)$, and $(4,\ 5)$&nbsp; let the 2D&ndash;PDF&nbsp; $f_{xy} (x, y) = 0.25$.
-*Au&szlig;erhalb dieses in der Grafik rot markierten Definitionsgebietes&nbsp; $D$&nbsp; gibt es keine Werte.
+*There are no values outside this definition area&nbsp; $D$&nbsp; marked in red in the graph.
-Weiterhin sind in der Grafik die beiden Randwahrscheinlichkeitsdichten bez&uuml;glich den Größen&nbsp; $x$&nbsp; und&nbsp; $y$&nbsp; eingezeichnet, die bereits in der Aufgabe 4.1 ermittelt wurden.
+Furthermore, the two marginal probability densities with respect to the quantities&nbsp; $x$&nbsp; and&nbsp; $y$&nbsp; are drawn in the graph, which have already been determined in Exercise 4.1.
-Daraus lassen sich mit den Gleichungen des Kapitels&nbsp; [[Theory_of_Stochastic_Signals/Erwartungswerte_und_Momente|Erwartungswerte und Momente]]&nbsp; die Kenngrößen der beiden Zufallsgr&ouml;&szlig;en bestimmen:
+From this, the equations of chapter&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|Expected Values and Moments]]&nbsp; can be used to determine the characteristics of the two random variables:
 :$$m_x=8/3 ,\hspace{0.5cm} \sigma_x=\sqrt{8/9},$$
 :$$ m_y= 3,\hspace{0.95cm} \sigma_y = \sqrt{\rm 2/3}.$$
-Aufgrund der Tatsache, dass das Definitionsgebiet&nbsp; $D$&nbsp; durch zwei Gerade&nbsp; $y_1(x)$&nbsp; und&nbsp; $y_2(x)$&nbsp; begrenzt ist, kann hier das gemeinsame Moment erster Ordnung wie folgt berechnet werden.
+Due to the fact that the domain of definition&nbsp; $D$&nbsp; is bounded by two straight lines&nbsp; $y_1(x)$&nbsp; and&nbsp; $y_2(x)$&nbsp;, the first order joint moment can be calculated here as follows.
 :$$m_{xy}={\rm E}\big[x\cdot y\big]=\int_{x_{1}}^{x_{2}}x\cdot \int_{y_{1}(x)}^{y_{2}(x)}y \cdot f_{xy}(x,y) \, \,{\rm d}y\, {\rm d}x.$$
@@ Line 25: / Line 25: @@
-''Hinweise:''
+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]].
-*Bezug genommen wird auch auf das Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Erwartungswerte_und_Momente|Erwartungswerte und Momente]].
+*Reference is also made to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|Expected Values and Moments]].
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Wie lauten die Grenzgeraden des inneren Integrals zur&nbsp; $m_{xy}$&ndash;Berechnung?
+{What are the limit lines of the inner integral for&nbsp; $m_{xy}$&ndash;calculation?
 |type="()"}
 - $y_1(x) = x+1, $ &nbsp; &nbsp; $y_2(x) = 2x+1.$
@@ Line 42: / Line 42: @@
-{Berechnen Sie das gemeinsame Moment&nbsp; $m_{xy}$&nbsp; gem&auml;&szlig; dem Doppelintegral auf der Angabenseite. &nbsp;<i>Hinweis</i>: &nbsp;Setzen Sie&nbsp; $x_1 = 0$&nbsp; und&nbsp; $x_2 = 4$.
+{Calculate the joint moment&nbsp; $m_{xy}$&nbsp; according to the double integral on the statement page. &nbsp;<i>Note</i>: &nbsp;Set&nbsp; $x_1 = 0$&nbsp; and&nbsp; $x_2 = 4$.
 |type="{}"}
-$m_{xy} \ = \ $  { 8.667 3% }
+$m_{xy} \ = \ $ { 8.667 3% }
-{Welcher Wert ergibt sich f&uuml;r die Kovarianz&nbsp; $\mu_{xy}$ ?
+{What is the value of the covariance&nbsp; $\mu_{xy}$ ?
 |type="{}"}
 $\mu_{xy}\ = \ $ { 0.667 3% }
-{Wie gro&szlig; ist der Korrelationskoeffizient&nbsp; $\rho_{xy}$?
+{What is the value of the correlation coefficient $\rho_{xy}$?
 |type="{}"}
 $\rho_{xy}\ = \ $ { 0.866 3% }
-{Wie lautet die Gleichung der Korrelationsgeraden&nbsp; $y = K(x)$&nbsp;? An welcher Stelle&nbsp; $y_0$&nbsp; schneidet die Gerade die&nbsp; $y$-Achse? <br>Zeigen Sie, dass die Korrelationsgerade auch durch den Punkt&nbsp; $(m_x, m_y)$&nbsp; geht.
+{What is the equation of the correlation line&nbsp; $y = K(x)$&nbsp;? At what point&nbsp; $y_0$&nbsp; does the straight line intersect the&nbsp; $y$ axis? <br>Show that the correlation line also passes through the point&nbsp; $(m_x, m_y)$&nbsp;.
 |type="{}"}
 $y_0\ = \ $ { 1 3% }
@@ Line 65: / Line 65: @@
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Richtig ist der <u>mittlere Vorschlag</u>:
+'''(1)'''&nbsp; Correct is the <u>middle proposition</u>:
-*Sowohl&nbsp; $y_1(x)$&nbsp; als auch&nbsp; $y_2(x)$&nbsp;  schneiden die&nbsp; $y$-Achse bei&nbsp; $y= 1$.
