Aufgaben:Exercise 4.1: PDF, CDF and Probability - Revision history

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

2021-10-11T08:27:33Z

Text replacement - "value-discrete" to "discrete"

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

2021-10-11T08:27:22Z

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

Noah at 10:37, 27 September 2021

2021-09-27T10:37:54Z

Guenter at 15:30, 24 September 2021

2021-09-24T15:30:50Z

Guenter at 15:09, 24 September 2021

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Guenter at 15:07, 24 September 2021

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Guenter at 13:05, 24 September 2021

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Guenter at 12:45, 24 September 2021

2021-09-24T12:45:03Z

Noah at 21:14, 23 September 2021

2021-09-23T21:14:27Z

Show changes

Noah at 20:51, 23 September 2021

2021-09-23T20:51:49Z

@@ Line 9: / Line 9: @@
-The upper plot shows the cumulative distribution function&nbsp; $F_X(x)$&nbsp; of a  value-discrete random variable&nbsp;  $X$.&nbsp; The corresponding probability density function&nbsp; $f_X(x)$&nbsp; has to be determined in subtask&nbsp; '''(1)'''.
+The upper plot shows the cumulative distribution function&nbsp; $F_X(x)$&nbsp; of a  discrete random variable&nbsp;  $X$.&nbsp; The corresponding probability density function&nbsp; $f_X(x)$&nbsp; has to be determined in subtask&nbsp; '''(1)'''.
 The equation
@@ Line 20: / Line 20: @@
 y < -2 \hspace{0.1cm}{\rm und}\hspace{0.1cm}y > +2 \\ \end{array}$$
 of a continuous random variable&nbsp; $Y$,&nbsp; which is restricted to the range&nbsp; $|Y| \le 2$&nbsp;.&nbsp;
-In principle, the same relationship between PDF, CDF and probabilities exists for the continuous random variable&nbsp; $Y$&nbsp; as for a value-discrete random variable.&nbsp; Nevertheless, you will notice some differences in details.
+In principle, the same relationship between PDF, CDF and probabilities exists for the continuous random variable&nbsp; $Y$&nbsp; as for a discrete random variable.&nbsp; Nevertheless, you will notice some differences in details.
 For example, for the continuous random variable&nbsp; $Y$,&nbsp; the boundary transition can be omitted in the above equation, and we obtain simplified:
@@ Line 46: / Line 46: @@
 <quiz display=simple>
-{Determine the PDF&nbsp; $f_X(x)$&nbsp; of the  value-discrete random variable&nbsp; $X$.&nbsp; Which of the following statements are true?
+{Determine the PDF&nbsp; $f_X(x)$&nbsp; of the  discrete random variable&nbsp; $X$.&nbsp; Which of the following statements are true?
 |type="[]"}
 + The PDF is composed of five Dirac functions.
@@ Line 81: / Line 81: @@
 ===Solution===
 {{ML-Kopf}}
-[[File:P_ID2857__Inf_A_4_1a_neu.png|right|frame|PDF and CDF of the value-discrete random variable&nbsp; $X$]]
+[[File:P_ID2857__Inf_A_4_1a_neu.png|right|frame|PDF and CDF of the discrete random variable&nbsp; $X$]]
 '''(1)'''&nbsp; <u>Proposed solutions 1 and 2</u> are correct:
 *The cumulative distribution function&nbsp; $F_X(x)$&nbsp; is obtained from the probability density function&nbsp; $f_X(x)$&nbsp; by integration over the (renamed) random variable in the range from&nbsp; $- \infty$&nbsp; to&nbsp; $x$.
@@ Line 105: / Line 105: @@
 \hspace{0.15cm}\underline {= 0.8}\hspace{0.05cm}.$$
-The same result is obtained using the CDF.&nbsp; Here the general equation, which is equally valid for value-discrete and continuous random variables, is:
+The same result is obtained using the CDF.&nbsp; Here the general equation, which is equally valid for discrete and continuous random variables, is:
 :$${\rm Pr}(A < X \le B) =F_X(B) - F_X(A) \hspace{0.05cm}.$$
@@ Line 155: / Line 155: @@
 \hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}.$$
 *Thus, for the symmetric continuous random variable&nbsp; $Y$&nbsp;  holds indeed as expected: &nbsp;${\rm Pr}( Y > 0) = 1/2$.
-*Although the value-discrete random variable&nbsp; $X$&nbsp; is also symmetrical about&nbsp;$x= 0$ &nbsp; &rArr; &nbsp; ${\rm Pr}( X > 0)  = 0.3$&nbsp; was determined in subtask&nbsp; '''(3)''', on the other hand.
+*Although the discrete random variable&nbsp; $X$&nbsp; is also symmetrical about&nbsp;$x= 0$ &nbsp; &rArr; &nbsp; ${\rm Pr}( X > 0)  = 0.3$&nbsp; was determined in subtask&nbsp; '''(3)''', on the other hand.
 *Further, with &nbsp;$A = -1$&nbsp; and &nbsp;$B = +1$,&nbsp; one obtains because of&nbsp;$F_Y(-1) = 1- F_Y(+1)$:
 :$${\rm Pr}( |Y| \le 1)  =  {\rm Pr}(-1 \le Y \le +1)

