Aufgaben:Exercise 4.5: Two-dimensional Examination Evaluation - Revision history

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	[[Category:Theory of Stochastic Signals: Exercises\|^4.2 Gaussian 2D Random Variables^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.2 Gaussian 2D Random Variables^]]
	[[de:Aufgaben:4.~~5_2D~~-Prüfungsauswertung]]		[[de:Aufgaben:Aufgabe 4.5: 2D-Prüfungsauswertung]]

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	[[Category:Theory of Stochastic Signals: Exercises\|^4.2 Gaussian 2D Random Variables^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.2 Gaussian 2D Random Variables^]]
			[[de:Aufgaben:4.5_2D-Prüfungsauswertung]]

Guenter at 15:24, 24 February 2022

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Guenter at 15:06, 24 February 2022

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	[[File:P_ID267__Sto_A_4_5.png\|right\|frame\|Applied Gaussian 2D PDF  $f_{tp}(t,p)$]]		[[File:P_ID267__Sto_A_4_5.png\|right\|frame\|Applied Gaussian 2D–PDF  $f_{tp}(t,\hspace{0.08cm} p)$]]
	In a study, the master craftsman examinations were investigated, which always consist of a theoretical and additionally a practical part. In the graph denoted		In a study,  the master craftsman examinations were investigated,  which always consist of a theoretical and additionally a practical part.  In the graph denoted:
	* $t$  ~~denotes~~ the score in the theoretical test,		* $t$  is the score in the theoretical test,
	* $p$  the score in the practical test.		* $p$  is the score in the practical test.


	Both random variables  $(t$  and  $p)$  are normalized to the maximum ~~scores~~ and can therefore only take values between  $0$  and  $1$~~ ~~.		Both random variables  $(t$  and  $p)$  are normalized to the maximum and can therefore only take values between  $0$  and  $1$.

	Moreover, both random variables are to be interpreted as continuous random variables, i.e.:   $t$  and  $p$  are not restricted to discrete numerical values.		Moreover,  both random variables are to be interpreted as continuous random variables, i.e.:   $t$  and  $p$  are not restricted to discrete numerical values.

	*The graph shows the PDF  $f_{tp}(t, p)$  of the two-dimensional random variable  $(t, p)$,  which was published after evaluating a total of  $N = 10\hspace{0.08cm}000$  final papers.		*The graph shows the PDF  $f_{tp}(t, p)$  of the two-dimensional random variable  $(t,\hspace{0.08cm} p)$,  which was published after evaluating a total of  $N = 10\hspace{0.08cm}000$  final papers.
	*This function was empirically approximated using an evaluation program as follows:		*This function was empirically approximated using an evaluation program as follows:
	:$$f_{tp}(t,p) = \rm 13.263\cdot \rm exp \Bigg\{-\frac{(\it t - \rm 0.5)^{\rm 2}}{\rm 0.0288}-\frac{(\it p-\rm 0.7)^{\rm 2}}{\rm 0.0072} + \frac{(\it t-\rm 0.5)(\it p-\rm 0.7)}{\rm 0.0090}\Bigg\}.$$		:$$f_{tp}(t,\hspace{0.08cm} p) = \rm 13.263\cdot \rm exp \Bigg\{-\frac{(\it t - \rm 0.5)^{\rm 2}}{\rm 0.0288}-\frac{(\it p-\rm 0.7)^{\rm 2}}{\rm 0.0072} + \frac{(\it t-\rm 0.5)(\it p-\rm 0.7)}{\rm 0.0090}\Bigg\}.$$





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	Hints:		Hints:
	*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Two-Dimensional_Gaussian_Random_Variables\|Two-dimensional Gaussian Random Variables]].		*The exercise belongs to the chapter  [[Theory_of_Stochastic_Signals/Two-Dimensional_Gaussian_Random_Variables\|Two-dimensional Gaussian Random Variables]].
			*Reference is also made to the chapter  [[Theory_of_Stochastic_Signals/Gaussian_Distributed_Random_Variables\|Gaussian distributed random variables]].
	*More information on this topic is provided in the learning video  [[Gaußsche_2D-Zufallsgrößen_(Lernvideo)\|~~Gaussian~~ 2D ~~random variables~~]]:		*More information on this topic is provided in the  (German language) learning video  [[Gaußsche_2D-Zufallsgrößen_(Lernvideo)\|"Gaußsche 2D-Zufallsgrößen"]]:
	::Part 1:   Gaussian random variables without statistical bindings,		::Part 1:   Gaussian random variables without statistical bindings,
	::Part 2:   Gaussian random variables with statistical bindings.		::Part 2:   Gaussian random variables with statistical bindings.
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	<quiz display=simple>		<quiz display=simple>
	{What is the mean value  $m_t$  of the results obtained in the ~~theory~~ part?		{What is the mean value  $m_t$  of the results obtained in the theoretical part?
	\|type="{}"}		\|type="{}"}
	$m_t \ = \ $ { 0.5 3% }		$m_t \ = \ $ { 0.5 3% }


