Aufgaben:Exercise 2.5Z: Flower Meadow - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:53:37Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:53, 16 March 2026
Line 97:		Line 97:

	[[Category:Theory of Stochastic Signals: Exercises\|^2.4 Poisson Distribution^]]		[[Category:Theory of Stochastic Signals: Exercises\|^2.4 Poisson Distribution^]]
	[[de:~~Zusatzaufgaben~~:2.~~5_Blumenwiese~~]]		[[de:Aufgaben:Aufgabe 2.5Z: Blumenwiese]]

Maintenance script: Add German interlanguage link

2026-03-16T12:29:10Z

Add German interlanguage link

← Older revision		Revision as of 14:29, 16 March 2026
Line 97:		Line 97:

	[[Category:Theory of Stochastic Signals: Exercises\|^2.4 Poisson Distribution^]]		[[Category:Theory of Stochastic Signals: Exercises\|^2.4 Poisson Distribution^]]
			[[de:Zusatzaufgaben:2.5_Blumenwiese]]

Hwang at 12:20, 18 January 2023

2023-01-18T12:20:21Z

← Older revision		Revision as of 14:20, 18 January 2023
Line 66:		Line 66:


	'''(2)'''  For the ~~quadratic mean~~ of the random variable  $z$  applies accordingly:		'''(2)'''  For the second moment of the random variable  $z$  applies accordingly:
	:$$m_{\rm 2\it z}=\frac{1}{10}\cdot (0^2+1^2+ 2\cdot 2^2+ 2\cdot 3^2+2\cdot 4^2+ 5^2+6^2)=12.$$		:$$m_{\rm 2\it z}=\frac{1}{10}\cdot (0^2+1^2+ 2\cdot 2^2+ 2\cdot 3^2+2\cdot 4^2+ 5^2+6^2)=12.$$

Bene: Text replacement - " rms " to " standard deviation "

2022-02-17T11:34:31Z

Text replacement - " rms " to " standard deviation "

← Older revision		Revision as of 13:34, 17 February 2022
Line 76:		Line 76:

	'''(3)'''  Correct are  <u>solutions 1, 2, and 4</u>:		'''(3)'''  Correct are  <u>solutions 1, 2, and 4</u>:
	*Mean and ~~rms~~ agree here.  This is indicative of the Poisson distribution with rate  $\lambda = 3$  <br>(equal to the mean and equal to the variance,  not equal to the standard deviation).		*Mean and standard deviation agree here.  This is indicative of the Poisson distribution with rate  $\lambda = 3$  <br>(equal to the mean and equal to the variance,  not equal to the standard deviation).
	*Naturally,  it is questionable to make this statement on the basis of only ten values.		*Naturally,  it is questionable to make this statement on the basis of only ten values.
	*However,  in the case of moments,  a smaller sample number is less serious than,  for example,  in the case of probabilities.		*However,  in the case of moments,  a smaller sample number is less serious than,  for example,  in the case of probabilities.

Bene: Text replacement - "rms value" to "standard deviation"

2022-02-17T11:11:29Z

Text replacement - "rms value" to "standard deviation"

← Older revision		Revision as of 13:11, 17 February 2022
Line 34:		Line 34:


	{Determine the ~~rms value~~ of the random variable  $z$.		{Determine the standard deviation of the random variable  $z$.
	\|type="{}"}		\|type="{}"}
	$\sigma_z\ = \ $ { 1.732 3% }		$\sigma_z\ = \ $ { 1.732 3% }
Line 43:		Line 43:
	+ Actually,  one would have to use considerably more than ten random numbers  (squares)  for the moment calculation.		+ Actually,  one would have to use considerably more than ten random numbers  (squares)  for the moment calculation.
	+ The random variable  $z$  is in fact Poisson distributed.		+ The random variable  $z$  is in fact Poisson distributed.
	- The rate  $\lambda$  of the Poisson distribution is equal to the ~~rms value~~  $\sigma_z$.		- The rate  $\lambda$  of the Poisson distribution is equal to the standard deviation  $\sigma_z$.
	+ The rate  $\lambda$  of the Poisson distribution is equal to the mean  $m_z$.		+ The rate  $\lambda$  of the Poisson distribution is equal to the mean  $m_z$.