@@ Line 19: / Line 19: @@
 :$$ f_Y(y) = \left\{ \begin{array}{c} \hspace{0.1cm}1/2 \cdot \cos^2(\pi/4 \cdot y) \\ \hspace{0.1cm} 0 \\  \end{array} \right.\quad \begin{array}{*{20}c}   {\rm{f\ddot{u}r}}  \\     {\rm{f\ddot{u}r}}  \\ \end{array}\begin{array}{*{20}l}   | y| \le 2, \\
 y < -2 \hspace{0.1cm}{\rm und}\hspace{0.1cm}y > +2 \\ \end{array}$$
-of a value-continuous random variable&nbsp; $Y$,&nbsp; which is restricted to the range&nbsp; $|Y| \le 2$&nbsp;.&nbsp;
+of a continuous random variable&nbsp; $Y$,&nbsp; which is restricted to the range&nbsp; $|Y| \le 2$&nbsp;.&nbsp;
-In principle, the same relationship between PDF, CDF and probabilities exists for the value-continuous random variable&nbsp; $Y$&nbsp; as for a value-discrete random variable.&nbsp; Nevertheless, you will notice some differences in details.
+In principle, the same relationship between PDF, CDF and probabilities exists for the continuous random variable&nbsp; $Y$&nbsp; as for a value-discrete random variable.&nbsp; Nevertheless, you will notice some differences in details.
-For example, for the value-continuous random variable&nbsp; $Y$,&nbsp; the boundary transition can be omitted in the above equation, and we obtain simplified:
+For example, for the continuous random variable&nbsp; $Y$,&nbsp; the boundary transition can be omitted in the above equation, and we obtain simplified:
 :$${\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A) =\int_{A}^{B} \hspace{-0.01cm}  f_Y(y)
 \hspace{0.1cm}{\rm d}y\hspace{0.05cm}.$$
@@ Line 36: / Line 36: @@
 Hints:
 *The exercise belongs to the chapter&nbsp; [[Information_Theory/Differentielle_Entropie|Differential Entropy]].
-*Useful hints for solving this problem and further information on value-continuous random variables can be found in the third chapter&nbsp; "Continuous Random Variables"&nbsp; of the book&nbsp;  [[Theory of Stochastic Signals]].
+*Useful hints for solving this problem and further information on continuous random variables can be found in the third chapter&nbsp; "Continuous Random Variables"&nbsp; of the book&nbsp;  [[Theory of Stochastic Signals]].
 *Given also is the following indefinite integral:
@@ Line 58: / Line 58: @@
 ${\rm Pr}(|X| ≤ 1) \ =  \ $ { 0.8 3% }
-{What are the values of the cumulative distribution function&nbsp; $F_Y(y)  ={\rm Pr}(Y \le y)$&nbsp; of the  value-continuous random variable&nbsp; $Y$,&nbsp; in particular:
+{What are the values of the cumulative distribution function&nbsp; $F_Y(y)  ={\rm Pr}(Y \le y)$&nbsp; of the  continuous random variable&nbsp; $Y$,&nbsp; in particular:
 |type="{}"}
 $F_Y(y = 0) \ =  \ $ { 0.5 3% }
@@ Line 105: / Line 105: @@
 \hspace{0.15cm}\underline {= 0.8}\hspace{0.05cm}.$$
-The same result is obtained using the CDF.&nbsp; Here the general equation, which is equally valid for value-discrete and value-continuous random variables, is:
+The same result is obtained using the CDF.&nbsp; Here the general equation, which is equally valid for value-discrete and continuous random variables, is:
 :$${\rm Pr}(A < X \le B) =F_X(B) - F_X(A) \hspace{0.05cm}.$$
@@ Line 115: / Line 115: @@
-[[File:P_ID2858__Inf_A_4_1c_neu.png|right|frame|PDF and CDF of the value-continuous random variable&nbsp; $Y$]]
+[[File:P_ID2858__Inf_A_4_1c_neu.png|right|frame|PDF and CDF of the continuous random variable&nbsp; $Y$]]
 '''(3)'''&nbsp; The cumulative distribution function&nbsp; $F_Y(y)$&nbsp; is obtained from the (renamed) WDF&nbsp; $f_Y(\eta)$&nbsp; by integrating&nbsp; $- \infty$&nbsp; to&nbsp; $x$.&nbsp; Due to symmetry, this can be written in the range&nbsp; $0 \le y \le +2$:
 :$$F_Y(y) = \int_{-\infty}^{\hspace{0.05cm}y} \hspace{-0.1cm}f_Y(\eta) \hspace{0.1cm}{\rm d}\eta ={1}/{2}+\int_{0}^{\hspace{0.05cm}y} \hspace{-0.1cm}f_Y(\eta) \hspace{0.1cm}{\rm d}\eta$$
@@ Line 126: / Line 126: @@
-'''(4)'''&nbsp; The probability that the value-continuous random variable&nbsp; $Y$&nbsp; lies in the range from&nbsp; $-\varepsilon$&nbsp; to&nbsp; $+\varepsilon$&nbsp; can be calculated using the given equation as follows:
+'''(4)'''&nbsp; The probability that the continuous random variable&nbsp; $Y$&nbsp; lies in the range from&nbsp; $-\varepsilon$&nbsp; to&nbsp; $+\varepsilon$&nbsp; can be calculated using the given equation as follows:
 :$${\rm Pr}(-\varepsilon \le Y \le +\varepsilon) = F_Y(+\varepsilon) - F_Y(-\varepsilon) \hspace{0.05cm}.$$
@@ Line 135: / Line 135: @@
      F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{+}) - F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{-})\hspace{0.05cm}.$$
-*Since for a value-continuous random variable the two limits are equal, $\underline{{\rm Pr}(Y = 0) = 0}$.
+*Since for a continuous random variable the two limits are equal, $\underline{{\rm Pr}(Y = 0) = 0}$.
-'''In general''': &nbsp; The probability&nbsp; ${\rm Pr}(Y = y_0)$&nbsp; that a value-continuous  random variable&nbsp; $Y$&nbsp; takes a fixed value&nbsp; $y_0$,&nbsp; is always zero.
+'''In general''': &nbsp; The probability&nbsp; ${\rm Pr}(Y = y_0)$&nbsp; that a continuous  random variable&nbsp; $Y$&nbsp; takes a fixed value&nbsp; $y_0$,&nbsp; is always zero.
@@ Line 149: / Line 149: @@
-'''(6)'''&nbsp; We again assume the equation&nbsp; $ {\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A)$&nbsp; &nbsp; valid for the value-continuous random quantity &nbsp;$Y$:
+'''(6)'''&nbsp; We again assume the equation&nbsp; $ {\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A)$&nbsp; &nbsp; valid for the continuous random quantity &nbsp;$Y$:
 *With&nbsp; $A = 0$&nbsp; and&nbsp; $B \to \infty$&nbsp; $($or&nbsp; $B = 2)$&nbsp; we obtain:
 :$${\rm Pr}( Y > 0) = {\rm Pr}(0 \le Y \le \infty)