	{What is the mean  $m_p$  of the results obtained in the ~~practice~~ part?  Give also the PDF of the zero mean random variable  $(t\hspace{0.05cm}', p\hspace{0.05cm}')$~~ ~~ .		{What is the mean  $m_p$  of the results obtained in the practical part?  Give also the PDF of the zero mean random variable  $(t\hspace{0.05cm}', p\hspace{0.05cm}')$.
	\|type="{}"}		\|type="{}"}
	$m_p \ = \ $ { 0.7 3% }		$m_p \ = \ $ { 0.7 3% }
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	\|type="[]"}		\|type="[]"}
	+ The Gaussian approach is only an approximation for this problem.		+ The Gaussian approach is only an approximation for this problem.
	- If an examinee was above average in the theory part, it is to be expected that he is rather bad in practice.		- If an examinee was above average in the theory part,  it is to be expected that he is rather bad in practice.
	{What is the probability that a participant scored between  $49\%$  and  $51\%$  of the points in the theory~~–~~ and the practice~~–~~exam respectively?
			{What is the probability that a participant scored between  $49\%$  and  $51\%$  of the points in the theory and the practice exam respectively?
	\|type="{}"}		\|type="{}"}
	${\rm Pr}\big [(0.49 ≤ t ≤0.51)∩(0.49≤ p ≤0.51)\big]\ = \ $ { 2 3% } $\ \cdot 10^{-5}$		${\rm Pr}\big [(0.49 ≤ t ≤0.51)∩(0.49≤ p ≤0.51)\big]\ = \ $ { 2 3% } $\ \cdot 10^{-5}$

Noah at 21:56, 23 January 2022

2022-01-23T21:56:49Z

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Guenter at 10:58, 12 January 2022

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	[[Category:Theory of Stochastic Signals: Exercises\|^4.2 ~~Gaußsche~~ 2D~~-Zufallsgrößen~~^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.2 Gaussian 2D Random Variables^]]

Guenter: Guenter moved page Aufgaben:Aufgabe 4.5: 2D-Prüfungsauswertung to Aufgaben:Exercise 4.5: Two-dimensional Examination Evaluation

2022-01-12T10:55:54Z

Guenter moved page Aufgaben:Aufgabe 4.5: 2D-Prüfungsauswertung to Aufgaben:Exercise 4.5: Two-dimensional Examination Evaluation

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Javier: Text replacement - "Category:Aufgaben zu Stochastische Signaltheorie" to "Category:Theory of Stochastic Signals: Exercises"

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	[[Category:~~Aufgaben zu Stochastische Signaltheorie~~\|^4.2 Gaußsche 2D-Zufallsgrößen^]]		[[Category:Theory of Stochastic Signals: Exercises\|^4.2 Gaußsche 2D-Zufallsgrößen^]]

Javier: Text replacement - "[[Stochastische_Signaltheorie/" to "[[Theory_of_Stochastic_Signals/"

2020-07-09T09:52:08Z

Text replacement - "[[Stochastische_Signaltheorie/" to "[[Theory_of_Stochastic_Signals/"

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	''Hinweise:''		''Hinweise:''
	*Die Aufgabe gehört zum Kapitel  [[~~Stochastische_Signaltheorie~~/Zweidimensionale_Gaußsche_Zufallsgrößen\|Zweidimensionale Gaußsche Zufallsgrößen]].		*Die Aufgabe gehört zum Kapitel  [[Theory_of_Stochastic_Signals/Zweidimensionale_Gaußsche_Zufallsgrößen\|Zweidimensionale Gaußsche Zufallsgrößen]].

	*Weitere Informationen zu dieser Thematik liefert das Lernvideo  [[Gaußsche_2D-Zufallsgrößen_(Lernvideo)\|Gaußsche 2D-Zufallsgrößen]]:		*Weitere Informationen zu dieser Thematik liefert das Lernvideo  [[Gaußsche_2D-Zufallsgrößen_(Lernvideo)\|Gaußsche 2D-Zufallsgrößen]]:

