Aufgaben:Exercise 4.2: Triangle Area again - Revision history

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	[[Category:Theory of Stochastic Signals: Exercises\|^4.1 Two-Dimensional Random Variables^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.1 Two-Dimensional Random Variables^]]
	[[de:Aufgaben:4.~~2_Wieder_Dreieckgebiet~~]]		[[de:Aufgaben:Aufgabe 4.2: Wieder Dreieckgebiet]]

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	[[Category:Theory of Stochastic Signals: Exercises\|^4.1 Two-Dimensional Random Variables^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.1 Two-Dimensional Random Variables^]]
			[[de:Aufgaben:4.2_Wieder_Dreieckgebiet]]

Guenter at 15:43, 13 March 2023

2023-03-13T15:43:56Z

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	{What is the equation of the ~~correlation~~ line  $y = K(x)$?  At what point  $y_0$  does the straight line intersect the  $y$–axis? <br>Show that the correlation line also passes through the point  $(m_x, m_y)$.		{What is the equation of the regression line  $y = K(x)$?  At what point  $y_0$  does the straight line intersect the  $y$–axis? <br>Show that the correlation line also passes through the point  $(m_x, m_y)$.
	\|type="{}"}		\|type="{}"}
	$y_0\ = \ $ { 1 3% }		$y_0\ = \ $ { 1 3% }
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	'''(3)'''  Since both random variables each have a nonzero mean,  it follows for the covariance:		'''(3)'''  Since both random variables each have a nonzero mean,  it follows for the covariance:
	[[File:EN_Sto_Z_4_2d.png\|right\|frame\|~~Correlation~~ line ~~'''KORREKTUR'''~~]]		[[File:EN_Sto_Z_4_2d.png\|right\|frame\|Regression line]]
	:$$\it \mu_{xy}=\it m_{xy}-m_{x}\cdot m_{y}=\frac{\rm 26}{\rm 3}-\frac{\rm 8}{\rm 3}\cdot\rm 3={2}/{3} \hspace{0.15cm}\underline{=0.667}.$$		:$$\it \mu_{xy}=\it m_{xy}-m_{x}\cdot m_{y}=\frac{\rm 26}{\rm 3}-\frac{\rm 8}{\rm 3}\cdot\rm 3={2}/{3} \hspace{0.15cm}\underline{=0.667}.$$

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	:$$\rho_{xy}=\frac{\mu_{xy}}{\sigma_{x}\cdot\sigma_{y}}=\frac{{\rm 2}/{\rm 3}}{\sqrt{{\rm 8}/{\rm 9}}\cdot\sqrt{{\rm 2}/{\rm 3}}}=\sqrt{0.75}\hspace{0.15cm}\underline{=\rm 0.866}.$$		:$$\rho_{xy}=\frac{\mu_{xy}}{\sigma_{x}\cdot\sigma_{y}}=\frac{{\rm 2}/{\rm 3}}{\sqrt{{\rm 8}/{\rm 9}}\cdot\sqrt{{\rm 2}/{\rm 3}}}=\sqrt{0.75}\hspace{0.15cm}\underline{=\rm 0.866}.$$

	'''(5)'''  For the ~~correlation~~ line ("KG"  in the graph),  in general:		'''(5)'''  For the regression line ("RL"  in the graph),  in general:
	:$$ y-m_{y}=\rho_{xy}\cdot\frac{\sigma_{y}}{\sigma_ {x}}\cdot(x-m_{x}).$$		:$$ y-m_{y}=\rho_{xy}\cdot\frac{\sigma_{y}}{\sigma_ {x}}\cdot(x-m_{x}).$$
	*Using the numerical values calculated above,  we obtain		*Using the numerical values calculated above,  we obtain

