Aufgaben:Exercise 4.7Z: Generation of a Joint PDF - Revision history

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	[[Category:Theory of Stochastic Signals: Exercises\|^4.3 Linear Combinations^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.3 Linear Combinations^]]
	[[de:~~Zusatzaufgaben~~:4.~~7_Erzeugung_einer_2D–WDF~~]]		[[de:Aufgaben:Aufgabe 4.7Z: Erzeugung einer 2D–WDF]]

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	[[Category:Theory of Stochastic Signals: Exercises\|^4.3 Linear Combinations^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.3 Linear Combinations^]]
			[[de:Zusatzaufgaben:4.7_Erzeugung_einer_2D–WDF]]

Guenter at 16:45, 25 February 2022

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Guenter at 16:39, 25 February 2022

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	}}		}}

	[[File:P_ID423__Sto_Z_4_7.png\|right\|frame\|Requirements for the generation of a <br>2D random variable]]		[[File:P_ID423__Sto_Z_4_7.png\|right\|frame\|Requirements for the generation of a <br>two-dimensional random variable]]
	Given statistically independent quantities  $u$  and  $v$, both of which are uniformly distributed between  $-1$  and  $+1$  and thus each have variance  $\sigma^2 = 2/3$ , generate a 2D random variable  $(x, y)$  where for the components:		Given statistically independent quantities  $u$  and  $v$,
			*both of which are uniformly distributed between  $-1$  and  $+1$,  and
			*thus each have variance  $\sigma^2 = 2/3$,


			generate a two-dimensional random variable  $(x,\hspace{0.08cm} y)$  where for the components:
	:$$x = A \cdot u + B \cdot v + C,$$		:$$x = A \cdot u + B \cdot v + C,$$
	:$$y= D \cdot u + E \cdot v + F.$$		:$$y= D \cdot u + E \cdot v + F.$$

	The ~~2D–~~random variable  $(x, y)$  to be generated should have the following statistical properties:		The two-dimensional random variable  $(x,\hspace{0.08cm} y)$  to be generated should have the following statistical properties:
	* Let the variances be  $\sigma_x^2 = 4$  and  $\sigma_y^2 = 10$.		* Let the variances be  $\sigma_x^2 = 4$  and  $\sigma_y^2 = 10$.
	* Let the random variable  $x$  be mean-free  $(m_x =0)$.		* Let the random variable  $x$  be mean-free  $(m_x =0)$.
	* For the mean of  $y$  let  $m_y = 1$ hold.		* For the mean of  $y$  let  $m_y = 1$  hold.
	* The correlation coefficient between  $x$  and  $y$  is  $\rho_{xy} = \sqrt{0.9} = 0.949.$		* The correlation coefficient between  $x$  and  $y$  is  $\rho_{xy} = \sqrt{0.9} = 0.949.$
	* The random variable  $x$  possess a triangular PDF $f_x(x)$  corresponding to the above graph.		* The random variable  $x$  possess a triangular PDF $f_x(x)$  corresponding to the above graph.
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			Hints:


	''Hints:''
	*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Linear_Combinations_of_Random_Variables\|Linear Combinations of Random Variables]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Linear_Combinations_of_Random_Variables\|Linear Combinations of Random Variables]].
	*In particular, reference is made to the page  [Theory_of_Stochastic_Signals/Linear_Combinations_of_Random_Variables#Generation_of_correlated_random_variables\|Generation of correlated random variables]].		*In particular,  reference is made to the page  [[Theory_of_Stochastic_Signals/Linear_Combinations_of_Random_Variables#Generation_of_correlated_random_variables\|Generation of correlated random variables]].
	*To avoid ambiguity, it is specified that all coefficients  $A$, ... , $F$  should be non-negative.		*To avoid ambiguity,  it is specified that all coefficients  $A$, ... , $F$  should be non-negative.


