Aufgaben:Exercise 3.12: Cauchy Distribution - Revision history

Guenter at 12:24, 3 February 2022

2022-02-03T12:24:13Z

Guenter at 12:12, 3 February 2022

2022-02-03T12:12:46Z

Noah at 09:02, 13 January 2022

2022-01-13T09:02:33Z

Noah at 09:00, 13 January 2022

2022-01-13T09:00:49Z

Noah at 21:57, 12 January 2022

2022-01-12T21:57:24Z

Noah at 21:57, 12 January 2022

2022-01-12T21:57:10Z

Guenter at 10:32, 20 December 2021

2021-12-20T10:32:57Z

Guenter: Guenter moved page Aufgaben:Aufgabe 3.12: Cauchyverteilung to Aufgaben:Exercise 3.12: Cauchy Distribution

2021-12-20T10:30:54Z

Guenter moved page Aufgabe 3.12: Cauchyverteilung to Exercise 3.12: Cauchy Distribution

Javier: Text replacement - "„" to """

2021-05-28T14:37:39Z

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Javier: Text replacement - "”" to """

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 ===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Comparing the given PDF with the general equation in the theory part, we see that the parameter&nbsp; $\lambda= 2$&nbsp; is.
+'''(1)'''&nbsp; Comparing the given PDF with the general equation in the theory part,&nbsp; we see that the parameter is&nbsp; $\lambda= 2$.
 *From this follows&nbsp; (after integration over the PDF):
 :$$F_x ( r ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(\it r/\rm 2).$$
@@ Line 50: / Line 50: @@
 :$$F_x ( r = -2 ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(-1)=\frac{1}{2} - \frac{\rm 1}{\rm \pi} \cdot \frac{\rm \pi}{4 }=0.25.$$
-*The probability we are looking for is given by the difference of.
+*The probability we are looking for is given by the difference:
 :$${\rm Pr} (|x| < 2) = 0.75 - 0.25 \hspace{0.15cm}\underline{=50\%}.$$
-'''(2)'''&nbsp; According to the result of the subtask&nbsp; '''(1)'''&nbsp; is&nbsp; $F_x ( r = 4 ) = 0.5 + 1/\pi = 0.852$.
+'''(2)'''&nbsp; According to the result of the subtask&nbsp; '''(1)'''&nbsp; &rArr; &nbsp; $F_x ( r = 4 ) = 0.5 + 1/\pi = 0.852$.
-*Thus, for the "complementary" probability&nbsp; ${\rm Pr} (x > 4)= 0.148$.
+*Thus,&nbsp; for the&nbsp; "complementary"&nbsp; probability:&nbsp; ${\rm Pr} (x > 4)= 0.148$.
-*For symmetry reasons, the probability we are looking for is twice as large:
+*For symmetry reasons,&nbsp; the probability we are looking for is twice as large:
 :$${\rm Pr} (|x| >4) \hspace{0.15cm}\underline{ = 29.6\%}.$$
@@ Line 65: / Line 65: @@
 \hspace{-0.15cm}
 \frac{\it x^{\rm 2}}{\rm 1+(\it x/\rm 2)^{\rm 2}} \,\,{\rm d}x.$$
-*For gro&szlig;e&nbsp; $x$&nbsp; the integrand yields the constant value&nbsp; $4$. Therefore the integral diverges.
+*For large&nbsp; $x$&nbsp; the integrand yields the constant value&nbsp; $4$. Therefore the integral diverges.
-*With&nbsp; $\sigma_x \to \infty$&nbsp; however, even Chebyshev's inequality does not provide an evaluable bound.
+*Chebyshev's inequality does not provide an evaluable bound,&nbsp;  even with&nbsp; $\sigma_x \to \infty$.
-*"Natural" random variables (physically interpretable) can never be cauchy distributed, otherwise they would have to have infinite power.
+*"Natural" random variables&nbsp; (physically interpretable)&nbsp; can never be cauchy distributed,&nbsp; otherwise they would have an infinite power.
-*On the other hand, an "artificial" (or mathematical) random variable (example: the quotient of two zero mean quantities) is not subject to this restriction.
+*On the other hand,&nbsp; an&nbsp; "artificial"&nbsp; (or mathematical)&nbsp; random variable  is not subject to this restriction. &nbsp; Example: '''The quotient of two zero mean quantities'''.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID207__Sto_A_3_12.png|right|frame|PDF of a Cauchy distribution]]
+[[File:P_ID207__Sto_A_3_12.png|right|frame|Cauchy PDF]]
-The probability density function of the Cauchy distribution is given as follows:
+The probability density function&nbsp; $\rm (PDF)$&nbsp; of the Cauchy distribution is given as follows:
 :$$f_x(x)=\frac{\rm 1}{\rm 2 \pi}\cdot \frac{\rm 1}{\rm 1+ (\it x/\rm 2)^{\rm 2}}.$$
 From the graph you can already see the extremely slow decay of the PDF course.
 Hints:
-*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Further_Distributions|Further Distributions]].
+*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Further_Distributions|"Further Distributions"]].
-*In particular, reference is made to the page&nbsp;  [[Theory_of_Stochastic_Signals/Further_Distributions#Cauchy_PDF|Cauchy PDF]]&nbsp;.
+*In particular, reference is made to the section&nbsp;  [[Theory_of_Stochastic_Signals/Further_Distributions#Cauchy_PDF|"Cauchy PDF"]].
@@ Line 25: / Line 21: @@
 <quiz display=simple>
-{What is the distribution function&nbsp; $F_x(r)$?&nbsp; What is the probability that&nbsp; $x$&nbsp; is smaller than&nbsp; $2$ by a small magnitude?
+{What is the cumulative distribution function&nbsp; $\rm (CDF)$&nbsp; $F_x(r)$?&nbsp; What is the probability that&nbsp; $|x|<2$?
 |type="{}"}
 ${\rm Pr} (|x| < 2) \ = \ $ { 50 3% } $ \ \%$
-{What is the probability that&nbsp; $x$&nbsp; is greater in magnitude than&nbsp; $4$?
+{What is the probability that&nbsp; $|x|>4$?
 |type="{}"}
 ${\rm Pr} (|x| > 4) \ = \ $ { 29.6 3% } $ \ \%$

