# 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  ("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"):
$$\sigma^2=m_2-m_1^2.$$
• The  standard deviation  $σ$  is the square root of the variance:
$$\sigma=\sqrt{m_2-m_1^2}.$$

$\text{Notes on units:}$

1. For a random signal  $x(t)$   ⇒   $m_2$  gives the total power  (DC power plus AC power)  related to the resistance  $1 \hspace{0.03cm} Ω$.
2. If  $x(t)$  describes a voltage,  accordingly  $m_2$  has the unit  ${\rm V}^2$  and the rms value  (  "root mean square")  $x_{\rm eff}=\sqrt{m_2}$   has the unit  ${\rm V}$.  The total power for any reference resistance  $R$  is calculated to   $P=m_2/R$  and accordingly  has the unit  $\rm V^2/(V/A) = W$.
3. If  $x(t)$  describes a current waveform,  then  $m_2$  has the unit  ${\rm A}^2$  and the rms value  $x_{\rm eff}=\sqrt{m_2}$  has the unit  ${\rm A}$.   The total power for any reference resistance  $R$  is calculated to   $P=m_2\cdot R$  and accordingly  has the unit  $\rm A^2 \cdot(V/A) = W$.
4. Only in the special case  $m_1=0$   ⇒   the variance is  $σ^2=m_2$.  Then the standard deviation   $σ$  coincides also with the rms value  $x_{\rm eff}$ .

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

$\text{Example 2:}$  For a binary signal  $x(t)$  with the amplitude values.

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

and the probabilities of occurrence  $p_{\rm L} = 0.2$  resp.  $p_{\rm H} = 0.8$  results for the second moment:

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

The rms value  $x_{\rm eff}=\sqrt{m_2}=2.72\,{\rm V}$  is independent of the reference resistance  $R$  unlike the total power. For the latter, with  $R=1 \hspace{0.1cm} Ω$  the value  $P=7.4 \hspace{0.1cm}{\rm W}$,  with  $R=50 \hspace{0.1cm} Ω$  on the other hand, only  $P=0.148 \hspace{0.1cm}{\rm W}$.

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 standard deviation  $σ = 0.8 \hspace{0.05cm} \rm V$.

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

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