Guenter at 16:19, 31 March 2022

2022-03-31T16:19:02Z

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	'''(3)'''  Since both random variables each have a nonzero mean,  it follows for the covariance:		'''(3)'''  Since both random variables each have a nonzero mean,  it follows for the covariance:
	[[File:~~P_ID223__Sto_A_4_2_d~~.png\|right\|frame\|Correlation line]]		[[File:EN_Sto_Z_4_2d.png\|right\|frame\|Correlation line '''KORREKTUR''']]
	:$$\it \mu_{xy}=\it m_{xy}-m_{x}\cdot m_{y}=\frac{\rm 26}{\rm 3}-\frac{\rm 8}{\rm 3}\cdot\rm 3={2}/{3} \hspace{0.15cm}\underline{=0.667}.$$		:$$\it \mu_{xy}=\it m_{xy}-m_{x}\cdot m_{y}=\frac{\rm 26}{\rm 3}-\frac{\rm 8}{\rm 3}\cdot\rm 3={2}/{3} \hspace{0.15cm}\underline{=0.667}.$$

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

2022-02-17T11:11:39Z

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

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	:$$\it \mu_{xy}=\it m_{xy}-m_{x}\cdot m_{y}=\frac{\rm 26}{\rm 3}-\frac{\rm 8}{\rm 3}\cdot\rm 3={2}/{3} \hspace{0.15cm}\underline{=0.667}.$$		:$$\it \mu_{xy}=\it m_{xy}-m_{x}\cdot m_{y}=\frac{\rm 26}{\rm 3}-\frac{\rm 8}{\rm 3}\cdot\rm 3={2}/{3} \hspace{0.15cm}\underline{=0.667}.$$

	'''(4)'''  With the given ~~rms values~~ we obtain:		'''(4)'''  With the given standard deviations we obtain:
	:$$\rho_{xy}=\frac{\mu_{xy}}{\sigma_{x}\cdot\sigma_{y}}=\frac{{\rm 2}/{\rm 3}}{\sqrt{{\rm 8}/{\rm 9}}\cdot\sqrt{{\rm 2}/{\rm 3}}}=\sqrt{0.75}\hspace{0.15cm}\underline{=\rm 0.866}.$$		:$$\rho_{xy}=\frac{\mu_{xy}}{\sigma_{x}\cdot\sigma_{y}}=\frac{{\rm 2}/{\rm 3}}{\sqrt{{\rm 8}/{\rm 9}}\cdot\sqrt{{\rm 2}/{\rm 3}}}=\sqrt{0.75}\hspace{0.15cm}\underline{=\rm 0.866}.$$

Guenter at 12:44, 7 February 2022

2022-02-07T12:44:19Z

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	===Solution===		===Solution===
	{{ML-Kopf}}		{{ML-Kopf}}
	'''(1)'''  Correct is the <u>middle proposition</u>:		'''(1)'''  Correct is the  <u>middle proposition</u>:
	*Both  $y_1(x)$  and  $y_2(x)$  intersect the  $y$-axis at  $y= 1$.		*Both  $y_1(x)$  and  $y_2(x)$  intersect the  $y$-axis at  $y= 1$.
	*The lower boundary line has slope  $0.5$, the upper has slope  $1$.		*The lower boundary line has slope  $0.5$,  the upper has slope  $1$.


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	*This leads to the integral or final result:		*This leads to the integral or final result:
	:$$m_{xy}={1}/{8}\int_{\rm 0}^{\rm 4}(\frac{3}{4}\cdot x^{3}{\rm +} x^2\,{\rm d}x = \rm \frac{1}{8} \cdot (\frac{3}{16}\cdot 4^4+\rm \frac{4^3}{3})=\frac{26}{3}\hspace{0.15cm}\underline{ \approx 8.667}.$$		:$$m_{xy}={1}/{8}\int_{\rm 0}^{\rm 4}(\frac{3}{4}\cdot x^{3}{\rm +} x^2\,{\rm d}x = \rm \frac{1}{8} \cdot (\frac{3}{16}\cdot 4^4+\rm \frac{4^3}{3})=\frac{26}{3}\hspace{0.15cm}\underline{ \approx 8.667}.$$
	'''(3)'''  Since both random variables each have a nonzero mean, it follows ~~für~~ the covariance:
	[[File:P_ID223__Sto_A_4_2_d.png\|right\|frame\|~~correlation~~ line]]		'''(3)'''  Since both random variables each have a nonzero mean,  it follows for the covariance:
			[[File:P_ID223__Sto_A_4_2_d.png\|right\|frame\|Correlation line]]
	:$$\it \mu_{xy}=\it m_{xy}-m_{x}\cdot m_{y}=\frac{\rm 26}{\rm 3}-\frac{\rm 8}{\rm 3}\cdot\rm 3={2}/{3} \hspace{0.15cm}\underline{=0.667}.$$		:$$\it \mu_{xy}=\it m_{xy}-m_{x}\cdot m_{y}=\frac{\rm 26}{\rm 3}-\frac{\rm 8}{\rm 3}\cdot\rm 3={2}/{3} \hspace{0.15cm}\underline{=0.667}.$$

