Exercise 3.4: Characteristic Function

Rectangular and trapezoidal PDF

Given here are three random variables $x$, $y$ and $z$, mostly by their respective probability density functions:

Nothing else is known about the random variable $x$: This can be both a discrete or a continuous random variable, and can have any PDF $f_x(x)$ The mean is generally equal $m_x$.
The continuous random variable $y$ can take values in the range between $1$ to $3$ with equal probability. Mean: $m_y = 2.$
The random variable $z$ has the following characteristic function:

$$C_z ({\it \Omega} ) = {\mathop{\rm si}\nolimits}( {3{\it \Omega}} ) \cdot {\mathop{\rm si}\nolimits} ( {2{\it \Omega} } ).$$

Besides, the qualitative course of the WDF $f_z(z)$ according to the blue sketch is assumed to be known. To be determined are the PDF parameters $a$, $b$, $c$ of this PDF.

Hints:

This exercise belongs to the chapter Expected values and moments.
Reference is made to the section Charakteristic funcion .

The characteristic function of a between $\pm a$ uniformly distributed random variable $z$ is:

$$C ( {\it \Omega} ) = {\mathop{\rm si}\nolimits} ( {a {\it \Omega} } )\quad {\rm{with}}\quad {\mathop{\rm si}\nolimits}( x ) = \sin ( x )/x.$$

Questions

	$C_x ( {\it \Omega} )$ is the Fourier transform of $f_x(x)$.
	The real part of $C_x ( {\it \Omega} )$ is an even function in ${\it \Omega}$.
	The imaginary part of $C_x ( {\it \Omega} )$ is an odd function in ${\it \Omega}$.
	The value at location ${\it \Omega} = 0$ is always $C_x ( {\it \Omega} ) = 1$.
	For a zero mean random variable $(m_x = 0)$ ⇒ $C_x ( {\it \Omega} )$ is always real.

${\rm Re}\big[C_y(\Omega\ =\ \pi/2)\big] \ = \ $

${\rm Im}\big[C_y(\Omega\ =\ \pi/2)\big] \ = \ $

$a \ = \ $

$b \ = \ $

$c \ = \ $

Solutions

Solution

(1) Correct are the proposed solutions 2, 3 and 4:

$C_x( {\it \Omega} )$ is not the Fourier transform to $f_x(x)$, but the inverse Fourier transform:

$$C_x( {\it \Omega } ) = \int_{ - \infty }^{ + \infty } {f_x }( x )\cdot {\rm{e}}^{\hspace{0.03cm}{\rm{j}}\hspace{0.03cm}{\it \Omega x}} \hspace{0.1cm}{\rm{d}}x .$$

Also for this, the real part is always even and the imaginary part odd. For ${\it \Omega} = 0$ holds:

$$C_x( {\it \Omega} = 0 ) = \int_{ - \infty }^{ + \infty } {f_x }( x ) \hspace{0.1cm}{\rm{d}}x = 1.$$

The last alternative does not always hold: A two-point distributed random variable $x \in \{-1, +3\}$ with probabilities $0.75$ and $0.25$ is zero mean $(m_x = 0)$, but has still a complex characteristic function.

(2) According to the general definition:

$$C_y( {\it \Omega } ) = \int_{ - \infty }^{ + \infty } {f_y }( y )\cdot {\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega\hspace{0.01cm}\hspace{0.05cm}\cdot \hspace{0.05cm} y}} \hspace{0.1cm}{\rm{d}}y = 0.5\int_1^3 {{\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}\Omega\hspace{0.05cm}\cdot \hspace{0.05cm} y} \hspace{0.1cm}{\rm{d}}y.} $$

After solving this integral, we get:

$$C_y ( {\it \Omega } ) = \frac{{{\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}3{\it \Omega } } - {\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega } } }}{{2{\rm{j}}{\it \Omega } }} = = \frac{{{\rm{e}}^{{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega } } - {\rm{e}}^{{\rm{ - j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega }} }}{{2{\rm{j}}{\it \Omega } }} \cdot {\rm{e}}^{{\rm{j\hspace{0.05cm}\cdot \hspace{0.05cm}2}}{\it \Omega } } .$$

Using Euler's theorem, this can also be written:

$$C_y ( {\it \Omega } ) = \frac{{\sin ( {\it \Omega } )}}{{\it \Omega } } \cdot {\rm{e}}^{{\rm{j2}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega } } = {\rm si} ( {\it \Omega } ) \cdot {\rm{e}}^{{\rm{j2}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega } }.$$

For ${\it \Omega} = \pi/2$ we thus obtain a purely real numerical value:

$${\rm Re}[C_y ({\it \Omega} = {\rm{\pi }}/2 )] = \frac{{\sin( {{\rm{\pi }}/2})}}{{{\rm{\pi }}/2}} \cdot {\rm{e}}^{{\rm{j\pi }}} = - \frac{2}{{\rm{\pi }}} \hspace{0.15cm}\underline{\approx -0.637}, \hspace{0.5cm} {\rm Im}[C_y ({\it \Omega} = {\rm{\pi }}/2 )] \hspace{0.15cm}\underline{= 0} .$$

(3) From the given correspondence it can be read that ${\rm si}(3 {\it \Omega} )$ is due to an between $\pm 3$ equally distributed random variable and ${\rm si}(2 {\it \Omega} )$ gives the transform of a uniform distribution between $\pm 2$.

Construction of the trapezoidal PDF

In the characteristic function, these two proportions are multiplicatively linked.
Thus, the resulting PDF $f_z(z)$ is the convolution of these two rectangular functions.
The three PDF parameters are thus:

$$\hspace{0.15cm}\underline{a = 1},\quad \hspace{0.15cm}\underline{b = 5}, \quad c = 1/6 \hspace{0.15cm}\underline{= 0.167}.$$