Aufgaben:Exercise 2.2Z: Discrete Random Variables

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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:

• The exercise belongs to the chapter  Moments of a Discrete Random Variable.

• • The topic of this chapter is illustrated with examples in the  (German language)  learning video
    Momentenberechnung bei diskreten Zufallsgrößen  $\Rightarrow$   Calculating Moments for Discrete-Valued Random Variables

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

Solution

(1)  Aufgrund der Symmetrie gilt:

$$\rm \it m_{\it a}=\rm 0; \hspace{0.5cm}\it m_{\rm 2\it a}=\rm 0.5\cdot (-1)^2 + 0.5\cdot (1)^2{ = 1}.$$

• Daraus erhält man mit dem Satz von Steiner:

$$\it\sigma_a^{\rm 2} = \rm\sqrt{1-0^2}=1 \hspace{0.5cm}bzw. \hspace{0.5cm}\it\sigma_a\hspace{0.15cm} \underline{=\rm 1}.$$

(2)  Allgemein gilt für das Moment  $k$–ter Ordnung:

$$ m_{k}=(1-p)\cdot 0^{ k} + p\cdot 1^{k}= p.$$

• Daraus folgt mit  $p = 1/4$:

$$m_{b}= m_{2b}= p, \hspace{0.5cm} \sigma_{\it b}=\sqrt{p\cdot (1- p)}\hspace{0.15cm} \underline{=\rm 0.433} .$$

(3)  Für die Zufallsgröße  $c$  gilt:

$$m_{\it c} = 0\hspace{0.3cm} ({\rm symmetrisch\hspace{0.1cm}um\hspace{0.1cm}0)},$$

$$ m_{2\it c}= {1}/{4}\cdot(-1)^2+{1}/{2}\cdot 0^2+{1}/{4}\cdot (1)^2={1}/{2} \hspace{0.5cm}$$

$$\Rightarrow \hspace{0.5cm}\sigma_{\it c}=\rm \sqrt{1/2}\hspace{0.15cm} \underline{=0.707}.$$

(4)  Nach den allgemeinen Regeln für Erwartungswerte gilt mit  $p = 0.25$:

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

(5)  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:

$${\rm E}\big[a\cdot b\big] = {\rm E}\big[ a\big] \cdot {\rm E}\big[ b\big]= m_{ a}\cdot m_{ b} = 0, \hspace{0.2cm} {\rm da}\hspace{0.2cm} m_{ a}=\rm 0.$$

• Gleiches gilt für die anderen gemischten Terme.  Daher erhält man mit  $p = 0.25$:

$$ m_{2 d}=m_{2 a}+4\cdot m_{ 2 b}+m_{ 2 c}=1+4\cdot p+0.5\hspace{0.15cm} \underline{=\rm 2.5}.$$

(6)  Für allgemeines  $p$  bzw.  für  $p = 0.25$  ergibt sich:

$$\sigma_{\it d}^{\rm 2}=1.5+4\cdot p - 4 \cdot p^{\rm 2}=2.25 \hspace{0.5cm}\Rightarrow \hspace{0.5cm} \sigma_{d}\hspace{0.15cm} \underline{=\rm 1.5}.$$

• Die maximale Varianz ergäbe sich für  $p = 0.50$   zu  $\sigma_{\it d}^{\rm 2}=2.50$.