Advanced Signals and Systems - Stochastic Processes II

 

5. Bivariate probability densitiy functions.

Task

Consider the following joint pdf of two stationary random variables \(v_1\), \(v_2\):

\begin{eqnarray*} f_{v_1v_2}(V_1,V_2,\kappa) = \left\{\begin{array}{r@{,\qquad}l} P_0, & \sqrt{V_1^2 + V_2^2} \leq \hat{V} \\ 0, & \hspace{0.6cm} \mbox{otherwise} \end{array} \right. \end{eqnarray*}

  1. Sketch \(f_{v_1v_2}(V_1,V_2,\kappa)\) and determine \(P_0\).
    What kind of distribution is \(f_{v_1v_2}(V_1,V_2,\kappa)\)?
  2. Calculate the marginal probability density functions \(f_{v_1}(V_1)\) and \(f_{v_2}(V_2)\).
  3. Are the random variables \(v_1\) and \(v_2\)
    1. independent
    2. orthogonal
    3. uncorrelated?
  4. Specify a uniform joint pdf \(f_{v_1v_2}(V_1,V_2,\kappa)\) of two independent, uncorrelated but not orthogonal random signals \(v_1(n)\), \(v_2(n)\).

Amount and difficulty

  • Working time: approx. xx minutes
  • Difficulty: ?

Additional Material

  1. Joint pdf of the two stationary random variables \(v_1\) and \(v_2\): \begin{equation} f_{v_1v_2}(V_1,V_2,\kappa)= \begin{cases} \quad P_0, \qquad &\sqrt{V_1^2+V_2^2} \le \hat{V} \notag \\ \quad 0, \qquad &\text{otherwise} \end{cases} \end{equation}

    Joint pdf \(f_{v_1v_2}(V_1,V_2,\kappa)\):


    Product of the two marginal pdfs \(f_{v_1}(V_1)\cdot f_{v_2}(V_2)\):

  2. The following joint pdf fulfills the conditions:

    \(\Rightarrow\) \(v_1(n)\) and \(v_2(n)\) are independent, as \(f_{v_1v_2}(V_1,V_2,\kappa) = f_{v_1}(V_1)\cdot f_{v_2}(V_2)\)

Solution

  1. The following picture shows the two-dimensional uniform distribution with circular base which was defined by the pdf given in the problem.

    The value for \(P_0\) can be derived by \begin{equation*} F_{v_1v_2}(\infty, \infty, \kappa) = \int \limits_{-\infty}^{\infty} \int \limits_{-\infty}^{\infty} f_{v_1v_2} (V_1,V_2,\kappa) \; dV_1 \;dV_2 = 1 \ \ \ \Longrightarrow \ \ \ P_0 = \frac{1}{\pi \hat{V}^2} \end{equation*}
  2. The marginal probability density function is defined by \begin{equation*} f_{v_1}(V_1) = \int \limits_{V_2 = -\infty}^{\infty} f_{v_1v_2}(V_1, V_2, \kappa) \; dV_2 \end{equation*} Solving this equation, the two marginal probability density function are defined by: \begin{align*} f_{v_1}(V_1) &= 2 \cdot P_0 \sqrt{\hat{V}^2-V_1^2}\\ f_{v_2}(V_2) &= 2 \cdot P_0 \sqrt{\hat{V}^2-V_2^2}\\ \end{align*}
  3. The variables are
    1. independent, if \begin{equation*} f_{v_1v_2} (V_1,V_2,\kappa) = f_{v_1} (V_1) \cdot f_{v_2} (V_2) \text{ .} \end{equation*} \(\longrightarrow\) \(v_1\) and \(v_2\) are not independent
    2. orthogonal, if \begin{equation*} s_{v_1v_2}(\kappa) = \int \limits_{-\infty}^{\infty} \int \limits_{-\infty}^{\infty} V_1^* V_2 \cdot f_{v_1v_2} (V_1,V_2,\kappa) \; dV_1 \; dV_2 = 0 \text{ .} \end{equation*} \(\longrightarrow v_1\) and \(v_2\) are orthogonal
    3. uncorrelated, if \begin{equation*} \psi_{v_1v_2}(\kappa) = s_{v_1v_2}(\kappa) - \mu_{v_1}^* \mu_{v_2} = 0 \text{ .} \end{equation*} As both marginal pdfs are symmetric about their origin, their mean values are equal to zero.
      \(\longrightarrow v_1\) and \(v_2\) are uncorrelated
  4. The following joint pdf fulfills the conditions:

    \(\Rightarrow\) The shown pdf is uniform as all possible combinations of \(V_1\) and \(V_2\) are equally likely to be observed. 
    \(\Rightarrow\) \(v_1(n)\) and \(v_2(n)\) are independent, as \(f_{v_1v_2}(V_1,V_2,\kappa) = f_{v_1}(V_1)\cdot f_{v_2}(V_2)\).
    \(\Rightarrow\) As \(v_1(n)\) and \(v_2(n)\) are independent, they are also uncorrelated.
    \(\Rightarrow\) \(v_1(n)\) and \(v_2(n)\) are not orthogonal, as \(\mu_{v_1} \neq 0\) and \(\mu_{v_2} \neq 0\).

