Theory of Stochastic Signals/Linear Combinations of Random Variables - Revision history

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2026-03-16T14:15:43Z

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	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Conclusion:}$ After evaluating this equation in an analogous manner to above,  it follows:		$\text{Conclusion:}$ After evaluating this equation in an analogous manner to above,  it follows:
	:$$\mu_{xy } = (A \cdot D + B \cdot E) \cdot \sigma^{\rm 2 }		:$$\mu_{xy } = (A \cdot D + B \cdot E) \cdot \sigma^{\rm 2 }\hspace{0.3cm}\Rightarrow\hspace{0.3cm}\rho_{xy } = \frac{\rho_{xy } }{\sigma_x \cdot \sigma_y} = \frac {A \cdot D + B \cdot E}{\sqrt{(A^{\rm 2}+B^{\rm 2})(D^{\rm 2}+E^{\rm 2} ) } }. $$}}
	\hspace{0.3cm}\Rightarrow\hspace{0.3cm}
	\rho_{xy } = \frac{\rho_{xy } }{\sigma_x \cdot \sigma_y} = \frac {A \cdot D + B \cdot E}{\sqrt{(A^{\rm 2}+B^{\rm 2})(D^{\rm 2}+E^{\rm 2} ) } }. $$}}

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2026-03-16T12:28:48Z

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	{{Display}}		{{Display}}
			[[de:Stochastische_Signaltheorie/Linearkombinationen_von_Zufallsgrößen]]

Hwang at 14:02, 21 December 2022

2022-12-21T14:02:08Z

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	:$$m_x =A \cdot m_u + B \cdot m_v + C =C,$$		:$$m_x =A \cdot m_u + B \cdot m_v + C =C,$$
	:$$m_y =D \cdot m_u + E \cdot m_v + F =F.$$		:$$m_y =D \cdot m_u + E \cdot m_v + F =F.$$
	Thus,  the coefficients  $C$  and  $F$  give only the mean values of  $x$  and  $y$.  Both are always set to zero on the following ~~pages~~.		Thus,  the coefficients  $C$  and  $F$  give only the mean values of  $x$  and  $y$.  Both are always set to zero in the following sections.
	==Resulting correlation coefficient==		==Resulting correlation coefficient==
	<br>		<br>

Hwang at 13:58, 21 December 2022

2022-12-21T13:58:07Z

Guenter at 16:27, 31 March 2022

2022-03-31T16:27:59Z

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	$(1)$   To generate a two-dimensional random variable with desired characteristics   $σ_x =1$,  $σ_y = 1.55$  and  $ρ_{xy} = -0.8$   the parameter set is suitable, for example:		$(1)$   To generate a two-dimensional random variable with desired characteristics   $σ_x =1$,  $σ_y = 1.55$  and  $ρ_{xy} = -0.8$   the parameter set is suitable, for example:
	[[File:~~P_ID419__Sto_T_4_3_S3_neu~~.png\|right \|frame\| Two-dimensional random variables generated by linear combination]]		[[File:EN_Sto_T_4_3_S3_v1.png\|right \|frame\| Two-dimensional random variables generated by linear combination]]
	:$$A = -0.8, \; B = 0.6, \; D = 1.55, \; E = 0.$$		:$$A = -0.8, \; B = 0.6, \; D = 1.55, \; E = 0.$$
	*This parameter set underlies the left graph.		*This parameter set underlies the left graph.

Guenter at 12:51, 25 February 2022

2022-02-25T12:51:03Z

← Older revision		Revision as of 14:51, 25 February 2022
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	[[Aufgaben:Exercise_4.7Z:_Generation_of_a_joint_PDF\|Exercise 4.7Z: Generation of a Joint PDF]]		[[Aufgaben:Exercise_4.7Z:_Generation_of_a_joint_PDF\|Exercise 4.7Z: Generation of a Joint PDF]]

	[[Aufgaben:Exercise_4.8:~~_Diamond_shaped_joint_PDF~~\|Exercise 4.8: Diamond shaped Joint PDF]]		[[Aufgaben:Exercise_4.8:_Diamond-shaped_Joint_PDF\|Exercise 4.8: Diamond-shaped Joint PDF]]

	[[Aufgaben:Exercise_4.8Z:_AWGN_Channel\|Exercise 4.8Z: AWGN Channel]]		[[Aufgaben:Exercise_4.8Z:_AWGN_Channel\|Exercise 4.8Z: AWGN Channel]]

