Aufgaben:Exercise 3.3Z: Moments for Triangular PDF - Revision history

Hwang at 12:19, 18 January 2023

2023-01-18T12:19:32Z

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

2022-02-17T11:34:31Z

Text replacement - " rms " to " standard deviation "

Guenter at 14:49, 5 January 2022

2022-01-05T14:49:43Z

Guenter at 14:13, 5 January 2022

2022-01-05T14:13:43Z

Noah at 01:20, 28 December 2021

2021-12-28T01:20:37Z

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Noah at 21:29, 26 December 2021

2021-12-26T21:29:57Z

Guenter at 10:06, 20 December 2021

2021-12-20T10:06:39Z

Guenter: Guenter moved page Aufgaben:Aufgabe 3.3Z: Momente bei Dreieck-WDF to Aufgaben:Exercise 3.3Z: Moments for Triangular PDF

2021-12-20T10:00:00Z

Guenter moved page Aufgabe 3.3Z: Momente bei Dreieck-WDF to Exercise 3.3Z: Moments for Triangular PDF

Javier: Text replacement - "”" to """

2021-05-28T14:31:06Z

Text replacement - "”" to """

Javier: Text replacement - "„" to """

2021-05-28T14:28:47Z

Text replacement - "„" to """

@@ Line 43: / Line 43: @@
-{What is the quadratic mean and the rms&nbsp; $\sigma_x$&nbsp; of the random variable&nbsp; $x$?
+{What is the second moment and the rms&nbsp; $\sigma_x$&nbsp; of the random variable&nbsp; $x$?
 |type="{}"}
 $\sigma_x\ = \ $ { 0.943 3% }

@@ Line 61: / Line 61: @@
-{Calculate the rms of the random variable&nbsp; $y$.
+{Calculate the standard deviation of the random variable&nbsp; $y$.
 |type="{}"}
 $\sigma_y \ = \ $ { 1.633 3% }

@@ Line 87: / Line 87: @@
-'''(2)'''&nbsp; The &nbsp; $(k= 2)$&nbsp; is&nbsp; $m_2 = 8/3$.
+'''(2)'''&nbsp; The&nbsp; $(k= 2)$&nbsp; is&nbsp; $m_2 = 8/3$.
-*From this follows with <i>Steiner's theorem</i>:
+*From this follows with&nbsp; "Steiner's theorem":
 :$$\sigma_x^{\rm 2}={8}/{3}-({4}/{3})^2=\rm {8}/{9}\hspace{0.5cm}\Rightarrow\hspace{0.5cm} \sigma_x\hspace{0.15cm}\underline{\approx \rm 0.943}.$$
-'''(3)'''&nbsp; With &nbsp;$m_1 = 4/3$, &nbsp;$m_2 = 8/3$&nbsp; and &nbsp;$m_3 = 32/5$&nbsp;, the given equation for the third order central moment gives:  &nbsp; $\mu_3 = 64/135 \approx 0.474$.
+'''(3)'''&nbsp; With &nbsp;$m_1 = 4/3$, &nbsp;$m_2 = 8/3$&nbsp; and &nbsp;$m_3 = 32/5$,&nbsp; the given equation for the third order central moment gives:  &nbsp; $\mu_3 = 64/135 \approx 0.474$.
-*From this follows for&nbsp; <i>Charlier's skewness</i>:
+*From this follows for the&nbsp; "Charlier's skewness":
 :$$S_x=\rm \frac{64/135}{\Big(\sqrt {8/9}\Big)^3}=\frac{\sqrt{8}}{5}\hspace{0.15cm}\underline{\approx 0.566}.$$
-'''(4)'''&nbsp; Correct are <u>the proposed solutions 1, 3 and 4</u>:
+'''(4)'''&nbsp; Correct are&nbsp; <u>the proposed solutions 1, 3 and 4</u>:
-*For symmetric PDF, all odd moments are zero, including the mean&nbsp; $m_y$.
+*For symmetric PDF,&nbsp; all odd moments are zero,&nbsp; including the mean&nbsp; $m_y$.
-*Therefore, in terms of randomness&nbsp; $y$&nbsp; there is no difference between the moments&nbsp; $m_k$&nbsp; and the central moments&nbsp; $\mu_k$.
+*Therefore,&nbsp; according&nbsp; $y$:&nbsp; There is no difference between the moments&nbsp; $m_k$&nbsp; and the central moments&nbsp; $\mu_k$.
-*The moments&nbsp; $m_k$&nbsp; with even&nbsp; $k$&nbsp; are the same for the random variables&nbsp; $x$&nbsp; and&nbsp; $y$&nbsp; . This is evident from the time averages.
+*The moments&nbsp; $m_k$&nbsp; with even&nbsp; $k$&nbsp; are the same for the random variables&nbsp; $x$&nbsp; and&nbsp; $y$.&nbsp; This is evident from the time averages:
-*Since&nbsp; $x^2(t) = y^2(t)$, for&nbsp; $k = 2n$&nbsp; the moments are also equal:
+*Since&nbsp; $x^2(t) = y^2(t)$,&nbsp; for&nbsp; $k = 2n$&nbsp; the moments are equal too:
 :$$m_k=m_{2 n}=\ \text{...}\int [x^2(t)]^n \hspace{0.1cm}{\rm d} x=\ \text{...}\int [y^2(t)]^n \hspace{0.1cm}{\rm d} y.$$
@@ Line 114: / Line 114: @@
-'''(6)'''&nbsp; The fourth-order central moment is equal to the moment&nbsp; $m_4$ for symmetric WDF.
+'''(6)'''&nbsp;  For symmetrical PDF,&nbsp; the fourth-order central moment is equal to the moment&nbsp; $m_4$.
 *From the general equation calculated in subtask&nbsp; '''(1)'''&nbsp; one obtains&nbsp; $\mu_4 = 256/15.$
 *From this follows for the kurtosis:
 :$$K_y=\frac{\mu_{\rm 4}}{\sigma_y^{\rm 4}}=\rm \frac{256/15}{(8/3)^2}\hspace{0.15cm}\underline{=2.4}.$$
-::''Note:'' &nbsp; This numerical value is valid for the triangle WDF in general and lies between the kurtosis values of uniform distribution&nbsp; $(K = 1.8)$&nbsp; and Gaussian distribution&nbsp; $(K = 3)$. This is a quantitative evaluation of the fact that here
+::<u>Note:</u> &nbsp; This numerical value is valid for the triangle PDF in general and lies between the kurtosis values of the uniform distribution&nbsp; $(K = 1.8)$&nbsp; and the Gaussian distribution&nbsp; $(K = 3)$. This is a quantitative evaluation of the fact that here
 ::*the outliers are more pronounced than in the case of a uniformly distributed random size,
 ::*but due to the limitation less pronounced than with Gaussian sizes.
-Then we will prove that the asymmetric triangular WDF $f_x(x)$ also has the same kurtosis as shown in the upper sketch on the data sheet:
+*Then we will prove that the asymmetric triangular PDF&nbsp; $f_x(x)$&nbsp; has the same kurtosis as shown in the upper sketch on the data sheet:
 : $$\mu_{ 4} = m_{\rm 4}- 4\cdot m_{\rm 3}\cdot m_{\rm 1}+ 6\cdot m_{\rm 2}\cdot m_{\rm 1}^{\rm 2}- 3\cdot m_{\rm 1}^{\rm 4}= \frac{256}{15} - 4 \cdot \frac{32}{5}\cdot \frac{4}{3} + 6 \cdot \frac{8}{3}\cdot \left(\frac{4}{3}\right)^2 -3 \cdot \left(\frac{4}{3}\right)^4 =\frac{256}{15 \cdot 9}$$

