Theory of Stochastic Signals/Moments of a Discrete Random Variable

1. Theory of Stochastic Signals
2. Discrete Random Variables
3. Moments of a Discrete Random Variable

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Calculation as ensemble average or time average

The probabilities and the relative frequencies provide extensive information about a discrete random variable. Reduced information is obtained by the so-called moments  $m_k$, where  $k$  represents a natural number.

$\text{Two alternative ways of calculation:}$ 

Under the  Ergodicity  implied here, there are two different calculation possibilities for the moment  $k$-th order:

• the  ensemble averaging'  or  expected value formation   ⇒  averaging over all possible values  $\{ z_\mu\}$  with the index  $\mu = 1 , \hspace{0.1cm}\text{ ...} \hspace{0.1cm} , M$:

$$m_k = {\rm E} \big[z^k \big] = \sum_{\mu = 1}^{M}p_\mu \cdot z_\mu^k \hspace{2cm} \rm with \hspace{0.1cm} {\rm E\big[\text{ ...} \big]\hspace{-0.1cm}:} \hspace{0.3cm} \rm expected\hspace{0.1cm}value ;$$

• the  time averaging  over the random sequence  $\langle z_ν\rangle$  with the index  $ν = 1 , \hspace{0.1cm}\text{ ...} \hspace{0.1cm} , N$:

$$m_k=\overline{z_\nu^k}=\hspace{0.01cm}\lim_{N\to\infty}\frac{1}{N}\sum_{\nu=\rm 1}^{\it N}z_\nu^k\hspace{1.7cm}\rm with\hspace{0.1cm}horizontal\hspace{0.1cm}line\hspace{-0.1cm}:\hspace{0.1cm}time\hspace{0.1cm}average.$$

Note:

• Both types of calculations lead to the same asymptotic result for sufficiently large values of  $N$  .
• For finite  $N$ , a comparable error results as when the probability is approximated by the relative frequency.

Linear mean - DC component

$\text{Definition:}$  With  $k = 1$  we obtain from the general equation for moments the  linear mean:

$$m_1 =\sum_{\mu=1}^{M}p_\mu\cdot z_\mu =\lim_{N\to\infty}\frac{1}{N}\sum_{\nu=1}^{N}z_\nu.$$

• The left part of this equation describes the ensemble averaging  (over all possible values),
• while the right equation gives the determination as time average.
• In the context of signals, this quantity is also referred to as the  DC component 

DC component  $m_1$  of a binary signal

$\text{Example 1:}$  A binary signal  $x(t)$  with the two possible amplitude values.

• $1\hspace{0.03cm}\rm V$  $($for the symbol  $\rm L)$,
• $3\hspace{0.03cm}\rm V$  $($for the symbol  $\rm H)$

as well as the occurrence probabilities  $p_{\rm L} = 0.2$  respectively  $p_{\rm H} = 0.8$  has the linear mean (DC)

$$m_1 = 0.2 \cdot 1\,{\rm V}+ 0.8 \cdot 3\,{\rm V}= 2.6 \,{\rm V}. $$

This is drawn as a red line in the graph.
If we determine this parameter by time averaging over the displayed  $N = 12$  signal values, we obtain a slightly smaller value:

$$m_1\hspace{0.01cm}' = 4/12 \cdot 1\,{\rm V}+ 8/12 \cdot 3\,{\rm V}= 2.33 \,{\rm V}. $$

Here, the probabilities of occurrence  $p_{\rm L} = 0.2$  and  $p_{\rm H} = 0.8$  were replaced by the corresponding frequencies  $h_{\rm L} = 4/12$  and  $h_{\rm H} = 8/12$  respectively. The relative error due to insufficient sequence length  $N$  is greater than  $10\%$ in the example.

Note about our (admittedly somewhat unusual) nomenclature:.

We denote binary symbols here as in circuit theory with  $\rm L$  (Low) and  $\rm H$  (High) to avoid confusion.

• In coding theory, it is useful to map  $\{ \text{L, H}\}$  to  $\{0, 1\}$  to take advantage of the possibilities of modulo algebra.
• In contrast, to describe modulation with bipolar (antipodal) signals, one better chooses the mapping  $\{ \text{L, H}\}$ ⇔ $ \{-1, +1\}$.

