Difference between revisions of "Aufgaben:Exercise 2.2Z: Discrete Random Variables"

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===Solution===
 
===Solution===
 
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'''(1)'''  Aufgrund der Symmetrie gilt:
+
'''(1)'''  Due to the symmetry holds:
 
:$$\rm \it m_{\it a}=\rm 0; \hspace{0.5cm}\it m_{\rm 2\it a}=\rm 0.5\cdot (-1)^2 + 0.5\cdot (1)^2{ = 1}.$$
 
:$$\rm \it m_{\it a}=\rm 0; \hspace{0.5cm}\it m_{\rm 2\it a}=\rm 0.5\cdot (-1)^2 + 0.5\cdot (1)^2{ = 1}.$$
  
*Daraus erhält man mit dem Satz von Steiner:
+
*From this one obtains with Steiner's theorem:
:$$\it\sigma_a^{\rm 2} = \rm\sqrt{1-0^2}=1 \hspace{0.5cm}bzw. \hspace{0.5cm}\it\sigma_a\hspace{0.15cm} \underline{=\rm 1}.$$
+
:$$\it\sigma_a^{\rm 2} = \rm\sqrt{1-0^2}=1 \hspace{0.5cm}or \hspace{0.5cm}\it\sigma_a\hspace{0.15cm} \underline{=\rm 1}.$$
  
  
  
'''(2)'''  Allgemein gilt für das Moment  $k$–ter Ordnung:
+
'''(2)'''  In general, for the  $k$–th order moment:
 
:$$ m_{k}=(1-p)\cdot 0^{ k} + p\cdot 1^{k}= p.$$
 
:$$ m_{k}=(1-p)\cdot 0^{ k} + p\cdot 1^{k}= p.$$
  
*Daraus folgt mit  $p = 1/4$:
+
*From this follows with  $p = 1/4$:
 
:$$m_{b}= m_{2b}= p, \hspace{0.5cm} \sigma_{\it b}=\sqrt{p\cdot (1- p)}\hspace{0.15cm} \underline{=\rm 0.433} .$$
 
:$$m_{b}= m_{2b}= p, \hspace{0.5cm} \sigma_{\it b}=\sqrt{p\cdot (1- p)}\hspace{0.15cm} \underline{=\rm 0.433} .$$
  
  
  
'''(3)'''  Für die Zufallsgröße  $c$  gilt:
+
'''(3)'''  For the random variable  $c$  holds:
:$$m_{\it c} = 0\hspace{0.3cm} ({\rm symmetrisch\hspace{0.1cm}um\hspace{0.1cm}0)},$$
+
:$$m_{\it c} = 0\hspace{0.3cm} ({\rm symmetric\hspace{0.1cm}um\hspace{0.1cm}0)},$$
 
:$$ m_{2\it c}= {1}/{4}\cdot(-1)^2+{1}/{2}\cdot 0^2+{1}/{4}\cdot (1)^2={1}/{2} \hspace{0.5cm}$$
 
:$$ m_{2\it c}= {1}/{4}\cdot(-1)^2+{1}/{2}\cdot 0^2+{1}/{4}\cdot (1)^2={1}/{2} \hspace{0.5cm}$$
 
:$$\Rightarrow \hspace{0.5cm}\sigma_{\it c}=\rm \sqrt{1/2}\hspace{0.15cm} \underline{=0.707}.$$
 
:$$\Rightarrow \hspace{0.5cm}\sigma_{\it c}=\rm \sqrt{1/2}\hspace{0.15cm} \underline{=0.707}.$$
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'''(4)'''  Nach den allgemeinen Regeln für Erwartungswerte gilt mit  $p = 0.25$:
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'''(4)'''  According to the general rules for expected values, with  $p = 0.25$:
 