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID142__Sto_Z_3_3.png|right|frame|Two triangular PDF]]
+[[File:P_ID142__Sto_Z_3_3.png|right|frame|Two triangular PDFs]]
-We consider in this task two random signals&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; each with triangular PDF, namely
+We consider in this exercise two random signals&nbsp; $x(t)$&nbsp; and&nbsp; $y(t)$&nbsp; each with triangular PDF,&nbsp; namely
 * the one-sided triangular PDF according to the upper graph:
-:$$f_x(x)=\left\{ \begin{array}{*{4}{c}} 0.5 \cdot (1-{ x}/{\rm 4}) & \rm f\ddot{u}r\hspace{0.2cm}{\rm 0 \le {\it x} \le 4},\rm 0 & \rm else. \end{array} \right.$$
+:$$f_x(x)=\left\{ \begin{array}{*{4}{c}}  0.5 \cdot (1-{ x}/{\rm 4}) & \rm for\hspace{0.2cm}{\rm 0 \le {\it x} \le  4},\\\rm 0 & \rm else. \end{array} \right.$$
 * the two-sided triangular PDF according to the graph below:
-:$$ f_y(y)=\left\{ \begin{array}{*{4}{c}} 0.25 \cdot (1-{ |y|}/{\rm 4}) & \rm f\ddot{u}r\hspace{0.2cm}{ -4 \le {\it y}  \le \rm 4},\rm 0 & \rm else. \end{array} \right.$$
+:$$ f_y(y)=\left\{ \begin{array}{*{4}{c}} 0.25 \cdot (1-{ |y|}/{\rm 4}) & \rm for\hspace{0.2cm}{ -4 \le {\it y}  \le \rm 4},\\\rm 0 & \rm else. \end{array} \right.$$
-To solve this problem, consider the equation for the central moments:
+To solve this problem,&nbsp; consider the equation for the central moments:
 :$$\mu_k=\sum\limits_{\kappa = \rm 0}^{\it k}\left({k} \atop {\kappa}\right)\cdot m_k\cdot(-m_{\rm 1})^{k - \kappa}.$$
-Specifically, this equation yields the following results:
+Specifically,&nbsp; this equation yields the following results:
 :$$\mu_{\rm 1}=0,\hspace{0.5cm}\mu_{\rm 2}=\it m_{\rm 2}-\it m_{\rm 1}^{\rm 2},\hspace{0.5cm}\mu_{\rm 3}=\it m_{\rm 3}-\rm 3\cdot\it m_{\rm 2}\cdot \it m_{\rm 1} {\rm +}\rm 2\cdot\it m_{\rm 1}^{\rm 3},$$
 :$$\mu_{\rm 4}=\it m_{\rm 4}-\rm 4\cdot\it m_{\rm 3}\cdot \it m_{\rm 1}\rm +6\cdot\it m_{\rm 2}\cdot\it m_{\rm 1}^{\rm 2}-\rm 3\cdot\it m_{\rm 1}^{\rm 4}.$$
-From the central moments of higher order one can derive, among others:
+From the central moments of higher order one can derive among others:
@@ Line 32: / Line 29: @@
 Hints:
-*This exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Erwartungswerte_und_Momente|Erwartungswerte und Momente]].
+*This exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments|Expected Values and Moments]].
-*Reference is made to the page&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments#Some_common_central_moments|Some common central moments]].
+*Reference is made to the section&nbsp; [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments#Some_common_central_moments|Some common central moments]].
@@ Line 41: / Line 38: @@
 <quiz display=simple>
-{Calculate from the present PDF&nbsp; $f_x(x)$&nbsp; the moment&nbsp; $k$-th order. <br>What value results for the linear mean&nbsp; $m_x = m_1$?
+{Calculate from the present PDF&nbsp; $f_x(x)$&nbsp; the&nbsp; $k$-th order moment.&nbsp; What value results for the linear mean&nbsp; $m_x = m_1$?
 |type="{}"}
 $m_x \ = \ $ { 1.333 3% }
@@ Line 51: / Line 48: @@
-{For random variable&nbsp; $x$&nbsp; how large is Charlier's skewness&nbsp; $S_x = \mu_3/\sigma_x^3$? &nbsp; Why is&nbsp; $S_x \ne 0$?
+{For random variable&nbsp; $x$:&nbsp; What is the Charlier's skewness&nbsp; $S_x = \mu_3/\sigma_x^3$?&nbsp; Why is&nbsp; $S_x \ne 0$?
 |type="{}"}
 $S_x \ = \ $ { 0.566 3% }
-{Which statements are true for the symmetrically distributed random variable&nbsp; $y$&nbsp;?
+{Which statements are true for the symmetrically distributed random variable&nbsp; $y$?
 |type="[]"}
 + All moments with odd&nbsp; $k$&nbsp; are&nbsp; $m_k =0$.
 - All moments with even&nbsp; $k$&nbsp; are&nbsp; $m_k =0$.
-+ All moments&nbsp; $m_k$&nbsp; with even&nbsp; $k$&nbsp; are calculated as in subtask&nbsp; '''(1)'''&nbsp;.
++ All moments&nbsp; $m_k$&nbsp; with even&nbsp; $k$&nbsp; are calculated as in subtask&nbsp; '''(1)'''.
 + The central moments &nbsp; $\mu_k$&nbsp; are equal to the non-centered moments&nbsp; $m_k$.
@@ Line 69: / Line 66: @@
-{What value is obtained for the kurtosis&nbsp; $K_y$&nbsp; of the random variable&nbsp; $y$? &nbsp; Interpret the result.
+{What is the kurtosis&nbsp; $K_y$&nbsp; of the random variable&nbsp; $y$?&nbsp; Interpret the result.
 |type="{}"}
 $K_y \ = \ $ { 2.4 3% }

