Aufgaben:Exercise 3.12: Cauchy Distribution - Revision history

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	[[Category:Theory of Stochastic Signals: Exercises\|^3.7 Further Distributions^]]		[[Category:Theory of Stochastic Signals: Exercises\|^3.7 Further Distributions^]]
	[[de:Aufgaben:3.~~12_Cauchyverteilung~~]]		[[de:Aufgaben:Aufgabe 3.12: Cauchyverteilung]]

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	'''(3)'''  <u>All proposed solutions</u> are true:		'''(3)'''  <u>All proposed solutions</u> are true:
	*For the variance of the Cauchy distribution holds namely:		*For the variance of the Cauchy distribution holds namely:
	:$$\sigma_x^{\rm 2}=\frac{1}{2\pi}\int_{-\infty}^{+\infty}		:$$\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.$$
	\hspace{-0.15cm}
	\frac{\it x^{\rm 2}}{\rm 1+(\it x/\rm 2)^{\rm 2}} \,\,{\rm d}x.$$
	*For large  $x$  the integrand yields the constant value  $4$. Therefore the integral diverges.		*For large  $x$  the integrand yields the constant value  $4$. Therefore the integral diverges.
	*Chebyshev's inequality does not provide an evaluable bound,  even with  $\sigma_x \to \infty$.		*Chebyshev's inequality does not provide an evaluable bound,  even with  $\sigma_x \to \infty$.

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	[[Category:Theory of Stochastic Signals: Exercises\|^3.7 Further Distributions^]]		[[Category:Theory of Stochastic Signals: Exercises\|^3.7 Further Distributions^]]
			[[de:Aufgaben:3.12_Cauchyverteilung]]

Guenter at 12:24, 3 February 2022

2022-02-03T12:24:13Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Comparing the given PDF with the general equation in the theory part, we see that the parameter  $\lambda= 2$~~  is~~.		'''(1)'''  Comparing the given PDF with the general equation in the theory part,  we see that the parameter is  $\lambda= 2$.
	*From this follows  (after integration over the PDF):		*From this follows  (after integration over the PDF):
	:$$F_x ( r ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(\it r/\rm 2).$$		:$$F_x ( r ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(\it r/\rm 2).$$
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	:$$F_x ( r = -2 ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(-1)=\frac{1}{2} - \frac{\rm 1}{\rm \pi} \cdot \frac{\rm \pi}{4 }=0.25.$$		:$$F_x ( r = -2 ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(-1)=\frac{1}{2} - \frac{\rm 1}{\rm \pi} \cdot \frac{\rm \pi}{4 }=0.25.$$

	*The probability we are looking for is given by the difference ~~of.~~		*The probability we are looking for is given by the difference:
	:$${\rm Pr} (\|x\| < 2) = 0.75 - 0.25 \hspace{0.15cm}\underline{=50\%}.$$		:$${\rm Pr} (\|x\| < 2) = 0.75 - 0.25 \hspace{0.15cm}\underline{=50\%}.$$


	'''(2)'''  According to the result of the subtask  '''(1)'''  is  $F_x ( r = 4 ) = 0.5 + 1/\pi = 0.852$.		'''(2)'''  According to the result of the subtask  '''(1)'''  ⇒   $F_x ( r = 4 ) = 0.5 + 1/\pi = 0.852$.
	*Thus, for the "complementary" probability  ${\rm Pr} (x > 4)= 0.148$.		*Thus,  for the  "complementary"  probability:  ${\rm Pr} (x > 4)= 0.148$.
	*For symmetry reasons, the probability we are looking for is twice as large:		*For symmetry reasons,  the probability we are looking for is twice as large:
	:$${\rm Pr} (\|x\| >4) \hspace{0.15cm}\underline{ = 29.6\%}.$$		:$${\rm Pr} (\|x\| >4) \hspace{0.15cm}\underline{ = 29.6\%}.$$

