Aufgaben:Exercise 3.11: Chebyshev's Inequality - Revision history

Maintenance script: Fix interlanguage link: resolve redirect chain

2026-03-16T15:54:06Z

Fix interlanguage link: resolve redirect chain

← Older revision		Revision as of 17:54, 16 March 2026
Line 94:		Line 94:

	[[Category:Theory of Stochastic Signals: Exercises\|^3.7 Further Distributions^]]		[[Category:Theory of Stochastic Signals: Exercises\|^3.7 Further Distributions^]]
	[[de:Aufgaben:3.~~11_Tschebyscheff~~]]		[[de:Aufgaben:Aufgabe 3.11: Tschebyscheffsche Ungleichung]]

Maintenance script: Join split $$ formula lines for correct rendering

2026-03-16T13:48:00Z

Join split $$ formula lines for correct rendering

← Older revision		Revision as of 15:48, 16 March 2026
Line 82:		Line 82:

	*Since in this special case the random variable is always  $x >0$,  it further holds:		*Since in this special case the random variable is always  $x >0$,  it further holds:
	:$$p_k= {\rm Pr}( x \ge k+1)=\int_{k+\rm 1}^{\infty}\hspace{-0.15cm}		:$$p_k= {\rm Pr}( x \ge k+1)=\int_{k+\rm 1}^{\infty}\hspace{-0.15cm}{\rm e}^{-x}\, {\rm d} x={\rm e}^{-( k + 1)}.$$
	{\rm e}^{-x}\, {\rm d} x={\rm e}^{-( k + 1)}.$$

	*This yields the following numerical values for the exponential distribution:		*This yields the following numerical values for the exponential distribution:

Maintenance script: Add German interlanguage link

2026-03-16T12:31:59Z

Add German interlanguage link

← Older revision		Revision as of 14:31, 16 March 2026
Line 95:		Line 95:

	[[Category:Theory of Stochastic Signals: Exercises\|^3.7 Further Distributions^]]		[[Category:Theory of Stochastic Signals: Exercises\|^3.7 Further Distributions^]]
			[[de:Aufgaben:3.11_Tschebyscheff]]

Guenter at 15:39, 13 March 2023

2023-03-13T15:39:05Z

← Older revision		Revision as of 17:39, 13 March 2023
Line 3:		Line 3:
	}}		}}

	[[File:EN_Sto_A_3_11_neu.png\|right\|frame\|Exemplary Chebyshev's bound ~~'''Korrektur''' Chebyshev~~]]		[[File:EN_Sto_A_3_11_neu.png\|right\|frame\|Exemplary Chebyshev bound]]

	[[File:P_ID921__Sto_A_3_11_b.png\|frame\|Values of the  "complementary Gaussian error function"]]		[[File:P_ID921__Sto_A_3_11_b.png\|frame\|Values of the  "complementary Gaussian error function"]]

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 20:		Line 20:


	From this plot it can be seen that the  "Chebyshev's Inequality"  is only a very rough bound.  <br>It should be used only if really only the mean and the ~~rms~~ are known from the random size.		From this plot it can be seen that the  "Chebyshev's Inequality"  is only a very rough bound.  <br>It should be used only if really only the mean and the standard deviation are known from the random size.

Bene: Text replacement - "root mean square" to "standard deviation"

2022-02-17T11:01:02Z

Text replacement - "root mean square" to "standard deviation"

← Older revision		Revision as of 13:01, 17 February 2022
Line 8:		Line 8:
	If nothing else is known about a random variable  $x$  than only		If nothing else is known about a random variable  $x$  than only
	*the mean value  $m_x$,  and		*the mean value  $m_x$,  and
	*the ~~root mean square~~  $\sigma_x$,		*the standard deviation  $\sigma_x$,

Guenter at 16:01, 8 February 2022

2022-02-08T16:01:41Z

← Older revision		Revision as of 18:01, 8 February 2022
Line 3:		Line 3:
	}}		}}

	[[File:~~EN_Sto_A_3_11~~.png\|right\|frame\|Exemplary Chebyshev's bound '''Korrektur''' Chebyshev]]		[[File:EN_Sto_A_3_11_neu.png\|right\|frame\|Exemplary Chebyshev's bound '''Korrektur''' Chebyshev]]

	[[File:P_ID921__Sto_A_3_11_b.png\|frame\|Values of the  "complementary Gaussian error function"]]		[[File:P_ID921__Sto_A_3_11_b.png\|frame\|Values of the  "complementary Gaussian error function"]]