← 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: @@
 ===Solution===
 {{ML-Kopf}}
-[[File:P_ID2857__Inf_A_4_1a_neu.png|right|frame|PDF and CDF of the discrete random variable&nbsp; $X$]]
+[[File:P_ID2857__Inf_A_4_1a_neu.png|right|frame|PDF and CDF of the value-discrete random variable&nbsp; $X$]]
 '''(1)'''&nbsp; <u>Proposed solutions 1 and 2</u> are correct:
-*The cumulative distribution function (CDF)&nbsp; $F_X(x)$&nbsp; is obtained from the probability density function&nbsp; $f_X(x)$&nbsp; by integration over the (renamed) random variable in the range from&nbsp; $- \infty$&nbsp; to&nbsp; $x$.
+*The cumulative distribution function&nbsp; $F_X(x)$&nbsp; is obtained from the probability density function&nbsp; $f_X(x)$&nbsp; by integration over the (renamed) random variable in the range from&nbsp; $- \infty$&nbsp; to&nbsp; $x$.
-*The inverse is: &nbsp; given the CDF, obtain the PDF by differentiation.
+*The inverse is: &nbsp; Given the CDF, obtain the PDF by differentiation.
 *The given CDF contains five discontinuity points, which after differentiation lead to five Dirac functions:
 :$$f_X(x) =  0.1 \cdot {\rm \delta}( x+2)
@@ Line 90: / Line 90: @@
   +  0.4 \cdot {\rm \delta}( x) + 0.2 \cdot {\rm \delta}( x-1)
    + 0.1 \cdot {\rm \delta}( x-2)\hspace{0.05cm}.$$
-*The Dirac weights give the occurrence probabilities of the random variable&nbsp; $X = \{-2,\ -1,\ 0,\ +1,\ +2\}$ an, <br>for example:
+*The Dirac weights give the occurrence probabilities of the random variable&nbsp; $X = \{-2,\ -1,\ 0,\ +1,\ +2\}$&nbsp;, e.g.:
 :$${\rm Pr}(X = 0) = F_X(x \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{+}) - F_X(x \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{-}) =
 .7 - 0.3 = 0.4\hspace{0.05cm}.$$
@@ Line 105: / Line 105: @@
 \hspace{0.15cm}\underline {= 0.8}\hspace{0.05cm}.$$
-The same result is obtained using the distribution function.&nbsp; Here the general equation, which is equally valid for discrete-value and continuous-value random variables, is:
+The same result is obtained using the CDF.&nbsp; Here the general equation, which is equally valid for value-discrete and value-continuous random variables, is:
 :$${\rm Pr}(A < X \le B) =F_X(B) - F_X(A) \hspace{0.05cm}.$$
 * Thus, with&nbsp; $A= 0$&nbsp; and&nbsp; $B = +2$&nbsp; we obtain:
 :$${\rm Pr}(0 < X \le +2) = {\rm Pr}(X >0)= F_X(+2) - F_X(0) = 1 - 0.7 \hspace{0.15cm}\underline {= 0.3} \hspace{0.05cm}.$$
-*Setting $A=-2$ and $B = +1$, we get:
+*Setting&nbsp; $A=-2$&nbsp; and&nbsp; $B = +1$,&nbsp; we get:
 :$${\rm Pr}(-2 < X \le +1) = {\rm Pr}(|X|  \le 1)= F_X(+1) - F_X(-2) = 0.9 - 0.1 \hspace{0.15cm}\underline {= 0.8} \hspace{0.05cm}.$$
-[[File:P_ID2858__Inf_A_4_1c_neu.png|right|frame|PDF and CDF of the continuous random variable&nbsp; $Y$]]