Guenter at 11:03, 27 November 2019

2019-11-27T11:03:28Z

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	'''(1)'''  and '''(2)'''		'''(1)'''  and '''(2)'''
	*The mean values  $m_t\hspace{0.15cm}\underline{= 0.5}$  and  $m_p\hspace{0.15cm}\underline{= 0.7}$  can be estimated from the sketch and obtained exactly from the given equation.		*The mean values  $m_t\hspace{0.15cm}\underline{= 0.5}$  and  $m_p\hspace{0.15cm}\underline{= 0.7}$  can be estimated from the sketch and obtained exactly from the given equation.
	*The 2D PDF of the zero mean variable is:		*The 2D–PDF of the zero mean variable is:
	:$$f_{\it t\hspace{0.05cm}'\hspace{0.05cm}p\hspace{0.05cm}'}(\it t\hspace{0.05cm}', \it p\hspace{0.05cm}'{\rm )} = \rm 13.263\cdot \rm exp\Bigg (-\frac{\it {\rm (}t\hspace{0.05cm}'{\rm )}^{\rm 2}}{\rm 0.0288} - \frac{\it {\rm (}p\hspace{0.05cm}'{\rm )}^{\rm 2}}{\rm 0.0072}+\frac{\it t\hspace{0.05cm}'\cdot p\hspace{0.05cm}'}{\rm 0.0090}\Bigg ). $$		:$$f_{\it t\hspace{0.05cm}'\hspace{0.05cm}p\hspace{0.05cm}'}(\it t\hspace{0.05cm}',\hspace{0.08cm} \it p\hspace{0.05cm}'{\rm )} = \rm 13.263\cdot \rm exp\Bigg (-\frac{\it {\rm (}t\hspace{0.05cm}'{\rm )}^{\rm 2}}{\rm 0.0288} - \frac{\it {\rm (}p\hspace{0.05cm}'{\rm )}^{\rm 2}}{\rm 0.0072}+\frac{\it t\hspace{0.05cm}'\cdot p\hspace{0.05cm}'}{\rm 0.0090}\Bigg ). $$
	*For simplicity, the apostrophe is omitted below to denote zero mean variables.		*For simplicity,  the apostrophe is omitted below to denote zero mean variables.
	*Both  $t$  and  $p$  are to be understood as zero mean up to and including the subtask  '''(4)'''~~' ~~ .		*Both  $t$  and  $p$  are to be understood as zero mean up to and including the subtask  '''(4)'''.



	'''(3)'''  The general equation of a zero mean 2D random variable is:		'''(3)'''  The general equation of a zero mean two-dimensional random variable is:
	:$$f_{\it tp}(\it t, \it p)=\frac{\rm 1}{\rm 2\it \pi \cdot \sigma_{\it t}\cdot \sigma_{\it p} \cdot\sqrt{\rm 1- \it\rho^{\rm 2}}}\hspace{0.1cm}\cdot \hspace{0.1cm}\rm exp\Bigg\{-\hspace{0.1cm}\frac{\it t^{\rm 2}}{\rm 2\cdot (\rm 1-\rho^{\rm 2})\cdot \sigma_{\it t}^{\rm 2}} -\hspace{0.1cm}\frac{\it p^{\rm 2}}{\rm 2\cdot (\rm 1-\it\rho^{\rm 2}{\rm )}\cdot \sigma_{\it p}^{\rm 2}}+\hspace{0.1cm}\frac{\rho\cdot \it t\cdot \it p}{ (\rm 1-\it \rho^{\rm 2}{\rm )}\cdot\sigma_{\it t}\cdot\sigma_{\it p}}\Bigg\}.$$		:$$f_{\it tp}(\it t,\hspace{0.08cm} \it p)=\frac{\rm 1}{\rm 2\it \pi \cdot \sigma_{\it t}\cdot \sigma_{\it p} \cdot\sqrt{\rm 1- \it\rho^{\rm 2}}}\hspace{0.1cm}\cdot \hspace{0.1cm}\rm exp\Bigg\{-\hspace{0.1cm}\frac{\it t^{\rm 2}}{\rm 2\cdot (\rm 1-\rho^{\rm 2})\cdot \sigma_{\it t}^{\rm 2}} -\hspace{0.1cm}\frac{\it p^{\rm 2}}{\rm 2\cdot (\rm 1-\it\rho^{\rm 2}{\rm )}\cdot \sigma_{\it p}^{\rm 2}}+\hspace{0.1cm}\frac{\rho\cdot \it t\cdot \it p}{ (\rm 1-\it \rho^{\rm 2}{\rm )}\cdot\sigma_{\it t}\cdot\sigma_{\it p}}\Bigg\}.$$