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	\hspace{-0.15cm}		\hspace{-0.15cm}
	\frac{\it x^{\rm 2}}{\rm 1+(\it x/\rm 2)^{\rm 2}} \,\,{\rm d}x.$$		\frac{\it x^{\rm 2}}{\rm 1+(\it x/\rm 2)^{\rm 2}} \,\,{\rm d}x.$$
	*For ~~große~~  $x$  the integrand yields the constant value  $4$. Therefore the integral diverges.		*For large  $x$  the integrand yields the constant value  $4$. Therefore the integral diverges.
	*~~With~~  $\sigma_x \to \infty$~~  however, even Chebyshev's inequality does not provide an evaluable bound~~.		*Chebyshev's inequality does not provide an evaluable bound,  even with  $\sigma_x \to \infty$.
	*"Natural" random variables (physically interpretable) can never be cauchy distributed, otherwise they would have ~~to have~~ infinite power.		*"Natural" random variables  (physically interpretable)  can never be cauchy distributed,  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,  an  "artificial"  (or mathematical)  random variable is not subject to this restriction.   Example: '''The quotient of two zero mean quantities'''.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 12:12, 3 February 2022

2022-02-03T12:12:46Z

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	}}		}}

	[[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  $\rm (PDF)$  of the Cauchy distribution is given as follows:
	:$$f_x(x)=\frac{\rm 1}{\rm 2 \pi}\cdot \frac{\rm 1}{\rm 1+ (\it x/\rm 2)^{\rm 2}}.$$		:$$f_x(x)=\frac{\rm 1}{\rm 2 \pi}\cdot \frac{\rm 1}{\rm 1+ (\it x/\rm 2)^{\rm 2}}.$$

	From the graph you can already see the extremely slow decay of the PDF course.		From the graph you can already see the extremely slow decay of the PDF course.







	Hints:		Hints:
	*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Further_Distributions\|Further Distributions]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Further_Distributions\|"Further Distributions"]].
	*In particular, reference is made to the ~~page~~  [[Theory_of_Stochastic_Signals/Further_Distributions#Cauchy_PDF\|Cauchy PDF]]~~ ~~.		*In particular, reference is made to the section  [[Theory_of_Stochastic_Signals/Further_Distributions#Cauchy_PDF\|"Cauchy PDF"]].


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	<quiz display=simple>		<quiz display=simple>
	{What is the distribution function  $F_x(r)$?  What is the probability that  $x~~$  is smaller than  $~~2$ ~~by a small magnitude~~?		{What is the cumulative distribution function  $\rm (CDF)$  $F_x(r)$?  What is the probability that  $\|x\|<2$?
	\|type="{}"}		\|type="{}"}
	${\rm Pr} (\|x\| < 2) \ = \ $ { 50 3% } $ \ \%$		${\rm Pr} (\|x\| < 2) \ = \ $ { 50 3% } $ \ \%$


	{What is the probability that  $x~~$  is greater in magnitude than  $~~4$?		{What is the probability that  $\|x\|>4$?
	\|type="{}"}		\|type="{}"}
	${\rm Pr} (\|x\| > 4) \ = \ $ { 29.6 3% } $ \ \%$		${\rm Pr} (\|x\| > 4) \ = \ $ { 29.6 3% } $ \ \%$

Noah at 09:02, 13 January 2022

2022-01-13T09:02:33Z

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	<quiz display=simple>		<quiz display=simple>
	{What is the distribution function  $F_x(r)$?  What is the probability that  $x$  is smaller than  $2$ by a small ~~amount~~?		{What is the distribution function  $F_x(r)$?  What is the probability that  $x$  is smaller than  $2$ by a small magnitude?
	\|type="{}"}		\|type="{}"}
	${\rm Pr} (\|x\| < 2) \ = \ $ { 50 3% } $ \ \%$		${\rm Pr} (\|x\| < 2) \ = \ $ { 50 3% } $ \ \%$


	{What is the probability that  $x$  is greater in ~~amount~~ than  $4$?		{What is the probability that  $x$  is greater in magnitude than  $4$?
	\|type="{}"}		\|type="{}"}
	${\rm Pr} (\|x\| > 4) \ = \ $ { 29.6 3% } $ \ \%$		${\rm Pr} (\|x\| > 4) \ = \ $ { 29.6 3% } $ \ \%$

Noah at 09:00, 13 January 2022

2022-01-13T09:00:49Z

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	*For große  $x$  the integrand yields the constant value  $4$. Therefore the integral diverges.		*For große  $x$  the integrand yields the constant value  $4$. Therefore the integral diverges.
	*With  $\sigma_x \to \infty$  however, even Chebyshev's inequality does not provide an evaluable bound.		*With  $\sigma_x \to \infty$  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.		*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}}		{{ML-Fuß}}

