Difference between revisions of "Aufgaben:Exercise 4.3Z: Dirac-shaped "2D-PDF""

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{{quiz-Header|Buchseite=Stochastische Signaltheorie/Zweidimensionale Zufallsgrößen
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{{quiz-Header|Buchseite=Theory_of_Stochastic_Signals/Two-Dimensional_Random_Variables
 
}}
 
}}
  
[[File:P_ID257__Sto_Z_4_3.png|right|frame|Betrachtete diracförmige 2D-WDF]]
+
[[File:P_ID257__Sto_Z_4_3.png|right|frame|Dirac-shaped 2D– PDF]]
In der Grafik ist die zweidimensionale Wahrscheinlichkeitsdichtefunktion $f_{xy}(x, y)$ der zwei diskreten Zufallsgrößen $x$ und $y$ dargestellt.  
+
The graph shows the two-dimensional probability density function $f_{xy}(x, y)$  of two discrete random variables  $x$,   $y$.  
*Diese 2D–WDF besteht aus acht Diracpunkten, durch Kreuze markiert. Die Zahlenwerte geben die entsprechenden Wahrscheinlichkeiten an.
+
*This 2D–PDF consists of eight Dirac points,   marked by crosses. 
*Es ist zu erkennen, dass sowohl $x$ als auch $y$ alle ganzzahligen Werte zwischen den Grenzen $-2$ und $+2$ annehmen können.
+
*The numerical values indicate the corresponding probabilities.
*Die Varianzen der beiden Zufallsgrößen sind wie folgt gegeben:    $\sigma_x^2 = 2$,   $\sigma_y^2 = 1.4$.  
+
*It can be seen that both  $x$  and  $y$  can take all integer values between the limits  $-2$  and  $+2$.
<br>
+
*The variances of the two random variables are given as follows:  &nbsp; $\sigma_x^2 = 2$, &nbsp; $\sigma_y^2 = 1.4$.  
  
  
  
  
''Hinweise:''
+
 
*Die Aufgabe gehört zum  Kapitel [[Stochastische_Signaltheorie/Zweidimensionale_Zufallsgrößen|Zweidimensionale Zufallsgrößen]].
+
Hints:  
*Bezuig genommen wird auch auf das Kapitel [[Stochastische_Signaltheorie/Momente_einer_diskreten_Zufallsgröße|Momente einer diskreten Zufallsgröße]]
+
*The exercise belongs to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Two-Dimensional_Random_Variables|Two-Dimensional Random Variables]].
 +
*Reference is also made to the chapter&nbsp; [[Theory_of_Stochastic_Signals/Moments_of_a_Discrete_Random_Variable|Moments of a Discrete Random Variable]]
 
   
 
   
  
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===Fragebogen===
+
===Questions===
  
 
<quiz display=simple>
 
<quiz display=simple>
{Welche der folgenden Aussagen trefen hinsichtlich der Zufallsgr&ouml;&szlig;e $x$ zu?
+
{Which of the following statements are true regarding the random variable&nbsp; $x$?
 
|type="[]"}
 
|type="[]"}
+ Die Wahrscheinlichkeiten für $-2$, $-1$, &nbsp; $0$, $+1$ und $+2$ sind gleich.
+
+ The probabilities for&nbsp; $-2$,&nbsp; $-1$,&nbsp; &nbsp; $0$,&nbsp; $+1$&nbsp; and&nbsp; $+2$&nbsp; are equal.
+ Die Zufallsgr&ouml;&szlig;e $x$ ist mittelwertfrei $(m_x = 0)$.  
+
+ The random variable&nbsp; $x$&nbsp; is mean-free&nbsp; $(m_x = 0)$.  
- Die Wahrscheinlichkeit ${\rm Pr}(x \le 1)$ ist $0.9$.
+
- The probability&nbsp; ${\rm Pr}(x \le 1)=0.9$.
  
  
{Welche der folgenden Aussagen treffen hinsichtlich der Zufallsgr&ouml;&szlig;e $y$ zu?
+
{Which of the following statements are true with respect to the random variable&nbsp; $y$?
 
|type="[]"}
 
|type="[]"}
- Die Wahrscheinlichkeiten für $-2$, $-1$, &nbsp; $0$, $+1$ und $+2$ sind gleich.
+
- The probabilities for&nbsp; $-2$,&nbsp; $-1$,&nbsp; &nbsp; $0$,&nbsp; $+1$&nbsp; and&nbsp; $+2$&nbsp; are equal.
+ Die Zufallsgr&ouml;&szlig;e $y$ ist mittelwertfrei $(m_y = 0)$.  
+
+ The random variable&nbsp; $y$&nbsp; is mean-free&nbsp; $(m_y = 0)$.  
+ Die Wahrscheinlichkeit ${\rm Pr}(y \le 1)$ ist $0.9$.
+
+ The probability&nbsp; ${\rm Pr}(y \le 1)=0.9$.
  