:$$m_{\it d} = {\rm E}\big[a-2 b+c\big]= {\rm E}\big[a\big] \hspace{0.1cm} -\hspace{0.1cm}\rm 2 \hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[ b\big]\hspace{0.1cm}+\hspace{0.1cm} {\rm E}\big[ c\big] =  m_{ a}\hspace{0.1cm}-\hspace{0.1cm}2\hspace{0.05cm}\cdot\hspace{0.05cm} m_{\it b}\hspace{0.1cm}+\hspace{0.1cm} m_{\it c} =    0-2\hspace{0.05cm}\cdot\hspace{0.05cm} p + 0 \hspace{0.15cm} \underline{= -0.5}.$$
 
:$$m_{\it d} = {\rm E}\big[a-2 b+c\big]= {\rm E}\big[a\big] \hspace{0.1cm} -\hspace{0.1cm}\rm 2 \hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[ b\big]\hspace{0.1cm}+\hspace{0.1cm} {\rm E}\big[ c\big] =  m_{ a}\hspace{0.1cm}-\hspace{0.1cm}2\hspace{0.05cm}\cdot\hspace{0.05cm} m_{\it b}\hspace{0.1cm}+\hspace{0.1cm} m_{\it c} =    0-2\hspace{0.05cm}\cdot\hspace{0.05cm} p + 0 \hspace{0.15cm} \underline{= -0.5}.$$
  
  
 +
'''(5)'''  Analogous to the subtask  '''(4)'''  we obtain for the root mean square:
 +
:$$m_{2d}= {\rm E}\big[( a-2b+c)^{\rm 2}\big] =  {\rm E}\big[a^{\rm 2}\big]\hspace{0.1cm}+\hspace{0.1cm}4\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[ b^{\rm 2}\big]\hspace{0.1cm}+\hspace{0.1cm} {\rm E}\big[c^{\rm 2}\big]\hspace{0.1cm}  -  \hspace{0.1cm}4\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[a\hspace{0.05cm}\cdot \hspace{0.05cm}b\big]\hspace{0.1cm}+\hspace{0.1cm} 2\hspace{0.05cm}\cdot\hspace{0.05cm}{\rm E}\big[ a\hspace{0.05cm}\cdot \hspace{0.05cm}c\big]\hspace{0.1cm}-\hspace{0.1cm} 4\hspace{0.05cm}\cdot\hspace{0.05cm}{\rm E}\big[ b\hspace{0.05cm}\cdot \hspace{0.05cm}c\big].$$
  
'''(5)'''  Analog zur Teilaufgabe  '''(4)'''  erhält man für den quadratischen Mittelwert:
 
:$$m_{2d}= {\rm E}\big[( a-2b+c)^{\rm 2}\big] =  {\rm E}\big[a^{\rm 2}\big]\hspace{0.1cm}+\hspace{0.1cm}4\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[ b^{\rm 2}\big]\hspace{0.1cm}+\hspace{0.1cm} {\rm E}\big[c^{\rm 2}\big]\hspace{0.1cm}  -  \hspace{0.1cm}4\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[a\hspace{0.05cm}\cdot \hspace{0.05cm}b\big]\hspace{0.1cm}+\hspace{0.1cm} 2\hspace{0.05cm}\cdot\hspace{0.05cm}{\rm E}\big[ a\hspace{0.05cm}\cdot \hspace{0.05cm}c\big]\hspace{0.1cm}-\hspace{0.1cm} 4\hspace{0.05cm}\cdot\hspace{0.05cm}{\rm E}\big[ b\hspace{0.05cm}\cdot \hspace{0.05cm}c\big].$$
 
  
*Da aber  $a$  und  $b$  statistisch voneinander unabhängig sind,  gilt auch:
+
*But since  $a$  and  $b$  are statistically independent of each other,  also holds:
:$${\rm E}\big[a\cdot b\big] = {\rm E}\big[ a\big] \cdot {\rm E}\big[ b\big]= m_{ a}\cdot m_{ b} = 0, \hspace{0.2cm} {\rm da}\hspace{0.2cm} m_{ a}=\rm 0.$$
+
:$${\rm E}\big[a\cdot b\big] = {\rm E}\big[ a\big] \cdot {\rm E}\big[ b\big]= m_{ a}\cdot m_{ b} = 0, \hspace{0.2cm} {\rm da}\hspace{0.2cm} m_{ a}=\rm 0.$$
  