Line 71:		Line 71:
	*According to Steiner's theorem, the variance is:		*According to Steiner's theorem, the variance is:
	:$$\sigma_z^2 =12 -3^2 = 3,$$		:$$\sigma_z^2 =12 -3^2 = 3,$$
	:and thus the ~~rms value~~:		:and thus the standard deviation:
	:$$\underline{\sigma_z \approx 1.732}.$$		:$$\underline{\sigma_z \approx 1.732}.$$


	'''(3)'''  Correct are  <u>solutions 1, 2, and 4</u>:		'''(3)'''  Correct are  <u>solutions 1, 2, and 4</u>:
	*Mean and rms agree here.  This is indicative of the Poisson distribution with rate  $\lambda = 3$  <br>(equal to the mean and equal to the variance,  not equal to the ~~rms value~~).		*Mean and rms agree here.  This is indicative of the Poisson distribution with rate  $\lambda = 3$  <br>(equal to the mean and equal to the variance,  not equal to the standard deviation).
	*Naturally,  it is questionable to make this statement on the basis of only ten values.		*Naturally,  it is questionable to make this statement on the basis of only ten values.
	*However,  in the case of moments,  a smaller sample number is less serious than,  for example,  in the case of probabilities.		*However,  in the case of moments,  a smaller sample number is less serious than,  for example,  in the case of probabilities.

Guenter at 14:26, 18 December 2021

2021-12-18T14:26:39Z

← Older revision		Revision as of 16:26, 18 December 2021
Line 17:		Line 17:

	Hints:		Hints:
	*This exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Poisson_Distribution\|Poisson ~~distribution~~]].		*This exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Poisson_Distribution\|Poisson Distribution]].
	*Reference is also made to the chapter  [[Theory_of_Stochastic_Signals/Moments_of_a_Discrete_Random_Variable\|Moments of a Discrete Random Variable]].		*Reference is also made to the chapter  [[Theory_of_Stochastic_Signals/Moments_of_a_Discrete_Random_Variable\|Moments of a Discrete Random Variable]].

Line 60:		Line 60:
	</quiz>		</quiz>

	===~~Musterlösung~~===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  The linear mean of these ten numbers gives  $\underline{m_z = 3}$.		'''(1)'''  The linear mean of these ten numbers gives
			:$$\underline{m_z = 3}.$$


Line 68:		Line 69:
	:$$m_{\rm 2\it z}=\frac{1}{10}\cdot (0^2+1^2+ 2\cdot 2^2+ 2\cdot 3^2+2\cdot 4^2+ 5^2+6^2)=12.$$		:$$m_{\rm 2\it z}=\frac{1}{10}\cdot (0^2+1^2+ 2\cdot 2^2+ 2\cdot 3^2+2\cdot 4^2+ 5^2+6^2)=12.$$

	*According to Steiner's theorem, the variance is ~~thus equal to.~~		*According to Steiner's theorem, the variance is:
	:$$\sigma_z^2 =12 -3^2 = 3$$		:$$\sigma_z^2 =12 -3^2 = 3,$$
	:and ~~accordingly~~ the rms		:and thus the rms value:
	:$$\underline{\sigma_z \approx 1.732}.$$		:$$\underline{\sigma_z \approx 1.732}.$$


			'''(3)'''  Correct are  <u>solutions 1, 2, and 4</u>:
			*Mean and rms agree here.  This is indicative of the Poisson distribution with rate  $\lambda = 3$  <br>(equal to the mean and equal to the variance,  not equal to the rms value).
			*Naturally,  it is questionable to make this statement on the basis of only ten values.
			*However,  in the case of moments,  a smaller sample number is less serious than,  for example,  in the case of probabilities.

	~~'''(3)'''  Correct <u>solutions 1, 2, and 4</u>:~~
	*Mean and rms agree here.  This is indicative of the Poisson distribution with rate  $\lambda = 3$  (equal to the mean and equal to the variance, not equal to the rms).
	*Naturally, it is questionable to make this statement on the basis of only ten values. However, in the case of moments, a smaller sample number is less serious than, for example, in the case of probabilities.