	*The standard deviations  $\sigma_t$  and  $\sigma_p$  as well as the correlation coefficient  $\rho$  can be obtained by coefficient comparison:		*The standard deviations  $\sigma_t$  and  $\sigma_p$  as well as the correlation coefficient  $\rho$  can be obtained by coefficient comparison:
	*A comparison of the first two terms in the exponent shows that  $\sigma_t = 2 \cdot \sigma_p$  must hold.  Thus the PDF is:		*A comparison of the first two terms in the exponent shows that  $\sigma_t = 2 \cdot \sigma_p$  must hold.  Thus the PDF is:
	:$$f_{\it tp}(\it t, \it p)=\frac{\rm 1}{\rm 4\it \pi \cdot \sigma_{\it p}^{\rm 2} \cdot\sqrt{\rm 1- \it\rho^{\rm 2}}}\hspace{0.1cm}\cdot \hspace{0.1cm}\rm exp\Bigg\{-\hspace{0.1cm}\frac{\it t^{\rm 2}}{\rm 8\cdot (\rm 1-\rho^{\rm 2})\cdot \sigma_{\it p}^{\rm 2}} -\hspace{0.1cm}\frac{\it p^{\rm 2}}{\rm 2\cdot (\rm 1-\it\rho^{\rm 2}{\rm )}\cdot \sigma_{\it p}^{\rm 2}}+\hspace{0.1cm}\frac{\rho\cdot \it t\cdot \it p}{\rm 2\cdot (\rm 1-\it \rho^{\rm 2}{\rm )}\cdot\sigma_{\it p}^{\rm 2}}\Bigg\}.$$		:$$f_{\it tp}(\it t,\hspace{0.08cm} \it p)=\frac{\rm 1}{\rm 4\it \pi \cdot \sigma_{\it p}^{\rm 2} \cdot\sqrt{\rm 1- \it\rho^{\rm 2}}}\hspace{0.1cm}\cdot \hspace{0.1cm}\rm exp\Bigg\{-\hspace{0.1cm}\frac{\it t^{\rm 2}}{\rm 8\cdot (\rm 1-\rho^{\rm 2})\cdot \sigma_{\it p}^{\rm 2}} -\hspace{0.1cm}\frac{\it p^{\rm 2}}{\rm 2\cdot (\rm 1-\it\rho^{\rm 2}{\rm )}\cdot \sigma_{\it p}^{\rm 2}}+\hspace{0.1cm}\frac{\rho\cdot \it t\cdot \it p}{\rm 2\cdot (\rm 1-\it \rho^{\rm 2}{\rm )}\cdot\sigma_{\it p}^{\rm 2}}\Bigg\}.$$
	*From the second term of the exponent follows:		*From the second term of the exponent follows:
	:$$2\cdot(1-\rho^{\rm 2})\cdot\sigma_{p}^{ 2}=0.0072\hspace{0.5cm}\Rightarrow \hspace{0.5cm} \sigma_{p}^{2} = \frac{ 0.0036}{(1-\rho^{\rm 2})}.$$		:$$2\cdot(1-\rho^{\rm 2})\cdot\sigma_{p}^{ 2}=0.0072\hspace{0.5cm}\Rightarrow \hspace{0.5cm} \sigma_{p}^{2} = \frac{ 0.0036}{(1-\rho^{\rm 2})}.$$
	*The factor  $K = 13.263$  now gives the result.		*The factor  $K = 13.263$  now gives the result:
	:$$K = \frac{\sqrt{\rm 1-\it\rho^{\rm 2}}}{\rm 4\it\pi\cdot \rm 0.0036}=\rm 13.263 \hspace{0.5cm}\Rightarrow \hspace{0.5cm}\sqrt{\rm 1-\it\rho^{\rm 2}}=\rm 0.6 \hspace{0.5cm}\Rightarrow \hspace{0.5cm}\hspace{0.15cm}\underline{ \rm \rho = \rm 0.8}.$$		:$$K = \frac{\sqrt{\rm 1-\it\rho^{\rm 2}}}{\rm 4\it\pi\cdot \rm 0.0036}=\rm 13.263 \hspace{0.5cm}\Rightarrow \hspace{0.5cm}\sqrt{\rm 1-\it\rho^{\rm 2}}=\rm 0.6 \hspace{0.5cm}\Rightarrow \hspace{0.5cm}\hspace{0.15cm}\underline{ \rm \rho = \rm 0.8}.$$
	*From this we get the standard deviations to  $\sigma_t\hspace{0.15cm}\underline{= 0.2}$  and  $\sigma_p\hspace{0.15cm}\underline{= 0.1}$.		*From this we get the standard deviations to  $\sigma_t\hspace{0.15cm}\underline{= 0.2}$  and  $\sigma_p\hspace{0.15cm}\underline{= 0.1}$.
			*For control,  we use the last term of the exponent:
	*For control, we use the last term of the exponent:
	:$$\frac{(1 - \rho^{2})\cdot \sigma_{\it t}\cdot\sigma_{\it p}}{\it \rho} = \frac{0.36\cdot 0.1\cdot 0.2}{0.8} = \rm 0.009.$$		:$$\frac{(1 - \rho^{2})\cdot \sigma_{\it t}\cdot\sigma_{\it p}}{\it \rho} = \frac{0.36\cdot 0.1\cdot 0.2}{0.8} = \rm 0.009.$$
	*This agrees with the given value.		*This agrees with the given value.