Guenter at 12:49, 25 February 2022

2022-02-25T12:49:13Z

← Older revision		Revision as of 14:49, 25 February 2022
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	[[Aufgaben:Exercise_4.7:_Weighted_Sum_and_Difference\|Exercise 4.7: Weighted Sum and Difference]]		[[Aufgaben:Exercise_4.7:_Weighted_Sum_and_Difference\|Exercise 4.7: Weighted Sum and Difference]]

	[[Aufgaben:Exercise_4.7Z:_Generation_of_a_joint_PDF\|Exercise 4.7Z: Generation of a ~~joint~~ PDF]]		[[Aufgaben:Exercise_4.7Z:_Generation_of_a_joint_PDF\|Exercise 4.7Z: Generation of a Joint PDF]]

	[[Aufgaben:Exercise_4.8:_Diamond_shaped_joint_PDF\|Exercise 4.8: Diamond shaped ~~joint~~ PDF]]		[[Aufgaben:Exercise_4.8:_Diamond_shaped_joint_PDF\|Exercise 4.8: Diamond shaped Joint PDF]]

	[[Aufgaben:Exercise_4.8Z:_AWGN_Channel\|Exercise 4.8Z: AWGN Channel]]		[[Aufgaben:Exercise_4.8Z:_AWGN_Channel\|Exercise 4.8Z: AWGN Channel]]

Guenter at 12:44, 25 February 2022

2022-02-25T12:44:09Z

← Older revision		Revision as of 14:44, 25 February 2022
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	[[Aufgaben:Exercise_4.7:_Weighted_Sum_and_Difference\|Exercise 4.7: Weighted Sum and Difference]]		[[Aufgaben:Exercise_4.7:_Weighted_Sum_and_Difference\|Exercise 4.7: Weighted Sum and Difference]]

	[[Aufgaben:Exercise_4.7Z:~~_Generation_of_a_"2D-PDF"~~\|Exercise 4.7Z: Generation of a joint PDF]]		[[Aufgaben:Exercise_4.7Z:_Generation_of_a_joint_PDF\|Exercise 4.7Z: Generation of a joint PDF]]

	[[Aufgaben:Exercise_4.8:~~_Diamond_shaped_"2D-PDF"~~\|Exercise 4.8: Diamond shaped joint PDF]]		[[Aufgaben:Exercise_4.8:_Diamond_shaped_joint_PDF\|Exercise 4.8: Diamond shaped joint PDF]]

	[[Aufgaben:Exercise_4.8Z:_AWGN_Channel\|Exercise 4.8Z: AWGN Channel]]		[[Aufgaben:Exercise_4.8Z:_AWGN_Channel\|Exercise 4.8Z: AWGN Channel]]

Guenter at 12:42, 25 February 2022

2022-02-25T12:42:14Z

← Older revision		Revision as of 14:42, 25 February 2022
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	[[Aufgaben:Exercise_4.7:_Weighted_Sum_and_Difference\|Exercise 4.7: Weighted Sum and Difference]]		[[Aufgaben:Exercise_4.7:_Weighted_Sum_and_Difference\|Exercise 4.7: Weighted Sum and Difference]]

	[[Aufgaben:Exercise_4.7Z:_Generation_of_a_"2D-PDF"\|Exercise 4.7Z: Generation of a ~~"2D-~~PDF"]]		[[Aufgaben:Exercise_4.7Z:_Generation_of_a_"2D-PDF"\|Exercise 4.7Z: Generation of a joint PDF]]

	[[Aufgaben:Exercise_4.8:_Diamond_shaped_"2D-PDF"\|Exercise 4.8: Diamond shaped ~~"2D-~~PDF"]]		[[Aufgaben:Exercise_4.8:_Diamond_shaped_"2D-PDF"\|Exercise 4.8: Diamond shaped joint PDF]]

	[[Aufgaben:Exercise_4.8Z:_AWGN_Channel\|Exercise 4.8Z: AWGN Channel]]		[[Aufgaben:Exercise_4.8Z:_AWGN_Channel\|Exercise 4.8Z: AWGN Channel]]

