Exercise 3.4: Characteristic Function
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Rectangular PDF and trapezoidal PDF
Given here are the 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 and 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 only take values in the range between $1$ to $3$ with equal probability. The mean is $$m_y = 2.$$
- 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 $a$, $b$ and $c$ of this PDF.
Hints:
- This exercise belongs to the chapter expected values and moments.
- Reference is made to the page charakteristic funcion .
- The characteristic function of a between $\pm a$ uniformly distributed random variable $z$ is:
- $$C_z ( {\it \Omega} ) = {\mathop{\rm si}\nolimits} ( {a {\it \Omega} } )\quad {\rm{with}}\quad {\mathop{\rm si}\nolimits}( x ) = \sin ( x )/x.$$
Questions
Solutions
(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 Fourier retransform:
- $$C_x( {\it \Omega } ) = \int_{ - \infty }^{ + \infty } {f_x }( x )\cdot {\rm{e}}^{\hspace{0.03cm}{\rm{j}}\hspace{0.03cm}{\it \Omega x}} \hspace{0.1cm}{\rm{d}}x .$$
- 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 still has a complex characteristic function.
(2) According to the general definition:
- $$C_y( {\it \Omega } ) = \int_{ - \infty }^{ + \infty } {f_y }( y )\cdot {\rm{e}}^{{\rm{j}}{\it \Omega y}} \hspace{0.1cm}{\rm{d}}y = 0.5\int_1^3 {{\rm{e}}^{{\rm{j}}\Omega y} \hspace{0.1cm}{\rm{d}}y.} $$
- After solving this integral, we get:
- $$C_y ( {\it \Omega } ) = \frac{{{\rm{e}}^{{\rm{j}}3{\it \Omega } } - {\rm{e}}^{{\rm{j}}{\it \Omega } } }}{{2{\rm{j}}{\it \Omega } }} = \frac{{{\rm{e}}^{{\rm{j}}{\it \Omega } } - {\rm{e}}^{{\rm{ - j}}{\it \Omega }} }}{{2{\rm{j}}{\it \Omega } }} \cdot {\rm{e}}^{{\rm{j2}}{\it \Omega } } .$$
- Using Euler's theorem, this can also be written:
- $$C_y ( {\it \Omega } ) = \frac{{\sin ( {\it \Omega } )}}{{\it \Omega } } \cdot {\rm{e}}^{{\rm{j2}}{\it \Omega } } .$$
- 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 $\pm 2$ .
- 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:
Construction of trapezoidal PDF
- The three PDF parameters are thus:
- $$\hspace{0.15cm}\underline{a = 1},\quad \hspace{0.15cm}\underline{b = 5}, \quad c = 1/6 \hspace{0.15cm}\underline{= 0.167}.$$