Noah at 21:57, 12 January 2022

2022-01-12T21:57:24Z

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	}}		}}

	[[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:		The probability density function of the Cauchy distribution is given as follows:
	:$$f_x(x)=\frac{\rm 1}{\rm 2 \pi}\cdot \frac{\rm 1}{\rm 1+ (\it x/\rm 2)^{\rm 2}}.$$		:$$f_x(x)=\frac{\rm 1}{\rm 2 \pi}\cdot \frac{\rm 1}{\rm 1+ (\it x/\rm 2)^{\rm 2}}.$$

Noah at 21:57, 12 January 2022

2022-01-12T21:57:10Z

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	{{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:
	:$$f_x(x)=\frac{\rm 1}{\rm 2 \pi}\cdot \frac{\rm 1}{\rm 1+ (\it x/\rm 2)^{\rm 2}}.$$		:$$f_x(x)=\frac{\rm 1}{\rm 2 \pi}\cdot \frac{\rm 1}{\rm 1+ (\it x/\rm 2)^{\rm 2}}.$$

	~~Aus der Grafik ist bereits der extrem langsame Abfall des WDF–Verlaufs zu erkennen~~.		From the graph you can already see the extremely slow decay of the PDF course.


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	~~''Hinweise~~:''		Hints:
	*~~Die Aufgabe gehört zum Kapitel~~  [[Theory_of_Stochastic_Signals/~~Weitere_Verteilungen~~\|~~Weitere Verteilungen~~]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Further_Distributions\|Further Distributions]].
	*~~Insbesondere wird auf die Seite~~  [[Theory_of_Stochastic_Signals/~~Weitere_Verteilungen~~#~~Cauchyverteilung~~\|~~Cauchyverteilung~~]]  ~~Bezug genommen~~.		*In particular, reference is made to the page  [[Theory_of_Stochastic_Signals/Further_Distributions#Cauchy_PDF\|Cauchy PDF]] .




	===~~Fragebogen~~===		===Questions===

	<quiz display=simple>		<quiz display=simple>
	{~~Wie lautet die Verteilungsfunktion~~  $F_x(r)$?  ~~Mit welcher Wahrscheinlichkeit ist~~  $x$  ~~betragsmäßig kleiner als~~  $2$?		{What is the distribution function  $F_x(r)$?  What is the probability that  $x$  is smaller than  $2$ by a small amount?
	\|type="{}"}		\|type="{}"}
	${\rm Pr} (\|x\| < 2) \ = \ $ { 50 3% } $ \ \%$		${\rm Pr} (\|x\| < 2) \ = \ $ { 50 3% } $ \ \%$


	{~~Mit welcher Wahrscheinlichkeit ist~~  $x$  ~~betragsmäßig größer als~~  $4$?		{What is the probability that  $x$  is greater in amount than  $4$?
	\|type="{}"}		\|type="{}"}
	${\rm Pr} (\|x\| > 4) \ = \ $ { 29.6 3% } $ \ \%$		${\rm Pr} (\|x\| > 4) \ = \ $ { 29.6 3% } $ \ \%$


	{~~Welche der folgenden Aussagen treffen für die Cauchyverteilung zu~~?		{Which of the following statements are true for the Cauchy distribution?
	\|type="[]"}		\|type="[]"}
	+ ~~Die Cauchyverteilung besitzt eine unendlich große Varianz~~.		+ The Cauchy distribution has an infinitely large variance.
	+ ~~Die Tschebyscheff–Ungleichung macht hier keinen Sinn~~.		+ The Chebyshev inequality makes no sense here.
	+ ~~Eine~~ in ~~der Natur messbare Zufallsgröße ist nie cauchyverteilt~~.		+ A random variable that can be measured in nature is never Cauchy distributed.


	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Vergleicht man die vorgegebene WDF mit der allgemeinen Gleichung im Theorieteil~~, ~~so erkennt man, dass der Parameter~~  $\lambda= 2$  ~~ist~~.		'''(1)'''  Comparing the given PDF with the general equation in the theory part, we see that the parameter  $\lambda= 2$  is.
	*~~Daraus folgt~~  (~~nach Integration über die WDF~~):		*From this follows  (after integration over the PDF):
	:$$F_x ( r ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(\it r/\rm 2).$$		:$$F_x ( r ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(\it r/\rm 2).$$