@@ Line 25: / Line 25: @@
 <quiz display=simple>
-{What is the distribution function&nbsp; $F_x(r)$?&nbsp; What is the probability that&nbsp; $x$&nbsp; is smaller than&nbsp; $2$ by a small amount?
+{What is the distribution function&nbsp; $F_x(r)$?&nbsp; What is the probability that&nbsp; $x$&nbsp; is smaller than&nbsp; $2$ by a small magnitude?
 |type="{}"}
 ${\rm Pr} (|x| < 2) \ = \ $ { 50 3% } $ \ \%$
-{What is the probability that&nbsp; $x$&nbsp; is greater in amount than&nbsp; $4$?
+{What is the probability that&nbsp; $x$&nbsp; is greater in magnitude than&nbsp; $4$?
 |type="{}"}
 ${\rm Pr} (|x| > 4) \ = \ $ { 29.6 3% } $ \ \%$

@@ Line 71: / Line 71: @@
 *For gro&szlig;e&nbsp; $x$&nbsp; the integrand yields the constant value&nbsp; $4$. Therefore the integral diverges.
 *With&nbsp; $\sigma_x \to \infty$&nbsp; however, even Chebyshev's inequality does not provide an evaluable bound.
-*Natural" random variables (physically interpretable) can never be cauchy distributed, otherwise they would have to have infinite power.
+*"Natural" random variables (physically interpretable) can never be cauchy distributed, otherwise they would have to have infinite power.
 *On the other hand, an "artificial" (or mathematical) random variable (example: the quotient of two zero mean quantities) is not subject to this restriction.
 {{ML-Fuß}}

@@ Line 3: / Line 3: @@
 }}
-[[File:P_ID207__Sto_A_3_12.png|right|frame|PDF of Cauchy distribution]]
+[[File:P_ID207__Sto_A_3_12.png|right|frame|PDF of a Cauchy distribution]]
 The probability density function of the Cauchy distribution is given as follows:
 :$$f_x(x)=\frac{\rm 1}{\rm 2 \pi}\cdot \frac{\rm 1}{\rm 1+ (\it x/\rm 2)^{\rm 2}}.$$