			'''(4)'''  In total,  there  are $80000$  such squares,  each with three flowers in the mean.
	'''(4)'''  In total, there  are $80000$  such squares, each with three flowers in the mean.
	*This suggests a total of   $\underline{B = 240}$  thousand flowers.		*This suggests a total of   $\underline{B = 240}$  thousand flowers.



	'''(5)'''  According to the Poisson distribution, this probability results in		'''(5)'''  According to the Poisson distribution,  this probability results in
	$${\rm Pr}(z = 0) = \frac{3^0}{0!} \cdot{\rm e}^{-3}\hspace{0.15cm}\underline{\approx 5\%}.$$		:$${\rm Pr}(z = 0) = \frac{3^0}{0!} \cdot{\rm e}^{-3}\hspace{0.15cm}\underline{\approx 5\%}.$$

	*However, the small sample size  $N = 10$  on which this task was based would have indicated probability  ${\rm Pr}(z = 0) = { 10\%}$  since only in a single square no single flower was counted.		*However,  the small sample size  $N = 10$  on which this task was based would have indicated probability  ${\rm Pr}(z = 0) = { 10\%}$  <br>since only in a single square no single flower was counted.

	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 14:17, 18 December 2021

2021-12-18T14:17:17Z

← Older revision		Revision as of 16:17, 18 December 2021
Line 3:		Line 3:
	}}		}}

	[[File:P_ID124__Sto_Z_2_5.gif\|right\|frame\|~~flower~~ meadow – another ~~<br>~~example of the Poisson distribution]]		[[File:P_ID124__Sto_Z_2_5.gif\|right\|frame\|Flower meadow – another example of the Poisson distribution]]
	A farmer is happy about the splendor of flowers on his land and wants to know how many dandelions are currently blooming on his meadow.		A farmer is happy about the splendor of flowers on his land and wants to know how many dandelions are currently blooming on his meadow.
	*He knows that the meadow has an area of $5000$  square meters and he also knows from the agricultural school that the number of flowers in a small area is always ~~poisson~~ distributed.		*He knows that the meadow has an area of  $5000$  square meters and he also knows from the agricultural school that the number of flowers in a small area is always Poisson distributed.
	*He stakes out ten squares, each with an edge length of   $\text{25 cm}$ , randomly distributed over the entire meadow and counts the flowers in each of these squares:		*He stakes out ten squares,  each with an edge length of  $\text{25 cm}$,  randomly distributed over the entire meadow and counts the flowers in each of these squares:
	::$$\rm 3, \ 4, \ 1, \ 5, \ 0, \ 3, \ 2, \ 4, \ 2, \ 6.$$		::$$\rm 3, \ 4, \ 1, \ 5, \ 0, \ 3, \ 2, \ 4, \ 2, \ 6.$$

	Consider these numerical values as random results of the discrete random variable  $z$.		Consider these numerical values as random results of the discrete random variable  $z$.

	It is obvious that the sample size ~~is very small at~~  $10$  but – this much is revealed – the farmer is lucky.  First consider how you would proceed to solve this task, and then answer the following questions.		It is obvious that the sample size  $(10)$  is very small but – this much is revealed – the farmer is lucky.  First consider how you would proceed to solve this task,  and then answer the following questions.






Line 33:		Line 29:

	<quiz display=simple>		<quiz display=simple>
	{Find the mean of  $z$,  that is, the mean number of flowers counted in the ten squares.		{Find the mean of  $z$,  that is,  the mean number of flowers counted in each of the ten squares.
	\|type="{}"}		\|type="{}"}
	$m_z \ =$ { 3 3% }		$m_z \ =$ { 3 3% }


	{Determine the rms of the random variable  $z$.		{Determine the rms value of the random variable  $z$.
	\|type="{}"}		\|type="{}"}
	$\sigma_z\ = \ $ { 1.732 3% }		$\sigma_z\ = \ $ { 1.732 3% }
Line 45:		Line 41:
	{Which of the following statements are true?		{Which of the following statements are true?
	\|type="[]"}		\|type="[]"}
	+ Actually, one would have to use considerably more than ten random numbers (squares) for the moment calculation.		+ Actually,  one would have to use considerably more than ten random numbers  (squares)  for the moment calculation.
	+ The random ~~size~~ $z$  is in fact Poisson distributed.		+ The random variable  $z$  is in fact Poisson distributed.
	- The rate  $\lambda$  of the Poisson distribution is equal to the rms  $\sigma_z$.		- The rate  $\lambda$  of the Poisson distribution is equal to the rms value  $\sigma_z$.
	+ The rate  $\lambda$  of the Poisson distribution is equal to the mean  $m_z$.		+ The rate  $\lambda$  of the Poisson distribution is equal to the mean  $m_z$.