Guenter at 12:36, 25 February 2022

2022-02-25T12:36:56Z

← Older revision		Revision as of 14:51, 25 February 2022
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	[[Aufgaben:Exercise_4.7Z:_Generation_of_a_joint_PDF\|Exercise 4.7Z: Generation of a Joint PDF]]		[[Aufgaben:Exercise_4.7Z:_Generation_of_a_joint_PDF\|Exercise 4.7Z: Generation of a Joint PDF]]

	[[Aufgaben:Exercise_4.8:~~_Diamond_shaped_joint_PDF~~\|Exercise 4.8: Diamond shaped Joint PDF]]		[[Aufgaben:Exercise_4.8:_Diamond-shaped_Joint_PDF\|Exercise 4.8: Diamond-shaped Joint PDF]]

	[[Aufgaben:Exercise_4.8Z:_AWGN_Channel\|Exercise 4.8Z: AWGN Channel]]		[[Aufgaben:Exercise_4.8Z:_AWGN_Channel\|Exercise 4.8Z: AWGN Channel]]

← Older revision		Revision as of 14:49, 25 February 2022
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	[[Aufgaben:Exercise_4.7:_Weighted_Sum_and_Difference\|Exercise 4.7: Weighted Sum and Difference]]		[[Aufgaben:Exercise_4.7:_Weighted_Sum_and_Difference\|Exercise 4.7: Weighted Sum and Difference]]

	[[Aufgaben:Exercise_4.7Z:_Generation_of_a_joint_PDF\|Exercise 4.7Z: Generation of a ~~joint~~ PDF]]		[[Aufgaben:Exercise_4.7Z:_Generation_of_a_joint_PDF\|Exercise 4.7Z: Generation of a Joint PDF]]

	[[Aufgaben:Exercise_4.8:_Diamond_shaped_joint_PDF\|Exercise 4.8: Diamond shaped ~~joint~~ PDF]]		[[Aufgaben:Exercise_4.8:_Diamond_shaped_joint_PDF\|Exercise 4.8: Diamond shaped Joint PDF]]

	[[Aufgaben:Exercise_4.8Z:_AWGN_Channel\|Exercise 4.8Z: AWGN Channel]]		[[Aufgaben:Exercise_4.8Z:_AWGN_Channel\|Exercise 4.8Z: AWGN Channel]]

← Older revision		Revision as of 14:44, 25 February 2022
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	[[Aufgaben:Exercise_4.7:_Weighted_Sum_and_Difference\|Exercise 4.7: Weighted Sum and Difference]]		[[Aufgaben:Exercise_4.7:_Weighted_Sum_and_Difference\|Exercise 4.7: Weighted Sum and Difference]]

	[[Aufgaben:Exercise_4.7Z:~~_Generation_of_a_"2D-PDF"~~\|Exercise 4.7Z: Generation of a joint PDF]]		[[Aufgaben:Exercise_4.7Z:_Generation_of_a_joint_PDF\|Exercise 4.7Z: Generation of a joint PDF]]

	[[Aufgaben:Exercise_4.8:~~_Diamond_shaped_"2D-PDF"~~\|Exercise 4.8: Diamond shaped joint PDF]]		[[Aufgaben:Exercise_4.8:_Diamond_shaped_joint_PDF\|Exercise 4.8: Diamond shaped joint PDF]]

	[[Aufgaben:Exercise_4.8Z:_AWGN_Channel\|Exercise 4.8Z: AWGN Channel]]		[[Aufgaben:Exercise_4.8Z:_AWGN_Channel\|Exercise 4.8Z: AWGN Channel]]

← Older revision		Revision as of 14:42, 25 February 2022
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	[[Aufgaben:Exercise_4.7:_Weighted_Sum_and_Difference\|Exercise 4.7: Weighted Sum and Difference]]		[[Aufgaben:Exercise_4.7:_Weighted_Sum_and_Difference\|Exercise 4.7: Weighted Sum and Difference]]

	[[Aufgaben:Exercise_4.7Z:_Generation_of_a_"2D-PDF"\|Exercise 4.7Z: Generation of a ~~"2D-~~PDF"]]		[[Aufgaben:Exercise_4.7Z:_Generation_of_a_"2D-PDF"\|Exercise 4.7Z: Generation of a joint PDF]]

	[[Aufgaben:Exercise_4.8:_Diamond_shaped_"2D-PDF"\|Exercise 4.8: Diamond shaped ~~"2D-~~PDF"]]		[[Aufgaben:Exercise_4.8:_Diamond_shaped_"2D-PDF"\|Exercise 4.8: Diamond shaped joint PDF]]

