Aufgaben:Exercise 4.1Z: Calculation of Moments - Revision history

Hwang at 12:19, 18 January 2023

2023-01-18T12:19:45Z

Hwang at 05:07, 18 September 2022

2022-09-18T05:07:41Z

Guenter at 15:26, 17 February 2022

2022-02-17T15:26:53Z

Javier: Text replacement - "value-continuous" to "continuous"

2021-10-11T08:27:14Z

Text replacement - "value-continuous" to "continuous"

Javier: Text replacement - "continuous-valued" to "continuous"

2021-10-11T08:20:00Z

Text replacement - "continuous-valued" to "continuous"

Noah at 10:37, 27 September 2021

2021-09-27T10:37:59Z

Guenter at 15:58, 24 September 2021

2021-09-24T15:58:01Z

Guenter at 15:46, 24 September 2021

2021-09-24T15:46:57Z

Noah at 07:45, 24 September 2021

2021-09-24T07:45:37Z

Noah at 07:40, 24 September 2021

2021-09-24T07:40:31Z

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@@ Line 114: / Line 114: @@
 [[File:EN_Inf_A_4_3.png|right|frame|To illustrate the sample solution to problem&nbsp; '''(5)''']]
 '''(5)'''&nbsp; Only the <u>proposed solution 2</u> is correct:
-*The quadratic mean of&nbsp; "Laplace"&nbsp; is the same as for the exponential distribution because of the symmetric PDF:
+*The second moment of&nbsp; "Laplace"&nbsp; is the same as for the exponential distribution because of the symmetric PDF:
 :$$m_2 = \frac{\lambda}{2} \cdot \int_{-\infty}^{\infty} \hspace{-0.01cm} y^2 \cdot  {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}|y|}\hspace{0.1cm}{\rm d}y = \lambda \cdot\int_{0}^{\infty} \hspace{-0.01cm} y^2 \cdot  {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}y}\hspace{0.1cm}{\rm d}y = {2}/{\lambda^2} \hspace{0.05cm}.$$

@@ Line 67: / Line 67: @@
 - The variance is&nbsp; $\sigma^2 = 1/\lambda^2$.
-{With what probabilities does the random variable&nbsp; $(X$&nbsp; or &nbsp; $Y)$&nbsp; differ from the respective mean in amount by more than the dispersion &nbsp;$\sigma$?
+{With what probabilities does the random variable&nbsp; $(X$&nbsp; or &nbsp; $Y)$&nbsp; differ from the respective mean in magnitude by more than the dispersion &nbsp;$\sigma$?
 |type="{}"}
 $\text{Exponential:}\; \;{\rm Pr}( |X   -   m_X| > \sigma_X) \ = \ $  { 0.135 3% }

@@ Line 58: / Line 58: @@
 |type="[]"}
 + The linear mean is&nbsp; $m_1 = 1/\lambda$.
-+ The root mean square is&nbsp; $m_2 = 2/\lambda^2$.
++ The second order moment is&nbsp; $m_2 = 2/\lambda^2$.
 + The variance is&nbsp; $\sigma^2 = 1/\lambda^2$.
@@ Line 64: / Line 64: @@
 |type="[]"}
 - The linear mean is&nbsp; $m_1 = 1/\lambda$.
-+ The root mean square is&nbsp; $m_2 = 2/\lambda^2$.
++ The second order moment is&nbsp; $m_2 = 2/\lambda^2$.
 - The variance is&nbsp; $\sigma^2 = 1/\lambda^2$.
@@ Line 102: / Line 102: @@
 * the linear mean (first order moment):
 :$$m_1 = \lambda \cdot\int_{0}^{\infty} \hspace{-0.01cm} x \cdot  {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x = \lambda \cdot \left [\frac{{\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}}{(-\lambda)^2}\cdot(-\lambda \cdot x-1)\right ]_{0}^{\infty}= {1}/{\lambda} \hspace{0.05cm},$$
-* the root mean square (second order moment):
+* the second order moment:
 :$$m_2 = \lambda \cdot\int_{0}^{\infty} \hspace{-0.01cm} x^2 \cdot  {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x = \lambda \cdot\left [ {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\cdot
 (\frac{x^2}{-\lambda} - \frac{2x}{\lambda^2} + \frac{2}{\lambda^3})

@@ Line 9: / Line 9: @@
 :$$f_Y(y) =  A_{ Y} \cdot {\rm e}^{-\lambda \hspace{0.05cm} \cdot \hspace{0.05cm} |\hspace{0.03cm}y\hspace{0.03cm}|}\hspace{0.05cm}.$$
-The two value-continuous random variables&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; are to be compared with respect to the following characteristics:
+The two continuous random variables&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; are to be compared with respect to the following characteristics:
 *The linear mean&nbsp; $m_1$&nbsp; (first order moment),
 *the second order moment &nbsp; &#8658; &nbsp; $m_2$,