*Gleiches gilt für die anderen gemischten Terme.  Daher erhält man mit  $p = 0.25$:
+
*The same holds for the other mixed terms.  Therefore, using  $p = 0.25$, we obtain:
:$$ m_{2 d}=m_{2 a}+4\cdot m_{ 2 b}+m_{ 2 c}=1+4\cdot p+0.5\hspace{0.15cm} \underline{=\rm 2.5}.$$
+
:$$ m_{2 d}=m_{2 a}+4\cdot m_{ 2 b}+m_{ 2 c}=1+4\cdot p+0.5\hspace{0.15cm} \underline{=\rm 2.5}.$$
  
  
  
'''(6)'''  Für allgemeines  $p$  bzw.  für  $p = 0.25$  ergibt sich:
+
'''(6)'''  For general  $p$  resp.  for  $p = 0.25$  results:
 
:$$\sigma_{\it d}^{\rm 2}=1.5+4\cdot p - 4 \cdot p^{\rm 2}=2.25 \hspace{0.5cm}\Rightarrow \hspace{0.5cm}  \sigma_{d}\hspace{0.15cm} \underline{=\rm 1.5}.$$
 
:$$\sigma_{\it d}^{\rm 2}=1.5+4\cdot p - 4 \cdot p^{\rm 2}=2.25 \hspace{0.5cm}\Rightarrow \hspace{0.5cm}  \sigma_{d}\hspace{0.15cm} \underline{=\rm 1.5}.$$
  
*Die maximale Varianz ergäbe sich für  $p = 0.50$   zu  $\sigma_{\it d}^{\rm 2}=2.50$.
+
*The maximum variance for  $p = 0.50$ adds up to  $\sigma_{\it d}^{\rm 2}=2.50$.
 +
 
 
{{ML-Fuß}}
 
{{ML-Fuß}}
  

Revision as of 21:32, 3 December 2021


Different rectangular signals

Let be given three discrete random variables  $a$,  $b$  and  $c$,  which are defined as the current values of the represented signals.  These have the following properties:

  • The random variable  $a$  can take the values  $+1$  and  $-1$  with equal probability.
  • The random variable  $b$  is also two-point distributed, but with  ${\rm Pr}(b = 1) = p$  and  ${\rm Pr}(b = 0) = 1 - p$.
  • The probabilities of  $c$  be  ${\rm Pr}(c = 0) = 1/2$  and  ${\rm Pr}(c = +1) = Pr(c = -1) =1/4$.
  • There are no statistical dependencies between the three random variables  $a$,  $b$  and  $c$  .
  • Another random variable  $d=a$,  $b$  and  $c$  is formed from the random variables  $d=a-2 b+c$ .


The graph shows sections of these four random variables.  It can be seen that  $d$  can take all integer values between  $-4$  and  $+2$ .




Hints:


Questions

1

What is the dispersion (standard deviation) of the random variable $a$?

$\sigma_a \ = \ $

2

What is the dispersion of the random variable  $b$?  Set  $p = 0.25$.

$\sigma_b \ = \ $

3

What is the spread of the random variable $c$?

$\sigma_c \ = \ $

4

Calculate the mean  $m_d$  of the random variable  $d$  for $p = 0.25$.

$m_d\ = \ $

5

What is the root mean square value  $m_{2d}$  of this random variable?

$m_{2d}\ = \ $

6

What is the dispersion  $\sigma_d$?