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	{Determine the coefficients  $D$  and  $E$, where  $D > E$  should hold.		{Determine the coefficients  $D$  and  $E$,  where  $D > E$  should hold.
	\|type="{}"}		\|type="{}"}
	$D \ = \ $ { 3.464 3% }		$D \ = \ $ { 3.464 3% }
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	{Specify the maximum values for  $x$  and  $y$~~ ~~.		{Specify the maximum values for  $x$  and  $y$.
	\|type="{}"}		\|type="{}"}
	$x_\text{max}\ = \ $ { 3.464 3% }		$x_\text{max}\ = \ $ { 3.464 3% }

Guenter: Guenter moved page Aufgaben:Exercise 4.7Z: Generation of a joint PDF to Aufgaben:Exercise 4.7Z: Generation of a Joint PDF

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Guenter moved page Aufgaben:Exercise 4.7Z: Generation of a joint PDF to Aufgaben:Exercise 4.7Z: Generation of a Joint PDF

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Guenter: Guenter moved page Aufgaben:Exercise 4.7Z: Generation of a two-dimensional PDF to Aufgaben:Exercise 4.7Z: Generation of a joint PDF

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Guenter moved page Aufgaben:Exercise 4.7Z: Generation of a two-dimensional PDF to Aufgaben:Exercise 4.7Z: Generation of a joint PDF

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Guenter: Guenter moved page Aufgaben:Exercise 4.7Z: Generation of a "2D-PDF" to Aufgaben:Exercise 4.7Z: Generation of a two-dimensional PDF

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Guenter moved page Aufgaben:Exercise 4.7Z: Generation of a "2D-PDF" to Aufgaben:Exercise 4.7Z: Generation of a two-dimensional PDF

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Noah at 16:11, 24 February 2022

2022-02-24T16:11:41Z

← Older revision		Revision as of 18:11, 24 February 2022
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	[[File:P_ID424__Sto_Z_4_7_d.png\|right\|frame\|Rhombic ~~two dimensional~~ PDF]]		[[File:P_ID424__Sto_Z_4_7_d.png\|right\|frame\|Rhombic joint PDF]]
	'''(3)'''  With  $ A = B = \sqrt{3}$  corresponding to the last subtask, two equations of determination remain for  $D$  and  $E$:		'''(3)'''  With  $ A = B = \sqrt{3}$  corresponding to the last subtask, two equations of determination remain for  $D$  and  $E$:
	:$$\sigma_y^2 = \sigma^2 \cdot ( D^2 + E^2)= 10 \hspace{0.5cm} \Rightarrow \hspace{0.5cm} D^2 + E^2 = \frac {\sigma_y^2}{\sigma^2} = \frac {10}{2/3} \stackrel{!}{=}15,$$		:$$\sigma_y^2 = \sigma^2 \cdot ( D^2 + E^2)= 10 \hspace{0.5cm} \Rightarrow \hspace{0.5cm} D^2 + E^2 = \frac {\sigma_y^2}{\sigma^2} = \frac {10}{2/3} \stackrel{!}{=}15,$$

Noah at 16:11, 24 February 2022

2022-02-24T16:11:12Z

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Guenter at 17:54, 24 January 2022

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← Older revision		Revision as of 19:54, 24 January 2022
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	[[Category:Theory of Stochastic Signals: Exercises\|^4.3 ~~Linearkombinationen~~^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.3 Linear Combinations^]]

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Given the mean values ~~given~~, it must hold:		'''(1)'''  Given the mean values,  it must hold:
	:$$ C = m_x\hspace{0.15cm}\underline{ = 0},$$		:$$ C = m_x\hspace{0.15cm}\underline{ = 0},$$
	:$$ F = m_y\hspace{0.15cm}\underline{ = 1}.$$		:$$ F = m_y\hspace{0.15cm}\underline{ = 1}.$$
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	:$$\sigma_x^2 = \sigma^2 \cdot ( A^2 + B^2)= {2}/{3} \cdot ( A^2 + B^2) .$$		:$$\sigma_x^2 = \sigma^2 \cdot ( A^2 + B^2)= {2}/{3} \cdot ( A^2 + B^2) .$$

	*Because  $\sigma_x^2 = 4$  it follows  $A^2 + B^2= 6$.		*Because of  $\sigma_x^2 = 4$  it follows  $A^2 + B^2= 6$.
	*A triangular PDF means that  $A = \pm B$  must hold.		*A triangular PDF means that  $A = \pm B$  must hold.
	*Thus, since negative coefficients have been excluded, we obtain:		*Thus,  since negative coefficients have been excluded,  we obtain:
	:$$ A = B = \sqrt{3}\hspace{0.15cm}\underline{ = 1.732}.$$		:$$ A = B = \sqrt{3}\hspace{0.15cm}\underline{ = 1.732}.$$


