Aufgaben:Exercise 4.2: Triangular PDF - Revision history

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

2021-10-11T08:27:01Z

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

Noah at 10:38, 27 September 2021

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Guenter at 09:47, 25 September 2021

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Guenter at 09:45, 25 September 2021

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Guenter at 09:36, 25 September 2021

2021-09-25T09:36:21Z

Noah at 07:54, 24 September 2021

2021-09-24T07:54:20Z

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Guenter at 09:15, 16 August 2021

2021-08-16T09:15:46Z

Guenter: Guenter moved page Aufgaben:Aufgabe 4.2: Dreieckförmige WDF to Aufgaben:Exercise 4.2: Triangular PDF

2021-08-16T09:02:30Z

Guenter moved page Aufgabe 4.2: Dreieckförmige WDF to Exercise 4.2: Triangular PDF

Javier: Text replacement - "„" to """

2021-05-28T14:36:57Z

Text replacement - "„" to """

Javier: Text replacement - "”" to """

2021-05-28T14:34:50Z

Text replacement - "”" to """

@@ Line 34: / Line 34: @@
 Hints:
 *The task belongs to the chapter&nbsp; [[Information_Theory/Differentielle_Entropie|Differential Entropy]].
-*Useful hints for solving this task and further information on value-continuous 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]].
 *Given the following indefinite integral:

← Older revision		Revision as of 10:38, 27 September 2021
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−	{{quiz-Header\|Buchseite=~~Informationstheorie~~/~~Differentielle Entropie~~	+	{{quiz-Header\|Buchseite=Information_Theory/Differential_Entropy
	}}		}}

@@ Line 118: / Line 118: @@
 *The first integral for the range&nbsp; $-1 \le y \le 0$&nbsp; is identical in form to that of subtask&nbsp; $(1)$&nbsp; and only shifted with respect to it, which does not affect the result.
-*Now the height&nbsp; $C = 1$&nbsp; instead of&nbsp; $C = 2$&nbsp; has to be considered::
+*Now the height&nbsp; $C = 1$&nbsp; instead of&nbsp; $C = 2$&nbsp; has to be considered:
 :$$I_{\rm neg} =- C/2 \cdot
 \big [ {\rm ln} \hspace{0.1cm} (C) - 1/2
@@ Line 126: / Line 126: @@
 \hspace{0.05cm}.$$
-*The second integrand is identical to the first except for a shift and reflection.&nbsp; Moreover, the integration intervals do not overla &nbsp; &#8658; &nbsp; $I_{\rm pos} = I_{\rm neg}$:
+*The second integrand is identical to the first except for a shift and reflection.&nbsp; Moreover, the integration intervals do not overlap &nbsp; &#8658; &nbsp; $I_{\rm pos} = I_{\rm neg}$:
 :$$h_{\rm nat}(Y)  = 2 \cdot I_{\rm neg} =  1/2 \cdot
 {\rm ln} \hspace{0.1cm} ({\rm e}) = {\rm ln} \hspace{0.1cm} (\sqrt{\rm e}) \hspace{0.3cm}
@@ Line 137: / Line 137: @@
 '''(4)'''&nbsp; For the differential entropy of the random variable&nbsp; $Z = A \cdot Y$&nbsp; holds in general:
 :$$h(Z)  = h(A \cdot Y)  = h(Y) + {\rm log}_2 \hspace{0.1cm} (A) \hspace{0.05cm}.$$
-*Thus, from the requiremen&nbsp; $h(Z) = 1 \ \rm bit$&nbsp; and the result of subta&nbsp; $(3)$&nbsp; follows:
+*Thus, from the requiremen&nbsp; $h(Z) = 1 \ \rm bit$&nbsp; and the result of subtask&nbsp; $(3)$&nbsp; follows:
 :$${\rm log}_2 \hspace{0.1cm} (A) = 1\,{\rm bit} - 0.721 \,{\rm bit} = 0.279 \,{\rm bit}
 \hspace{0.3cm} \Rightarrow\hspace{0.3cm} A = 2^{0.279}\hspace{0.15cm}\underline