Noah at 21:46, 15 December 2021

2021-12-15T21:46:45Z

← Older revision		Revision as of 23:46, 15 December 2021
Line 38:		Line 38:


	{Determine the ~~dispersion~~ of the random variable  $z$.		{Determine the rms of the random variable  $z$.
	\|type="{}"}		\|type="{}"}
	$\sigma_z\ = \ $ { 1.732 3% }		$\sigma_z\ = \ $ { 1.732 3% }
Line 47:		Line 47:
	+ Actually, one would have to use considerably more than ten random numbers (squares) for the moment calculation.		+ Actually, one would have to use considerably more than ten random numbers (squares) for the moment calculation.
	+ The random size $z$  is in fact Poisson distributed.		+ The random size $z$  is in fact Poisson distributed.
	- The rate  $\lambda$  of the Poisson distribution is equal to the ~~dispersion~~  $\sigma_z$.		- The rate  $\lambda$  of the Poisson distribution is equal to the rms  $\sigma_z$.
	+ The rate  $\lambda$  of the Poisson distribution is equal to the mean  $m_z$.		+ The rate  $\lambda$  of the Poisson distribution is equal to the mean  $m_z$.

Line 66:		Line 66:
	===Musterlösung===		===Musterlösung===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  ~~Der lineare Mittelwert dieser zehn Zahlen ergibt~~  $\underline{m_z = 3}$.		'''(1)'''  The linear mean of these ten numbers gives  $\underline{m_z = 3}$.


	'''(2)'''  ~~Für den quadratischen Mittelwert der Zufallsgröße~~  $z$  ~~gilt entsprechend~~:		'''(2)'''  For the quadratic mean of the random variable  $z$  applies accordingly:
	:$$m_{\rm 2\it z}=\frac{1}{10}\cdot (0^2+1^2+ 2\cdot 2^2+ 2\cdot 3^2+2\cdot 4^2+ 5^2+6^2)=12.$$		:$$m_{\rm 2\it z}=\frac{1}{10}\cdot (0^2+1^2+ 2\cdot 2^2+ 2\cdot 3^2+2\cdot 4^2+ 5^2+6^2)=12.$$

	*~~Die Varianz ist nach dem Satz von~~ Steiner ~~somit gleich~~		*According to Steiner's theorem, the variance is thus equal to.
	:$$\sigma_z^2 =12 -3^2 = 3$$		:$$\sigma_z^2 =12 -3^2 = 3$$
	:~~und dementsprechend die Streuung~~		:and accordingly the rms
	:$$\underline{\sigma_z \approx 1.732}.$$		:$$\underline{\sigma_z \approx 1.732}.$$



	'''(3)'''  ~~Richtig sind die~~ <u>~~Lösungsvorschläge~~ 1, 2 ~~und~~ 4</u>:		'''(3)'''  Correct <u>solutions 1, 2, and 4</u>:
	*~~Mittelwert und Streuung stimmen hier überein~~.  ~~Dies ist ein Indiz für die Poissonverteilung mit der Rate~~  $\lambda = 3$  (~~gleich dem Mittelwert und gleich der Varianz~~, ~~nicht gleich der Streuung~~).		*Mean and rms agree here.  This is indicative of the Poisson distribution with rate  $\lambda = 3$  (equal to the mean and equal to the variance, not equal to the rms).
	*~~Natürlich ist es fragwürdig~~, ~~diese Aussage auf der Basis von nur zehn Werten zu treffen~~. ~~Bei den Momenten ist eine geringere Stichprobenanzahl aber weniger gravierend als beispielsweise bei den Wahrscheinlichkeiten~~.		*Naturally, it is questionable to make this statement on the basis of only ten values. However, in the case of moments, a smaller sample number is less serious than, for example, in the case of probabilities.



