Aufgaben:Exercise 3.3: Moments for Cosine-square PDF - Revision history

Hwang at 12:20, 18 January 2023

2023-01-18T12:20:33Z

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

2022-02-17T11:34:20Z

Text replacement - " rms " to " standard deviation "

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

2022-02-17T11:11:21Z

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

Guenter: Guenter moved page Aufgaben:Exercise 3.3: Moments for cos²-PDF to Aufgaben:Exercise 3.3: Moments for Cosine-square PDF

2022-02-02T15:42:07Z

Guenter moved page Exercise 3.3: Moments for cos²-PDF to Exercise 3.3: Moments for Cosine-square PDF

Guenter at 13:49, 5 January 2022

2022-01-05T13:49:49Z

Guenter at 13:29, 5 January 2022

2022-01-05T13:29:41Z

Guenter: Guenter moved page Aufgaben:Exercise 3.3: Moments at cos²-PDF to Aufgaben:Exercise 3.3: Moments for cos²-PDF

2022-01-05T13:22:39Z

Guenter moved page Exercise 3.3: Moments at cos²-PDF to Exercise 3.3: Moments for cos²-PDF

Noah at 23:47, 27 December 2021

2021-12-27T23:47:13Z

Noah at 17:44, 27 December 2021

2021-12-27T17:44:19Z

Noah at 17:11, 27 December 2021

2021-12-27T17:11:43Z

@@ Line 29: / Line 29: @@
 <quiz display=simple>
-{Which of the following statements are true for any given PDF&nbsp; $f_x(x)$&nbsp;?&nbsp; Used are the following quantities: &nbsp; linear mean&nbsp; $m_1$,&nbsp; quadratic mean&nbsp; $m_2$,&nbsp; variance&nbsp; $\sigma^2$.
+{Which of the following statements are true for any given PDF&nbsp; $f_x(x)$&nbsp;?&nbsp; Used are the following quantities: &nbsp; linear mean&nbsp; $m_1$,&nbsp; second moment&nbsp; $m_2$,&nbsp; variance&nbsp; $\sigma^2$.
 |type="[]"}
 - $m_2 = 0$,&nbsp;&nbsp;if &nbsp; $m_1 \ne 0$.
@@ Line 85: / Line 85: @@
 '''(3)'''&nbsp; The standard deviation of the signal&nbsp; $x(t)$&nbsp; is equal to the standard deviation &nbsp; $\sigma_x$&nbsp; or equal to the root of the variance&nbsp; $\sigma_x^2$.
-*Since the random variable&nbsp; $x$&nbsp; has mean&nbsp; $m_x {= 0}$,&nbsp; the variance is equal to the quadratic mean according to Steiner's theorem.
+*Since the random variable&nbsp; $x$&nbsp; has mean&nbsp; $m_x {= 0}$,&nbsp; the variance is equal to the second moment according to Steiner's theorem.
 *This is also referred to as the power&nbsp; $($with respect to&nbsp; $1 \ \rm \Omega)$&nbsp; in the context of signals. Thus:
 :$$\sigma_x^{\rm 2}=\int_{-\infty}^{+\infty}x^{\rm 2}\cdot f_x(x)\hspace{0.1cm}{\rm d}x=2 \cdot \int_{\rm 0}^{\rm 2} x^2/2 \cdot \cos^2({\pi}/4\cdot\it x)\hspace{0.1cm} {\rm d}x.$$

@@ Line 84: / Line 84: @@
-'''(3)'''&nbsp; The standard deviation of the signal&nbsp; $x(t)$&nbsp; is equal to the rms &nbsp; $\sigma_x$&nbsp; or equal to the root of the variance&nbsp; $\sigma_x^2$.
+'''(3)'''&nbsp; The standard deviation of the signal&nbsp; $x(t)$&nbsp; is equal to the standard deviation &nbsp; $\sigma_x$&nbsp; or equal to the root of the variance&nbsp; $\sigma_x^2$.
 *Since the random variable&nbsp; $x$&nbsp; has mean&nbsp; $m_x {= 0}$,&nbsp; the variance is equal to the quadratic mean according to Steiner's theorem.
 *This is also referred to as the power&nbsp; $($with respect to&nbsp; $1 \ \rm \Omega)$&nbsp; in the context of signals. Thus:
@@ Line 107: / Line 107: @@
 '''(6)'''&nbsp; The mean value does not change the variance and rms.
-*By compressing by the factor&nbsp; $2$&nbsp; the rms becomes smaller compared to subtask&nbsp; '''(3)'''&nbsp; also by this factor:
+*By compressing by the factor&nbsp; $2$&nbsp; the standard deviation becomes smaller compared to subtask&nbsp; '''(3)'''&nbsp; also by this factor:
 :$$\sigma_y=\sigma_x/\rm 2\hspace{0.15cm}\underline{\approx 0.361}.$$
 {{ML-Fuß}}