@@ Line 16: / Line 16: @@
 :$$h(X) =
 \hspace{0.1cm} - \hspace{-0.45cm} \int\limits_{{\rm supp}\hspace{0.03cm}(\hspace{-0.03cm}f_X)} \hspace{-0.35cm}  f_X(x) \cdot {\rm log} \hspace{0.1cm} \big [ f_X(x) \big ] \hspace{0.1cm}{\rm d}x
-\hspace{0.6cm}{\rm mit}\hspace{0.6cm} {\rm supp}(f_X) = \{ x\text{:} \  f_X(x) > 0 \}
+\hspace{0.6cm}{\rm with}\hspace{0.6cm} {\rm supp}(f_X) = \{ x\text{:} \  f_X(x) > 0 \}
 \hspace{0.05cm}.$$
 *If the&nbsp;  "natural logarithm",&nbsp; the pseudo-unit&nbsp;  "nat"&nbsp;  must be added.
@@ Line 70: / Line 70: @@
 :$$f_X(x) = 2x = C \cdot x
 \hspace{0.05cm}.$$
-*Here we have replaced "2" by&nbsp; $C$&nbsp; &nbsp; &#8658; &nbsp; generalization in order to be able to use the following calculation again in subtask&nbsp; $(3)$&nbsp;.
+*Here we have replaced&nbsp; "2"&nbsp; by&nbsp; $C$&nbsp; &nbsp; &#8658; &nbsp; generalization in order to be able to use the following calculation again in subtask&nbsp; $(3)$&nbsp;.
-*Since the differential entropy is sought in "nat", we use the natural logarithm.&nbsp; With the substitution&nbsp; $\xi = C \cdot x$&nbsp; we obtain:
+*Since the differential entropy is sought in&nbsp; "nat",&nbsp; we use the natural logarithm.&nbsp; With the substitution&nbsp; $\xi = C \cdot x$&nbsp; we obtain:
 :$$h_{\rm nat}(X) = \hspace{0.1cm} - \int_{0}^{1} \hspace{0.1cm}   C \cdot x \cdot {\rm ln} \hspace{0.1cm} \big[ C \cdot x \big] \hspace{0.1cm}{\rm d}x =
 \hspace{0.1cm} - \hspace{0.1cm}\frac{1}{C} \cdot \int_{0}^{C} \hspace{0.1cm}   \xi \cdot {\rm ln} \hspace{0.1cm} [ \xi ] \hspace{0.1cm}{\rm d}\xi
@@ Line 79: / Line 79: @@
 \frac{1}{4}\right ]_{\xi = 0}^{\xi = C}
 \hspace{0.05cm}$$
-*Here the indefinite integral given in the front was used.&nbsp; After inserting the limits, considering&nbsp; $C=2$, we obtain::
+*Here the indefinite integral given in the front was used.&nbsp; After inserting the limits, considering&nbsp; $C=2$,&nbsp; we obtain::
 :$$h_{\rm nat}(X) =
 - C/2 \cdot
@@ Line 96: / Line 96: @@
 '''(2)'''&nbsp; In general:
 :$$h_{\rm bit}(X) = \frac{h_{\rm nat}(X)}{{\rm ln} \hspace{0.1cm} (2)\,{\rm nat/bit}} = - 0.279
 \hspace{0.3cm} \Rightarrow\hspace{0.3cm}
@@ Line 101: / Line 103: @@
 \hspace{0.15cm}\underline {= - 0.279\,{\rm bit}}
 \hspace{0.05cm}.$$
-*You can save this conversion if you directly replace&nbsp; $(1)$&nbsp; direkt "ln" by "log<sub>2</sub>" already in the analytical result of subtask:
+*You can save this conversion if you directly replace&nbsp; $(1)$&nbsp; direct&nbsp; "ln"&nbsp; by&nbsp; "log<sub>2</sub>"&nbsp; already in the analytical result of subtask:
 :$$h(X) = \  {\rm log}_2 \hspace{0.1cm} (\sqrt{\rm e}/2)\hspace{0.05cm}, \hspace{1.3cm}
 {\rm pseudo-unit\hspace{-0.1cm}:\hspace{0.15cm} bit}
@@ Line 108: / Line 111: @@
 '''(3)'''&nbsp; We again use the natural logarithm and divide the integral into two partial integrals:
 :$$h(Y) =
@@ Line 116: / Line 118: @@
 *The first integral for the range&nbsp; $-1 \le y \le 0$&nbsp; is identical in form to that of subtask&nbsp; $(1)$&nbsp; and only shifted with respect to it, which does not affect the result.
-*Now the heighte&nbsp; $C = 1$&nbsp; instead of&nbsp; $C = 2$ has to be considered::
+*Now the height&nbsp; $C = 1$&nbsp; instead of&nbsp; $C = 2$&nbsp; has to be considered::
 :$$I_{\rm neg} =- C/2 \cdot
 \big [ {\rm ln} \hspace{0.1cm} (C) - 1/2