	'''(4)'''  The <u>proposed solution 1</u> is correct.		'''(4)'''  The  <u>proposed solution 1</u>  is correct.
	*Basically,  $(t, p)$  is not a true Gaussian random variable,  since both components are bounded.		*Basically,  $(t,\hspace{0.08cm} p)$  is not a true Gaussian random variable,  since both components are bounded.
	*The probabilities for the events  $t < 0$,     $t >1$,     $p < 0$ and  $p >1$  are therefore zero.		*The probabilities for the events  $t < 0$,     $t >1$,     $p < 0$     and   $p >1$  are therefore zero.
	*However, for Gaussian variables with the mean values and standard deviations present here, we get.		*However,  for Gaussian variables with the mean values and standard deviations present here, we get:
	:$$\rm Pr(\it t < \rm 0) = \rm Pr(\it t > \rm 1) = \rm Q(2.5)\approx 6\cdot 10^{-3},$$		:$$\rm Pr(\it t < \rm 0) = \rm Pr(\it t > \rm 1) = \rm Q(2.5)\approx 6\cdot 10^{-3},$$
	:$$\rm Pr(\it p > \rm 1) = \rm Q(3)\approx 1.3\cdot 10^{-3},$$		:$$\rm Pr(\it p > \rm 1) = \rm Q(3)\approx 1.3\cdot 10^{-3},$$
	:$$\rm Pr(\it p < \rm 0) = \rm Q(7)\approx 10^{-12}.$$		:$$\rm Pr(\it p < \rm 0) = \rm Q(7)\approx 10^{-12}.$$

	*The correlation coefficient  $\rho = 0.8$  is positive here. If the examinee has done rather well in the theory part, it is to be expected (at least for this exercise) that the practical part will also go well.		*The correlation coefficient  $\rho = 0.8$  is positive here.  If the examinee has done rather well in the theory part,  it is to be expected  (at least for this exercise)  that the practical part will also go well.
	*Suggestion 2 is therefore wrong ~~here~~. In practice, this is certainly not always the case.		*Suggestion 2 is therefore wrong.  In practice,  this is certainly not always the case.



	'''(5)'''  For this probability, $\Delta t = \Delta p = 0.02$:		'''(5)'''  For this probability with  $\Delta t = \Delta p = 0.02$:
	:$$\rm Pr\left [( \rm 0.5-\frac{\rm\Delta\it t}{\rm 2}\le \it t \le \rm 0.5+\frac{\rm\Delta\it t}{\rm 2})\cap(\rm 0. 5-\frac{\rm\~~delta~~\it p}{\rm 2}\le \it p \le \rm 0.5+\frac{\rm\~~delta~~\it p}{\rm 2})\right ] \approx \rm\Delta\it t\cdot\rm\Delta\it p\cdot \it f_{tp}{\rm (}t=\rm 0.5, \it p = \rm 0.5).$$		:$$\rm Pr\left [( \rm 0.5-\frac{\rm\Delta\it t}{\rm 2}\le \it t \le \rm 0.5+\frac{\rm\Delta\it t}{\rm 2})\cap(\rm 0. 5-\frac{\rm\Delta\it p}{\rm 2}\le \it p \le \rm 0.5+\frac{\rm\Delta\it p}{\rm 2})\right ] \approx \rm\Delta\it t\cdot\rm\Delta\it p\cdot \it f_{tp}{\rm (}t=\rm 0.5, \it p = \rm 0.5).$$

	*For the 2D PDF, taking into account the mean values  $m_t{= 0.5}$  and  $m_p{= 0.7}$:		*For the 2D–PDF,  taking into account the mean values  $m_t{= 0.5}$  and  $m_p{= 0.7}$:
	:$$f_{tp}(\it t=\rm 0.5, \it p=\rm 0.5) = \rm 13.263\cdot {\rm e}^{-(-0.2)^2/0.0072}\approx 0.0513.$$		:$$f_{tp}(\it t=\rm 0.5,\hspace{0.08cm} \it p=\rm 0.5) = \rm 13.263\cdot {\rm e}^{-(-0.2)^2/0.0072}\approx 0.0513.$$

	*Thus, the probability we are looking for is given by		*Thus,  the probability we are looking for is given by
	:$${\rm Pr}\big[(0.49 ≤ t ≤0.51)∩(0.49≤ p ≤0.51)\big] =0.02 \cdot 0.0513\hspace{0.15cm}\underline{\approx 2 · 10^{-5}}.$$		:$${\rm Pr}\big[(0.49 ≤ t ≤0.51)∩(0.49≤ p ≤0.51)\big] =0.02 \cdot 0.0513\hspace{0.15cm}\underline{\approx 2 · 10^{-5}}.$$
	{{ML-Fuß}}		{{ML-Fuß}}

