Moments of a Discrete Random Variable

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 condition "Ergodicity" implicitly assumed here, there are two different calculation possibilities for the $k$-th order moment:

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.

First order moment – linear mean – DC component

$\text{Definition:}$ With $k = 1$ we obtain from the general equation the first order moment ⇒ 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 "direct current" $\rm (DC)$ component.

DC component $m_1$ of a binary signal

$\text{Example 1:}$ A binary signal $x(t)$ with the two possible 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 $p_{\rm L} = 0.2$ and $p_{\rm H} = 0.8$ has the linear mean ("DC component")

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

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

$\text{Note about our (admittedly somewhat unusual) nomenclature:}$

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\}$.

Second order moment – power – variance – standard deviation

$\text{Definitions:}$

Analogous to the linear mean, $k = 2$ obtains the second order moment):

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

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

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

The standard deviation $σ$ is the square root of the variance:

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

$\text{Hinweise zu den Einheiten:}$

Bei einem Nachrichtensignal $x(t)$ gibt $m_2$ die gesamte Leistung (Gleichleistung plus Wechselleitung) eines Zufallssignals an, bezogen auf den Widerstand $1 \hspace{0.03cm} Ω$.
Beschreibt $x(t)$ eine Spannung, so besitzt dementsprechend $m_2$ die Einheit ${\rm V}^2$ und der Effektivwert (englisch: "root mean square") $x_{\rm eff}=\sqrt{m_2}$ hat die Einheit ${\rm V}$. Die Gesamtleistung für beliebigen Bezugswiderstand $R$ berechnet sich zu $P=m_2/R$ und besitzt dementsprechend die Einheit $\rm V^2/(V/A) = W$.
Beschreibt $x(t)$ einen Stromverlauf, so besitzt $m_2$ die Einheit ${\rm A}^2$ und der Effektivwert $x_{\rm eff}=\sqrt{m_2}$ hat die Einheit ${\rm A}$. Die Gesamtleistung für beliebigen Bezugswiderstand $R$ berechnet sich zu $P=m_2\cdot R$ und besitzt dementsprechend die Einheit $\rm A^2 \cdot(V/A) = W$.
Nur im Sonderfall $m_1=0$ ist die Varianz $σ^2=m_2$. Dann stimmt auch die Standardabweichung $σ$ mit dem Effektivwert $x_{\rm eff}$ überein.

The following (German language) learning video illustrates the defined quantities using the example of a digital signal:
Momentenberechnung bei diskreten Zufallsgrößen ⇒ "Moment Calculation for Discrete Random Variables".

"Standard deviation" of a binary signal

$\text{Example 2:}$ Bei einem Binärsignal $x(t)$ 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. $p_{\rm H} = 0.8$ ergibt sich für das zweite Moment:

$$m_2 = 0.2 \cdot (1\,{\rm V})^2+ 0.8 \cdot (3\,{\rm V})^2 = 7.4 \hspace{0.1cm}{\rm V}^2,$$

Der Effektivwert $x_{\rm eff}=\sqrt{m_2}=2.72\,{\rm V}$ ist unabhängig vom Bezugswiderstand $R$ im Gegensatz zur Gesamtleistung. Für diese ergibt sich mit $R=1 \hspace{0.1cm} Ω$ der Wert $P=7.4 \hspace{0.1cm}{\rm W}$, mit $R=50 \hspace{0.1cm} Ω$ dagegen nur $P=0.148 \hspace{0.1cm}{\rm W}$.

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

die Varianz $ 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$,
die Standardabweichung (Streuung) $σ = 0.8 \hspace{0.05cm} \rm V$.

Die gleiche Varianz $ σ^2 = 0.64\hspace{0.05cm} {\rm V}^2$ und die gleiche Standardabweichung $σ = 0.8 \hspace{0.05cm} \rm V$ 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:

$$m_1 = 1.6 \hspace{0.05cm}{\rm V}, \hspace{0.5cm}P = {m_1}^2 +\sigma^2 = 3.2 \hspace{0.05cm}{\rm V}^2.$$

Muss anschließend noch gelöscht werden

$\text{Definitions:}$

Analogous to the linear mean, $k = 2$ is obtained for the second order moment ⇒ the quadratic mean:

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

$m_2$ indicates also the total power $P$ (DC power plus AC power) of a random signal, referenced to $1 \hspace{0.03cm} Ω$ resistance. The square root $\sqrt{P}$ is called "standard deviation".

Together with the DC component $m_1$, the variance $σ^2$ can be determined from the second order moment as a further parameter ("Steiner's theorem"):

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

The square root $σ$ of the variance is called "standard deviation":

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

$\text{Notes on units:}$

For message signals, $m_2$ indicates the total power (DC power plus AC power) of a random signal, referenced to $1 \hspace{0.03cm} Ω$ resistance.
If $z$ describes a voltage, $P=m_2$ accordingly has the unit ${\rm V}^2$, and the "standard deviation" has the unit ${\rm V}$.
The variance $σ^2$ of a random signal corresponds physically to the "alternating power" or "AC power".
These definitions are based on the reference resistance $1 \hspace{0.03cm} Ω$.

The following (German language) learning video illustrates the defined quantities using the example of a digital signal:
Momentenberechnung bei diskreten Zufallsgrößen ⇒ "Moment Calculation for Discrete Random Variables".

"Standard deviation" of a binary signal

$\text{Example 2:}$ A binary signal $x(t)$ with the two possible 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 $p_{\rm L} = 0.2$ and $p_{\rm H} = 0.8$ has the total signal power

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

if one assumes the reference resistance $R = 1 \hspace{0.05cm} Ω$ .

With the DC component $m_1 = 2.6 \hspace{0.05cm}\rm V$ $($see $\text{Example 1})$ it follows for

the variance $σ^2 = 7.4 \hspace{0.05cm}{\rm V}^2 - \big [2.6 \hspace{0.05cm}\rm V\big ]^2 = 0.64\hspace{0.05cm} {\rm V}^2$,
the alternating power $P_{\rm AC} = 0.64\hspace{0.05cm} {\rm W}$ ⇒ same numerical value as $σ^2$, but different unit,
the standard deviation $s_{\rm eff} = σ = 0.8 \hspace{0.05cm} \rm V$.

Insertion: With other reference resistance ⇒ $R \ne 1 \hspace{0.1cm} Ω$, not all these calculations apply. For example, with $R = 50 \hspace{0.1cm} Ω$, the power $P_{\rm total} $, the alternating power $P_{\rm AC}$, and the standard deviation $s_{\rm eff}$ have the following physical values:

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

The same variance $σ^2 = 0.64\hspace{0.05cm} {\rm V}^2$ and the same standard deviation $s_{\rm eff}= 0.8 \hspace{0.05cm} \rm V$ are obtained for amplitudes $0\hspace{0.05cm}\rm V$ $($for symbol $\rm L)$ and $2\hspace{0.05cm}\rm V$ $($for symbol $\rm H)$, provided that the probabilities $p_{\rm L} = 0.2$ and $p_{\rm H} = 0.8$ remain the same. Only the DC component and the total power change:

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

Exercises for the chapter

Exercise 2.2: Multi-Level Signals

Exercise 2.2Z: Discrete Random Variables