	[[Aufgaben:Exercise_4.8Z:_AWGN_Channel\|Exercise 4.8Z: AWGN Channel]]		[[Aufgaben:Exercise_4.8Z:_AWGN_Channel\|Exercise 4.8Z: AWGN Channel]]

← Older revision		Revision as of 15:58, 21 December 2022
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	==Resulting correlation coefficient==		==Resulting correlation coefficient==
	<br>		<br>
	Let us now consider the  '''variances'''  of the random variables according to the linear combinations.		Let us now consider the  »'''variances'''«  of the random variables according to the linear combinations.
	*For the random variable  $x$  holds independently of the parameter  $C$:		*For the random variable  $x$  holds independently of the parameter  $C$:
	:$$\sigma _x ^2 = {\rm E}\big[x ^{\rm 2}\big] = A^{\rm 2} \cdot {\rm E}\big[u^{\rm 2}\big] + B^{\rm 2} \cdot {\rm E}\big[v^{\rm 2}\big] + {\rm 2} \cdot A \cdot B \cdot {\rm E}\big[u \cdot v\big].$$		:$$\sigma _x ^2 = {\rm E}\big[x ^{\rm 2}\big] = A^{\rm 2} \cdot {\rm E}\big[u^{\rm 2}\big] + B^{\rm 2} \cdot {\rm E}\big[v^{\rm 2}\big] + {\rm 2} \cdot A \cdot B \cdot {\rm E}\big[u \cdot v\big].$$
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	:$$\sigma _y ^2 =(D^2 + E^2) \cdot \sigma ^2.$$		:$$\sigma _y ^2 =(D^2 + E^2) \cdot \sigma ^2.$$

	The  '''covariance'''  $μ_{xy}$  is identical to the joint moment  $m_{xy}$ for zero mean random variables  $x$  and  $y$   ⇒   $C = F = 0$:		The  »'''covariance'''«  $μ_{xy}$  is identical to the joint moment  $m_{xy}$ for zero mean random variables  $x$  and  $y$   ⇒   $C = F = 0$:
	:$$\mu_{xy } = m_{xy } = {\rm E}\big[x \cdot y\big] = {\rm E}\big[(A \cdot u + B \cdot v)\cdot (D \cdot u + E \cdot v)\big].$$		:$$\mu_{xy } = m_{xy } = {\rm E}\big[x \cdot y\big] = {\rm E}\big[(A \cdot u + B \cdot v)\cdot (D \cdot u + E \cdot v)\big].$$
	Note here that  ${\rm E}\big[ \text{...} \big]$  denotes an expected value,  while  $E$  describes a coefficient.		Note here that  ${\rm E}\big[ \text{...} \big]$  denotes an expected value,  while  $E$  describes a coefficient.
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	==Generation of correlated random variables==		==Generation of correlated random variables==
	<br>		<br>
	The  [[Theory_of_Stochastic_Signals/Linear_Combinations_of_Random_Variables#Resulting_correlation_coefficient\|previously used equations]]  can be used to generate a two-dimensional random variable  $(x,\hspace{0.08cm} y)$  with given characteristics  $σ_x$,  $σ_y$  and  $ρ_{xy}$.		The  [[Theory_of_Stochastic_Signals/Linear_Combinations_of_Random_Variables#Resulting_correlation_coefficient\|$\text{previously used equations}$]]  can be used to generate a two-dimensional random variable  $(x,\hspace{0.08cm} y)$  with given characteristics  $σ_x$,  $σ_y$  and  $ρ_{xy}$.

	*If no further preconditions are met other than these three nominal values, one of the four coefficients  $A, \ B, \ D$  and  $E$  is arbitrary.		*If no further preconditions are met other than these three nominal values, one of the four coefficients  $A, \ B, \ D$  and  $E$  is arbitrary.