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	'''(1)'''  und '''(2)'''		'''(1)'''  und '''(2)'''
	*Die Mittelwerte $m_t\hspace{0.15cm}\underline{= 0.5}$ und $m_p\hspace{0.15cm}\underline{= 0.7}$ können aus der Skizze abgeschätzt und aus der angegebenen Gleichung exakt ermittelt werden.		*Die Mittelwerte  $m_t\hspace{0.15cm}\underline{= 0.5}$  und  $m_p\hspace{0.15cm}\underline{= 0.7}$  können aus der Skizze abgeschätzt und aus der angegebenen Gleichung exakt ermittelt werden.
	*Die 2D–WDF der mittelwertfreien Größe lautet:		*Die 2D–WDF der mittelwertfreien Größe lautet:
	:$$f_{\it t\hspace{0.05cm}'\hspace{0.05cm}p\hspace{0.05cm}'}(\it t\hspace{0.05cm}', \it p\hspace{0.05cm}'{\rm )} = \rm 13.263\cdot \rm exp\Bigg (-\frac{\it {\rm (}t\hspace{0.05cm}'{\rm )}^{\rm 2}}{\rm 0.0288} - \frac{\it {\rm (}p\hspace{0.05cm}'{\rm )}^{\rm 2}}{\rm 0.0072}+\frac{\it t\hspace{0.05cm}'\cdot p\hspace{0.05cm}'}{\rm 0.0090}\Bigg ). $$		:$$f_{\it t\hspace{0.05cm}'\hspace{0.05cm}p\hspace{0.05cm}'}(\it t\hspace{0.05cm}', \it p\hspace{0.05cm}'{\rm )} = \rm 13.263\cdot \rm exp\Bigg (-\frac{\it {\rm (}t\hspace{0.05cm}'{\rm )}^{\rm 2}}{\rm 0.0288} - \frac{\it {\rm (}p\hspace{0.05cm}'{\rm )}^{\rm 2}}{\rm 0.0072}+\frac{\it t\hspace{0.05cm}'\cdot p\hspace{0.05cm}'}{\rm 0.0090}\Bigg ). $$
	*Zur Vereinfachung wird im Folgenden auf den Apostroph zur Kennzeichnung mittelwertfreier Größen verzichtet. Sowohl $t$ als auch $p$ sind bis einschließlich der Teilaufgabe '''(4)''' als mittelwertfrei zu verstehen.		*Zur Vereinfachung wird im Folgenden auf den Apostroph zur Kennzeichnung mittelwertfreier Größen verzichtet.
			*Sowohl  $t$  als auch  $p$  sind bis einschließlich der Teilaufgabe  '''(4)'''  als mittelwertfrei zu verstehen.



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	:$$f_{\it tp}(\it t, \it p)=\frac{\rm 1}{\rm 2\it \pi \cdot \sigma_{\it t}\cdot \sigma_{\it p} \cdot\sqrt{\rm 1- \it\rho^{\rm 2}}}\hspace{0.1cm}\cdot \hspace{0.1cm}\rm exp\Bigg\{-\hspace{0.1cm}\frac{\it t^{\rm 2}}{\rm 2\cdot (\rm 1-\rho^{\rm 2})\cdot \sigma_{\it t}^{\rm 2}} -\hspace{0.1cm}\frac{\it p^{\rm 2}}{\rm 2\cdot (\rm 1-\it\rho^{\rm 2}{\rm )}\cdot \sigma_{\it p}^{\rm 2}}+\hspace{0.1cm}\frac{\rho\cdot \it t\cdot \it p}{ (\rm 1-\it \rho^{\rm 2}{\rm )}\cdot\sigma_{\it t}\cdot\sigma_{\it p}}\Bigg\}.$$		:$$f_{\it tp}(\it t, \it p)=\frac{\rm 1}{\rm 2\it \pi \cdot \sigma_{\it t}\cdot \sigma_{\it p} \cdot\sqrt{\rm 1- \it\rho^{\rm 2}}}\hspace{0.1cm}\cdot \hspace{0.1cm}\rm exp\Bigg\{-\hspace{0.1cm}\frac{\it t^{\rm 2}}{\rm 2\cdot (\rm 1-\rho^{\rm 2})\cdot \sigma_{\it t}^{\rm 2}} -\hspace{0.1cm}\frac{\it p^{\rm 2}}{\rm 2\cdot (\rm 1-\it\rho^{\rm 2}{\rm )}\cdot \sigma_{\it p}^{\rm 2}}+\hspace{0.1cm}\frac{\rho\cdot \it t\cdot \it p}{ (\rm 1-\it \rho^{\rm 2}{\rm )}\cdot\sigma_{\it t}\cdot\sigma_{\it p}}\Bigg\}.$$