$\sigma_d\ = \ $


Solution

(1)  Due to the symmetry holds:

$$\rm \it m_{\it a}=\rm 0; \hspace{0.5cm}\it m_{\rm 2\it a}=\rm 0.5\cdot (-1)^2 + 0.5\cdot (1)^2{ = 1}.$$
  • From this one obtains with Steiner's theorem:
$$\it\sigma_a^{\rm 2} = \rm\sqrt{1-0^2}=1 \hspace{0.5cm}or \hspace{0.5cm}\it\sigma_a\hspace{0.15cm} \underline{=\rm 1}.$$


(2)  In general, for the  $k$–th order moment:

$$ m_{k}=(1-p)\cdot 0^{ k} + p\cdot 1^{k}= p.$$
  • From this follows with  $p = 1/4$:
$$m_{b}= m_{2b}= p, \hspace{0.5cm} \sigma_{\it b}=\sqrt{p\cdot (1- p)}\hspace{0.15cm} \underline{=\rm 0.433} .$$


(3)  For the random variable  $c$  holds:

$$m_{\it c} = 0\hspace{0.3cm} ({\rm symmetric\hspace{0.1cm}um\hspace{0.1cm}0)},$$
$$ m_{2\it c}= {1}/{4}\cdot(-1)^2+{1}/{2}\cdot 0^2+{1}/{4}\cdot (1)^2={1}/{2} \hspace{0.5cm}$$
$$\Rightarrow \hspace{0.5cm}\sigma_{\it c}=\rm \sqrt{1/2}\hspace{0.15cm} \underline{=0.707}.$$


(4)  According to the general rules for expected values, with  $p = 0.25$:

$$m_{\it d} = {\rm E}\big[a-2 b+c\big]= {\rm E}\big[a\big] \hspace{0.1cm} -\hspace{0.1cm}\rm 2 \hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[ b\big]\hspace{0.1cm}+\hspace{0.1cm} {\rm E}\big[ c\big] = m_{ a}\hspace{0.1cm}-\hspace{0.1cm}2\hspace{0.05cm}\cdot\hspace{0.05cm} m_{\it b}\hspace{0.1cm}+\hspace{0.1cm} m_{\it c} = 0-2\hspace{0.05cm}\cdot\hspace{0.05cm} p + 0 \hspace{0.15cm} \underline{= -0.5}.$$


(5)  Analogous to the subtask  (4)  we obtain for the root mean square:

$$m_{2d}= {\rm E}\big[( a-2b+c)^{\rm 2}\big] = {\rm E}\big[a^{\rm 2}\big]\hspace{0.1cm}+\hspace{0.1cm}4\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[ b^{\rm 2}\big]\hspace{0.1cm}+\hspace{0.1cm} {\rm E}\big[c^{\rm 2}\big]\hspace{0.1cm} - \hspace{0.1cm}4\hspace{0.05cm}\cdot\hspace{0.05cm} {\rm E}\big[a\hspace{0.05cm}\cdot \hspace{0.05cm}b\big]\hspace{0.1cm}+\hspace{0.1cm} 2\hspace{0.05cm}\cdot\hspace{0.05cm}{\rm E}\big[ a\hspace{0.05cm}\cdot \hspace{0.05cm}c\big]\hspace{0.1cm}-\hspace{0.1cm} 4\hspace{0.05cm}\cdot\hspace{0.05cm}{\rm E}\big[ b\hspace{0.05cm}\cdot \hspace{0.05cm}c\big].$$


  • But since  $a$  and  $b$  are statistically independent of each other,  also holds:
$${\rm E}\big[a\cdot b\big] = {\rm E}\big[ a\big] \cdot {\rm E}\big[ b\big]= m_{ a}\cdot m_{ b} = 0, \hspace{0.2cm} {\rm da}\hspace{0.2cm} m_{ a}=\rm 0.$$
  • The same holds for the other mixed terms.  Therefore, using  $p = 0.25$, we obtain:
$$ m_{2 d}=m_{2 a}+4\cdot m_{ 2 b}+m_{ 2 c}=1+4\cdot p+0.5\hspace{0.15cm} \underline{=\rm 2.5}.$$


(6)  For general  $p$  resp.  for  $p = 0.25$  results:

$$\sigma_{\it d}^{\rm 2}=1.5+4\cdot p - 4 \cdot p^{\rm 2}=2.25 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} \sigma_{d}\hspace{0.15cm} \underline{=\rm 1.5}.$$
  • The maximum variance for  $p = 0.50$ adds up to  $\sigma_{\it d}^{\rm 2}=2.50$.