← Older revision		Revision as of 14:36, 25 February 2022
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	==Prerequisites and mean values==		==Prerequisites and mean values==
	<br>		<br>
	Throughout the chapter "Linear Combinations of Random Variables" we make the following assumptions:		Throughout the chapter  "Linear Combinations of Random Variables"  we make the following assumptions:
	*The random variables  $u$  and  $v$  are zero mean each   ⇒   $m_u = m_v = 0$  and also statistically independent of each other   ⇒   $ρ_{uv} = 0$.		*The random variables  $u$  and  $v$  are zero mean each   ⇒   $m_u = m_v = 0$  and also statistically independent of each other   ⇒   $ρ_{uv} = 0$.
	*The two random variables  $u$  and  $v$  each have equal standard deviation  $σ$.  No statement is made about the nature of the distribution.		*The two random variables  $u$  and  $v$  each have equal standard deviation  $σ$.  No statement is made about the nature of the distribution.
	*Let the two random variables  $x$  and  $y$  be linear combinations of  $u$  and  $v$, where:		*Let the two random variables  $x$  and  $y$  be linear combinations of  $u$  and  $v$,  where:
	:$$x=A \cdot u + B \cdot v + C,$$		:$$x=A \cdot u + B \cdot v + C,$$
	:$$y=D \cdot u + E \cdot v + F.$$		:$$y=D \cdot u + E \cdot v + F.$$

	Thus, for the (linear) mean values of the new random variables  $x$  and  $y$  we obtain according to the general rules of calculation for expected values:		Thus,  for the  (linear)  mean values of the new random variables  $x$  and  $y$  we obtain according to the general rules of calculation for expected values:
	:$$m_x =A \cdot m_u + B \cdot m_v + C =C,$$		:$$m_x =A \cdot m_u + B \cdot m_v + C =C,$$
	:$$m_y =D \cdot m_u + E \cdot m_v + F =F.$$		:$$m_y =D \cdot m_u + E \cdot m_v + F =F.$$
	Thus, the coefficients  $C$  and  $F$  give only the mean values of  $x$  and  $y$  Both are always set to zero in the following pages.		Thus,  the coefficients  $C$  and  $F$  give only the mean values of  $x$  and  $y$.  Both are always set to zero on the following pages.
	==Resulting correlation coefficient==		==Resulting correlation coefficient==
	<br>		<br>
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	:$$\sigma _x ^2 = {\rm E}\big[x ^{\rm 2}\big] = A^{\rm 2} \cdot {\rm E}\big[u^{\rm 2}\big] + B^{\rm 2} \cdot {\rm E}\big[v^{\rm 2}\big] + {\rm 2} \cdot A \cdot B \cdot {\rm E}\big[u \cdot v\big].$$		:$$\sigma _x ^2 = {\rm E}\big[x ^{\rm 2}\big] = A^{\rm 2} \cdot {\rm E}\big[u^{\rm 2}\big] + B^{\rm 2} \cdot {\rm E}\big[v^{\rm 2}\big] + {\rm 2} \cdot A \cdot B \cdot {\rm E}\big[u \cdot v\big].$$

	*The expected values of  $u^2$  and  $v^2$  are by definition equal to  $σ^2$, ~~respectively,~~ because  $u$  and  $v$  are zero mean.		*The expected values of  $u^2$  and  $v^2$  are by definition equal to  $σ^2$,  because  $u$  and  $v$  are zero mean.
	*Since  $u$  and  $v$  are moreover assumed to be statistically independent, one can ~~also~~ write for the expected value of the product:		*Since  $u$  and  $v$  are moreover assumed to be statistically independent,  one can write for the expected value of the product:
	:$${\rm E}\big[u \cdot v\big] = {\rm E}\big[u\big] \cdot {\rm E}\big[v\big] = m_u \cdot m_v = \rm 0.$$		:$${\rm E}\big[u \cdot v\big] = {\rm E}\big[u\big] \cdot {\rm E}\big[v\big] = m_u \cdot m_v = \rm 0.$$
	*Thus, for the variances of the random variables formed by linear combinations, we obtain:		*Thus,  for the variances of the random variables formed by linear combinations,  we obtain:
	:$$\sigma _x ^2 =(A^2 + B^2) \cdot \sigma ^2,$$		:$$\sigma _x ^2 =(A^2 + B^2) \cdot \sigma ^2,$$
	:$$\sigma _y ^2 =(D^2 + E^2) \cdot \sigma ^2.$$		:$$\sigma _y ^2 =(D^2 + E^2) \cdot \sigma ^2.$$

