Aufgaben:Exercise 4.6: Coordinate Rotation - Revision history

Guenter at 15:59, 24 February 2022

2022-02-24T15:59:11Z

Guenter at 15:49, 24 February 2022

2022-02-24T15:49:12Z

Noah at 21:24, 25 January 2022

2022-01-25T21:24:53Z

Guenter at 10:59, 12 January 2022

2022-01-12T10:59:13Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.6: Koordinatendrehung to Aufgaben:Exercise 4.6: Coordinate Rotation

2022-01-12T10:56:47Z

Guenter moved page Aufgabe 4.6: Koordinatendrehung to Exercise 4.6: Coordinate Rotation

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

2021-03-23T12:44:57Z

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

Javier: Text replacement - "[[Stochastische_Signaltheorie/" to "[[Theory_of_Stochastic_Signals/"

2020-07-09T09:52:19Z

Text replacement - "[[Stochastische_Signaltheorie/" to "[[Theory_of_Stochastic_Signals/"

Guenter at 11:25, 27 November 2019

2019-11-27T11:25:54Z

Guenter at 11:19, 27 November 2019

2019-11-27T11:19:38Z

Guenter at 07:49, 17 August 2018

2018-08-17T07:49:26Z

@@ Line 29: / Line 29: @@
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Two-Dimensional_Gaussian_Random_Variables|Two-dimensional Gaussian Random Variables]].
 *Reference is also made to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Two-Dimensional_Gaussian_Random_Variables#Rotation_of_the_coordinate_system|Rotation of the coordinate system]].
-*Given are the approximations&nbsp; ${\rm Q}(2.3) \approx 0.01$&nbsp; and&nbsp; ${\rm Q}(2.6) \approx 0.005$&nbsp; for the complementary Gaussian error integral.
+*Given are the approximations&nbsp; ${\rm Q}(2.3) \approx 0.01$&nbsp; and&nbsp; ${\rm Q}(2.6) \approx 0.005$&nbsp; for the complementary Gaussian error function.
 *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,
@@ Line 61: / Line 61: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; From &nbsp;$\xi = x + y$&nbsp; and &nbsp;$\eta = -x + y$&nbsp; it follows directly:
+'''(1)'''&nbsp; From &nbsp; $\xi = x + y$ &nbsp; and &nbsp; $\eta = -x + y$ &nbsp; it follows directly:
 :$$x = {1}/{2} \cdot ( \xi - \eta ) ,\hspace{0.5cm}y = {1}/{2}\cdot ( \xi +\eta ) .$$
-*Substituting these values for the negative exponent, we get:
+*Substituting these values for the negative exponent,&nbsp; we get:
 :$$\frac{x^2}{2} + \frac{y^2}{8} = \frac{1}{8} \cdot ( \xi - \eta )^2 + \frac{1}{32} \cdot ( \xi + \eta )^2.$$
-*Multiplied out, this gives:
+*After Multiplication,&nbsp; this gives:
 :$$\frac{5}{32} \cdot \xi^2 + \frac{5}{32} \cdot \eta^2 - \frac{3}{16} \cdot \xi \cdot \eta .$$
-*Since the coefficients on&nbsp; $\xi^2$&nbsp; and&nbsp; $\eta^2$&nbsp; are equal, then&nbsp; $\sigma_\xi = \sigma_\eta$.&nbsp; &nbsp; <u>The quotient we are looking for is therefore 1</u>.
+*Since the coefficients on&nbsp; $\xi^2$&nbsp; and&nbsp; $\eta^2$&nbsp; are equal &nbsp; &rArr; &nbsp; $\sigma_\xi = \sigma_\eta$.&nbsp; &nbsp; The quotient we are looking for is therefore
-'''(2)'''&nbsp; By comparing coefficients, we obtain for&nbsp; $\sigma_\xi = \sigma_\eta$&nbsp; the system of equations:
+'''(2)'''&nbsp; By comparing coefficients,&nbsp; we obtain for&nbsp; $\sigma_\xi = \sigma_\eta$&nbsp; the system of equations:
 :$$2 \cdot \sigma_\xi^2 \cdot (1 - \rho_{\xi\eta}^2)= \frac{32}{5},$$
 :$$\frac{\sigma_\xi^2 \cdot (1 - \rho_{\xi\eta}^2)}{\rho_{\xi\eta}}= \frac{16}{3}.$$
-*Substituting the first equation into the second, we get&nbsp; $\rho_{\xi\eta}\hspace{0.15cm}\underline {= 0.6}$&nbsp; and&nbsp; $\sigma_{\xi} = \sqrt{5}\hspace{0.15cm}\underline {= 2.236}$.
+*Substituting the first equation into the second,&nbsp; we get &nbsp; $\rho_{\xi\eta}\hspace{0.15cm}\underline {= 0.6}$ &nbsp; and &nbsp; $\sigma_{\xi} = \sqrt{5}\hspace{0.15cm}\underline {= 2.236}$.
-'''(3)'''&nbsp; After coordinate transformation, we can write f&uuml;r this probability:
+'''(3)'''&nbsp; After coordinate transformation,&nbsp; we can write f&uuml;r this probability:
 :$${\rm Pr} ( | x + y | \le C ) = {\rm Pr} ( | \xi | \le C ) = 1 - 2 \cdot {\rm Pr} ( \xi >C ).$$
-*With the complementary Gaussian error integral, it further follows:
+*With the complementary Gaussian error function,&nbsp; it further follows:
 :$${\rm Pr} ( | x + y | \le C ) = 1 - 2 \cdot {\rm Q} ( {C}/{\sigma_\xi}) = 0.99 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Q} ( {C}/{\sigma_\xi}) = 0.005.$$