← Older revision		Revision as of 21:29, 26 December 2021
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−	{{quiz-Header\|Buchseite=~~Stochastische Signaltheorie~~/~~Erwartungswerte und Momente~~	+	{{quiz-Header\|Buchseite=Theory_of_Stochastic_Signals/Expected_Values_and_Moments
	}}		}}

← Older revision		Revision as of 10:06, 20 December 2021
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−	[[Category:Theory of Stochastic Signals: Exercises\|^3.3 ~~Erwartungswerte und Momente~~^]]	+	[[Category:Theory of Stochastic Signals: Exercises\|^3.3 Expected Values and Moments^]]

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 Aus den Zentralmomenten höherer Ordnung kann man unter anderem ableiten:
-*die&nbsp; "Charliersche Schiefe</i>&rdquo;&nbsp; $S =  {\mu_3}/{\sigma^3}\hspace{0.05cm},$
+*die&nbsp; "Charliersche Schiefe</i>"&nbsp; $S =  {\mu_3}/{\sigma^3}\hspace{0.05cm},$
-*die&nbsp; "Kurtosis</i>&rdquo;&nbsp; $K =  {\mu_4}/{\sigma^4}\hspace{0.05cm}.$
+*die&nbsp; "Kurtosis</i>"&nbsp; $K =  {\mu_4}/{\sigma^4}\hspace{0.05cm}.$