	The  '''covariance'''  $μ_{xy}$  is identical to the joint moment  $m_{xy}$ for zero mean random variables  $x$  and  $y$   ⇒   $C = F = 0$~~ ~~ :		The  '''covariance'''  $μ_{xy}$  is identical to the joint moment  $m_{xy}$ for zero mean random variables  $x$  and  $y$   ⇒   $C = F = 0$:
	:$$\mu_{xy } = m_{xy } = {\rm E}\big[x \cdot y\big] = {\rm E}\big[(A \cdot u + B \cdot v)\cdot (D \cdot u + E \cdot v)\big].$$		:$$\mu_{xy } = m_{xy } = {\rm E}\big[x \cdot y\big] = {\rm E}\big[(A \cdot u + B \cdot v)\cdot (D \cdot u + E \cdot v)\big].$$
	Note here that  ${\rm E}\big[ \text{...} \big]$  denotes an expected value, while  $E$  describes a ~~variable~~.		Note here that  ${\rm E}\big[ \text{...} \big]$  denotes an expected value,  while  $E$  describes a coefficient.

	{{BlaueBox\|TEXT=		{{BlaueBox\|TEXT=
	$\text{Conclusion:}$ After evaluating this equation in an analogous manner to above, it follows:		$\text{Conclusion:}$ After evaluating this equation in an analogous manner to above,  it follows:
	:$$\mu_{xy } = (A \cdot D + B \cdot E) \cdot \sigma^{\rm 2 }		:$$\mu_{xy } = (A \cdot D + B \cdot E) \cdot \sigma^{\rm 2 }
	\hspace{0.3cm}\~~rightarrow~~\hspace{0.3cm}		\hspace{0.3cm}\Rightarrow\hspace{0.3cm}
	\rho_{xy } = \frac{\rho_{xy } }{\sigma_x \cdot \sigma_y} = \frac {A \cdot D + B \cdot E}{\sqrt{(A^{\rm 2}+B^{\rm 2})(D^{\rm 2}+E^{\rm 2} ) } }. $$}}		\rho_{xy } = \frac{\rho_{xy } }{\sigma_x \cdot \sigma_y} = \frac {A \cdot D + B \cdot E}{\sqrt{(A^{\rm 2}+B^{\rm 2})(D^{\rm 2}+E^{\rm 2} ) } }. $$}}


	We now exclude two special cases:		We now exclude two special cases:
	*$A = B = 0$  ⇒   $x ≡ 0$~~  as well as.~~		*$A = B = 0$  ⇒   $x ≡ 0$,
	*$D = E = 0$  ⇒   $y ≡ 0$.		*$D = E = 0$  ⇒   $y ≡ 0$.

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	$\text{Example 1:}$		$\text{Example 1:}$
	If we set  $A = E = 0$,  we get the correlation coefficient  $ρ_{xy} = 0$.  This result is insightful:		If we set  $A = E = 0$,  we get the correlation coefficient  $ρ_{xy} = 0$.  This result is insightful:
	*Now  $x$  depends only on  $v$  and  $y$  depends exclusively on  $u$~~ ~~ .		*Now  $x$  depends only on  $v$  and  $y$  depends exclusively on  $u$.
	*But since  $u$  and  $v$  were assumed to be statistically independent, there are also no relationships between  $x$  and  $y$.		*But since  $u$  and  $v$  were assumed to be statistically independent,  there are also no relationships between  $x$  and  $y$.
	*Similarly,  $ρ_{xy} = 0$  results for the combination  $B = D = 0$.}}		*Similarly,   $ρ_{xy} = 0$  results for the combination  $B = D = 0$.}}


	{{GraueBox\|TEXT=		{{GraueBox\|TEXT=
	$\text{Example 2:}$  The constellation  $B = E = 0$  leads to the fact that both  $x$  and  $y$  depend only on  $u$  Then the correlation coefficient is obtained:		$\text{Example 2:}$  The constellation  $B = E = 0$  leads to the fact that both  $x$  and  $y$  depend only on  $u$.  Then the following correlation coefficient is obtained:
	:$$\rho_{xy } = \frac {A \cdot D }{\sqrt{A^{\rm 2}\cdot D^{\rm 2} } } = \frac {A \cdot D }{\vert A\vert \cdot D\vert } =\pm 1. $$		:$$\rho_{xy } = \frac {A \cdot D }{\sqrt{A^{\rm 2}\cdot D^{\rm 2} } } = \frac {A \cdot D }{\vert A\vert \cdot D\vert } =\pm 1. $$