Quadratic Mean – Variance – Dispersion

$\text{Definitions:}$ 

• Analogous to the linear mean,   $k = 2$  is obtained for the  root mean square:

$$m_2 =\sum_{\mu=\rm 1}^{\it M}p_\mu\cdot z_\mu^2 =\lim_{N\to\infty}\frac{\rm 1}{\it N}\sum_{\nu=\rm 1}^{\it N}z_\nu^2.$$

• Together with the DC component  $m_1$ , the  variance  $σ^2$  can be determined from this as a further parameter (Steiner's Theorem):

$$\sigma^2=m_2-m_1^2.$$

• In statistics, the  dispersion  $σ$  is the square root of the variance; sometimes this quantity is also called  standard deviation :

$$\sigma=\sqrt{m_2-m_1^2}.$$

Notes on units:

• For message signals,  $m_2$  indicates the (average) power  power of a random signal, referenced to  $1 \hspace{0.03cm} Ω$ resistance.
• If  $z$  describes a voltage,  $m_2$  accordingly has the unit${\rm V}^2$.
• The variance  $σ^2$  of a random signal corresponds physically to the  alternating power  and the dispersion  $σ$  to therms (root mean square) value.
• These definitions are based on the reference resistance  $1 \hspace{0.03cm} Ω$  zugrunde.

The (German) learning video   Momentenberechnung bei diskreten Zufallsgrößen   $\Rightarrow$ Moment Calculation for Discrete Random Variables, illustrates the defined quantities using the example of a digital signal.

Standard deviation of a binary signals

$\text{Example 2:}$  A binary signal  $x(t)$  mit den Amplitudenwerten

• $1\hspace{0.03cm}\rm V$  $($für das Symbol  $\rm L)$,
• $3\hspace{0.03cm}\rm V$  $($für das Symbol  $\rm H)$

sowie den Auftrittswahrscheinlichkeiten  $p_{\rm L} = 0.2$  bzw.  $p_{\rm H} = 0.8$  besitzt die gesamte Signalleistung

$$P_{\rm Gesamt} = 0.2 \cdot (1\,{\rm V})^2+ 0.8 \cdot (3\,{\rm V})^2 = 7.4 \hspace{0.05cm}{\rm V}^2,$$

wenn man vom Bezugswiderstand  $R = 1 \hspace{0.05cm} Ω$  ausgeht.

Mit dem Gleichanteil  $m_1 = 2.6 \hspace{0.05cm}\rm V$  $($siehe  $\text{Beispiel 1})$  folgt daraus für

• die Wechselleistung (Varianz)  $P_{\rm W} = σ^2 = 7.4 \hspace{0.05cm}{\rm V}^2 - \big [2.6 \hspace{0.05cm}\rm V\big ]^2 = 0.64\hspace{0.05cm} {\rm V}^2$,
• den Effektivwert  $s_{\rm eff} = σ = 0.8 \hspace{0.05cm} \rm V$.

Einschub:   Bei anderem Bezugswiderstand   ⇒   $R \ne 1 \hspace{0.1cm} Ω$  gelten nicht alle diese Berechnungen.  Beispielsweise haben mit  $R = 50 \hspace{0.1cm} Ω$  die Leistung $P_{\rm Gesamt} $,  die Wechselleistung  $P_{\rm W}$  und der Effektivwert  $s_{\rm eff}$  folgende Werte:

$$P_{\rm Gesamt} \hspace{-0.05cm}= \hspace{-0.05cm} \frac{m_2}{R} \hspace{-0.05cm}= \hspace{-0.05cm} \frac{7.4\,{\rm V}^2}{50\,{\rm \Omega} } \hspace{-0.05cm}= \hspace{-0.05cm}0.148\,{\rm W},\hspace{0.5cm} P_{\rm W} \hspace{-0.05cm} = \hspace{-0.05cm} \frac{\sigma^2}{R} \hspace{-0.05cm}= \hspace{-0.05cm}12.8\,{\rm mW} \hspace{0.05cm},\hspace{0.5cm} s_{\rm eff} \hspace{-0.05cm} = \hspace{-0.05cm}\sqrt{R \cdot P_{\rm W} } \hspace{-0.05cm}= \hspace{-0.05cm} \sigma \hspace{-0.05cm}= \hspace{-0.05cm} 0.8\,{\rm V}.$$

Die gleiche Varianz und der gleiche Effektivwert  $s_{\rm eff}$  ergeben sich für die Amplituden  $0\hspace{0.05cm}\rm V$  $($für das Symbol  $\rm L)$  und $2\hspace{0.05cm}\rm V$  $($für das Symbol  $\rm H)$ , vorausgesetzt, die Auftrittswahrscheinlichkeiten  $p_{\rm L} = 0.2$  und  $p_{\rm H} = 0.8$  bleiben gleich.  Nur der Gleichanteil und die Gesamtleistung ändern sich:

$$m_1 = 1.6 \hspace{0.05cm}{\rm V}, \hspace{0.5cm}P_{\rm Gesamt} = P_{\rm W} + {m_1}^2 = 3.2 \hspace{0.05cm}{\rm V}^2.$$

Aufgaben zum Kapitel

Aufgabe 2.2: Mehrstufensignale

Aufgabe 2.2Z: Diskrete Zufallsgrößen