@@ Line 44: / Line 44: @@
-{What is the rms value of the signal&nbsp; $x(t)$?
+{What is the standard deviation of the signal&nbsp; $x(t)$?
 |type="{}"}
 $\sigma_x \ = \ $ { 0.722 3% }
@@ Line 61: / Line 61: @@
-{What is the rms value of the signal&nbsp; $y(t)$?
+{What is the standard deviation of the signal&nbsp; $y(t)$?
 |type="{}"}
 $\sigma_y\ = \ $ { 0.361 3% }

@@ Line 70: / Line 70: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Under all circumstances, <u>statements 3, 4, and 5</u> are correct:
+'''(1)'''&nbsp; Under all circumstances,&nbsp; <u>statements 3, 4, and 5</u>&nbsp; are correct:
-*The first statement is never true, as is evident from the&nbsp; <i>Steiner's theorem</i>&nbsp;.
+*The first statement is never true,&nbsp; as is evident from the&nbsp; "Steiner's theorem".
-*The second statement is only valid in the (trivial) special case&nbsp; $x = 0$.
+*The second statement is only valid in the&nbsp; (trivial)&nbsp; special case&nbsp; $x = 0$.
@@ Line 84: / Line 84: @@
-'''(3)'''&nbsp; The rms of the signal&nbsp; $x(t)$&nbsp; is equal to the rms &nbsp; $\sigma_x$&nbsp; or equal to the root of the variance&nbsp; $\sigma_x^2$.
+'''(3)'''&nbsp; The standard deviation of the signal&nbsp; $x(t)$&nbsp; is equal to the rms &nbsp; $\sigma_x$&nbsp; or equal to the root of the variance&nbsp; $\sigma_x^2$.
-*Since the random variable&nbsp; $x$&nbsp; has mean&nbsp; $m_x {= 0}$&nbsp;, the variance is equal to the root mean square according to <i>Steiner's theorem</i>.
+*Since the random variable&nbsp; $x$&nbsp; has mean&nbsp; $m_x {= 0}$,&nbsp; the variance is equal to the quadratic mean according to Steiner's theorem.
 *This is also referred to as the power&nbsp; $($with respect to&nbsp; $1 \ \rm \Omega)$&nbsp; in the context of signals. Thus:
 :$$\sigma_x^{\rm 2}=\int_{-\infty}^{+\infty}x^{\rm 2}\cdot f_x(x)\hspace{0.1cm}{\rm d}x=2 \cdot \int_{\rm 0}^{\rm 2} x^2/2 \cdot \cos^2({\pi}/4\cdot\it x)\hspace{0.1cm} {\rm d}x.$$
 *With the relation&nbsp; $\cos^2(\alpha) = 0.5 \cdot \big[1 + \cos(2\alpha)\big]$&nbsp; it follows:
 :$$\sigma_x^2=\int_{\rm 0}^{\rm 2}{x^{\rm 2}}/{\rm 2} \hspace{0.1cm}{\rm d}x + \int_{\rm 0}^{\rm 2}{x^{\rm 2}}/{2}\cdot \cos({\pi}/{\rm 2}\cdot\it x) \hspace{0.1cm} {\rm d}x.$$
 *These two standard integrals can be found in tables. One obtains with&nbsp; $a = \pi/2$:
 :$$\sigma_x^{\rm 2}=\left[\frac{x^{\rm 3}}{\rm 6} + \frac{x}{a^2}\cdot {\cos}(a x) + \left( \frac{x^{\rm2}}{{\rm2}a} - \frac{1}{a^3} \right) \cdot \sin(a \cdot x)\right]_{x=0}^{x=2} \hspace{0.5cm}
@@ Line 98: / Line 96: @@
-'''(4)'''&nbsp; Correct is the <u>first mentioned suggestion</u>:
+'''(4)'''&nbsp; Correct is the&nbsp; <u>first mentioned suggestion</u>:
-*The variant&nbsp; $y = 2x$&nbsp; would yield a random size distributed between&nbsp; $-4$&nbsp; and&nbsp; $+4$&nbsp; .
+*The variant&nbsp; $y = 2x$&nbsp; would yield a random variable distributed between&nbsp; $-4$&nbsp; and&nbsp; $+4$.
-*In the last proposition&nbsp; $y = x/2-1$&nbsp; the mean&nbsp; $m_y = -1$ would w&auml;re.
+*In the last proposition&nbsp; $y = x/2-1$&nbsp; the mean would be&nbsp; $m_y = -1$.