	*If  $A$  and  $D$  have the same sign, then  $ρ_{xy} = +1$.		*If  $A$  and  $D$  have the same sign,  then  $ρ_{xy} = +1$.
	*For different signs, the correlation coefficient is  $-1$.		*For different signs,  the correlation coefficient is  $-1$.
	*Also, for  $A = D = 0$  ~~the~~ coefficient  $ρ_{xy} = ±1$ if  $B \ne 0$  and  $E \ne 0$  holds. }}		*Also for  $A = D = 0:$  The coefficient  $ρ_{xy} = ±1$,  if  $B \ne 0$  and  $E \ne 0$  holds. }}
	==Generation of correlated random variables==		==Generation of correlated random variables==
	<br>		<br>
	The  [[Theory_of_Stochastic_Signals/Linear_Combinations_of_Random_Variables#Resulting_correlation_coefficient\|previously used equations]]  can be used to generate a two-dimensional random variable  $(x, y)$  with given characteristics  $σ_x$,  $σ_y$  and  $ρ_{xy}$~~ ~~ .		The  [[Theory_of_Stochastic_Signals/Linear_Combinations_of_Random_Variables#Resulting_correlation_coefficient\|previously used equations]]  can be used to generate a two-dimensional random variable  $(x,\hspace{0.08cm} y)$  with given characteristics  $σ_x$,  $σ_y$  and  $ρ_{xy}$.

	*If no further preconditions are met other than these three nominal values, one of the four coefficients  $A, \ B, \ D$  and  $E$  is arbitrary.		*If no further preconditions are met other than these three nominal values, one of the four coefficients  $A, \ B, \ D$  and  $E$  is arbitrary.
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	{{GraueBox\|TEXT=		{{GraueBox\|TEXT=
	$\text{Example 3:}$		$\text{Example 3:}$
	We always assume zero mean Gaussian random variables  $u$  and  $v$  Both have variance  $σ^2 = 1$.		We always assume zero mean Gaussian random variables  $u$  and  $v$.  Both have variance  $σ^2 = 1$.

	$(1)$   To generate a 2D random variable with ~~the~~ desired characteristics  $σ_x =1$,  $σ_y = 1.55$  and  $ρ_{xy} = -0.8$  the parameter set is suitable, for example.		$(1)$   To generate a two-dimensional random variable with desired characteristics   $σ_x =1$,  $σ_y = 1.55$  and  $ρ_{xy} = -0.8$   the parameter set is suitable, for example:
	[[File:P_ID419__Sto_T_4_3_S3_neu.png\|right \|frame\| 2D random variables generated by linear combination]]		[[File:P_ID419__Sto_T_4_3_S3_neu.png\|right \|frame\| Two-dimensional random variables generated by linear combination]]
	:$$A = -0.8, \; B = 0.6, \; D = 1.55, \; E = 0.$$		:$$A = -0.8, \; B = 0.6, \; D = 1.55, \; E = 0.$$
	*This set ~~of parameters~~ underlies the left graph.		*This parameter set underlies the left graph.
	*The regression line  $K(x)$  is shown in red.		*The regression line  $(\rm RL)$  is shown in red.
	*It runs at an angle of about  $-50^\circ$.		*It runs at an angle of about  $-50^\circ$.
	*Drawn in purple is the ellipse major axis, which lies slightly above the regression line.		*Drawn in purple is the ellipse major axis, which lies slightly above the regression line.
Line 91:		Line 91:
	$(2)$   The parameter set for the right graph is:		$(2)$   The parameter set for the right graph is:
	:$$A = -0.625,\; B = 0.781,\; D = 1.501,\; E = -0.390.$$		:$$A = -0.625,\; B = 0.781,\; D = 1.501,\; E = -0.390.$$
	*In a statistical sense, the same result is obtained even though the two point clouds differ in detail.		*In a statistical sense,  the same result is obtained even though the two point clouds differ in detail.
	*~~In particular, with~~ respect to regression line and ellipse major axis, there is no difference with respect to the parameter set  $(1)$. }}		*With respect to regression line  $(\rm RL)$  and ellipse major axis  $(\rm EA)$,  there is no difference with respect to the parameter set  $(1)$. }}

	==Exercises for the chapter==		==Exercises for the chapter==