	Die Standardabweichungen $\sigma_t$ und $\sigma_p$ sowie der Korrelationskoeffizient $\rho$ lassen sich durch Koeffizientenvergleich ermitteln:		*Die Standardabweichungen  $\sigma_t$  und  $\sigma_p$  sowie der Korrelationskoeffizient  $\rho$  lassen sich durch Koeffizientenvergleich ermitteln:
	*Ein Vergleich der beiden ersten Terme im Exponenten zeigt, dass $\sigma_t = 2 \cdot \sigma_p$ gelten muss. Damit lautet die WDF:		*Ein Vergleich der beiden ersten Terme im Exponenten zeigt, dass  $\sigma_t = 2 \cdot \sigma_p$  gelten muss.  Damit lautet die WDF:
	:$$f_{\it tp}(\it t, \it p)=\frac{\rm 1}{\rm 4\it \pi \cdot \sigma_{\it p}^{\rm 2} \cdot\sqrt{\rm 1- \it\rho^{\rm 2}}}\hspace{0.1cm}\cdot \hspace{0.1cm}\rm exp\Bigg\{-\hspace{0.1cm}\frac{\it t^{\rm 2}}{\rm 8\cdot (\rm 1-\rho^{\rm 2})\cdot \sigma_{\it p}^{\rm 2}} -\hspace{0.1cm}\frac{\it p^{\rm 2}}{\rm 2\cdot (\rm 1-\it\rho^{\rm 2}{\rm )}\cdot \sigma_{\it p}^{\rm 2}}+\hspace{0.1cm}\frac{\rho\cdot \it t\cdot \it p}{\rm 2\cdot (\rm 1-\it \rho^{\rm 2}{\rm )}\cdot\sigma_{\it p}^{\rm 2}}\Bigg\}.$$		:$$f_{\it tp}(\it t, \it p)=\frac{\rm 1}{\rm 4\it \pi \cdot \sigma_{\it p}^{\rm 2} \cdot\sqrt{\rm 1- \it\rho^{\rm 2}}}\hspace{0.1cm}\cdot \hspace{0.1cm}\rm exp\Bigg\{-\hspace{0.1cm}\frac{\it t^{\rm 2}}{\rm 8\cdot (\rm 1-\rho^{\rm 2})\cdot \sigma_{\it p}^{\rm 2}} -\hspace{0.1cm}\frac{\it p^{\rm 2}}{\rm 2\cdot (\rm 1-\it\rho^{\rm 2}{\rm )}\cdot \sigma_{\it p}^{\rm 2}}+\hspace{0.1cm}\frac{\rho\cdot \it t\cdot \it p}{\rm 2\cdot (\rm 1-\it \rho^{\rm 2}{\rm )}\cdot\sigma_{\it p}^{\rm 2}}\Bigg\}.$$
	*Aus dem zweiten Term des Exponenten folgt:		*Aus dem zweiten Term des Exponenten folgt:
	:$$2\cdot(1-\rho^{\rm 2})\cdot\sigma_{p}^{ 2}=0.0072\hspace{0.5cm}\Rightarrow \hspace{0.5cm} \sigma_{p}^{2} = \frac{ 0.0036}{(1-\rho^{\rm 2})}.$$		:$$2\cdot(1-\rho^{\rm 2})\cdot\sigma_{p}^{ 2}=0.0072\hspace{0.5cm}\Rightarrow \hspace{0.5cm} \sigma_{p}^{2} = \frac{ 0.0036}{(1-\rho^{\rm 2})}.$$
	*Der Faktor $K = 13.263$ liefert nun das Ergebnis		*Der Faktor  $K = 13.263$  liefert nun das Ergebnis
	:$$K = \frac{\sqrt{\rm 1-\it\rho^{\rm 2}}}{\rm 4\it\pi\cdot \rm 0.0036}=\rm 13.263 \hspace{0.5cm}\Rightarrow \hspace{0.5cm}\sqrt{\rm 1-\it\rho^{\rm 2}}=\rm 0.6 \hspace{0.5cm}\Rightarrow \hspace{0.5cm}\hspace{0.15cm}\underline{ \rm \rho = \rm 0.8}.$$		:$$K = \frac{\sqrt{\rm 1-\it\rho^{\rm 2}}}{\rm 4\it\pi\cdot \rm 0.0036}=\rm 13.263 \hspace{0.5cm}\Rightarrow \hspace{0.5cm}\sqrt{\rm 1-\it\rho^{\rm 2}}=\rm 0.6 \hspace{0.5cm}\Rightarrow \hspace{0.5cm}\hspace{0.15cm}\underline{ \rm \rho = \rm 0.8}.$$
	*Daraus ergeben sich die Streuungen zu $\sigma_t\hspace{0.15cm}\underline{= 0.2}$ und $\sigma_p\hspace{0.15cm}\underline{= 0.1}$.		*Daraus ergeben sich die Streuungen zu  $\sigma_t\hspace{0.15cm}\underline{= 0.2}$  und  $\sigma_p\hspace{0.15cm}\underline{= 0.1}$.

	*Zur Kontrolle verwenden wir den letzten Term des Exponenten:		*Zur Kontrolle verwenden wir den letzten Term des Exponenten:
	:$$\frac{(1 - \rho^{2})\cdot \sigma_{\it t}\cdot\sigma_{\it p}}{\it \rho} = \frac{0.36\cdot 0.1\cdot 0.2}{0.8} = \rm 0.009.$$		:$$\frac{(1 - \rho^{2})\cdot \sigma_{\it t}\cdot\sigma_{\it p}}{\it \rho} = \frac{0.36\cdot 0.1\cdot 0.2}{0.8} = \rm 0.009.$$
	*Dies stimmt mit dem vorgegebenen Wert überein.		*Dies stimmt mit dem vorgegebenen Wert überein.