@@ Line 7: / Line 7: @@
 :$$f_x(x)= {1}/{2}\cdot \cos^2({\pi}/{4}\cdot { x}).$$
-Next to this, we consider a second random variable&nbsp; $y$ that returns only values between&nbsp; $0$&nbsp; and&nbsp; $2$&nbsp; with the following PDF:
+Next to this,&nbsp; we consider a second random variable&nbsp; $y$&nbsp; that returns only values between&nbsp; $0$&nbsp; and&nbsp; $2$&nbsp; with the following PDF:
 :$$f_y(y)=\sin^2({\pi}/{2}\cdot y).$$
-*Both density functions are shown in the graph.
+*Both probability density functions are shown in the graph.
-*Outside the ranges&nbsp; $-2 < x < +2$ &nbsp;resp.&nbsp; $0 < x < +2$&nbsp;, the followings holds: &nbsp; $f_x(x) = 0$ &nbsp;resp.&nbsp; $f_y(y) = 0$.
+*Outside the ranges&nbsp; $-2 < x < +2$ &nbsp;resp.&nbsp; $0 < x < +2$,&nbsp; the followings holds: &nbsp; $f_x(x) = 0$ &nbsp;resp.&nbsp; $f_y(y) = 0$.
-*Both random variables can be taken as (normalized) instantaneous values of the associated random signals&nbsp; $x(t)$&nbsp; or&nbsp; $y(t)$&nbsp; respectively.
+*Both random variables can be taken as&nbsp; (normalized)&nbsp; instantaneous values of the associated random signals&nbsp; $x(t)$&nbsp; or&nbsp; $y(t)$&nbsp; resp.
@@ Line 23: / Line 20: @@
 Hints:
-*This exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|expected values and moments]].
+*This exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|Expected values and moments]].
-*To solve this problem, you can use the following indefinite integral:
+*To solve this problem,&nbsp; you can use the following indefinite integral:
 :$$\int x^{2}\cdot {\cos}(ax)\,{\rm d}x=\frac{2 x}{ a^{ 2}}\cdot \cos(ax)+ \left [\frac{x^{\rm 2}}{\it a} - \frac{\rm 2}{\it a^{\rm 3}} \right ]\cdot \rm sin(\it ax \rm ) .$$
@@ Line 32: / Line 29: @@
 <quiz display=simple>
-{Which of the following statements are true for any given PDF&nbsp; $f_x(x)$&nbsp;?
+{Which of the following statements are true for any given PDF&nbsp; $f_x(x)$&nbsp;?&nbsp; Used are the following quantities: &nbsp; linear mean&nbsp; $m_1$,&nbsp; quadratic mean&nbsp; $m_2$,&nbsp; variance&nbsp; $\sigma^2$.
 |type="[]"}
 - $m_2 = 0$,&nbsp;&nbsp;if &nbsp; $m_1 \ne 0$.
@@ Line 43: / Line 39: @@
-{How large is the DC component (linear mean) of the signal&nbsp; $x(t)$?
+{What is the DC component&nbsp; ("linear mean")&nbsp; of the signal&nbsp; $x(t)$?
 |type="{}"}
 $m_x \ = \ $ { 0. }
@@ Line 53: / Line 49: @@
-{The random variable&nbsp; $y$&nbsp; can be derived from&nbsp; $x$&nbsp;. Which assignment is valid?
+{The random variable&nbsp; $y$&nbsp; can be derived from&nbsp; $x$.&nbsp; Which assignment is valid?
 |type="()"}
 + $y = 1+x/2.$
@@ Line 60: / Line 56: @@
-{How large is the DC component of the signal $y(t)$?
+{What is the DC component of the signal&nbsp; $y(t)$?
 |type="{}"}
 $m_y\ = $ { 1 3% }