@@ Line 4: / Line 4: @@
 [[File:P_ID431__Sto_A_4_6_neu.png|right|frame|Coordinate rotation of a joint PDF]]
-In the exercise we consider a two-dimensional Gaussian random variable&nbsp; $(x, y)$&nbsp; with statistically independent components&nbsp; Let the standard deviations of the two components be&nbsp; $\sigma_x = 1$&nbsp; and&nbsp; $\sigma_y = 2$.
+In the exercise we consider a two-dimensional Gaussian random variable&nbsp; $(x,\hspace{0.08cm} y)$&nbsp; with statistically independent components.&nbsp; Let the standard deviations of the two components be&nbsp; $\sigma_x = 1$&nbsp; and&nbsp; $\sigma_y = 2$.
-We want to calculate the probability that the two-dimensional random variable $(x, y)$&nbsp; lies within the shaded area:
+We want to calculate the probability that the two-dimensional random variable&nbsp; $(x,\hspace{0.08cm} y)$&nbsp; lies within the shaded area:
 :$$-C \le x + y \le +C.$$
@@ Line 16: / Line 16: @@
 *From&nbsp; $x+y= \pm C$&nbsp; it thus follows&nbsp; $\xi=\pm C$.
 *The two two-dimensional density functions are then:
-:$$f_{xy} (x,y) = \frac{1}{4 \pi} \cdot \exp \left [ - ( x^2\hspace {-0.1cm} /2 + y^2\hspace {-0.1cm} /8) \right ] ,$$
+:$$f_{xy} (x,\hspace{0.08cm}y) = \frac{1}{4 \pi} \cdot \exp \left [ - ( x^2\hspace {-0.1cm} /2 + y^2\hspace {-0.1cm} /8) \right ] ,$$
-:$$f_{\xi\eta} (\xi, \eta) = \frac{1}{2 \pi \cdot \sigma_\xi \cdot \sigma_\eta \cdot \sqrt{1 - \rho_{\xi\eta}^2}} \cdot \exp \left [ - \frac{1}{2 \cdot (1 - \rho_{\xi\eta}^2)} \cdot ( \frac {\xi^2}{\sigma_\xi^2} + \frac {\eta^2}{\sigma_\eta^2 }- 2 \rho_{\xi\eta}\cdot \frac {\xi \cdot \eta}{\sigma_\xi \cdot \sigma_\eta}) \right ] .$$
+:$$f_{\xi\eta} (\xi,\hspace{0.08cm} \eta) = \frac{1}{2 \pi \cdot \sigma_\xi \cdot \sigma_\eta \cdot \sqrt{1 - \rho_{\xi\eta}^2}} \cdot \exp \left [ - \frac{1}{2 \cdot (1 - \rho_{\xi\eta}^2)} \cdot ( \frac {\xi^2}{\sigma_\xi^2} + \frac {\eta^2}{\sigma_\eta^2 }- 2 \rho_{\xi\eta}\cdot \frac {\xi \cdot \eta}{\sigma_\xi \cdot \sigma_\eta}) \right ] .$$
@@ Line 29: / Line 29: @@
 *The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Two-Dimensional_Gaussian_Random_Variables|Two-dimensional Gaussian Random Variables]].
 *Reference is also made to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Two-Dimensional_Gaussian_Random_Variables#Rotation_of_the_coordinate_system|Rotation of the coordinate system]].
 *Given are the approximations&nbsp; ${\rm Q}(2.3) \approx 0.01$&nbsp; and&nbsp; ${\rm Q}(2.6) \approx 0.005$&nbsp; for the complementary Gaussian error integral.
-*More information on this topic is provided in the learning video&nbsp; [[Gaußsche_2D-Zufallsgrößen_(Lernvideo)|Gaussian 2D 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 42: / Line 41: @@
 <quiz display=simple>
-{Determine the ratio of the two standard deviations of the new random variables by comparing coefficients $(\xi, \eta)$.
+{Determine the ratio of the two standard deviations of the new random variables by comparing coefficients $(\xi,\hspace{0.08cm} \eta)$.
 |type="{}"}
 $\sigma_\xi/\sigma_\eta \ = \ $ { 1 3% }
@@ Line 53: / Line 52: @@
-{Calculate the probability that&nbsp; $ |\hspace{0.05cm}x+y\hspace{0.05cm}| \le C$&nbsp; holds.&nbsp; How large should&nbsp; $C$&nbsp; be chosen so that&nbsp; $99\%$&nbsp; of all sizes are in the shaded area?
+{Calculate the probability that&nbsp; $ |\hspace{0.05cm}x+y\hspace{0.05cm}| \le C$&nbsp; holds.&nbsp; How large should&nbsp; $C$&nbsp; be chosen so that&nbsp; $99\%$&nbsp; of all quantities are in the shaded area?
 |type="{}"}
 $C_{99\%} \ = \ $ { 5.814 3% }