  
{Berechnen Sie den Wert der zweidimensionalen VTF an der Stelle $(+1, +1)$.
+
{Calculate the value of the two-dimensional cumulative distribution function&nbsp; $\rm (CDF)$&nbsp; at location&nbsp; $(+1, +1)$.
 
|type="{}"}
 
|type="{}"}
$F_{xy}(+1, +1) \ = \ $ { 0.8 3% }
+
$F_{xy}(+1, +1) \ = \ $ { 0.8 3% }
 
 
  
{Berechnen Sie die Wahrscheinlichkeit, dass $x \le 1$ gilt, unter der Bedingung, dass gleichzeitig $y \le 1$ ist.
+
{Calculate the probability that&nbsp; $x \le 1$&nbsp; holds,&nbsp; conditioned on&nbsp; $y \le 1$&nbsp; simultaneously.
 
|type="{}"}
 
|type="{}"}
 
${\rm Pr}(x ≤ 1\hspace{0.05cm} | \hspace{0.05cm}y ≤ 1)\ = \ $ { 0.889 3% }
 
${\rm Pr}(x ≤ 1\hspace{0.05cm} | \hspace{0.05cm}y ≤ 1)\ = \ $ { 0.889 3% }
  
  
{Berechnen Sie das gemeinsame Moment $m_{xy}$ der Zufallsgr&ouml;&szlig;en $x$ und $y$.
+
{Calculate the joint moment&nbsp; $m_{xy}$&nbsp; of the random variables&nbsp; $x$&nbsp; and&nbsp; $y$.
 
|type="{}"}
 
|type="{}"}
 
$m_{xy}\ = \ $ { 1.2 3% }
 
$m_{xy}\ = \ $ { 1.2 3% }
  
  
{Berechnen Sie den Korrelationskoeffizienten $\rho_{xy}$ und geben Sie die Gleichung der Korrelationsgeraden $K(x)$ an.
+
{Calculate the correlation coefficient&nbsp; $\rho_{xy}$.&nbsp; Give the equation of the correlation line&nbsp; $K(x)$&nbsp; What is its angle to the&nbsp; $x$&ndash;axis?
<br>Wie gro&szlig; ist deren Winkel zur $x$-Achse?
 
 
|type="{}"}
 
|type="{}"}
 
$\rho_{xy}\ = \ $ { 0.707 3% }
 
$\rho_{xy}\ = \ $ { 0.707 3% }
$\theta_{y\hspace{0.05cm}→\hspace{0.05cm} x}\ = \ $ { 31 3% } $\ \rm Grad$
+
$\theta_{y\hspace{0.05cm}→\hspace{0.05cm} x}\ = \ $ { 31 3% } $\ \rm degrees$
  
  
{Welche der nachfolgenden Aussagen sind zutreffend?
+
{Which of the following statements are true?
 
|type="[]"}
 
|type="[]"}
- Die Zufallsgr&ouml;&szlig;en $x$ und $y$ sind statistisch unabh&auml;ngig.
+
- The random variables&nbsp; $x$&nbsp; and&nbsp; $y$&nbsp; are statistically independent.
+ Man erkennt bereits aus der vorgegebenen 2D-WDF, dass $x$ und $y$ statistisch voneinander abh&auml;ngen.
+
+ It can already be seen from the given 2D&ndash;PDF that&nbsp; $x$&nbsp; and&nbsp; $y$&nbsp; are statistically dependent on each other.
+ Aus dem berechneten Korrelationskoeffizienten $\rho_{xy}$ kann man auf die statistische Abh&auml;ngigkeit zwischen $x$ und $y$ schließen.
+
+ From the calculated correlation coefficient&nbsp; $\rho_{xy}$&nbsp; one can conclude the statistical dependence between&nbsp; $x$&nbsp; and&nbsp; $y$&nbsp;.
  
