Expected Values and Moments

Moment calculation as ensemble average

The probability density function (PDF), like the distribution function (CDF), provides very extensive information about the random variable under consideration. Less, but more compact information is provided by the so-called expected values and moments.

Their calculation possibilities have already been given for discrete random variables in the chapter moments of a discrete random variable .
Now these integrative descriptive quantities "expected value" and "moment" are considered in the context of the probability density function (PDF) of continuous random variables and thus formulated more generally.

$\text{Definition:}$ The expected value with respect to any weighting function $g(x)$ can be calculated with the PDF $f_{\rm x}(x)$ in the following way:

$${\rm E}\big[g (x ) \big] = \int_{-\infty}^{+\infty} g(x)\cdot f_{x}(x) \,{\rm d}x.$$

Substituting into this equation for $g(x) = x^k$ we get the moment of $k$-th order:

$$m_k = {\rm E}\big[x^k \big] = \int_{-\infty}^{+\infty} x^k\cdot f_{x} (x ) \, {\rm d}x.$$

From this equation follows.

with $k = 1$ for the linear mean:

$$m_1 = {\rm E}\big[x \big] = \int_{-\infty}^{ \rm +\infty} x\cdot f_{x} (x ) \,{\rm d}x,$$

with $k = 2$ for the root mean square:

$$m_2 = {\rm E}\big[x^{\rm 2} \big] = \int_{-\infty}^{ \rm +\infty} x^{ 2}\cdot f_{ x} (x) \,{\rm d}x.$$

For a discrete, $M$–-level random variable, the formulas given here again yield the equations already given in the second chapter (calculation as a ensemble average):

$$m_1 = \sum\limits_{\mu=1}^{ M}\hspace{0.15cm}p_\mu\cdot x_\mu,$$

$$m_2 = \sum\limits_{\mu= 1}^{ M}\hspace{0.15cm}p_\mu\cdot x_\mu^2.$$

Here it is taken into account that the integral over the Dirac function $δ(x)$ is equal $1$ .

In connection with signals, the following terms are also common:

$m_1$ indicates the DC component ,
$m_2$ corresponds to the (referred to the unit resistance $1 \ Ω$ ) signal power.

For example, if $x$ denotes a voltage, then according to these equations

$m_1$ has the unit ${\rm V}$ and
$m_2$ the unit ${\rm V}^2$.

If one wants to indicate the power in "Watt" $\rm (W)$ , then $m_2$ must still be divided by the resistance value $R$ .

Central moments

$\text{Definition:}$ Especially important in statistics are the central moments, which, in contrast to the conventional moments, are each related to the mean $m_1$ :

$$\mu_k = {\rm E}\big[(x-m_{\rm 1})^k\big] = \int_{-\infty}^{+\infty} (x-m_{\rm 1})^k\cdot f_x(x) \,\rm d \it x.$$

The noncentered moments $m_k$ can be directly converted to the centered moments $\mu_k$ :

$$\mu_k = \sum\limits_{\kappa= 0}^{k} \left( \begin{array}{*{2}{c}} k \ \kappa \ \end{array} \right)\cdot m_\kappa \cdot (-m_1)^{k-\kappa}.$$

According to the general equations of last page the formal quantities $m_0 = 1$ and $\mu_0 = 1$ result. For the first order central moment, according to the above definition, always $\mu_1 = 0$ holds.

In the opposite direction, the following equations hold for $k = 1$, $k = 2$, and so on:

$$m_k = \sum\limits_{\kappa= 0}^{k} \left( \begin{array}{*{2}{c}} k \ \kappa \ \end{array} \right)\cdot \mu_\kappa \cdot {m_1}^{k-\kappa}.$$