@@ Line 1: / Line 1: @@
-{{quiz-Header|Buchseite=Stochastische Signaltheorie/Weitere Verteilungen
+{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Further_Distributions
 }}
-[[File:P_ID207__Sto_A_3_12.png|right|frame|WDF der Cauchyverteilung]]
+[[File:P_ID207__Sto_A_3_12.png|right|frame|PDF of Cauchy distribution]]
-Die Wahrscheinlichkeitsdichtefunktion der Cauchyverteilung ist wie folgt gegeben:
+The probability density function of the Cauchy distribution is given as follows:
 :$$f_x(x)=\frac{\rm 1}{\rm 2 \pi}\cdot \frac{\rm 1}{\rm 1+ (\it x/\rm 2)^{\rm 2}}.$$
-Aus der Grafik ist bereits der extrem langsame Abfall des WDF&ndash;Verlaufs zu erkennen.
+From the graph you can already see the extremely slow decay of the PDF course.
@@ Line 15: / Line 15: @@
-''Hinweise:''
+Hints:
-*Die Aufgabe gehört zum  Kapitel&nbsp; [[Theory_of_Stochastic_Signals/Weitere_Verteilungen|Weitere Verteilungen]].
+*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Further_Distributions|Further Distributions]].
-*Insbesondere wird auf die Seite&nbsp;  [[Theory_of_Stochastic_Signals/Weitere_Verteilungen#Cauchyverteilung|Cauchyverteilung]]&nbsp; Bezug genommen.
+*In particular, reference is made to the page&nbsp;  [[Theory_of_Stochastic_Signals/Further_Distributions#Cauchy_PDF|Cauchy PDF]]&nbsp;.
-===Fragebogen===
+===Questions===
 <quiz display=simple>
-{Wie lautet die Verteilungsfunktion&nbsp; $F_x(r)$?&nbsp; Mit welcher Wahrscheinlichkeit ist&nbsp; $x$&nbsp; betragsm&auml;&szlig;ig kleiner als&nbsp; $2$?
+{What is the distribution function&nbsp; $F_x(r)$?&nbsp; What is the probability that&nbsp; $x$&nbsp; is smaller than&nbsp; $2$ by a small amount?
 |type="{}"}
-${\rm Pr} (|x| < 2) \ = \ $  { 50 3% } $ \ \%$
+${\rm Pr} (|x| < 2) \ = \ $ { 50 3% } $ \ \%$
-{Mit welcher Wahrscheinlichkeit ist&nbsp; $x$&nbsp; betragsm&auml;&szlig;ig gr&ouml;&szlig;er als&nbsp; $4$?
+{What is the probability that&nbsp; $x$&nbsp; is greater in amount than&nbsp; $4$?
 |type="{}"}
 ${\rm Pr} (|x| > 4) \ = \ $ { 29.6 3% } $ \ \%$
-{Welche der folgenden Aussagen treffen f&uuml;r die Cauchyverteilung zu?
+{Which of the following statements are true for the Cauchy distribution?
 |type="[]"}
-+ Die Cauchyverteilung besitzt eine unendlich gro&szlig;e Varianz.
++ The Cauchy distribution has an infinitely large variance.
-+ Die Tschebyscheff&ndash;Ungleichung macht hier keinen Sinn.
++ The Chebyshev inequality makes no sense here.
-+ Eine in der Natur messbare Zufallsgr&ouml;&szlig;e ist nie cauchyverteilt.
++ A random variable that can be measured in nature is never Cauchy distributed.
 </quiz>
-===Musterlösung===
+===Solution===
 {{ML-Kopf}}
-'''(1)'''&nbsp; Vergleicht man die vorgegebene WDF mit der allgemeinen Gleichung im Theorieteil, so erkennt man, dass der Parameter&nbsp; $\lambda= 2$&nbsp; ist.
+'''(1)'''&nbsp; Comparing the given PDF with the general equation in the theory part, we see that the parameter&nbsp; $\lambda= 2$&nbsp; is.
