Aufgaben:Exercise 3.4: Characteristic Function

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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

1

Which statements are valid with respect to the characteristic function  $C_x ( {\it \Omega} )$  always – that is, at any PDF ?

	$C_x ( {\it \Omega} )$  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.

2

Calculate the characteristic function  $C_y( {\it \Omega} )$.  What are the real and imaginary parts at  ${\it \Omega} = \pi/2$?

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

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

3

Determine the characteristic parameters  $a$,  $b$  and  $c$  of the PDF  $f_z(z)$.

$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:

$$C_x( {\it \Omega } ) = \int_{ - \infty }^{ + \infty } {f_x }( x )\cdot {\rm{e}}^{\hspace{0.03cm}{\rm{j}}\hspace{0.05cm}\cdot \hspace{0.05cm}{\it \Omega\hspace{0.05cm}\cdot \hspace{0.05cm} x}} \hspace{0.1cm}{\rm{d}}x .$$

• 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:

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