$\text{Example 1:}$ All moments of a binary random variable with probabilities ${\rm Pr}(0) = 1 - p$ and ${\rm Pr}(1) = p$ are of equal value:

$$m_1 = m_2 = m_3 = m_4 = \hspace{0.05cm}\text{...} \hspace{0.05cm}= p.$$

Using the above equations, we then obtain for the first three central moments:

$$\mu_2 = m_2 - m_1^2 = p -p^2, $$

$$\mu_3 = m_3 - 3 \cdot m_2 \cdot m_1 + 2 \cdot m_1^3 = p - 3 \cdot p^2 + 2 \cdot p^3, $$

$$ \mu_4 = m_4 - 4 \cdot m_3 \cdot m_1 + 6 \cdot m_2 \cdot m_1^2 - 3 \cdot m_1^4 = p - 4 \cdot p^2 + 6 \cdot p^3- 3 \cdot p^4. $$

Some common central moments

From the last definition the following additional characteristics can be derived:

$\text{Definition:}$ The variance $σ^2$ of the considered random variable is the second order central moment ⇒ $\mu_2.$

The variance $σ^2$ physically corresponds to the alternating power and the dispersion $σ$ gives the rms value .
From the linear and the quadratic mean, the variance can be calculated according to Steiner's theorem in the following way:

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

$\text{Definition:}$ The Charlier's skewness' $S$ denotes the third central moment related to $σ^3$ .

For symmetric density function, this parameter is always $S=0$.
The larger $S = \mu_3/σ^3$ is, the more asymmetric is the WDF around the mean $m_1$.
For example, for the exponential distribution the (positive) skewness $S =2$, and this is independent of the distribution parameter $λ$.
For positive skewness $(S > 0)$ one speaks of "a right-skewed or left-sloping distribution"; this slopes flatter on the right side than on the left.
When the skewness is negative $(S < 0)$ there is a "left-skewed or right-steep distribution"; such a distribution falls flatter on the left side than on the right

. $\text{Definition:}$ The fourth-order central moment is also used for statistical analyses; The quotient $K = \mu_4/σ^4$ is called kurtosis .

For a Gaussian distributed random variable this always yields the value $K = 3$.
Using also the so-called excess $\gamma = K - 3$ , also known under the term "overkurtosis".
This parameter can be used, for example, to check whether a random variable at hand is approximately Gaussian: $\gamma \approx 0$.

$\text{Example 2:}$

If the PDF has fewer offshoots than the Gaussian distribution, the kurtosis $K < 3$. For example, for the uniformly distributed $K = 1.8$ ⇒ $\gamma = - 1.2$.
In contrast, $K > 3$ indicates that the spurs are more pronounced than for the Gaussian distribution. For example, for the exponential distribution $K = 9$.
For the Laplace distribution ⇒ two-sided exponential distribution results in a slightly smaller kurtosis $K = 6$ and the excess $\gamma = 3$.

Moment calculation as time average

The expected value calculation according to the previous equations of this section corresponds to a ensemble averaging, that is, averaging over all possible values $x_\mu$.

However, the moments $m_k$ can also be determined as time averages if the stochastic process generating the random variable is stationary and ergodic:

The exact definition for such a stationary and ergodic random process can be found in Chapter 4.4.
A time-averaging is always denoted by a sweeping line in the following.
For discrete time, the random signal $x(t)$ is replaced by the random sequence $〈x_ν〉$ .
For finite sequence length, these time averages are with $ν = 1, 2,\hspace{0.05cm}\text{...}\hspace{0.05cm} , N$:

$$m_k=\overline{x_{\nu}^{k}}=\frac{1}{N} \cdot \sum\limits_{\nu=1}^{N}x_{\nu}^{k},$$

$$m_1=\overline{x_{\nu}}=\frac{1}{N} \cdot \sum\limits_{\nu=1}^{N}x_{\nu},$$

$$m_2=\overline{x_{\nu}^{2}}=\frac{1}{N} \cdot \sum\limits_{\nu=1}^{N}x_{\nu}^{2}.$$

If the moments (or expected values) are to be determined by simulation, in practice this is usually done by time averaging. The corresponding computational algorithm differs only mariginally for discrete and continuous random variables.