@@ Line 24: / Line 24: @@
 Hints:
 *The task belongs to the chapter&nbsp; [[Information_Theory/Differentielle_Entropie|Differential Entropy]].
-*Useful hints for solving this task and further information on continuous-valued random variables can be found in the third chapter  "Continuous Random Variables" of the book&nbsp;  [[Theory of Stochastic Signals]].
+*Useful hints for solving this task and further information on continuous random variables can be found in the third chapter  "Continuous Random Variables" of the book&nbsp;  [[Theory of Stochastic Signals]].
 *Also given are the two indefinite integrals:

← Older revision		Revision as of 10:37, 27 September 2021
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−	{{quiz-Header\|Buchseite=~~Informationstheorie~~/~~Differentielle Entropie~~	+	{{quiz-Header\|Buchseite=Information_Theory/Differential_Entropy
	}}		}}

@@ Line 81: / Line 81: @@
 :$$A_{X} \cdot\int_{0}^{\infty} \hspace{-0.01cm}  {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\hspace{0.1cm}{\rm d}x = A_{X} \cdot (-1/\lambda)\cdot\big [{\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}x}\big ]_{0}^{\infty} = A_{X} \cdot (1/\lambda) \stackrel{!}{=} 1
   \hspace{0.3cm} \Rightarrow\hspace{0.3cm} A_{X} = \lambda \hspace{0.05cm}. $$
 '''(2)'''&nbsp; <u>Proposed solution 1</u> is correct:
-*From the graph on the information page, we can see that the heighz&nbsp; $A_Y$&nbsp; of the Laplace distribution is only half as large as the maximum of the exponential distribution:
+*From the graph on the information page, we can see that the height&nbsp; $A_Y$&nbsp; of the Laplace PDF is only half as large as the maximum of the exponential PDF:
 :$$A_Y = \lambda/2.$$
+'''(3)'''&nbsp; Correct is <u>YES</u>,&nbsp; although for&nbsp; $z \ne 0$&nbsp; always &nbsp;$f_X(z) = f_Y(z)$. &nbsp;  Let us now consider the special case&nbsp; $z= 0$:
-'''(3)'''&nbsp; Correct is <u>YES</u>, although for&nbsp; $z \ne 0$&nbsp; always &nbsp;$f_X(z) = f_Y(z)$&nbsp;.  Let us now consider the special case&nbsp; $z= 0$:
+* For the Laplace PDF:&nbsp; $f_Y(y = 0) = \lambda/2$.
-* For the Laplace distribution,&nbsp; $f_Y(y = 0) = \lambda/2$.
+* For the exponential PDF,&nbsp; the left-hand and right-hand limits differ for&nbsp; $x \to 0$.
-* For the exponential distribution, the left-hand and right-hand limits differ for&nbsp; $x \to 0$.
+*The PDF value at point&nbsp; $x= 0$&nbsp; is the average of these two limits:
 :$$f_X(0) = \frac{1}{2} \cdot \big [ 0 + \lambda \big] = \lambda/2 =  f_Y(0)\hspace{0.05cm}.$$
 '''(4)'''&nbsp; <u>All proposed solutions</u> are correct.&nbsp;
-For the exponential distribution, the&nbsp; $k$th order moment is generally calculated to be
+For the exponential distribution, the&nbsp; $k$&ndash;th order moment is generally calculated to be
 :$$m_k = \frac{k!}{\lambda^k} \hspace{0.3cm}\Rightarrow\hspace{0.3cm} m_1 = \frac{1}{\lambda}, \hspace{0.3cm} m_2 = \frac{2}{\lambda^2}, \hspace{0.3cm} m_3 = \frac{6}{\lambda^3}, \ \text{...}$$
 Thus one obtains for
@@ Line 115: / Line 112: @@
+[[File:EN_Inf_A_4_3.png|right|frame|To illustrate the sample solution to problem&nbsp; '''(5)''']]
 '''(5)'''&nbsp; Only the <u>proposed solution 2</u> is correct:
-*The quadratic mean of the Laplace distribution is exactly the same as for the exponential distribution because of the symmetric PDF:
+*The quadratic mean of&nbsp; "Laplace"&nbsp; is the same as for the exponential distribution because of the symmetric PDF:
 :$$m_2 = \frac{\lambda}{2} \cdot \int_{-\infty}^{\infty} \hspace{-0.01cm} y^2 \cdot  {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}|y|}\hspace{0.1cm}{\rm d}y = \lambda \cdot\int_{0}^{\infty} \hspace{-0.01cm} y^2 \cdot  {\rm e}^{-\lambda \hspace{0.05cm}\cdot \hspace{0.05cm}y}\hspace{0.1cm}{\rm d}y = {2}/{\lambda^2} \hspace{0.05cm}.$$
@@ Line 126: / Line 122: @@
 \hspace{0.3cm} \Rightarrow\hspace{0.3cm}
 \sigma = {\sqrt{2}}/{\lambda}\hspace{0.05cm}.$$
@@ Line 137: / Line 131: @@
 \right ]_{2/\lambda}^{\infty}
    =  {\rm e}^{-2} \hspace{0.15cm}\underline {\approx 0.135}.$$
-For the Laplace distribution (lower graph), with &nbsp;$m_Y = 0$&nbsp; and &nbsp;$\sigma_Y = \sqrt{2}/\lambda$ we obtain::
+For the Laplace distribution (lower graph),&nbsp; with &nbsp;$m_Y = 0$&nbsp; and &nbsp;$\sigma_Y = \sqrt{2}/\lambda$ we obtain::
 :$${\rm Pr}( |Y   -   m_Y| > \sigma_Y) =
 \cdot {\rm Pr}( Y > \sqrt{2}/\lambda)
@@ Line 146: / Line 140: @@
   = -  {\rm e}^{-\sqrt{2}} \hspace{0.15cm}\underline {\approx 0.243}\hspace{0.05cm}.$$
-A comparison of the shaded areas in the accompanying graph qualitatively confirms the result: <br> &nbsp; &nbsp;  The blue areas together are slightly larger than the red area.
+A comparison of the shaded areas in the accompanying graph qualitatively confirms the result: <br> &rArr; &nbsp;  The blue areas together are slightly larger than the red area.