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Stochastische Signaltheorie/Zweidimensionale Gaußsche Zufallsgrößen
+{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Two-Dimensional_Gaussian_Random_Variables
 }}
-[[File:P_ID431__Sto_A_4_6_neu.png|right|frame|Koordinatendrehung einer 2D-WDF]]
+[[File:P_ID431__Sto_A_4_6_neu.png|right|frame|Coordinate rotation of a joint PDF]]
-Wir betrachten in der Aufgabe eine zweidimensionale Gau&szlig;sche Zufallsgr&ouml;&szlig;e&nbsp; $(x, y)$&nbsp; mit statistisch unabh&auml;ngigen Komponenten.&nbsp;  Die Streuungen der beiden Komponenten seien&nbsp;  $\sigma_x = 1$&nbsp; und&nbsp; $\sigma_y = 2$.
+In the exercise we consider a two-dimensional Gaussian random variable&nbsp; $(x, y)$&nbsp; with statistically independent components&nbsp; Let the standard deviations of the two components be&nbsp; $\sigma_x = 1$&nbsp; and&nbsp; $\sigma_y = 2$.
-Berechnet werden soll die Wahrscheinlichkeit dafür, dass die zweidimensionale Zufallsgr&ouml;&szlig;e&nbsp; $(x, y)$&nbsp; innerhalb des gr&uuml;n schraffiert eingezeichneten Bereichs liegt:
+We want to calculate the probability that the two-dimensional random variable $(x, y)$&nbsp; lies within the shaded area:
 :$$-C \le x + y \le +C.$$
-Führen Sie zur L&ouml;sung eine Koordinatentransformation durch:
+Perform a coordinate transformation to solve:
 :$$\xi = \hspace{0.4cm} x +y,$$
 :$$\eta= -x +y .$$
-Dies entspricht einer Drehung des Koordinatensystems um&nbsp; $45^\circ$.
+This corresponds to a rotation of the coordinate system by&nbsp; $45^\circ$.
-*Aus&nbsp; $x+y= \pm C$&nbsp; folgt damit&nbsp; $\xi=\pm C$.
+*From&nbsp; $x+y= \pm C$&nbsp; it thus follows&nbsp; $\xi=\pm C$.
-*Die beiden zweidimensionalen Dichtefunktionen lauten dann:
+*The two two-dimensional density functions are then:
 :$$f_{xy} (x,y) = \frac{1}{4 \pi} \cdot \exp \left [ - ( x^2\hspace {-0.1cm} /2 + y^2\hspace {-0.1cm} /8) \right ] ,$$
-:$$f_{\xi\eta} (\xi, \eta) = \frac{1}{2 \pi \cdot \sigma_\xi \cdot \sigma_\eta \cdot \sqrt{1 - \rho_{\xi\eta}^2}} \cdot \exp \left [ - \frac{1}{2 \cdot (1 - \rho_{\xi\eta}^2)} \cdot ( \frac {\xi^2}{\sigma_\xi^2} +  \frac {\eta^2}{\sigma_\eta^2 }- 2 \rho_{\xi\eta}\cdot \frac {\xi \cdot \eta}{\sigma_\xi \cdot \sigma_\eta}) \right ] .$$