-*Daraus folgt&nbsp; (nach Integration &uuml;ber die WDF):
+*From this follows&nbsp; (after integration over the PDF):
 :$$F_x ( r ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(\it r/\rm 2).$$
-*Insbesondere sind
+*In particular.
 :$$F_x ( r = +2 ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(1)=\frac{1}{2} + \frac{\rm 1}{\rm \pi} \cdot \frac{\rm \pi}{4 }=0.75,$$
 :$$F_x ( r = -2 ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(-1)=\frac{1}{2} - \frac{\rm 1}{\rm \pi} \cdot \frac{\rm \pi}{4 }=0.25.$$
-*Die gesuchte Wahrscheinlichkeit ergibt sich als die Differenz zu
+*The probability we are looking for is given by the difference of.
 :$${\rm Pr} (|x| < 2) = 0.75 - 0.25 \hspace{0.15cm}\underline{=50\%}.$$
-'''(2)'''&nbsp; Nach dem Ergebnis der Teilaufgabe&nbsp; '''(1)'''&nbsp; ist&nbsp; $F_x ( r = 4 )  = 0.5 + 1/\pi = 0.852$.
+'''(2)'''&nbsp; According to the result of the subtask&nbsp; '''(1)'''&nbsp; is&nbsp; $F_x ( r = 4 ) = 0.5 + 1/\pi = 0.852$.
-*Damit gilt f&uuml;r die "komplementäre" Wahrscheinlichkeit&nbsp; ${\rm Pr} (x > 4)= 0.148$.
+*Thus, for the "complementary" probability&nbsp; ${\rm Pr} (x > 4)= 0.148$.
-*Die gesuchte Wahrscheinlichkeit ist aus Symmetriegründen doppelt so gro&szlig;:
+*For symmetry reasons, the probability we are looking for is twice as large:
 :$${\rm Pr} (|x| >4) \hspace{0.15cm}\underline{ = 29.6\%}.$$
-'''(3)'''&nbsp; <u>Alle Lösungsvorschläge</u> treffen zu:
+'''(3)'''&nbsp; <u>All proposed solutions</u> are true:
-*F&uuml;r die Varianz der Cauchyverteilung gilt nämlich:
+*For the variance of the Cauchy distribution holds namely:
 :$$\sigma_x^{\rm 2}=\frac{1}{2\pi}\int_{-\infty}^{+\infty}
 \hspace{-0.15cm}
 \frac{\it x^{\rm 2}}{\rm 1+(\it x/\rm 2)^{\rm 2}} \,\,{\rm d}x.$$
-*F&uuml;r gro&szlig;e&nbsp; $x$&nbsp; liefert der Integrand den konstanten Wert&nbsp; $4$. Deshalb divergiert das Integral.
+*For gro&szlig;e&nbsp; $x$&nbsp; the integrand yields the constant value&nbsp; $4$. Therefore the integral diverges.
-*Mit&nbsp; $\sigma_x \to \infty$&nbsp; liefert aber auch die Tschebyscheffsche Ungleichung keine auswertbare Schranke.
+*With&nbsp; $\sigma_x \to \infty$&nbsp; however, even Chebyshev's inequality does not provide an evaluable bound.
-*„Nat&uuml;rliche“ Zufallsgr&ouml;&szlig;en (physikalisch interpretierbar) k&ouml;nnen nie cauchyverteilt sein, da sie sonst eine unendlich gro&szlig;e Leistung besitzen m&uuml;ssten.
+*Natural" random variables (physically interpretable) can never be cauchy distributed, otherwise they would have to have infinite power.
-*Dagegen unterliegt eine „k&uuml;nstliche“ (oder mathematische) Zufallsgr&ouml;&szlig;e&nbsp; (Beispiel: &nbsp; der Quotient zweier mittelwertfreier Gau&szlig;gr&ouml;&szlig;en)&nbsp; nicht dieser Beschr&auml;nkung.
+*On the other hand, an "artificial" (or mathematical) random variable (example: the quotient of two zero mean quantities) is not subject to this restriction.
 {{ML-Fuß}}

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