  
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</quiz>
 
</quiz>
  
===Musterlösung===
+
===Solution===
 
{{ML-Kopf}}
 
{{ML-Kopf}}
'''(1)'''&nbsp; Richtig sind  <u>die beiden ersten Antworten</u>:
+
'''(1)'''&nbsp; Correct are the&nbsp; <u>first two answers</u>:
*Die Randwahrscheinlichkeitsdichtefunktion $f_{x}(x)$ erh&auml;lt man aus der 2D&ndash;WDF $f_{xy}(x, y)$ durch Integration über $y$.  
+
*The marginal probability density function&nbsp; $f_{x}(x)$&nbsp; is obtained from the 2D&ndash;PDF&nbsp; $f_{xy}(x, y)$&nbsp; by integration over&nbsp; $y$.  
*F&uuml;r alle m&ouml;glichen Werte $ x \in \{-2, -1, 0, +1, +2\}$ sind die Wahrscheinlichkeiten gleich $0.2$.
+
*For all possible values&nbsp; $ x \in \{-2, -1, \ 0, +1, +2\}$&nbsp; the probabilities are equal&nbsp; $0.2$.
*Es gilt ${\rm Pr}(x \le 1)= 0.8$ und der Mittelwert ist $m_x = 0$.  
+
*It holds&nbsp; ${\rm Pr}(x \le 1)= 0.8$.&nbsp; The mean is&nbsp; $m_x = 0$.  
 +
 
 +
 
 +
 
 +
[[File:P_ID258__Sto_Z_4_3_b.png|right|frame|Discrete marginal PDF&nbsp; $f_{y}(y)$]]
 +
'''(2)'''&nbsp; Correct are&nbsp; <u>the proposed solutions 2 and 3</u>:
 +
*By integration over&nbsp; $x$&nbsp; one obtains the PDF&nbsp; $f_{y}(y)$ sketched on the right.
 +
*Due to symmetry,&nbsp; the mean value&nbsp; $m_y = 0$&nbsp; is obtained.
 +
*The probability we are looking for is&nbsp; ${\rm Pr}(y \le 1)= 0.9$.
 +
 
 +
 
  
 +
'''(3)'''&nbsp; By definition:
 +
:$$F_{xy}(r_x, r_y) = {\rm Pr} \big [(x \le r_x)\cap(y\le r_y)\big ].$$
  
[[File:P_ID258__Sto_Z_4_3_b.png|right|Diskrete Rand-WDF]]
+
*For&nbsp; $r_x = r_y = 1$&nbsp; it follows:
'''(2)'''&nbsp; Richtig sind <u>die Lösungsvorschläge 2 und 3</u>:
+
:$$F_{xy}(+1, +1) = {\rm Pr}\big [(x \le 1)\cap(y\le 1)\big ].$$
*Durch Integration &uuml;ber $x$ erh&auml;lt man die rechts skizzierte WDF $f_{y}(y)$.
+
*As can be seen from the 2D&ndash;PDF on the information page,&nbsp; this probability is&nbsp; ${\rm Pr}\big [(x \le 1)\cap(y\le 1)\big ]\hspace{0.15cm}\underline{=0.8}$.
* Aufgrund der Symmetrie ergibt sich der Mittelwert $m_y = 0$.
 
*Die gesuchte Wahrscheinlichkeit ist ${\rm Pr}(y \le 1)= 0.9$.  
 
  
  
'''(3)'''&nbsp;  Definitionsgemäß gilt:
 
$$F_{xy}(r_x, r_y) = \rm Pr((\it x \le r_x)\cap(\it y\le r_y)).$$
 
  
F&uuml;r $r_x = r_y = 1$ folgt daraus:
+
'''(4)'''&nbsp; For this,&nbsp; Bayes' theorem can also be used to write:  
$$F_{xy}(+1, +1) = {\rm Pr}((x \le 1)\cap(y\le 1)).$$
+
:$$ \rm Pr(\it x \le \rm 1)\hspace{0.05cm} | \hspace{0.05cm} \it y \le \rm 1) = \frac{ \rm Pr\big [(\it x \le \rm 1)\cap(\it y\le \rm 1)\big ]}{ \rm Pr(\it y\le \rm 1)} = \it \frac{F_{xy} \rm (1, \rm 1)}{F_{y}\rm (1)}.$$
  
Wie aus der 2D&ndash;WDF auf der Angabenseite zu ersehen, ist diese Wahrscheinlichkeit ${\rm Pr}((x \le 1)\cap(y\le 1))\hspace{0.15cm}\underline{=0.8}$.
+
*With the results from&nbsp; '''(2)'''&nbsp; and&nbsp; '''(3)'''&nbsp; it follows&nbsp; $ \rm Pr(\it x \le \rm 1)\hspace{0.05cm} | \hspace{0.05cm} \it y \le \rm 1) = 0.8/0.9 = 8/9 \hspace{0.15cm}\underline{=0.889}$.
  