@@ Line 33: / Line 33: @@
 <quiz display=simple>
 {Which of the following statements are true for any given PDF&nbsp; $f_x(x)$&nbsp;?
-<br>Used are the following quantities: &nbsp; linear mean&nbsp; $m_1$,&nbsp; root mean square&nbsp; $m_2$,&nbsp; variance&nbsp; $\sigma^2$.
+<br>Used are the following quantities: &nbsp; linear mean&nbsp; $m_1$,&nbsp; quadratic mean&nbsp; $m_2$,&nbsp; variance&nbsp; $\sigma^2$.
 |type="[]"}
 - $m_2 = 0$,&nbsp;&nbsp;if &nbsp; $m_1 \ne 0$.
@@ Line 48: / Line 48: @@
-{What is the rms value (Effektivwert?) of the signal&nbsp; $x(t)$?
+{What is the rms value of the signal&nbsp; $x(t)$?
 |type="{}"}
 $\sigma_x \ = \ $ { 0.722 3% }
@@ Line 65: / Line 65: @@
-{What is the rms value (Effektivwert) of the signal&nbsp; $y(t)$?
+{What is the rms value of the signal&nbsp; $y(t)$?
 |type="{}"}
 $\sigma_y\ = \ $ { 0.361 3% }
@@ Line 74: / Line 74: @@
 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Unter allen Umst&auml;nden richtig sind <u>die Aussagen 3, 4 und 5</u>:
+'''(1)'''&nbsp; Under all circumstances, <u>statements 3, 4, and 5</u> are correct:
-*Die erste Aussage ist nie erf&uuml;llt, wie aus dem&nbsp; <i>Satz von Steiner</i>&nbsp; ersichtlich ist.
+*The first statement is never true, as is evident from the&nbsp; <i>Steiner's theorem</i>&nbsp;.
-*Die zweite Aussage gilt nur im (trivialen) Sonderfall&nbsp; $x = 0$.
+*The second statement is only valid in the (trivial) special case&nbsp; $x = 0$.
-Es gibt aber auch mittelwertfreie Zufallsgr&ouml;&szlig;en mit unsymmetrischer WDF.
+However, there are also zero-mean random variables with asymmetric PDF.
-*Das bedeutet: &nbsp;Die Aussage 6 trifft nicht immer zu.
+*This means: &nbsp;Statement 6 is not always true.
-'''(2)'''&nbsp; Aufgrund der WDF-Symmetrie bez&uuml;glich&nbsp; $x = 0$&nbsp; ergibt sich f&uuml;r den linearen Mittelwert&nbsp; $m_x \hspace{0.15cm}\underline{= 0}$.
+'''(2)'''&nbsp; Because of the PDF symmetry with respect to&nbsp; $x = 0$&nbsp; it follows that for the linear mean&nbsp; $m_x \hspace{0.15cm}\underline{= 0}$.
 '''(3)'''&nbsp; Der Effektivwert des Signals&nbsp; $x(t)$&nbsp; ist gleich der Streuung&nbsp; $\sigma_x$&nbsp; bzw. gleich der Wurzel aus der Varianz&nbsp; $\sigma_x^2$.
-*Da die Zufallsgr&ouml;&szlig;e&nbsp; $x$&nbsp; den Mittelwert&nbsp; $m_x {= 0}$&nbsp; aufweist, ist die Varianz nach dem <i>Satz von Steiner</i> gleich dem quadratischen Mittelwert.
+*Since the random variable&nbsp; $x$&nbsp; has mean&nbsp; $m_x {= 0}$&nbsp;, the variance is equal to the root mean square according to <i>Steiner's theorem</i>.
 *Dieser wird in Zusammenhang mit Signalen auch als die Leistung&nbsp; $($bezogen auf&nbsp; $1 \ \rm \Omega)$&nbsp; bezeichnet. Somit gilt:
 :$$\sigma_x^{\rm 2}=\int_{-\infty}^{+\infty}x^{\rm 2}\cdot f_x(x)\hspace{0.1cm}{\rm d}x=2 \cdot \int_{\rm 0}^{\rm 2} x^2/2 \cdot \cos^2({\pi}/4\cdot\it x)\hspace{0.1cm} {\rm d}x.$$
-*Mit der Beziehung&nbsp; $\cos^2(\alpha) = 0.5 \cdot \big[1 + \cos(2\alpha)\big]$&nbsp; folgt daraus:
+*With the relation&nbsp; $\cos^2(\alpha) = 0.5 \cdot \big[1 + \cos(2\alpha)\big]$&nbsp; it follows:
 :$$\sigma_x^2=\int_{\rm 0}^{\rm 2}{x^{\rm 2}}/{\rm 2} \hspace{0.1cm}{\rm d}x + \int_{\rm 0}^{\rm 2}{x^{\rm 2}}/{2}\cdot \cos({\pi}/{\rm 2}\cdot\it x) \hspace{0.1cm} {\rm d}x.$$
@@ Line 112: / Line 112: @@
-'''(6)'''&nbsp; The mean value &auml;does not change the variance and dispersion.
+'''(6)'''&nbsp; The mean value does not change the variance and dispersion.
 *By compressing by the factor&nbsp; $2$&nbsp; the scatter becomes smaller against&uuml;ber Teilaufgabe&nbsp; '''(3)'''&nbsp; also by this factor:
 :$$\sigma_y=\sigma_x/\rm 2\hspace{0.15cm}\underline{\approx 0.361}.$$