+[[File:P_ID2858__Inf_A_4_1c_neu.png|right|frame|PDF and CDF of the value-continuous random variable&nbsp; $Y$]]
-'''(3)'''&nbsp; The cumulative distribution function&nbsp; $F_Y(y)$&nbsp; is obtained from the (renamed) WDF&nbsp; $f_Y(\eta)$&nbsp; by integrating&nbsp; $- \infty$&nbsp; to&nbsp; $x$.&nbsp; Due to symmetry, this can be written in the range&nbsp; $0 \le y \le +2$&nbsp;:
+'''(3)'''&nbsp; The cumulative distribution function&nbsp; $F_Y(y)$&nbsp; is obtained from the (renamed) WDF&nbsp; $f_Y(\eta)$&nbsp; by integrating&nbsp; $- \infty$&nbsp; to&nbsp; $x$.&nbsp; Due to symmetry, this can be written in the range&nbsp; $0 \le y \le +2$:
 :$$F_Y(y) = \int_{-\infty}^{\hspace{0.05cm}y} \hspace{-0.1cm}f_Y(\eta) \hspace{0.1cm}{\rm d}\eta ={1}/{2}+\int_{0}^{\hspace{0.05cm}y} \hspace{-0.1cm}f_Y(\eta) \hspace{0.1cm}{\rm d}\eta$$
 :$$\Rightarrow \hspace{0.3cm}F_Y(y) = \frac{1}{2}+\int_{0}^{\hspace{0.05cm}y} \hspace{0.1cm}\frac{1}{2} \cdot \cos^2({\pi}/{4} \cdot \eta) \hspace{0.1cm}{\rm d}\eta = \frac{1}{2}+\frac{y}{4} + \frac{1}{2\pi} \cdot \sin({\pi}/{2} \cdot y).$$
@@ Line 126: / Line 126: @@
-'''(4)'''&nbsp; The probability that the continuous-value random variable&nbsp; $Y$&nbsp; lies in the range from&nbsp; $-\varepsilon$&nbsp; to&nbsp; $+\varepsilon$&nbsp; can be calculated using the given equation as follows:
+'''(4)'''&nbsp; The probability that the value-continuous random variable&nbsp; $Y$&nbsp; lies in the range from&nbsp; $-\varepsilon$&nbsp; to&nbsp; $+\varepsilon$&nbsp; can be calculated using the given equation as follows:
 :$${\rm Pr}(-\varepsilon \le Y \le +\varepsilon) = F_Y(+\varepsilon) - F_Y(-\varepsilon) \hspace{0.05cm}.$$
-*It was taken into account that for the continuous random variable&nbsp; $Y$&nbsp; the "<"sign can be replaced by the "&#8804;" sign without distortion.
+*It was taken into account that for the random variable&nbsp; $Y$&nbsp; the "<"sign can be replaced by the "&#8804;" sign without distortion.
-*With the boundary transition&nbsp; $\varepsilon \to 0$&nbsp;, the probability we are looking for is obtained:
+*With the boundary transition&nbsp; $\varepsilon \to 0$,&nbsp; the probability we are looking for is obtained:
 :$${\rm Pr}(Y = 0)  =\lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm}{\rm Pr}(-\varepsilon \le Y \le +\varepsilon) =
 \lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm} F_Y(+\varepsilon) - \lim_{\varepsilon\hspace{0.05cm}\rightarrow\hspace{0.05cm}0}\hspace{0.1cm} F_Y(-\varepsilon) =
      F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{+}) - F_Y(y \hspace{0.05cm}\rightarrow\hspace{0.05cm}0^{-})\hspace{0.05cm}.$$
-*Since for a continuous random variable the two limits are equal, $\underline{{\rm Pr}(Y = 0) = 0}$.
+*Since for a value-continuous random variable the two limits are equal, $\underline{{\rm Pr}(Y = 0) = 0}$.