	*~~Insbesondere sind~~		*In particular.
	:$$F_x ( r = +2 ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(1)=\frac{1}{2} + \frac{\rm 1}{\rm \pi} \cdot \frac{\rm \pi}{4 }=0.75,$$		:$$F_x ( r = +2 ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(1)=\frac{1}{2} + \frac{\rm 1}{\rm \pi} \cdot \frac{\rm \pi}{4 }=0.75,$$
	:$$F_x ( r = -2 ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(-1)=\frac{1}{2} - \frac{\rm 1}{\rm \pi} \cdot \frac{\rm \pi}{4 }=0.25.$$		:$$F_x ( r = -2 ) =\frac{1}{2} + \frac{\rm 1}{\rm \pi}\cdot \rm arctan(-1)=\frac{1}{2} - \frac{\rm 1}{\rm \pi} \cdot \frac{\rm \pi}{4 }=0.25.$$

	*~~Die gesuchte Wahrscheinlichkeit ergibt sich als die Differenz zu~~		*The probability we are looking for is given by the difference of.
	:$${\rm Pr} (\|x\| < 2) = 0.75 - 0.25 \hspace{0.15cm}\underline{=50\%}.$$		:$${\rm Pr} (\|x\| < 2) = 0.75 - 0.25 \hspace{0.15cm}\underline{=50\%}.$$


	'''(2)'''  ~~Nach dem Ergebnis der Teilaufgabe~~  '''(1)'''  ~~ist~~  $F_x ( r = 4 ) = 0.5 + 1/\pi = 0.852$.		'''(2)'''  According to the result of the subtask  '''(1)'''  is  $F_x ( r = 4 ) = 0.5 + 1/\pi = 0.852$.
	*~~Damit gilt für die~~ "~~komplementäre~~" ~~Wahrscheinlichkeit~~  ${\rm Pr} (x > 4)= 0.148$.		*Thus, for the "complementary" probability  ${\rm Pr} (x > 4)= 0.148$.
	*~~Die gesuchte Wahrscheinlichkeit ist aus Symmetriegründen doppelt so groß~~:		*For symmetry reasons, the probability we are looking for is twice as large:
	:$${\rm Pr} (\|x\| >4) \hspace{0.15cm}\underline{ = 29.6\%}.$$		:$${\rm Pr} (\|x\| >4) \hspace{0.15cm}\underline{ = 29.6\%}.$$


	'''(3)'''  <u>~~Alle Lösungsvorschläge~~</u> ~~treffen zu~~:		'''(3)'''  <u>All proposed solutions</u> are true:
	*~~Für die Varianz der Cauchyverteilung gilt nämlich~~:		*For the variance of the Cauchy distribution holds namely:
	:$$\sigma_x^{\rm 2}=\frac{1}{2\pi}\int_{-\infty}^{+\infty}		:$$\sigma_x^{\rm 2}=\frac{1}{2\pi}\int_{-\infty}^{+\infty}
	\hspace{-0.15cm}		\hspace{-0.15cm}
	\frac{\it x^{\rm 2}}{\rm 1+(\it x/\rm 2)^{\rm 2}} \,\,{\rm d}x.$$		\frac{\it x^{\rm 2}}{\rm 1+(\it x/\rm 2)^{\rm 2}} \,\,{\rm d}x.$$
	*~~Für~~ große  $x$  ~~liefert der Integrand den konstanten Wert~~  $4$. ~~Deshalb divergiert das Integral~~.		*For große  $x$  the integrand yields the constant value  $4$. Therefore the integral diverges.
	*~~Mit~~  $\sigma_x \to \infty$  ~~liefert aber auch die Tschebyscheffsche Ungleichung keine auswertbare Schranke~~.		*With  $\sigma_x \to \infty$  however, even Chebyshev's inequality does not provide an evaluable bound.
	*~~„Natürliche“ Zufallsgrößen~~ (~~physikalisch interpretierbar~~) ~~können nie cauchyverteilt sein~~, ~~da sie sonst eine unendlich große Leistung besitzen müssten~~.		*Natural" random variables (physically interpretable) can never be cauchy distributed, otherwise they would have to have infinite power.
	*~~Dagegen unterliegt eine „künstliche“~~ (~~oder mathematische~~) ~~Zufallsgröße ~~ (~~Beispiel~~: ~~  der Quotient zweier mittelwertfreier Gaußgrößen~~)~~  nicht dieser Beschränkung~~.		*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}}		{{ML-Fuß}}

Guenter at 10:32, 20 December 2021

2021-12-20T10:32:57Z

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	[[Category:Theory of Stochastic Signals: Exercises\|^3.7 ~~Weitere Verteilungen~~^]]		[[Category:Theory of Stochastic Signals: Exercises\|^3.7 Further Distributions^]]