@@ Line 2: / Line 2: @@
 }}
-[[File:P_ID2865__Inf_A_4_2.png|right|frame|Two triangular PDF]]
+[[File:P_ID2865__Inf_A_4_2.png|right|frame|Two triangular PDFs]]
-Two probability density functions (PDF) with a triangular shape are considered.
+Two probability density functions&nbsp; $\rm (PDF)$&nbsp; with triangular shapes are considered.
-* The random variable&nbsp;  $X$&nbsp;  is limited to the range of values from&nbsp;  $0$&nbsp;  to&nbsp;  $1$&nbsp;,&nbsp;  and it holds for the PDF (upper sketch):
+* The random variable&nbsp;  $X$&nbsp;  is limited to the range from&nbsp;  $0$&nbsp;  to&nbsp;  $1$&nbsp;,&nbsp;  and it holds for the PDF (upper sketch):
 :$$f_X(x) = \left\{ \begin{array}{c} 2x \\  0 \\  \end{array} \right. \begin{array}{*{20}c}   {\rm{f\ddot{u}r}} \hspace{0.1cm} 0 \le x \le 1 \\     {\rm else} \\ \end{array}
 \hspace{0.05cm}.$$
@@ Line 13: / Line 13: @@
 For both random variables, the&nbsp;  [[Information_Theory/Differentielle_Entropie|differential  entropy]]&nbsp; is to be determined in each case.
-For example, the corresponding equation for the random variable&nbsp;  $X$ is:
+For example, the corresponding equation for the random variable&nbsp;  $X$&nbsp; is:
 :$$h(X) =
 \hspace{0.1cm} - \hspace{-0.45cm} \int\limits_{{\rm supp}\hspace{0.03cm}(\hspace{-0.03cm}f_X)} \hspace{-0.35cm}  f_X(x) \cdot {\rm log} \hspace{0.1cm} \big [ f_X(x) \big ] \hspace{0.1cm}{\rm d}x
 \hspace{0.6cm}{\rm mit}\hspace{0.6cm} {\rm supp}(f_X) = \{ x\text{:} \  f_X(x) > 0 \}
 \hspace{0.05cm}.$$
-*If the&nbsp;  ''natural logarithm'', the pseudo-unit&nbsp;  "nat"&nbsp;  must be added.
+*If the&nbsp;  "natural logarithm",&nbsp; the pseudo-unit&nbsp;  "nat"&nbsp;  must be added.
-*If, on the other hand, the result is asked in&nbsp;  "bit"&nbsp; then the <i>dual logarithm</i> &nbsp; &#8658; &nbsp; "$\log_2$"&nbsp; is to be used.
+*If, on the other hand, the result is asked in&nbsp;  "bit"&nbsp; then the&nbsp;  "dual logarithm" &nbsp; &#8658; &nbsp; "$\log_2$"&nbsp; is to be used.
-In the fourth subtask, the new random variable &nbsp;$Z = A \cdot Y$&nbsp; is considered. Here, the PDF parameter&nbsp; $A$&nbsp; is to be determined in such a way that the differential entropy of the new random variable&nbsp; $Z$&nbsp; yields exactly&nbsp; $1$ bit :<br>
+In the fourth subtask, the new random variable &nbsp;$Z = A \cdot Y$&nbsp; is considered. Here,&nbsp; the PDF parameter&nbsp; $A$&nbsp; is to be determined in such a way that the differential entropy of the new random variable&nbsp; $Z$&nbsp; yields exactly&nbsp; $1$&nbsp; bit :<br>
-:$$h(Z) = h (A \cdot Y) =  h (Y)  + {\rm log}_2 \hspace{0.1cm} (A) = 1\,{\rm bit} \hspace{0.05cm}.$$
+:$$h(Z) = h (A \cdot Y) =  h (Y)  + {\rm log}_2 \hspace{0.1cm} (A) = 1\ {\rm bit} \hspace{0.05cm}.$$
@@ Line 34: / Line 34: @@
 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 value-continuous random variables can be found in the third chapter "Continuous Random Variables" of the book&nbsp;  [[Theory of Stochastic Signals]].
 *Given the following indefinite integral:

← Older revision		Revision as of 09:15, 16 August 2021
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−	[[Category:Information Theory: Exercises\|^4.1 ~~Differentielle Entropie~~^]]	+	[[Category:Information Theory: Exercises\|^4.1 Differential Entropy^]]

← Older revision	Revision as of 09:02, 16 August 2021
(No difference)