-'''In general''': &nbsp; The probability&nbsp; ${\rm Pr}(Y = y_0)$ that a continuous value random variable&nbsp; $Y$&nbsp; takes a fixed value&nbsp; $y_0$&nbsp; is always zero.
+'''In general''': &nbsp; The probability&nbsp; ${\rm Pr}(Y = y_0)$&nbsp; that a value-continuous  random variable&nbsp; $Y$&nbsp; takes a fixed value&nbsp; $y_0$,&nbsp; is always zero.
@@ Line 145: / Line 145: @@
 *Based on the PDF at hand, the result&nbsp; $Y=3$&nbsp; can be excluded.
 *The result&nbsp; $Y=0$&nbsp; on the other hand is quite possible, although&nbsp; ${\rm Pr}(Y = 0) = 0$&nbsp;.
-*For example, if one performs a random experiment&nbsp; $N \to \infty$&nbsp; times and obtains the result&nbsp; $Y= 0$ &nbsp; $N_0$&nbsp; times, then with finite &nbsp; $N_0$&nbsp; according to the classical definition of probability:
+*For example, if one performs a random experiment&nbsp; $N \to \infty$&nbsp; times and obtains the result&nbsp; $Y= 0$ &nbsp; for&nbsp; $N_0$&nbsp; times, then with finite &nbsp; $N_0$&nbsp; according to the classical definition of probability:
 :$${\rm Pr}(Y = 0) = \lim_{N\hspace{0.05cm}\rightarrow\hspace{0.05cm}\infty}\hspace{0.1cm}{N_0}/{N} = 0\hspace{0.05cm}.$$
+'''(6)'''&nbsp; We again assume the equation&nbsp; $ {\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A)$&nbsp; &nbsp; valid for the value-continuous random quantity &nbsp;$Y$:
-'''(6)'''&nbsp; We again assume the equation&nbsp; $Y$&nbsp; valid for the continuous random quantity &nbsp; $ {\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A)$&nbsp;:
+*With&nbsp; $A = 0$&nbsp; and&nbsp; $B \to \infty$&nbsp; $($or&nbsp; $B = 2)$&nbsp; we obtain:
 :$${\rm Pr}( Y > 0) = {\rm Pr}(0 \le Y \le \infty)
 = {\rm Pr}(0 \le Y \le 2) = F_Y(2) - F_Y(0)
 \hspace{0.15cm}\underline {= 0.5}\hspace{0.05cm}.$$
-*Thus, for the symmetric continuous random variable&nbsp; $Y$&nbsp;  is indeed as expected &nbsp;${\rm Pr}( Y > 0) = 1/2$.
+*Thus, for the symmetric continuous random variable&nbsp; $Y$&nbsp;  holds indeed as expected: &nbsp;${\rm Pr}( Y > 0) = 1/2$.
-*Although the discrete value random variable&nbsp; $X$&nbsp; is also symmetric about&nbsp;$x= 0$&nbsp;,&nbsp;${\rm Pr}( X > 0)  = 0.3$&nbsp; was determined in subtask&nbsp; '''(3)'''&nbsp;, on the other hand.
+*Although the value-discrete random variable&nbsp; $X$&nbsp; is also symmetrical about&nbsp;$x= 0$ &nbsp; &rArr; &nbsp; ${\rm Pr}( X > 0)  = 0.3$&nbsp; was determined in subtask&nbsp; '''(3)''', on the other hand.
-*Further, with &nbsp;$A = -1$&nbsp; and &nbsp;$B = +1$&nbsp;, one obtains because of&nbsp;$F_Y(-1) = 1- F_Y(+1)$:
+*Further, with &nbsp;$A = -1$&nbsp; and &nbsp;$B = +1$,&nbsp; one obtains because of&nbsp;$F_Y(-1) = 1- F_Y(+1)$:
 :$${\rm Pr}( |Y| \le 1)  =  {\rm Pr}(-1 \le Y \le +1)
 =  F_Y(+1) - F_Y(-1)   =   2 \cdot F_Y(+1) -1 = 2 \cdot 0.909 -1 \hspace{0.15cm}\underline {= 0.818}. $$