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID119__Sto_A_3_3.png|right|frame|Cosine&ndash;Rectangular PDF <br>and a similar PDF]]
+[[File:P_ID119__Sto_A_3_3.png|right|frame|Squared cosine PDF <br>and a similar PDF]]
 As in&nbsp; [[Aufgaben:Exercise_3.1:_cos²-PDF_and_PDF_with_Dirac_Functions|Exercise 3.1]]&nbsp; and&nbsp; [[Aufgaben:Exercise_3.2:_cos²-CDF_and_CDF_with_Step_Functions|Exercise 3. 2]]&nbsp; we consider the random variable restricted to the range of values from&nbsp; $-2$&nbsp; to&nbsp; $+2$&nbsp; with the following PDF in this section:
 :$$f_x(x)= {1}/{2}\cdot \cos^2({\pi}/{4}\cdot { x}).$$
@@ Line 11: / Line 11: @@
 *Both density functions are shown in the graph.
-*Outside the ranges&nbsp; $-2 < x < +2$ &nbsp;resp.&nbsp; $0 < x < +2$&nbsp; respectively&nbsp; $f_x(x) = 0$ &nbsp;resp.&nbsp; $f_y(y) = 0$ holds.
+*Outside the ranges&nbsp; $-2 < x < +2$ &nbsp;resp.&nbsp; $0 < x < +2$&nbsp;, the followings holds: &nbsp; $f_x(x) = 0$ &nbsp;resp.&nbsp; $f_y(y) = 0$.
-*Both random magnitudes&ouml;&ahead;can be taken as (normalized) instantaneous values of the associated random signals&nbsp; $x(t)$&nbsp; or&nbsp; $y(t)$&nbsp; respectively.
+*Both random variables can be taken as (normalized) instantaneous values of the associated random signals&nbsp; $x(t)$&nbsp; or&nbsp; $y(t)$&nbsp; respectively.
@@ Line 29: / Line 29: @@
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Welche der folgenden Aussagen treffen bei jeder beliebigen WDF&nbsp; $f_x(x)$&nbsp; zu?
+{Which of the following statements are true for any given PDF&nbsp; $f_x(x)$&nbsp;?
-<br>Verwendet sind folgende Größen: &nbsp; linearer Mittelwert&nbsp; $m_1$,&nbsp; quadratischer Mittelwert&nbsp; $m_2$,&nbsp; Varianz&nbsp; $\sigma^2$.
+<br>Used are the following quantities: &nbsp; linear mean&nbsp; $m_1$,&nbsp; root mean square&nbsp; $m_2$,&nbsp; variance&nbsp; $\sigma^2$.
 |type="[]"}
-- $m_2 = 0$,&nbsp;&nbsp;&nbsp;falls &nbsp; $m_1 \ne 0$.
+- $m_2 = 0$,&nbsp;&nbsp;if &nbsp; $m_1 \ne 0$.
-- $m_2 = 0$,&nbsp;&nbsp;&nbsp;falls &nbsp; $m_1 = 0$.
+- $m_2 = 0$,&nbsp;&nbsp;if &nbsp; $m_1 = 0$.
-+ $m_1 = 0$,&nbsp;&nbsp;&nbsp;falls &nbsp; $m_2 = 0$.
++ $m_1 = 0$,&nbsp;&nbsp;if &nbsp; $m_2 = 0$.