			'''(4)'''  With the given rms values we obtain:

	'''(4)'''  With the given rms we obtain:
	:$$\rho_{xy}=\frac{\mu_{xy}}{\sigma_{x}\cdot\sigma_{y}}=\frac{{\rm 2}/{\rm 3}}{\sqrt{{\rm 8}/{\rm 9}}\cdot\sqrt{{\rm 2}/{\rm 3}}}=\sqrt{0.75}\hspace{0.15cm}\underline{=\rm 0.866}.$$		:$$\rho_{xy}=\frac{\mu_{xy}}{\sigma_{x}\cdot\sigma_{y}}=\frac{{\rm 2}/{\rm 3}}{\sqrt{{\rm 8}/{\rm 9}}\cdot\sqrt{{\rm 2}/{\rm 3}}}=\sqrt{0.75}\hspace{0.15cm}\underline{=\rm 0.866}.$$

	'''(5)'''  For the correlation line (KG), in general:		'''(5)'''  For the correlation line ("KG"  in the graph),  in general:
	:$$ y-m_{y}=\rho_{xy}\cdot\frac{\sigma_{y}}{\sigma_ {x}}\cdot(x-m_{x}).$$		:$$ y-m_{y}=\rho_{xy}\cdot\frac{\sigma_{y}}{\sigma_ {x}}\cdot(x-m_{x}).$$
			*Using the numerical values calculated above,  we obtain
	*Using the numerical values calculated above, we obtain
	:$$y={\rm 3}/{\rm 4}\cdot x +\rm 1.$$		:$$y={\rm 3}/{\rm 4}\cdot x +\rm 1.$$
			*The correlation line intersects the  $y$-axis at  $\underline{y=1}$  and also passes through the point  $(4, 4)$.
	The correlation line intersects the  $y$-axis at $\underline{y=1}$ and also passes through the point  $(4, 4)$.  Any other result would ~~also~~ be impossible to interpret considering the definition area:		*Any other result would be impossible to interpret considering the definition area:
	*If one sets  $m_x = 8/3$ , one obtains  $y = m_y = 3$.		::If one sets  $m_x = 8/3$,  one obtains  $y = m_y = 3$.
	*This means:   The calculated correlation line actually passes through the point  $(m_x, m_y)$, as the theory says.		*This means:   The calculated correlation line actually passes through the point  $(m_x, m_y)$,  as the theory says.
	{{ML-Fuß}}		{{ML-Fuß}}

Guenter at 12:32, 7 February 2022

2022-02-07T12:32:09Z

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	[[File:P_ID226__Sto_A_4_2.png\|right\|frame\|Triangular 2D area and the two marginal probability densities]]		[[File:P_ID226__Sto_A_4_2.png\|right\|frame\|Triangular 2D area and the two marginal probability densities]]
	We consider the same random variable  $(x, \ y)$  as in ~~the~~  [[Aufgaben:Exercise_4.2:~~_Triangle_Area_again~~\|Exercise 4.1]]:		We consider the same random variable  $(x, \ y)$  as in  [[Aufgaben:Exercise_4.1:_Triangular_(x,_y)_Area\|Exercise 4.1]]:
	*In a domain defined by vertices  $(0,\ 1)$, $(4,\ 3)$, and $(4,\ 5)$  let the 2D–PDF  $f_{xy} (x, y) = 0.25$.		*In a domain defined by vertices  $(0,\ 1)$,  $(4,\ 3)$,  and $(4,\ 5)$  let the 2D–PDF  $f_{xy} (x, y) = 0.25$.
	*There are no values outside this definition area  $D$  marked in red in the graph.		*There are no values outside this definition area  $D$  marked in red in the graph.