+*Both&nbsp; $y_1(x)$&nbsp; and&nbsp; $y_2(x)$&nbsp; intersect the&nbsp; $y$-axis at&nbsp; $y= 1$.
-*Die untere Begrenzungslinie hat die Steigung&nbsp; $0.5$, die obere die Steigung&nbsp; $1$.
+*The lower boundary line has slope&nbsp; $0.5$, the upper has slope&nbsp; $1$.
-'''(2)'''&nbsp; Entsprechend den Hinweisen erhalten wir:
+'''(2)'''&nbsp; According to the clues we get:
-:$$m_{xy}=\int_{\rm 0}^{\rm 4} x \cdot \int_{\it x/\rm 2 +\rm 1}^{\it x+\rm 1} {1}/{4}\cdot  y \, \,{\rm d}y\,\, \, {\rm d}x = {1}/{8}\cdot \int_{\rm 0}^{\rm 4} x\cdot \big[( x+ 1)^{\rm 2}- ({ x}/{2}+1)^{\rm 2} \big]  \,\, {\rm d}x.$$
+: $$m_{xy}=\int_{\rm 0}^{\rm 4} x \cdot \int_{\it x/\rm 2 +\rm 1}^{\it x+\rm 1} {1}/{4}\cdot y \, \,{\rm d}y\,\, {\rm d}x = {1}/{8}\cdot \int_{\rm 0}^{\rm 4} x\cdot \big[( x+ 1)^{\rm 2}- ({ x}/{2}+1)^{\rm 2} \big] \,\, {\rm d}x. $$
-*Dies f&uuml;hrt zum Integral bzw. Endergebnis:
+*This leads to the integral or final result:
 :$$m_{xy}={1}/{8}\int_{\rm 0}^{\rm 4}(\frac{3}{4}\cdot x^{3}{\rm +} x^2\,{\rm d}x = \rm \frac{1}{8} \cdot (\frac{3}{16}\cdot 4^4+\rm \frac{4^3}{3})=\frac{26}{3}\hspace{0.15cm}\underline{ \approx 8.667}.$$
+'''(3)'''&nbsp; Since both random variables each have a nonzero mean, it follows f&uuml;r the covariance:
+[[File:P_ID223__Sto_A_4_2_d.png|right|frame|correlation line]]
 :$$\it \mu_{xy}=\it m_{xy}-m_{x}\cdot m_{y}=\frac{\rm 26}{\rm 3}-\frac{\rm 8}{\rm 3}\cdot\rm 3={2}/{3} \hspace{0.15cm}\underline{=0.667}.$$
-'''(4)'''&nbsp; Mit den angegebenen Streuungen erh&auml;lt man:
+'''(4)'''&nbsp; With the given rms we obtain:
 :$$\rho_{xy}=\frac{\mu_{xy}}{\sigma_{x}\cdot\sigma_{y}}=\frac{{\rm 2}/{\rm 3}}{\sqrt{{\rm 8}/{\rm 9}}\cdot\sqrt{{\rm 2}/{\rm 3}}}=\sqrt{0.75}\hspace{0.15cm}\underline{=\rm 0.866}.$$
+'''(5)'''&nbsp; For the correlation line (KG), in general:
 :$$ y-m_{y}=\rho_{xy}\cdot\frac{\sigma_{y}}{\sigma_ {x}}\cdot(x-m_{x}).$$
-*Mit den oben berechneten Zahlenwerten erh&auml;lt man
+*Using the numerical values calculated above, we obtain
-:$$y={\rm 3}/{\rm 4}\cdot  x +\rm 1.$$
+:$$y={\rm 3}/{\rm 4}\cdot x +\rm 1.$$
-Die Korrelationsgerade schneidet die&nbsp; $y$-Achse bei $\underline{y=1}$ und geht auch durch den Punkt&nbsp; $(4, 4)$.&nbsp; Jedes andere Ergebnis w&auml;re auch nicht zu interpretieren, wenn man das Definitionsgebiet betrachtet:
+The correlation line intersects the&nbsp; $y$-axis at $\underline{y=1}$ and also passes through the point&nbsp; $(4, 4)$.&nbsp; Any other result would also be impossible to interpret considering the definition area:
-*Setzt man&nbsp; $m_x = 8/3$&nbsp; ein, so erh&auml;lt man&nbsp; $y = m_y = 3$.
+*If one sets&nbsp; $m_x = 8/3$&nbsp;, one obtains&nbsp; $y = m_y = 3$.
-*Das heißt: &nbsp; Die berechnete Korrelationsgerade geht tats&auml;chlich durch den Punkt&nbsp; $(m_x, m_y)$, wie es die Theorie besagt.
+*This means: &nbsp; The calculated correlation line actually passes through the point&nbsp; $(m_x, m_y)$, as the theory says.
 {{ML-Fuß}}

← Older revision		Revision as of 19:15, 13 January 2022
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−	{{quiz-Header\|Buchseite=~~Stochastische Signaltheorie~~/~~Zweidimensionale Zufallsgrößen~~	+	{{quiz-Header\|Buchseite=Theory_of_Stochastic_Signals/Two-Dimensional_Random_Variables
	}}		}}

← Older revision		Revision as of 10:51, 12 January 2022
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−	[[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 10:46, 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~~\|^4.1 Zweidimensionale Zufallsgrößen^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^4.1 Zweidimensionale Zufallsgrößen^]]