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID2862__Inf_A_4_1_neu.png|right|frame|CDF (top) and PDF (bottom)]]
+[[File:P_ID2862__Inf_A_4_1_neu.png|right|frame|$\rm  (CDF)$&nbsp; (top),&nbsp; $\rm  (PDF)$&nbsp; (bottom)]]
 To repeat some important basics from the book "Theory of Stochastic Signals" we are dealing with
 * the&nbsp; [[Theory_of_Stochastic_Signals/Wahrscheinlichkeitsdichtefunktion|probability density function]]&nbsp;$\rm  (PDF)$,

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID2862__Inf_A_4_1_neu.png|right|frame|VTF (oben) und WDF (unten)]]
+[[File:P_ID2862__Inf_A_4_1_neu.png|right|frame|CDF (top) and PDF (bottom)]]
-Zur Wiederholung einiger wichtiger Grundlagen aus dem Buch "Stochastische Signaltheorie"
+To repeat some important basics from the book "Theory of Stochastic Signals"
 beschäftigen wir uns mit
-* der&nbsp; [[Theory_of_Stochastic_Signals/Wahrscheinlichkeitsdichtefunktion|Wahrscheinlichkeitsdichtefunktion]] (WDF),
+* the&nbsp; [[Theory_of_Stochastic_Signals/Wahrscheinlichkeitsdichtefunktion|probability density function]] (PDF),
-* der&nbsp; [[Theory_of_Stochastic_Signals/Verteilungsfunktion|Verteilungsfunktion]] (VTF).
+* the&nbsp; [[Theory_of_Stochastic_Signals/Verteilungsfunktion|cumulative distribution function]] (CDF).
-Die obere Darstellung zeigt die Verteilungsfunktion&nbsp; $F_X(x)$&nbsp; einer wertdiskreten Zufallsgröße&nbsp;  $X$.&nbsp; Die zugehörige WDF&nbsp; $f_X(x)$&nbsp; ist in der Teilaufgabe&nbsp; '''(1)'''&nbsp; zu bestimmen.
+The upper plot shows the distribution function&nbsp; $F_X(x)$&nbsp; of a discrete value random variable&nbsp;  $X$.&nbsp; The corresponding PDF&nbsp; $f_X(x)$&nbsp; has to be determined in subtask&nbsp; '''(1)'''&nbsp;.
-Die Gleichung
+The equation
 :$$ {\rm Pr}(A < X \le B) = F_X(B) - F_X(A) =  \lim_{\varepsilon \hspace{0.05cm}\rightarrow \hspace{0.05cm}0} \int_{A+\varepsilon}^{B+\varepsilon} \hspace{-0.15cm}  f_X(x) \hspace{0.1cm}{\rm d}x $$
-stellt zwei Möglichkeiten dar, um aus der VTF bzw. der WDF die Wahrscheinlichkeit für das Ereignis&nbsp; „Die Zufallsgröße&nbsp; $X$&nbsp; liegt in einem vorgegebenem Intervall”&nbsp;  zu berechnen.
+represents two ways to calculate the probability for the event&nbsp; „The random variable&nbsp; $X$&nbsp; lies in a given interval”&nbsp; from the CDF and the PDF, respectively.
-Die untere Grafik zeigt die Wahrscheinlichkeitsdichtefunktion
+The lower graph shows the probability density function
 :$$ f_Y(y) = \left\{ \begin{array}{c} \hspace{0.1cm}1/2 \cdot \cos^2(\pi/4 \cdot y) \\ \hspace{0.1cm} 0 \\  \end{array} \right.\quad \begin{array}{*{20}c}   {\rm{f\ddot{u}r}}  \\     {\rm{f\ddot{u}r}}  \\ \end{array}\begin{array}{*{20}l}   | y| \le 2, \\
 y < -2 \hspace{0.1cm}{\rm und}\hspace{0.1cm}y > +2 \\ \end{array}$$
-einer wertkontinuierlichen Zufallsgröße&nbsp; $Y$, die auf den Bereich&nbsp; $|Y| \le 2$&nbsp; begrenzt ist.
+of a continuous-valued random variable&nbsp; $Y$, which is restricted to the range&nbsp; $|Y| \le 2$&nbsp;.
-Prinzipiell besteht bei der kontinuierlichen Zufallsgröße&nbsp; $Y$&nbsp; der gleiche Zusammenhang zwischen WDF, VTF und Wahrscheinlichkeiten wie bei einer diskreten Zufallsgröße.&nbsp; Sie werden trotzdem einige Detailunterschiede feststellen.
+For example, for the continuous random variable&nbsp; $Y$&nbsp;, the boundary transition can be omitted in the above equation, and we obtain simplified:
 :$${\rm Pr}(A \le Y \le B) = F_Y(B) - F_Y(A) =\int_{A}^{B} \hspace{-0.01cm}  f_Y(y)
 \hspace{0.1cm}{\rm d}y\hspace{0.05cm}.$$
@@ Line 36: / Line 35: @@
-''Hinweise:''
+Hints:
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Information_Theory/Differentielle_Entropie|Differentielle Entropie]].
+*The task belongs to the chapter&nbsp; [[Information_Theory/Differentielle_Entropie|Differential Entropy]].
-*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]].
+*Useful hints for solving this problem 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]].
-*Gegeben ist zudem das folgende unbstimmte Integral:
+*Given also is the following indefinite integral:
 :$$\int \hspace{0.1cm} \cos^2(A \eta) \hspace{0.1cm}{\rm d}\eta =  \frac{\eta}{2} + \frac{1}{4A} \cdot \sin(2A  \eta).$$
-===Fragebogen===
+===Questions===
 <quiz display=simple>