	Furthermore, the two marginal probability densities with respect to the quantities  $x$  and  $y$  are drawn in the graph, which have already been determined in Exercise 4.1.		Furthermore,  the two marginal probability densities with respect to the quantities  $x$  and  $y$  are drawn in the graph,  which have already been determined in Exercise 4.1.

	From this, the equations of chapter  [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments\|Expected Values and Moments]]  can be used to determine the characteristics of the two random variables:		From this,  the equations of chapter  [[Theory_of_Stochastic_Signals/Expected_Values_and_Moments\|Expected Values and Moments]]  can be used to determine the characteristics of the two random variables:
	:$$m_x=8/3 ,\hspace{0.5cm} \sigma_x=\sqrt{8/9},$$		:$$m_x=8/3 ,\hspace{0.5cm} \sigma_x=\sqrt{8/9},$$
	:$$ m_y= 3,\hspace{0.95cm} \sigma_y = \sqrt{\rm 2/3}.$$		:$$ m_y= 3,\hspace{0.95cm} \sigma_y = \sqrt{\rm 2/3}.$$

	Due to the fact that the domain ~~of definition~~  $D$  is bounded by two straight lines  $y_1(x)$  and  $y_2(x)$ , the first order joint moment can be calculated here as follows.		Due to the fact that the definition domain  $D$  is bounded by two straight lines  $y_1(x)$  and  $y_2(x)$ , the first order joint moment can be calculated here as follows.
	:$$m_{xy}={\rm E}\big[x\cdot y\big]=\int_{x_{1}}^{x_{2}}x\cdot \int_{y_{1}(x)}^{y_{2}(x)}y \cdot f_{xy}(x,y) \, \,{\rm d}y\, {\rm d}x.$$		:$$m_{xy}={\rm E}\big[x\cdot y\big]=\int_{x_{1}}^{x_{2}}x\cdot \int_{y_{1}(x)}^{y_{2}(x)}y \cdot f_{xy}(x,y) \, \,{\rm d}y\, {\rm d}x.$$

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	<quiz display=simple>		<quiz display=simple>
	{What are the limit lines of the inner integral for  $m_{xy}$~~–~~calculation?		{What are the limit lines of the inner integral for the  $m_{xy}$ calculation?
	\|type="()"}		\|type="()"}
	- $y_1(x) = x+1, $     $y_2(x) = 2x+1.$		- $y_1(x) = x+1, $     $y_2(x) = 2x+1.$
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	{What is ~~the value of~~ the covariance  $\mu_{xy}$ ?		{What is the covariance  $\mu_{xy}$ ?
	\|type="{}"}		\|type="{}"}
	$\mu_{xy}\ = \ $ { 0.667 3% }		$\mu_{xy}\ = \ $ { 0.667 3% }


	{What is ~~the value of~~ the correlation coefficient $\rho_{xy}$?		{What is the correlation coefficient $\rho_{xy}$?
	\|type="{}"}		\|type="{}"}
	$\rho_{xy}\ = \ $ { 0.866 3% }		$\rho_{xy}\ = \ $ { 0.866 3% }


	{What is the equation of the correlation line  $y = K(x)$ ? At what point  $y_0$  does the straight line intersect the  $y$ axis? <br>Show that the correlation line also passes through the point  $(m_x, m_y)$~~ ~~.		{What is the equation of the correlation line  $y = K(x)$?  At what point  $y_0$  does the straight line intersect the  $y$–axis? <br>Show that the correlation line also passes through the point  $(m_x, m_y)$.
	\|type="{}"}		\|type="{}"}
	$y_0\ = \ $ { 1 3% }		$y_0\ = \ $ { 1 3% }

Noah at 18:18, 17 January 2022

2022-01-17T18:18:55Z

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Noah at 19:15, 13 January 2022

2022-01-13T19:15:09Z

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	{{quiz-Header\|Buchseite=~~Stochastische Signaltheorie~~/~~Zweidimensionale Zufallsgrößen~~		{{quiz-Header\|Buchseite=Theory_of_Stochastic_Signals/Two-Dimensional_Random_Variables
	}}		}}

Guenter at 10:51, 12 January 2022

2022-01-12T10:51:10Z

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	[[Category:Theory of Stochastic Signals: Exercises\|^4.1 ~~Zweidimensionale Zufallsgrößen~~^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.1 Two-Dimensional Random Variables^]]