	'''(4)'''  Der <u>Lösungsvorschlag 1</u> ist richtig.		'''(4)'''  Der <u>Lösungsvorschlag 1</u> ist richtig.
	*Im Grunde genommen ist $(t, p)$ keine echte Gaußsche Zufallsgröße, da beide Komponenten begrenzt sind.		*Im Grunde genommen ist  $(t, p)$  keine echte Gaußsche Zufallsgröße,  da beide Komponenten begrenzt sind.
	*Die Wahrscheinlichkeiten für die Ereignisse $t < 0$,     $t >1$,     $p < 0$ und $p >1$ sind somit Null.		*Die Wahrscheinlichkeiten für die Ereignisse  $t < 0$,     $t >1$,     $p < 0$ und  $p >1$  sind somit Null.
	*Bei Gaußschen Größen mit den hier vorliegenden Mittelwerten und Streuungen ergeben sich jedoch		*Bei Gaußschen Größen mit den hier vorliegenden Mittelwerten und Streuungen ergeben sich jedoch
	:$$\rm Pr(\it t < \rm 0) = \rm Pr(\it t > \rm 1) = \rm Q(2.5)\approx 6\cdot 10^{-3},$$		:$$\rm Pr(\it t < \rm 0) = \rm Pr(\it t > \rm 1) = \rm Q(2.5)\approx 6\cdot 10^{-3},$$
Line 100:		Line 103:
	:$$\rm Pr(\it p < \rm 0) = \rm Q(7)\approx 10^{-12}.$$		:$$\rm Pr(\it p < \rm 0) = \rm Q(7)\approx 10^{-12}.$$

	*Der Korrelationskoeffizient $\rho = 0.8$ ist hier positiv. Hat der Prüfling im Theorieteil eher gut abgeschnitten, so ist (zumindest bei dieser Aufgabe) zu erwarten, dass auch der praktische Teil gut läuft.		*Der Korrelationskoeffizient  $\rho = 0.8$  ist hier positiv. Hat der Prüfling im Theorieteil eher gut abgeschnitten, so ist (zumindest bei dieser Aufgabe) zu erwarten, dass auch der praktische Teil gut läuft.
	*Hier ist also der Lösungsvorschlag 2 falsch. In der Praxis ist das sicher nicht immer so.		*Hier ist also der Lösungsvorschlag 2 falsch. In der Praxis ist das sicher nicht immer so.


	'''(5)'''  Für diese Wahrscheinlichkeit gilt mit $\Delta t = \Delta p = 0.02$:
			'''(5)'''  Für diese Wahrscheinlichkeit gilt mit  $\Delta t = \Delta p = 0.02$:
	:$$\rm Pr\left [( \rm 0.5-\frac{\rm\Delta\it t}{\rm 2}\le \it t \le \rm 0.5+\frac{\rm\Delta\it t}{\rm 2})\cap(\rm 0.5-\frac{\rm\Delta\it p}{\rm 2}\le \it p \le \rm 0.5+\frac{\rm\Delta\it p}{\rm 2})\right ] \approx \rm\Delta\it t\cdot\rm\Delta\it p\cdot \it f_{tp}{\rm (}t=\rm 0.5, \it p = \rm 0.5).$$		:$$\rm Pr\left [( \rm 0.5-\frac{\rm\Delta\it t}{\rm 2}\le \it t \le \rm 0.5+\frac{\rm\Delta\it t}{\rm 2})\cap(\rm 0.5-\frac{\rm\Delta\it p}{\rm 2}\le \it p \le \rm 0.5+\frac{\rm\Delta\it p}{\rm 2})\right ] \approx \rm\Delta\it t\cdot\rm\Delta\it p\cdot \it f_{tp}{\rm (}t=\rm 0.5, \it p = \rm 0.5).$$

	Für die 2D-WDF gilt unter Berücksichtigung der Mittelwerte $m_t{= 0.5}$ und $m_p{= 0.7}$:		*Für die 2D-WDF gilt unter Berücksichtigung der Mittelwerte  $m_t{= 0.5}$  und  $m_p{= 0.7}$:
	:$$f_{tp}(\it t=\rm 0.5, \it p=\rm 0.5) = \rm 13.263\cdot {\rm e}^{-(-0.2)^2/0.0072}\approx 0.0513.$$		:$$f_{tp}(\it t=\rm 0.5, \it p=\rm 0.5) = \rm 13.263\cdot {\rm e}^{-(-0.2)^2/0.0072}\approx 0.0513.$$

	Damit ergibt sich die gesuchte Wahrscheinlichkeit zu		*Damit ergibt sich die gesuchte Wahrscheinlichkeit zu
	:$${\rm Pr}\big[(0.49 ≤ t ≤0.51)∩(0.49≤ p ≤0.51)\big] =0.02 \cdot 0.02 \cdot 0.0513\hspace{0.15cm}\underline{\approx 2 · 10^{-5}}.$$		:$${\rm Pr}\big[(0.49 ≤ t ≤0.51)∩(0.49≤ p ≤0.51)\big] =0.02 \cdot 0.02 \cdot 0.0513\hspace{0.15cm}\underline{\approx 2 · 10^{-5}}.$$
	{{ML-Fuß}}		{{ML-Fuß}}