-+ $m_2 > \sigma^2$,&nbsp;&nbsp;&nbsp;falls &nbsp; $m_1 \ne 0$.
++ $m_2 > \sigma^2$,&nbsp;&nbsp;if &nbsp; $m_1 \ne 0$.
-+ $m_1 = 0$,&nbsp;&nbsp;&nbsp;falls &nbsp; $f_x(-x) = f_x(x)$.
++ $m_1 = 0$,&nbsp;&nbsp;&nbsp;if &nbsp; $f_x(-x) = f_x(x)$.
-- $f_x(-x) = f_x(x)$,&nbsp;&nbsp;&nbsp;falls &nbsp; $m_1 = 0$.
+- $f_x(-x) = f_x(x)$,&nbsp;&nbsp;&nbsp;if &nbsp; $m_1 = 0$.
-{Wie gro&szlig; ist der Gleichanteil (lineare Mittelwert) des Signals&nbsp; $x(t)$?
+{How large is the DC component (linear mean) of the signal&nbsp; $x(t)$?
 |type="{}"}
 $m_x \ = \ $ { 0. }
-{Wie gro&szlig; ist der Effektivwert des Signals&nbsp; $x(t)$?
+{What is the rms value (Effektivwert?) of the signal&nbsp; $x(t)$?
 |type="{}"}
-$\sigma_x \ =  \ $ { 0.722 3% }
+$\sigma_x \ = \ $ { 0.722 3% }
-{Die Zufallsgr&ouml;&szlig;e&nbsp; $y$&nbsp; lässt sich aus&nbsp; $x$&nbsp; ableiten. Welche Zuordnung gilt?
+{The random variable&nbsp; $y$&nbsp; can be derived from&nbsp; $x$&nbsp;. Which assignment is valid?
 |type="()"}
 + $y = 1+x/2.$
@@ Line 60: / Line 60: @@
-{Wie gro&szlig; ist der Gleichanteil des Signals&nbsp; $y(t)$?
+{How large is the DC component of the signal $y(t)$?
 |type="{}"}
-$m_y\ =  \ $   { 1 3% }
+$m_y\ = $ { 1 3% }
-{Wie gro&szlig; ist der Effektivwert des Signals&nbsp; $y(t)$?
+{What is the rms value (Effektivwert) of the signal&nbsp; $y(t)$?
 |type="{}"}
-$\sigma_y\ =  \ $ { 0.361 3% }
+$\sigma_y\ = \ $ { 0.361 3% }
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
 '''(1)'''&nbsp; Unter allen Umst&auml;nden richtig sind <u>die Aussagen 3, 4 und 5</u>:
@@ Line 95: / Line 94: @@
 *Mit der Beziehung&nbsp; $\cos^2(\alpha) = 0.5 \cdot \big[1 + \cos(2\alpha)\big]$&nbsp; folgt daraus:
-:$$\sigma_x^2=\int_{\rm 0}^{\rm 2}{x^{\rm 2}}/{\rm 2} \hspace{0.1cm}{\rm d}x  +  \int_{\rm 0}^{\rm 2}{x^{\rm 2}}/{2}\cdot \cos({\pi}/{\rm 2}\cdot\it x) \hspace{0.1cm} {\rm d}x.$$
+:$$\sigma_x^2=\int_{\rm 0}^{\rm 2}{x^{\rm 2}}/{\rm 2} \hspace{0.1cm}{\rm d}x + \int_{\rm 0}^{\rm 2}{x^{\rm 2}}/{2}\cdot \cos({\pi}/{\rm 2}\cdot\it x) \hspace{0.1cm} {\rm d}x.$$
-*Diese beiden Standardintegrale findet man in Tabellen. Man erh&auml;lt mit&nbsp; $a = \pi/2$:
+*These two standard integrals can be found in tables. One obtains with&nbsp; $a = \pi/2$:
-:$$\sigma_x^{\rm 2}=\left[\frac{x^{\rm 3}}{\rm 6}  +  \frac{x}{a^2}\cdot {\cos}(a  x) + \left( \frac{x^{\rm2}}{{\rm2}a} - \frac{1}{a^3} \right) \cdot \sin(a \cdot  x)\right]_{x=0}^{x=2} \hspace{0.5cm}
+:$$\sigma_x^{\rm 2}=\left[\frac{x^{\rm 3}}{\rm 6} + \frac{x}{a^2}\cdot {\cos}(a x) + \left( \frac{x^{\rm2}}{{\rm2}a} - \frac{1}{a^3} \right) \cdot \sin(a \cdot x)\right]_{x=0}^{x=2} \hspace{0.5cm}
-\Rightarrow \hspace{0.5cm} \sigma_{x}^{\rm 2}=\frac{\rm 4}{\rm 3}-\frac{\rm 8}{\rm \pi^2}\approx 0.524\hspace{0.5cm} \Rightarrow \hspace{0.5cm}\sigma_x \hspace{0.15cm}\underline{\approx 0.722}.$$
+\rightarrow \hspace{0.5cm} \sigma_{x}^{\rm 2}=\frac{\rm 4}{\rm 3}-\frac{\rm 8}{\rm \pi^2}\approx 0.524\hspace{0.5cm} \Rightarrow \hspace{0.5cm}\sigma_x \hspace{0.15cm}\underline{\approx 0.722}.$$
-'''(4)'''&nbsp; Richtig ist der <u>erstgenannte Vorschlag</u>:
+'''(4)'''&nbsp; Correct is the <u>first mentioned suggestion</u>:
-* Die Variante&nbsp; $y = 2x$&nbsp; w&uuml;rde eine zwischen&nbsp; $-4$&nbsp; und&nbsp; $+4$&nbsp; verteilte Zufallsgr&ouml;&szlig;e liefern.
+*The variant&nbsp; $y = 2x$&nbsp; would yield a random size distributed between&nbsp; $-4$&nbsp; and&nbsp; $+4$&nbsp; .
-*Beim letzten Vorschlag&nbsp; $y = x/2-1$&nbsp; w&auml;re der Mittelwert&nbsp; $m_y = -1$.
+*In the last proposition&nbsp; $y = x/2-1$&nbsp; the mean&nbsp; $m_y = -1$ would w&auml;re.
-'''(5)'''&nbsp; Aus der Grafik auf dem Angabenblatt ist bereits offensichtlich, dass&nbsp; $m_y \hspace{0.15cm}\underline{=+1}$&nbsp; gelten muss.
+'''(5)'''&nbsp; From the graph on the specification sheet it is already obvious that&nbsp; $m_y \hspace{0.15cm}\underline{=+1}$&nbsp; must hold.
-'''(6)'''&nbsp; Der Mittelwert &auml;ndert nichts an der Varianz und an der Streuung.
+'''(6)'''&nbsp; The mean value &auml;does not change the variance and dispersion.
-*Durch die Stauchung um den Faktor&nbsp; $2$&nbsp; wird die Streuung gegen&uuml;ber Teilaufgabe&nbsp; '''(3)'''&nbsp; ebenfalls um diesen Faktor kleiner:
+*By compressing by the factor&nbsp; $2$&nbsp; the scatter becomes smaller against&uuml;ber Teilaufgabe&nbsp; '''(3)'''&nbsp; also by this factor:
 :$$\sigma_y=\sigma_x/\rm 2\hspace{0.15cm}\underline{\approx 0.361}.$$
 {{ML-Fuß}}