The topic of this chapter is illustrated with examples in the (German language) learning video Momentenberechnung bei diskreten Zufallsgrößen $\Rightarrow$ Moment calculation for discrete random variables.

Characteristic function

$\text{Definition:}$ Another special case of an expected value is the characteristic function', where here for the valuation function $g(x) = {\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.03cm}{\it Ω}\hspace{0.05cm}x}$ is to be set:

$$C_x({\it \Omega}) = {\rm E}\big[{\rm e}^{ {\rm j} \hspace{0.05cm} {\it \Omega} \hspace{0.05cm} x}\big] = \int_{-\infty}^{+\infty} {\rm e}^{ {\rm j} \hspace{0.05cm} {\it \Omega} \hspace{0.05cm} x}\cdot f_{\rm x}(x) \hspace{0.1cm}{\rm d}x.$$

A comparison with the chapter Fourier Transform and Fourier Back Transform in the book "Signal Representation" shows that the characteristic function can be interpreted as the Fourier back transform of the probability density function:

$$C_x ({\it \Omega}) \hspace{0.3cm} \circ \!\!-\!\!-\!\!-\!-\!\! \bullet \hspace{0.3cm} f_{x}(x).$$

If the random variable $x$ is dimensionless, then the argument $\it Ω$ of the characteristic function is also without unit.

The symbol $\it Ω$ was chosen because the argument here has some relation to the angular frequency in the second Fourier integral (compared to the representation in the $f$–domain, however, the factor $2\pi$ is missing in the exponent).
But it is again insistently pointed out that - if one wants to establish a relation to system theory - $C_x({\it Ω})$ would correspond to the "time function" and $f_{x}(x)$ to the "spectral function".

$\text{calculation possibility:}$ Developing the complex function ${\rm e}^{\hspace{0.03cm}{\rm j} \hspace{0.03cm}{\it Ω}\hspace{0.05cm}x}$ into a power series and interchanges expectation value formation and summation, the series representation of the characteristic function follows:

$$C_x ( {\it \Omega}) = 1 + \sum_{k=1}^{\infty}\hspace{0.2cm}\frac{m_k}{k!} \cdot ({\rm j} \hspace{0.01cm}{\it \Omega})^k .$$

The Exercise 3.4 shows more properties of the characteristic function.

$\text{Example 3:}$

For a symmetric binary (two-point distributed) random variable $x ∈ \{\pm1\}$ with probabilities ${\rm Pr}(-1) = {\rm Pr}(+1) = 1/2$ the characteristic function is cosine.
The analogue in systems theory is that the spectrum of a cosine signal with angular frequency ${\it Ω}_{\hspace{0.03cm}0}$ consists of two Dirac functions at $±{\it Ω}_{\hspace{0.03cm}0}$ .

$\text{Example 4:}$

A uniform distribution between $±y_0$ has the following characteristic function according to the Laws of Fourier Transform :

$$C_y({\it \Omega}) = \frac{1}{2 y_0} \cdot \int_{-y_0}^{+y_0} {\rm e}^{ {\rm j} \hspace{0.05cm} {\it \Omega} \hspace{0.05cm} y} \,{\rm d}y = \frac{ {\rm e}^{ {\rm j} \hspace{0.05cm} y_0 \hspace{0.05cm}{\it \Omega} } - {\rm e}^{ - {\rm j} \hspace{0.05cm} y_0 \hspace{0.05cm} {\it \Omega} } }{2 {\rm j} \cdot y_0 \cdot {\it \Omega} } = \frac{ {\rm sin}(y_0 \cdot {\it \Omega})}{ y_0 \cdot {\it \Omega} } = {\rm si}(y_0 \cdot {\it \Omega}). $$

We already know the function ${\rm si}(x) = \sin(x)/x$ from the book Signal Representation.

Exercises for the chapter

Exercise 3.3: Moments at cos²-PDF

Exercise 3.3Z: Moments for Triangular PDF

Exercise 3.4: Characteristic Function