@@ Line 3: / Line 3: @@
 }}
-[[File:EN_Inf_Z_4_1_vers2.png|right|frame|Exponential distribution (top) and Laplace distribution (bottom)]]
+[[File:EN_Inf_Z_4_1_vers2.png|right|frame|Exponential PDF (top), <br>Laplace PDF (bottom)]]
-The graph above shows the probability density function (PDF) of the&nbsp; [[Theory_of_Stochastic_Signals/Exponentialverteilte_Zufallsgrößen|exponential distribution]]:
+The upper graph shows the probability density function&nbsp; $\rm (PDF)$&nbsp; of the&nbsp; [[Theory_of_Stochastic_Signals/Exponentialverteilte_Zufallsgrößen|exponential distribution]]:
 :$$f_X(x) = \left\{ \begin{array}{c} A_{ X} \cdot {\rm e}^{-\lambda \hspace{0.05cm} \cdot \hspace{0.05cm}x} \\ A_{ X}/2 \\ 0 \\  \end{array} \right. \begin{array}{*{20}c}   {\rm{f\ddot{u}r}} \hspace{0.1cm}x>0, \\ {\rm{f\ddot{u}r}} \hspace{0.1cm}x=0,  \\    {\rm{f\ddot{u}r}} \hspace{0.1cm}x<0. \\ \end{array}$$
-Drawn below is the WDF of the&nbsp; [[Theory_of_Stochastic_Signals/Exponentialverteilte_Zufallsgrößen#Zweiseitige_Exponentialverteilung_.E2.80.93_Laplaceverteilung|Laplace distribution]], which can be specified for all&nbsp; $y$&ndash;values as follows:
+Drawn below is the PDF of the&nbsp; [[Theory_of_Stochastic_Signals/Exponentialverteilte_Zufallsgrößen#Zweiseitige_Exponentialverteilung_.E2.80.93_Laplaceverteilung|Laplace distribution]], which can be specified for all&nbsp; $y$&ndash;values as follows:
 :$$f_Y(y) =  A_{ Y} \cdot {\rm e}^{-\lambda \hspace{0.05cm} \cdot \hspace{0.05cm} |\hspace{0.03cm}y\hspace{0.03cm}|}\hspace{0.05cm}.$$
-The two continuous value random variables&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; are to be compared with respect to the following characteristics:
+The two value-continuous random variables&nbsp; $X$&nbsp; and&nbsp; $Y$&nbsp; are to be compared with respect to the following characteristics:
-*the linear mean&nbsp; $m_1$&nbsp; (first order moment),
+*The linear mean&nbsp; $m_1$&nbsp; (first order moment),
 *the second order moment &nbsp; &#8658; &nbsp; $m_2$,
 *the variance&nbsp; $\sigma^2 = m_2 - m_1^2$ &nbsp; &#8658; &nbsp; Steiner's theorem,
-*the dispersion&nbsp; $\sigma$.
+*the standard deviation&nbsp; $\sigma$.