	'''(4)'''  ~~Insgesamt gibt es~~  $80000$  ~~solcher Quadrate mit jeweils drei Blumen im Mittel~~.		'''(4)'''  In total, there  are $80000$  such squares, each with three flowers in the mean.
	*~~Dies lässt auf insgesamt~~  $\underline{B = 240}$  ~~Tausend Blumen schließen~~.		*This suggests a total of   $\underline{B = 240}$  thousand flowers.



	'''(5)'''  ~~Diese Wahrscheinlichkeit ergibt sich gemäß der Poissonverteilung zu~~		'''(5)'''  According to the Poisson distribution, this probability results in
	:$${\rm Pr}(z = 0) = \frac{3^0}{0!} \cdot{\rm e}^{-3}\hspace{0.15cm}\underline{\approx 5\%}.$$		$${\rm Pr}(z = 0) = \frac{3^0}{0!} \cdot{\rm e}^{-3}\hspace{0.15cm}\underline{\approx 5\%}.$$

			*However, the small sample size  $N = 10$  on which this task was based would have indicated probability  ${\rm Pr}(z = 0) = { 10\%}$  since only in a single square no single flower was counted.

	*Die dieser Aufgabe zugrunde gelegte kleine Stichprobenmenge  $N = 10$  hätte allerdings auf die Wahrscheinlichkeit  ${\rm Pr}(z = 0) = { 10\%}$  hingedeutet,  da nur in einem einzigen Quadrat keine einzige Blume gezählt wurde.
	{{ML-Fuß}}		{{ML-Fuß}}

Noah at 20:56, 15 December 2021

2021-12-15T20:56:14Z

← Older revision		Revision as of 22:56, 15 December 2021
Line 1:		Line 1:

	{{quiz-Header\|Buchseite=~~Stochastische Signaltheorie~~/~~Poissonverteilung~~		{{quiz-Header\|Buchseite=Theory_of_Stochastic_Signals/Poisson_Distribution
	}}		}}

	[[File:P_ID124__Sto_Z_2_5.gif\|right\|frame\|~~Blumenwiese~~ – ~~ein weiteres~~ <br>~~Beispiel der Poissonverteilung~~]]		[[File:P_ID124__Sto_Z_2_5.gif\|right\|frame\|flower meadow – another <br>example of the Poisson distribution]]
	~~Ein Bauer freut sich über die Blütenpracht auf seinem Grund und möchte wissen, wie viele Löwenzahn gerade auf seiner Wiese blühen~~.		A farmer is happy about the splendor of flowers on his land and wants to know how many dandelions are currently blooming on his meadow.
	*~~Er weiß, dass die Wiese eine Fläche von ~~ $5000$  ~~Quadratmeter hat und außerdem weiß er noch von der Landwirtschaftsschule, dass die Anzahl der Blumen~~ in ~~einem kleinen Gebiet stets  ''poissonverteilt''  ist~~.		*He knows that the meadow has an area of $5000$  square meters and he also knows from the agricultural school that the number of flowers in a small area is always poisson distributed.
	*~~Er steckt über der gesamten Wiese – zufällig verteilt – zehn Quadrate mit einer jeweiligen Kantenlänge von~~  $\text{25 cm}$  ~~ab und zählt~~ in ~~jedem dieser Quadrate die Blumen.  Dabei kommt er zu folgendem Ergebnis~~:		*He stakes out ten squares, each with an edge length of   $\text{25 cm}$ , randomly distributed over the entire meadow and counts the flowers in each of these squares:
	::$$\rm 3, \ 4, \ 1, \ 5, \ 0, \ 3, \ 2, \ 4, \ 2, \ 6.$$		::$$\rm 3, \ 4, \ 1, \ 5, \ 0, \ 3, \ 2, \ 4, \ 2, \ 6.$$

	~~Betrachten Sie diese Zahlenwerte als zufällige Ergebnisse der diskreten Zufallsgröße~~  $z$.		Consider these numerical values as random results of the discrete random variable  $z$.