+:$$f_{\xi\eta} (\xi, \eta) = \frac{1}{2 \pi \cdot \sigma_\xi \cdot \sigma_\eta \cdot \sqrt{1 - \rho_{\xi\eta}^2}} \cdot \exp \left [ - \frac{1}{2 \cdot (1 - \rho_{\xi\eta}^2)} \cdot ( \frac {\xi^2}{\sigma_\xi^2} + \frac {\eta^2}{\sigma_\eta^2 }- 2 \rho_{\xi\eta}\cdot \frac {\xi \cdot \eta}{\sigma_\xi \cdot \sigma_\eta}) \right ] .$$
@@ Line 26: / Line 26: @@
-''Hinweise:''
+Hints:
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Zweidimensionale_Gaußsche_Zufallsgrößen|Zweidimensionale Gaußsche Zufallsgrößen]].
+*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Two-Dimensional_Gaussian_Random_Variables|Two-dimensional Gaussian Random Variables]].
-*Bezug genommen wird insbesondere auf die Seite&nbsp; [[Theory_of_Stochastic_Signals/Zweidimensionale_Gaußsche_Zufallsgrößen#Drehung_des_Koordinatensystems|Drehung des Koordinatensystems]].
+*Reference is also made to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Two-Dimensional_Gaussian_Random_Variables#Rotation_of_the_coordinate_system|Rotation of the coordinate system]].
-*Gegeben sind die Näherungen&nbsp; ${\rm Q}(2.3) \approx 0.01$&nbsp; und&nbsp; ${\rm Q}(2.6) \approx 0.005$&nbsp; für das komplementäre Gaußsche Fehlerintegral.
+*Given are the approximations&nbsp; ${\rm Q}(2.3) \approx 0.01$&nbsp; and&nbsp; ${\rm Q}(2.6) \approx 0.005$&nbsp; for the complementary Gaussian error integral.
-*Weitere Informationen zu dieser Thematik liefert das Lernvideo&nbsp; [[Gaußsche_2D-Zufallsgrößen_(Lernvideo)|Gaußsche 2D-Zufallsgrößen]]:
+*More information on this topic is provided in the learning video&nbsp; [[Gaußsche_2D-Zufallsgrößen_(Lernvideo)|Gaussian 2D random variables]]:
-::Teil 1: &nbsp; Gaußsche Zufallsgrößen ohne statistische Bindungen,
+::Part 1: &nbsp; Gaussian random variables without statistical bindings,
-::Teil 2: &nbsp; Gaußsche Zufallsgrößen mit statistischen Bindungen.
+::Part 2: &nbsp; Gaussian random variables with statistical bindings.
@@ Line 39: / Line 39: @@
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Ermitteln Sie durch Koeffizientenvergleich das Verh&auml;ltnis der beiden Streuungen der neuen Zufallsgr&ouml;&szlig;e&nbsp; $(\xi, \eta)$.
+{Determine the ratio of the two standard deviations of the new random variables by comparing coefficients $(\xi, \eta)$.
 |type="{}"}
-$\sigma_\xi/\sigma_\eta \ = \ $  { 1 3% }
+$\sigma_\xi/\sigma_\eta \ = \ $ { 1 3% }