  
'''(4)'''&nbsp; Hierf&uuml;r kann mit dem Satz von Bayes auch geschrieben werden:
 
$$ \rm Pr(\it x \le \rm 1)\hspace{0.05cm} | \hspace{0.05cm} \it y \le \rm 1) = \frac{ \rm Pr((\it x \le \rm 1)\cap(\it y\le \rm 1))}{ \rm Pr(\it y\le \rm 1)} = \it \frac{F_{xy}(\rm 1, \rm 1)}{F_{y}(\rm 1)}.$$
 
  
Mit den Ergebnissen aus (2) und (3) folgt daraus $ \rm Pr(\it x \le \rm 1)\hspace{0.05cm} | \hspace{0.05cm} \it y \le \rm 1) = 0.8/0.9 = 8/9 \hspace{0.15cm}\underline{=0.889}$.
+
'''(5)'''&nbsp; According to the definition, the common moment is:
 +
:$$m_{xy} = {\rm E}\big[x\cdot y \big] = \sum\limits_{i} {\rm Pr}( x_i \cap y_i)\cdot x_i\cdot y_i. $$
  
 +
*There remain five Dirac delta functions with&nbsp; $x_i \cdot y_i \ne 0$:
 +
:$$m_{xy} = \rm 0.1\cdot (-2) (-1) + 0.2\cdot(-1) (-1)+ 0.2\cdot 1\cdot 1 + 0.1\cdot 2\cdot 1+ 0.1\cdot 2\cdot 2\hspace{0.15cm}\underline{=\rm 1.2}.$$
  
'''(5)'''&nbsp; Entsprechend der Definition gilt für das gemeinsame Moment:
 
$$m_{xy} = {\rm E}[x\cdot y] = \sum\limits_{i} {\rm Pr}( x_i \cap y_i)\cdot x_i\cdot y_i.  $$
 
  
Es verbleiben fünf Diracfunktionen mit $x_i \cdot y_i \ne 0$:
 
$$m_{xy} = \rm 0.1\cdot (-2) (-1) + 0.2\cdot(-1) (-1)+ 0.2\cdot 1\cdot 1 + 0.1\cdot 2\cdot 1+ 0.1\cdot 2\cdot 2\hspace{0.15cm}\underline{=\rm 1.2}.$$
 
  
 +
[[File:EN_Sto_Z_4_3_f.png|right|frame|2D&ndash;PDF and regression line&nbsp; $\rm (RL)$]]
 +
'''(6)'''&nbsp; For the correlation coefficient:
 +
:$$\rho_{xy} = \frac{\mu_{xy}}{\sigma_x\cdot \sigma_y} = \frac{1.2}{\sqrt{2}\cdot\sqrt{1.4}}\hspace{0.15cm}\underline{=0.717}.$$
  
[[File:P_ID259__Sto_Z_4_3_f.png|right|2D-WDF]]
+
*This takes into account that because&nbsp; $m_x = m_y = 0$&nbsp; the covariance&nbsp; $\mu_{xy}$&nbsp; is equal to the moment&nbsp; $m_{xy}$&nbsp; .
'''(6)'''&nbsp; F&uuml;r den Korrelationskoeffizienten gilt:
 
$$\rho_{xy} = \frac{\mu_{xy}}{\sigma_x\cdot \sigma_y} = \frac{1.2}{\sqrt{2}\cdot\sqrt{1.4}}\hspace{0.15cm}\underline{=0.717}.$$
 
  
Hier ist ber&uuml;cksichtigt, dass wegen $m_x = m_y = 0$ die Kovarianz $\mu_{xy}$ gleich dem Moment $m_{xy}$ ist.
+
*The equation of the correlation line is:
 +
:$$y=\frac{\sigma_y}{\sigma_x}\cdot \rho_{xy}\cdot x = \frac{\mu_{xy}}{\sigma_x^{\rm 2}}\cdot x = \rm 0.6\cdot \it x.$$
  