	~~Es ist offensichtlich, dass die Stichprobenmenge mit~~  $10$  ~~sehr klein ist, aber~~ – ~~soviel sei verraten~~ – ~~der Bauer hat Glück~~.  ~~Überlegen Sie sich zunächst, wie Sie zur Lösung dieser Aufgabe vorgehen würden~~, ~~und beantworten Sie dann die folgenden Fragen~~.		It is obvious that the sample size is very small at  $10$  but – this much is revealed – the farmer is lucky.  First consider how you would proceed to solve this task, and then answer the following questions.


Line 19:		Line 19:


	~~''Hinweise~~:''
	*~~Die Aufgabe gehört zum Kapitel~~  [[Theory_of_Stochastic_Signals/~~Poissonverteilung~~\|~~Poissonverteilung~~]].		Hints:
	*~~Bezug genommen wird auch auf das Kapitel~~  [[Theory_of_Stochastic_Signals/~~Momente_einer_diskreten_Zufallsgröße~~\|~~Momente einer diskreten Zufallsgröße~~]].		*This exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Poisson_Distribution\|Poisson distribution]].
			*Reference is also made to the chapter  [[Theory_of_Stochastic_Signals/Moments_of_a_Discrete_Random_Variable\|Moments of a Discrete Random Variable]].


Line 29:		Line 30:


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

	<quiz display=simple>		<quiz display=simple>
	{~~Ermitteln Sie den Mittelwert von~~  $z$,  ~~das heißt die mittlere Anzahl der~~ in ~~den zehn Quadraten abgezählten Blumen~~.		{Find the mean of  $z$,  that is, the mean number of flowers counted in the ten squares.
	\|type="{}"}		\|type="{}"}
	$m_z \ =$ { 3 3% }		$m_z \ =$ { 3 3% }


	{~~Bestimmen Sie die Streuung der Zufallsgröße~~  $z$.		{Determine the dispersion of the random variable  $z$.
	\|type="{}"}		\|type="{}"}
	$\sigma_z\ = \ $ { 1.732 3% }		$\sigma_z\ = \ $ { 1.732 3% }


	{~~Welche der folgenden Aussagen sind zutreffend~~?		{Which of the following statements are true?
	\|type="[]"}		\|type="[]"}
	+ ~~Eigentlich müsste man deutlich mehr als zehn Zufallszahlen~~ (~~Quadrate~~) ~~zur Momentenberechnung heranziehen~~.		+ Actually, one would have to use considerably more than ten random numbers (squares) for the moment calculation.
	+ ~~Die Zufallsgröße ~~ $z$  ~~ist tatsächlich poissonverteilt~~.		+ The random size $z$  is in fact Poisson distributed.
	- ~~Die Rate~~  $\lambda$  ~~der Poissonverteilung ist gleich der Streuung~~  $\sigma_z$.		- The rate  $\lambda$  of the Poisson distribution is equal to the dispersion  $\sigma_z$.
	+ ~~Die Rate~~  $\lambda$  ~~der Poissonverteilung ist gleich dem Mittelwert~~  $m_z$.		+ The rate  $\lambda$  of the Poisson distribution is equal to the mean  $m_z$.


	{~~Sagen Sie die Gesamtzahl~~  $B$  ~~aller Blumen auf der Wiese voraus~~.		{Predict the total number  $B$  of all flowers in the meadow.
	\|type="{}"}		\|type="{}"}
	$B\ = \ $ { 240 3% } $\ \text{~~Tausend~~}$		$B\ = \ $ { 240 3% } $\ \text{thousand}$


	{~~Wie groß ist die Wahrscheinlichkeit für ein Quadrat ganz ohne Blumen~~?		{What is the probability of a square without any flowers?
	\|type="{}"}		\|type="{}"}
	${\rm Pr}(z = 0) \ = \ $ { 5 3% } $\ \%$		${\rm Pr}(z = 0) \ = \ $ { 5 3% } $\ \%$

Guenter at 10:08, 1 December 2021

2021-12-01T10:08:50Z

← Older revision		Revision as of 12:08, 1 December 2021
Line 97:		Line 97:


	[[Category:Theory of Stochastic Signals: Exercises\|^2.4 ~~Poissonverteilung~~^]]		[[Category:Theory of Stochastic Signals: Exercises\|^2.4 Poisson Distribution^]]