-{Berechnen Sie die Streuung&nbsp; $\sigma_\xi$&nbsp; und den Korrelationskoeffizienten&nbsp; $\rho_{\xi\eta}$&nbsp; zwischen den neuen Zufallsgrößen&nbsp; $\xi$&nbsp; und&nbsp; $\eta$.
+{Calculate the standard deviation&nbsp; $\sigma_\xi$&nbsp; and correlation coefficient&nbsp; $\rho_{\xi\eta}$&nbsp; between the new random variables&nbsp; $\xi$&nbsp; and&nbsp; $\eta$.
 |type="{}"}
-$\sigma_\xi \ =  \ $ { 2.236 3% }
+$\sigma_\xi \ = \ $ { 2.236 3% }
-$\rho_{\xi\eta} \ =  \ $ { 0.6 3% }
+$\rho_{\xi\eta} \ = \ $ { 0.6 3% }
-{Berechnen Sie die Wahrscheinlichkeit, dass&nbsp; $ |\hspace{0.05cm}x+y\hspace{0.05cm}| \le C$&nbsp; gilt.&nbsp; Wie gro&szlig; ist&nbsp; $C$&nbsp; zu w&auml;hlen, damit&nbsp; $99\%$&nbsp; aller Gr&ouml;&szlig;en im schraffierten Bereich liegen?
+{Calculate the probability that&nbsp; $ |\hspace{0.05cm}x+y\hspace{0.05cm}| \le C$&nbsp; holds.&nbsp; How large should&nbsp; $C$&nbsp; be chosen so that&nbsp; $99\%$&nbsp; of all sizes are in the shaded area?
 |type="{}"}
-$C_{99\%} \ =  \ $ { 5.814 3% }
+$C_{99\%} \ = \ $ { 5.814 3% }
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Aus &nbsp;$\xi = x + y$&nbsp; und &nbsp;$\eta = -x + y$&nbsp; folgt direkt:
+'''(1)'''&nbsp; From &nbsp;$\xi = x + y$&nbsp; and &nbsp;$\eta = -x + y$&nbsp; it follows directly:
-:$$x = {1}/{2} \cdot  ( \xi - \eta ) ,\hspace{0.5cm}y = {1}/{2}\cdot ( \xi +\eta ) .$$
+:$$x = {1}/{2} \cdot ( \xi - \eta ) ,\hspace{0.5cm}y = {1}/{2}\cdot ( \xi +\eta ) .$$
-*Setzt man diese Werte f&uuml;r den negativen Exponenten ein, so erh&auml;lt man:
+*Substituting these values for the negative exponent, we get:
-:$$\frac{x^2}{2}  + \frac{y^2}{8} = \frac{1}{8} \cdot ( \xi - \eta )^2 +  \frac{1}{32} \cdot ( \xi + \eta )^2.$$
+:$$\frac{x^2}{2} + \frac{y^2}{8} = \frac{1}{8} \cdot ( \xi - \eta )^2 + \frac{1}{32} \cdot ( \xi + \eta )^2.$$
-*Ausmultipliziert ergibt dies:
+*Multiplied out, this gives:
-:$$\frac{5}{32} \cdot  \xi^2 +  \frac{5}{32} \cdot  \eta^2 - \frac{3}{16} \cdot  \xi \cdot  \eta .$$
+:$$\frac{5}{32} \cdot \xi^2 + \frac{5}{32} \cdot \eta^2 - \frac{3}{16} \cdot \xi \cdot \eta .$$
-*Da die Koeffizienten bei&nbsp; $\xi^2$&nbsp; und&nbsp; $\eta^2$&nbsp; gleich sind, gilt&nbsp; $\sigma_\xi = \sigma_\eta$.&nbsp; &nbsp;  <u>Der gesuchte Quotient ist somit 1</u>.
+*Since the coefficients on&nbsp; $\xi^2$&nbsp; and&nbsp; $\eta^2$&nbsp; are equal, then&nbsp; $\sigma_\xi = \sigma_\eta$.&nbsp; &nbsp; <u>The quotient we are looking for is therefore 1</u>.