Die Gleichung der Korrelationsgeraden lautet:
+
*See sketch on the right.&nbsp; The angle between the regression line&nbsp; $\rm (RL)$&nbsp;  and the&nbsp; $x$-axis is
$$y=\frac{\sigma_y}{\sigma_x}\cdot \rho_{xy}\cdot x = \frac{\mu_{xy}}{\sigma_x^{\rm 2}}\cdot x = \rm 0.6\cdot \it x.$$
+
:$$\theta_{y\hspace{0.05cm}\hspace{0.05cm} x} = \arctan(0.6) \hspace{0.15cm}\underline{=31^\circ}.$$
  
Im Bild ist die Gerade $y = K(x)$ eingezeichnet. Der Winkel zwischen Korrelationsgerade und  $x$-Achse betr&auml;gt $\theta_{y\hspace{0.05cm}→\hspace{0.05cm} x} = \arctan(0.6) \hspace{0.15cm}\underline{=31^\circ}.$
 
  
'''(7)'''&nbsp; Richtig sind die <u>Lösungsvorschläge 2 und 3</u>:
+
'''(7)'''&nbsp; The correct solutions are&nbsp; <u>solutions 2 and 3</u>:
*Bei statistischer Unabh&auml;ngigkeit m&uuml;sste $f_{xy}(x, y) = f_{x}(x) \cdot f_{y}(y)$ gelten, was hier nicht erf&uuml;llt ist.  
+
*If statistically independent,&nbsp; $f_{xy}(x, y) = f_{x}(x) \cdot f_{y}(y)$&nbsp; should hold,&nbsp; which is not done here.  
*Aus der Korreliertheit (folgt aus $\rho_{xy} \ne 0$) kann direkt auf die statistische Abh&auml;ngigkeit geschlossen werden, denn Korrelation bedeutet eine Sonderform der statistischen Abh&auml;ngigkeit, nämlich die  lineare statistische Abh&auml;ngigkeit.  
+
*From correlatedness&nbsp; $($follows from&nbsp; $\rho_{xy} \ne 0)$&nbsp; it is possible to directly infer statistical dependence,
 +
*because correlation means a special form of statistical dependence,&nbspnamely linear statistical dependence.  
  
 
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[[Category:Aufgaben zu Stochastische Signaltheorie|^4.1 Zweidimensionale Zufallsgrößen^]]
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[[Category:Theory of Stochastic Signals: Exercises|^4.1 Two-Dimensional Random Variables^]]

Latest revision as of 16:41, 10 April 2022

Dirac-shaped 2D– PDF

The graph shows the two-dimensional probability density function $f_{xy}(x, y)$  of two discrete random variables  $x$,  $y$.

  • This 2D–PDF consists of eight Dirac points,  marked by crosses. 
  • The numerical values indicate the corresponding probabilities.
  • It can be seen that both  $x$  and  $y$  can take all integer values between the limits  $-2$  and  $+2$.
  • The variances of the two random variables are given as follows:   $\sigma_x^2 = 2$,   $\sigma_y^2 = 1.4$.



Hints:




Questions

1

Which of the following statements are true regarding the random variable  $x$?

The probabilities for  $-2$,  $-1$,    $0$,  $+1$  and  $+2$  are equal.
The random variable  $x$  is mean-free  $(m_x = 0)$.
The probability  ${\rm Pr}(x \le 1)=0.9$.

2

Which of the following statements are true with respect to the random variable  $y$?

The probabilities for  $-2$,  $-1$,    $0$,  $+1$  and  $+2$  are equal.
The random variable  $y$  is mean-free  $(m_y = 0)$.
The probability  ${\rm Pr}(y \le 1)=0.9$.

3

Calculate the value of the two-dimensional cumulative distribution function  $\rm (CDF)$  at location  $(+1, +1)$.

$F_{xy}(+1, +1) \ = \ $

4

Calculate the probability that  $x \le 1$  holds,  conditioned on  $y \le 1$  simultaneously.

${\rm Pr}(x ≤ 1\hspace{0.05cm} | \hspace{0.05cm}y ≤ 1)\ = \ $

5

Calculate the joint moment  $m_{xy}$  of the random variables  $x$  and  $y$.

$m_{xy}\ = \ $

6

Calculate the correlation coefficient  $\rho_{xy}$.  Give the equation of the correlation line  $K(x)$  What is its angle to the  $x$–axis?

$\rho_{xy}\ = \ $

$\theta_{y\hspace{0.05cm}→\hspace{0.05cm} x}\ = \ $

$\ \rm degrees$

7

Which of the following statements are true?