Guenter at 11:59, 3 February 2022

2022-02-03T11:59:59Z

← Older revision		Revision as of 13:59, 3 February 2022
Line 3:		Line 3:
	}}		}}

	[[File:EN_Sto_A_3_11.png\|right\|frame\|Exemplary Chebyshev's bound]]		[[File:EN_Sto_A_3_11.png\|right\|frame\|Exemplary Chebyshev's bound '''Korrektur''' Chebyshev]]

	[[File:P_ID921__Sto_A_3_11_b.png\|frame\|Values of the  "complementary Gaussian error function"]]		[[File:P_ID921__Sto_A_3_11_b.png\|frame\|Values of the  "complementary Gaussian error function"]]
Line 64:		Line 64:
	*For  $\varepsilon < \sigma_x$   ⇒   Chebyshev yields a probability greater  $1$.  This information is useless.		*For  $\varepsilon < \sigma_x$   ⇒   Chebyshev yields a probability greater  $1$.  This information is useless.
	*The last statement is true.  For example,  with the uniform distribution:		*The last statement is true.  For example,  with the uniform distribution:
	:$${\rm Pr}(\| x- m_x \| \ge \varepsilon)=\left\{ \begin{array}{*{4}{c}} 1-{\varepsilon}/{\varepsilon_{\rm 0}} & \rm ~~f\ddot{u}r~~\hspace{0.1cm}{\it \varepsilon<\varepsilon_{\rm 0}=\sqrt{\rm 3}\cdot\sigma_x},\rm 0 & \rm else. \end{array} \right. $$		:$${\rm Pr}(\| x- m_x \| \ge \varepsilon)=\left\{ \begin{array}{*{4}{c}} 1-{\varepsilon}/{\varepsilon_{\rm 0}} & \rm for\hspace{0.1cm}{\it \varepsilon<\varepsilon_{\rm 0}=\sqrt{\rm 3}\cdot\sigma_x},\\\rm 0 & \rm else. \end{array} \right. $$


Line 88:		Line 88:
	:$$k= 1\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge \sigma_{x})= \rm e^{-2}= \rm 13.53\%,$$		:$$k= 1\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge \sigma_{x})= \rm e^{-2}= \rm 13.53\%,$$
	:$$k= 2\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge 2 \cdot \sigma_{x})= \rm \rm e^{-3}=\rm 4.97\% ,$$		:$$k= 2\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge 2 \cdot \sigma_{x})= \rm \rm e^{-3}=\rm 4.97\% ,$$
	:$$k= 3~~\text~~\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge 3 \cdot\sigma_{x})= \rm \rm e^{-4}\hspace{0.15cm}\underline{ =\rm 1.83\% },$$		:$$k= 3\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge 3 \cdot\sigma_{x})= \rm \rm e^{-4}\hspace{0.15cm}\underline{ =\rm 1.83\% },$$
	:$$k= 4\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge 4 \cdot \sigma_{x}) = \rm e^{-5}= \rm 0.67\%.$$		:$$k= 4\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge 4 \cdot \sigma_{x}) = \rm e^{-5}= \rm 0.67\%.$$
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 11:41, 3 February 2022

2022-02-03T11:41:37Z

Guenter at 11:31, 3 February 2022

2022-02-03T11:31:07Z

← Older revision		Revision as of 13:31, 3 February 2022
Line 3:		Line 3:
	}}		}}

	[[File:EN_Sto_A_3_11.png\|right\|frame\|Exemplary ~~Chebyshevsch barrier~~]]		[[File:EN_Sto_A_3_11.png\|right\|frame\|Exemplary Chebyshev's bound]]
	If nothing else is known about a random ~~size~~  $x$  than only
			[[File:P_ID921__Sto_A_3_11_b.png\|frame\|Values of the  "complementary Gaussian error function"]]
			If nothing else is known about a random variable  $x$  than only
	*the mean value  $m_x$  and		*the mean value  $m_x$  and
	*the ~~rms~~  $\sigma_x$,		*the root mean square  $\sigma_x$,


	so the  ~~<i>~~Chebyshev's Inequality~~</i>~~  an upper bound on the probability that $x$  deviates by more than ~~one value from its mean~~ value $\varepsilon$  ~~is given~~.		so the  "Chebyshev's Inequality"  gives an upper bound on the probability that $x$  deviates by more than a value  $\varepsilon$  from its mean.  This bound is:

	This bound is:
	:$${\rm Pr}(\|x-m_x\|\ge \varepsilon) \le {\sigma_x^{\rm 2}}/{\varepsilon^{\rm 2}}.$$		:$${\rm Pr}(\|x-m_x\|\ge \varepsilon) \le {\sigma_x^{\rm 2}}/{\varepsilon^{\rm 2}}.$$

	To explain:		To explain:
	*In the graph, this upper bound is drawn in red.		*In the graph,  this upper bound is drawn in red.
	*The ~~larger~~ curve shows the actual probability in the uniform distribution.		*The green curve shows the actual probability for the uniform distribution.
	*The blue points are for the exponential distribution.		*The blue points are for the exponential distribution.


	From this plot it can be seen that the  ~~<i>~~Chebyshev's Inequality~~</i>~~  is only a very rough bound.  <br>It should be used only if really only the mean and the rms are known from the random size.		From this plot it can be seen that the  "Chebyshev's Inequality"  is only a very rough bound.  <br>It should be used only if really only the mean and the rms are known from the random size.


	~~[[File:P_ID921__Sto_A_3_11_b.png\|frame\|Values of the complementary Gaussian error function]]~~
	<br>		<br>
	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#Chebyshev.27s_inequality\|Chebyshev's inequality]] .		*In particular,  reference is made to the section  [[Theory_of_Stochastic_Signals/Further_Distributions#Chebyshev.27s_inequality\|"Chebyshev's inequality"]] .

	*On the right, values of the complementary Gaussian error function  ${\rm Q}(x)$  are given.		*On the right,  values of the complementary Gaussian error function  ${\rm Q}(x)$  are given.


Line 39:		Line 39:
	{Which of the following statements are true?		{Which of the following statements are true?
	\|type="[]"}		\|type="[]"}
	- Conceivably, a random variable with  ${\rm Pr}(\|x -m_x> \| \ge 3\sigma_x) = 1/4$.		- Conceivably,  a random variable with  ${\rm Pr}(\|x -m_x \| \ge 3\sigma_x) = 1/4$.
	+ "Chebyshev" ~~yields f~~&~~uuml~~;r  $\varepsilon < \sigma_x$  no information.		+ "Chebyshev"  yields for  $\varepsilon < \sigma_x$  no information.
	+ ${\rm Pr}(\|x -m_x> \| \ge \sigma_x)$  is identically zero for large  $\varepsilon$  if  $x$  is bounded.		+ ${\rm Pr}(\|x -m_x \| \ge \sigma_x)$  is identically zero for large  $\varepsilon$  if  $x$  is bounded.

← Older revision		Revision as of 13:41, 3 February 2022
Line 7:		Line 7:
	[[File:P_ID921__Sto_A_3_11_b.png\|frame\|Values of the  "complementary Gaussian error function"]]		[[File:P_ID921__Sto_A_3_11_b.png\|frame\|Values of the  "complementary Gaussian error function"]]
	If nothing else is known about a random variable  $x$  than only		If nothing else is known about a random variable  $x$  than only
	*the mean value  $m_x$  and		*the mean value  $m_x$,  and
	*the root mean square  $\sigma_x$,		*the root mean square  $\sigma_x$,

Line 59:		Line 59:
	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Correct are <u>the proposed solutions 2 and 3</u>:		'''(1)'''  Correct are  <u>the proposed solutions 2 and 3</u>:
	*The first statement is false. Chebyshev's inequality provides the bound ~~here~~  $1/9$.		*The first statement is false.  Here,  Chebyshev's inequality provides the bound  $1/9$.
	*For no distribution ~~can~~ the probability considered here be equal  $1/4$~~ ~~.		*For no distribution the probability considered here can be equal  $1/4$.
	*For  $\varepsilon < \sigma_x$  Chebyshev yields a probability greater  $1$.  This information is useless.		*For  $\varepsilon < \sigma_x$   ⇒   Chebyshev yields a probability greater  $1$.  This information is useless.
	*The last statement is true.  For example, in the uniform distribution:		*The last statement is true.  For example,  with the uniform distribution:
	:$${\rm Pr}(\| x- m_x \| \ge \varepsilon)=\left\{ \begin{array}{*{4}{c}} 1-{\varepsilon}/{\varepsilon_{\rm 0}} & \rm f\ddot{u}r\hspace{0.1cm}{\it \varepsilon<\varepsilon_{\rm 0}=\sqrt{\rm 3}\cdot\sigma_x},\rm 0 & \rm else. \end{array} \right. $$		:$${\rm Pr}(\| x- m_x \| \ge \varepsilon)=\left\{ \begin{array}{*{4}{c}} 1-{\varepsilon}/{\varepsilon_{\rm 0}} & \rm f\ddot{u}r\hspace{0.1cm}{\it \varepsilon<\varepsilon_{\rm 0}=\sqrt{\rm 3}\cdot\sigma_x},\rm 0 & \rm else. \end{array} \right. $$