	[[File:P_ID424__Sto_Z_4_7_d.png\|right\|frame\|Rhombic joint PDF]]		[[File:P_ID424__Sto_Z_4_7_d.png\|right\|frame\|Rhombic joint PDF]]
	'''(3)'''  With  $ A = B = \sqrt{3}$  corresponding to the last subtask, two equations of determination remain for  $D$  and  $E$:		'''(3)'''  With  $ A = B = \sqrt{3}$  corresponding to the last subtask,  two equations of determination remain for  $D$  and  $E$:
	:$$\sigma_y^2 = \sigma^2 \cdot ( D^2 + E^2)= 10 \hspace{0.5cm} \Rightarrow \hspace{0.5cm} D^2 + E^2 = \frac {\sigma_y^2}{\sigma^2} = \frac {10}{2/3} \stackrel{!}{=}15,$$		:$$\sigma_y^2 = \sigma^2 \cdot ( D^2 + E^2)= 10 \hspace{0.5cm} \Rightarrow \hspace{0.5cm} D^2 + E^2 = \frac {\sigma_y^2}{\sigma^2} = \frac {10}{2/3} \stackrel{!}{=}15,$$
	:$$\rho_{xy} = \frac{A \cdot D + B \cdot E}{\sqrt{(A^2 + B^2)(D^2 + E^2)}} = \frac{\sqrt{3} \cdot (D + E)}{\sqrt{6 \cdot (D^2 + E^2)}} \stackrel{!}{=} \sqrt{0.9}.$$		:$$\rho_{xy} = \frac{A \cdot D + B \cdot E}{\sqrt{(A^2 + B^2)(D^2 + E^2)}} = \frac{\sqrt{3} \cdot (D + E)}{\sqrt{6 \cdot (D^2 + E^2)}} \stackrel{!}{=} \sqrt{0.9}.$$

	*From this it further follows:  $D + E = \sqrt{1.8 \cdot ( D^2 + E^2)} = \sqrt{27} = 3 \cdot \sqrt{3}.$		*From this it further follows:  $D + E = \sqrt{1.8 \cdot ( D^2 + E^2)} = \sqrt{27} = 3 \cdot \sqrt{3}.$
	*The equation, in conjunction with  $D^2 + E^2 = 15$  and the constraint  $(D>E)$  leads to the result:		*The equation,  in conjunction with  $D^2 + E^2 = 15$  and the constraint  $(D>E)$  leads to the result:
	:$$ D= 2 \cdot \sqrt{3}\hspace{0.15cm}\underline{ = 3.464}, \hspace{0.5cm}E= \sqrt{3} \hspace{0.15cm}\underline{= 1.732}.$$		:$$ D= 2 \cdot \sqrt{3}\hspace{0.15cm}\underline{ = 3.464}, \hspace{0.5cm}E= \sqrt{3} \hspace{0.15cm}\underline{= 1.732}.$$


	'''(4)'''  The random variables  $x$  and  $y$  ~~respectively,~~ take their maximum values when ~~respectively~~  $u= +1$ and  $v= +1$  holds:		'''(4)'''  The random variables  $x$  and  $y$  resp. take their maximum values when  $u= +1$ and  $v= +1$  holds:
	:$$ x_\text{max}= A+B \hspace{0.15cm}\underline{ = +3.464}, \hspace{0.5cm} x_\text{min} = - A - B= -3.464.$$		:$$ x_\text{max}= A+B \hspace{0.15cm}\underline{ = +3.464}, \hspace{0.5cm} x_\text{min} = - A - B= -3.464.$$
	:$$ y_\text{max}= D+E+F \hspace{0.15cm}\underline{ = +6.196}, \hspace{0.5cm} y_\text{min} = -D-E+F= -4.196.$$		:$$ y_\text{max}= D+E+F \hspace{0.15cm}\underline{ = +6.196}, \hspace{0.5cm} y_\text{min} = -D-E+F= -4.196.$$