The random variables  $x$  and  $y$  are statistically independent.
It can already be seen from the given 2D–PDF that  $x$  and  $y$  are statistically dependent on each other.
From the calculated correlation coefficient  $\rho_{xy}$  one can conclude the statistical dependence between  $x$  and  $y$ .


Solution

(1)  Correct are the  first two answers:

  • The marginal probability density function  $f_{x}(x)$  is obtained from the 2D–PDF  $f_{xy}(x, y)$  by integration over  $y$.
  • For all possible values  $ x \in \{-2, -1, \ 0, +1, +2\}$  the probabilities are equal  $0.2$.
  • It holds  ${\rm Pr}(x \le 1)= 0.8$.  The mean is  $m_x = 0$.


Discrete marginal PDF  $f_{y}(y)$

(2)  Correct are  the proposed solutions 2 and 3:

  • By integration over  $x$  one obtains the PDF  $f_{y}(y)$ sketched on the right.
  • Due to symmetry,  the mean value  $m_y = 0$  is obtained.
  • The probability we are looking for is  ${\rm Pr}(y \le 1)= 0.9$.


(3)  By definition:

$$F_{xy}(r_x, r_y) = {\rm Pr} \big [(x \le r_x)\cap(y\le r_y)\big ].$$
  • For  $r_x = r_y = 1$  it follows:
$$F_{xy}(+1, +1) = {\rm Pr}\big [(x \le 1)\cap(y\le 1)\big ].$$
  • As can be seen from the 2D–PDF on the information page,  this probability is  ${\rm Pr}\big [(x \le 1)\cap(y\le 1)\big ]\hspace{0.15cm}\underline{=0.8}$.


(4)  For this,  Bayes' theorem can also be used to write:

$$ \rm Pr(\it x \le \rm 1)\hspace{0.05cm} | \hspace{0.05cm} \it y \le \rm 1) = \frac{ \rm Pr\big [(\it x \le \rm 1)\cap(\it y\le \rm 1)\big ]}{ \rm Pr(\it y\le \rm 1)} = \it \frac{F_{xy} \rm (1, \rm 1)}{F_{y}\rm (1)}.$$
  • With the results from  (2)  and  (3)  it follows  $ \rm Pr(\it x \le \rm 1)\hspace{0.05cm} | \hspace{0.05cm} \it y \le \rm 1) = 0.8/0.9 = 8/9 \hspace{0.15cm}\underline{=0.889}$.


(5)  According to the definition, the common moment is:

$$m_{xy} = {\rm E}\big[x\cdot y \big] = \sum\limits_{i} {\rm Pr}( x_i \cap y_i)\cdot x_i\cdot y_i. $$
  • There remain five Dirac delta functions with  $x_i \cdot y_i \ne 0$:
$$m_{xy} = \rm 0.1\cdot (-2) (-1) + 0.2\cdot(-1) (-1)+ 0.2\cdot 1\cdot 1 + 0.1\cdot 2\cdot 1+ 0.1\cdot 2\cdot 2\hspace{0.15cm}\underline{=\rm 1.2}.$$


2D–PDF and regression line  $\rm (RL)$

(6)  For the correlation coefficient:

$$\rho_{xy} = \frac{\mu_{xy}}{\sigma_x\cdot \sigma_y} = \frac{1.2}{\sqrt{2}\cdot\sqrt{1.4}}\hspace{0.15cm}\underline{=0.717}.$$
  • This takes into account that because  $m_x = m_y = 0$  the covariance  $\mu_{xy}$  is equal to the moment  $m_{xy}$  .
  • The equation of the correlation line is:
$$y=\frac{\sigma_y}{\sigma_x}\cdot \rho_{xy}\cdot x = \frac{\mu_{xy}}{\sigma_x^{\rm 2}}\cdot x = \rm 0.6\cdot \it x.$$
  • See sketch on the right.  The angle between the regression line  $\rm (RL)$  and the  $x$-axis is
$$\theta_{y\hspace{0.05cm}→\hspace{0.05cm} x} = \arctan(0.6) \hspace{0.15cm}\underline{=31^\circ}.$$


(7)  The correct solutions are  solutions 2 and 3:

  • If statistically independent,  $f_{xy}(x, y) = f_{x}(x) \cdot f_{y}(y)$  should hold,  which is not done here.
  • From correlatedness  $($follows from  $\rho_{xy} \ne 0)$  it is possible to directly infer statistical dependence,
  • because correlation means a special form of statistical dependence,  namely linear statistical dependence.