-'''(2)'''&nbsp; Durch Koeffizientenvergleich erh&auml;lt man für&nbsp; $\sigma_\xi = \sigma_\eta$&nbsp; das Gleichungssystem:
+'''(2)'''&nbsp; By comparing coefficients, we obtain for&nbsp; $\sigma_\xi = \sigma_\eta$&nbsp; the system of equations:
-:$$2 \cdot \sigma_\xi^2 \cdot  (1 - \rho_{\xi\eta}^2)=  \frac{32}{5},$$
+:$$2 \cdot \sigma_\xi^2 \cdot (1 - \rho_{\xi\eta}^2)= \frac{32}{5},$$
-:$$\frac{\sigma_\xi^2 \cdot  (1 - \rho_{\xi\eta}^2)}{\rho_{\xi\eta}}=  \frac{16}{3}.$$
+:$$\frac{\sigma_\xi^2 \cdot (1 - \rho_{\xi\eta}^2)}{\rho_{\xi\eta}}= \frac{16}{3}.$$
-*Setzt man die erste Gleichung in die zweite ein, so ergibt sich&nbsp; $\rho_{\xi\eta}\hspace{0.15cm}\underline {= 0.6}$&nbsp;  und&nbsp; $\sigma_{\xi} = \sqrt{5}\hspace{0.15cm}\underline {= 2.236}$.
+*Substituting the first equation into the second, we get&nbsp; $\rho_{\xi\eta}\hspace{0.15cm}\underline {= 0.6}$&nbsp; and&nbsp; $\sigma_{\xi} = \sqrt{5}\hspace{0.15cm}\underline {= 2.236}$.
-'''(3)'''&nbsp; Nach Koordinatentransformation kann man f&uuml;r diese Wahrscheinlichkeit schreiben:
+'''(3)'''&nbsp; After coordinate transformation, we can write f&uuml;r this probability:
-:$${\rm Pr} ( | x + y | \le C ) = {\rm Pr} ( | \xi | \le C ) = 1 - 2 \cdot {\rm Pr} (  \xi >C ).$$
+:$${\rm Pr} ( | x + y | \le C ) = {\rm Pr} ( | \xi | \le C ) = 1 - 2 \cdot {\rm Pr} ( \xi >C ).$$
-*Mit dem komplement&auml;ren Gau&szlig;schen Fehlerintegral folgt daraus weiter:
+*With the complementary Gaussian error integral, it further follows:
-:$${\rm Pr} ( | x + y | \le C ) = 1 - 2 \cdot {\rm Q} (  {C}/{\sigma_\xi}) = 0.99 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Q} (  {C}/{\sigma_\xi}) = 0.005.$$
+:$${\rm Pr} ( | x + y | \le C ) = 1 - 2 \cdot {\rm Q} ( {C}/{\sigma_\xi}) = 0.99 \hspace{0.3cm} \Rightarrow \hspace{0.3cm} {\rm Q} ( {C}/{\sigma_\xi}) = 0.005.$$
-*Mit dem angegebenen Wert&nbsp; ${\rm Q}(2.6) \approx 0.005$&nbsp; erh&auml;lt man somit das Ergebnis:
+*With the given value&nbsp; ${\rm Q}(2.6) \approx 0.005$&nbsp; we thus obtain the result:
 :$$C \approx 2.6 \cdot \sigma_{\xi}\hspace{0.15cm}\underline {= 5.814}.$$
 {{ML-Fuß}}

← Older revision		Revision as of 10:59, 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 12:44, 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^]]

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