@@ Line 18: / Line 18: @@
 \hspace{0.6cm}{\rm mit}\hspace{0.6cm} {\rm supp}(f_X) = \{ x\text{:} \  f_X(x) > 0 \}
 \hspace{0.05cm}.$$
-*Verwendet man den&nbsp;  ''natürlichen Logarithmus'', so ist die Pseudo&ndash;Einheit&nbsp;  &bdquo;nat"&nbsp;  anzufügen.
+*Verwendet man den&nbsp;  ''natürlichen Logarithmus'', so ist die Pseudo&ndash;Einheit&nbsp;  "nat"&nbsp;  anzufügen.
-*Ist das Ergebnis dagegen in&nbsp;  &bdquo;bit"&nbsp;  gefragt, so ist der <i>Logarithmus dualis</i> &nbsp; &#8658; &nbsp; &bdquo;$\log_2$"&nbsp; zu verwenden.
+*Ist das Ergebnis dagegen in&nbsp;  "bit"&nbsp;  gefragt, so ist der <i>Logarithmus dualis</i> &nbsp; &#8658; &nbsp; "$\log_2$"&nbsp; zu verwenden.
@@ Line 34: / Line 34: @@
 ''Hinweise:''
 *Die Aufgabe gehört zum  Kapitel&nbsp; [[Information_Theory/Differentielle_Entropie|Differentielle Entropie]].
-*Nützliche Hinweise zur Lösung dieser Aufgabe und weitere Informationen zu den wertkontinuierlichen Zufallsgrößen finden Sie im dritten Kapitel &bdquo;Kontinuierliche Zufallsgrößen" des Buches&nbsp;  [[Stochastische Signaltheorie]].
+*Nützliche Hinweise zur Lösung dieser Aufgabe und weitere Informationen zu den wertkontinuierlichen Zufallsgrößen finden Sie im dritten Kapitel "Kontinuierliche Zufallsgrößen" des Buches&nbsp;  [[Stochastische Signaltheorie]].
 *Vorgegeben ist das folgende unbestimmte Integral:
@@ Line 45: / Line 45: @@
 ===Fragebogen===
 <quiz display=simple>
-{Berechnen Sie die differentielle Entropie der Zufallsgröße&nbsp; $X$&nbsp; in&nbsp; &bdquo;nat".
+{Berechnen Sie die differentielle Entropie der Zufallsgröße&nbsp; $X$&nbsp; in&nbsp; "nat".
 |type="{}"}
 $h(X) \ = \ $ { -0.199--0.187 } $\ \rm nat$
-{Welches Ergebnis erhält man mit der Pseudoeinheit&nbsp; &bdquo;bit"?
+{Welches Ergebnis erhält man mit der Pseudoeinheit&nbsp; "bit"?
 |type="{}"}
 $h(X) \ = \ $ { -0.288--0.270 } $\ \rm bit$
@@ Line 70: / Line 70: @@
 :$$f_X(x) = 2x = C \cdot x
 \hspace{0.05cm}.$$
-*Wir haben hierbei &bdquo;2" durch&nbsp; $C$&nbsp; ersetzt &nbsp; &#8658; &nbsp; Verallgemeinerung, um in der Teilaufgabe&nbsp; $(3)$&nbsp; die nachfolgende Berechnung nochmals nutzen zu können.
+*Wir haben hierbei "2" durch&nbsp; $C$&nbsp; ersetzt &nbsp; &#8658; &nbsp; Verallgemeinerung, um in der Teilaufgabe&nbsp; $(3)$&nbsp; die nachfolgende Berechnung nochmals nutzen zu können.
-*Da die differentielle Entropie in &bdquo;nat" gesucht ist, verwenden wir den natürlichen Logarithmus.&nbsp; Mit der Substitution&nbsp; $\xi = C \cdot x$&nbsp; erhalten wir:
+*Da die differentielle Entropie in "nat" gesucht ist, verwenden wir den natürlichen Logarithmus.&nbsp; Mit der Substitution&nbsp; $\xi = C \cdot x$&nbsp; erhalten wir:
 :$$h_{\rm nat}(X) = \hspace{0.1cm} - \int_{0}^{1} \hspace{0.1cm}   C \cdot x \cdot {\rm ln} \hspace{0.1cm} \big[ C \cdot x \big] \hspace{0.1cm}{\rm d}x =
 \hspace{0.1cm} - \hspace{0.1cm}\frac{1}{C} \cdot \int_{0}^{C} \hspace{0.1cm}   \xi \cdot {\rm ln} \hspace{0.1cm} [ \xi ] \hspace{0.1cm}{\rm d}\xi
@@ Line 101: / Line 101: @@
 \hspace{0.15cm}\underline {= - 0.279\,{\rm bit}}
 \hspace{0.05cm}.$$
-*Diese Umrechnung kann man sich sparen, wenn man bereits im analytischen Ergebnis der Teilaufgabe&nbsp; $(1)$&nbsp; direkt &bdquo;ln" durch &bdquo;log<sub>2</sub>" ersetzt:
+*Diese Umrechnung kann man sich sparen, wenn man bereits im analytischen Ergebnis der Teilaufgabe&nbsp; $(1)$&nbsp; direkt "ln" durch "log<sub>2</sub>" ersetzt:
 :$$h(X) = \  {\rm log}_2 \hspace{0.1cm} (\sqrt{\rm e}/2)\hspace{0.05cm}, \hspace{1.3cm}
 {\rm Pseudo-Einheit\hspace{-0.1cm}:\hspace{0.15cm} bit}