	'''(2)'''  For the Gaussian distribution holds:		'''(2)'''  For the Gaussian distribution holds:
	:$$p_k={\rm Pr}(\| x-m_x\| \ge k\cdot\sigma_{x})=\rm 2\cdot \rm Q(\it k).$$		:$$p_k={\rm Pr}(\| x-m_x\| \ge k\cdot\sigma_{x})=\rm 2\cdot \rm Q(\it k).$$

	*This results in the following numerical values  (in brackets:   bound according to Chebyshev):		*This results in the following numerical values  $($in brackets:   bound according to Chebyshev$)$:
	:$$k= 1\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge \sigma_{x}) = 31.7 \% \hspace{0.3cm}(100 \%),$$		:$$k= 1\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge \sigma_{x}) = 31.7 \% \hspace{0.3cm}(100 \%),$$
	:$$k= 2\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge 2 \cdot \sigma_{x}) = 4.54 \% \hspace{0.3cm}(25 \%),$$		:$$k= 2\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge 2 \cdot \sigma_{x}) = 4.54 \% \hspace{0.3cm}(25 \%),$$
Line 75:		Line 76:
	:$$k= 4\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge 4 \cdot \sigma_{x}) = 0.0064 \% \hspace{0.3cm}(6.25 \%).$$		:$$k= 4\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge 4 \cdot \sigma_{x}) = 0.0064 \% \hspace{0.3cm}(6.25 \%).$$

	'''(3)'''  Without restricting generality, we set  $\lambda = 1$
			'''(3)'''  Without restricting generality,  we set  $\lambda = 1$
	⇒   $m_x = \sigma_x = 1$.  Then holds:		⇒   $m_x = \sigma_x = 1$.  Then holds:
	:$${\rm Pr}(\|x - m_x\| \ge k\cdot\sigma_{x}) = {\rm Pr}(\| x-1\| \ge k).$$		:$${\rm Pr}(\|x - m_x\| \ge k\cdot\sigma_{x}) = {\rm Pr}(\| x-1\| \ge k).$$

	*Since in this special case the random variable is always  $x >0$ , it further holds:		*Since in this special case the random variable is always  $x >0$,  it further holds:
	:$$p_k= {\rm Pr}( x \ge k+1)=\int_{k+\rm 1}^{\infty}\hspace{-0.15cm}		:$$p_k= {\rm Pr}( x \ge k+1)=\int_{k+\rm 1}^{\infty}\hspace{-0.15cm}
	{\rm e}^{-x}\, {\rm d} x={\rm e}^{-( k + 1)}.$$		{\rm e}^{-x}\, {\rm d} x={\rm e}^{-( k + 1)}.$$

	*This yields the following numerical values for the exponential distribution:		*This yields the following numerical values for the exponential distribution:
	:$$k= 1\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge \sigma_{x}) \rm e^{-2}= \rm 13.53\%,$$		:$$k= 1\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge \sigma_{x})= \rm e^{-2}= \rm 13.53\%,$$
	:$$k= 2\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge 2 \cdot \sigma_{x})= \rm \rm e^{-3}=\rm 4.97\% ,$$		:$$k= 2\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge 2 \cdot \sigma_{x})= \rm \rm e^{-3}=\rm 4.97\% ,$$
	:$$k= 3\text\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge 3 \cdot\sigma_{x})= \rm \rm e^{-4}\hspace{0.15cm}\underline{ =\rm 1.83\% },$$		:$$k= 3\text\text{:}\hspace{0.5cm} {\rm Pr}(\|x-m_x\| \ge 3 \cdot\sigma_{x})= \rm \rm e^{-4}\hspace{0.15cm}\underline{ =\rm 1.83\% },$$