     

    6. ACF of a sum process.

    Task

    Given two stationary random processes \(v_1(n)\) and \(v_2(n)\), determine the auto-correlation function (ACF) \(s_{vv}\) of the sum process \(v(n,\kappa)=v_1(n)+v_2(n+\kappa)\).

    Amount and difficulty

        • Working time: approx. xx minutes
        • Difficulty: ?

    Solution

    Given: The sum process \begin{equation*} v(n,\kappa )=v_1(n)+v_2(n+\kappa ), \qquad \text{where } v_1(n) \text{ and } v_2(n) \text{ are stationary} \end{equation*}

    Wanted: The auto-correlation function (ACF) \(s_{vv}(\eta,\kappa )\) Definition of the ACF \(s_{vv}(\eta,\kappa )\): \begin{align*} s_{vv}(\eta,\kappa )&=E\{v(n,\kappa )\cdot v(n+\eta,\kappa ) \} \\ &= \cdots \\ s_{vv}(\eta,\kappa )&= s_{v_1v_1}(\eta) + s_{v_1v_2}(\kappa +\eta) + s_{v_1v_2}(\kappa -\eta)+ s_{v_2v_2}(\eta) \end{align*} \(\rightarrow\) \(s_{vv}(\eta,\kappa )\) is dependent on \(\kappa\) despite stationarity of \(v_1(n)\), \(v_2(n)\) and \(\mu_v\)!

 

7. PDF of a sum process.

Task

Let \(v_1(n)\) and \(v_2(n)\) be two independent random processes with following probability density functions:

  1. Determine the mean and the variance of the sum process \(v(n)=v_1(n)+v_2(n)\).
  2. Determine and sketch the pdf \(f_v(V)\) of the sum process.

Amount and difficulty

    • Working time: approx. xx minutes
    • Difficulty: ?

Solution

  1. The mean value of a sum process, for which the processes are independent, is defined by \begin{equation*} \mu_v = \mu_{v_1} + \mu_{v_1} = \cdots = \frac{2}{3} \text{ .} \end{equation*}

    The variance of a sum process, for which the processes are independent, is defined by \begin{equation*} \sigma^2 _v = \sigma^2_{v_1} + \sigma^2_{v_1} + 2 \psi_{v_1v_2}(\kappa) = \cdots = \frac{2}{9} \text{ .} \end{equation*}

    Hint: The problem says two independent random processes are given, we know that independence includes uncorrelatedness <\(\rightarrow \psi_{v_1v_2}(\kappa) = 0 \).

  2. In general the pdf of a sum process is defined by the convolution \begin{equation*} f_v(V) = f_{v_1}(V) * f_{v_2}(V) = \int \limits_{X=-\infty}^{\infty} f_{v_1}(X) \cdot f_{v_2}(V-X) \; dX\text{ .} \end{equation*}

    For solving this integral a case differentiation is carried out and the following result can be achieved: \begin{equation*} f_v(V) = \begin{cases} -\frac{1}{8}V^2+\frac{1}{4}V+\frac{3}{8} & \text{, } -1\leq V < 1\\ \frac{1}{8}V^2-\frac{3}{4}V+\frac{9}{8} & \text{, } 1\leq V < 3\\ 0 & \text{, otherwise.} \end{cases} \end{equation*}

 

8. Mapping of a random process.

Task

Consider a stationary, uniformly distributed random process \(v(n)\) mapped to \begin{equation*} y(n) = a \cdot v(n) + b \text{ .} \end{equation*} Determine the mean \(\mu_y\), the variance \(\sigma_x^2\) and the pdf \(f_y(Y)\) of the resulting random process \(y(n)\).

Amount and difficulty

    • Working time: approx. xx minutes
    • Difficulty: ?

Solution

The random variable is linearly mapped to \(y(n)\), so \(\rightarrow g(V) = a \cdot V +b = Y\). Remember that \(v(n)\) is uniformly distributed random process.

  1. Find the pdf of this new random variable \(Y\) \begin{equation*} f_y(Y) = \left. \frac{f_v(V)}{|dg(V)/dV|} \right|_{V=g^{-1}[Y]} \ \ \ \text{with } V=g^{-1}[Y]= \frac{Y-b}{a} \text{ , } \frac{dg(V)}{dV}=a \end{equation*} \begin{equation*} f_y(Y) = \begin{cases} \frac{1}{|a|} \frac{1}{V_{max}-V_{min}} &\text{, } Y \in \left[ a \cdot V_{min} + b, a \cdot V_{max} + b \right] \\ 0 &\text{, otherwise}\end{cases} \end{equation*}

  2. Find the mean value \begin{equation*} \mu_y = \frac{1}{2} \left[ (a \cdot V_{max} + b) + (a \cdot V_{min} + b) \right] = \cdots = a \cdot \mu_v + b \end{equation*}

  3. Find the variance \begin{equation*} \sigma^2_y = \frac{1}{12} \left[ a\cdot (V_{max}-V_{min}) \right]^2 = \cdots = a^2 \cdot \sigma^2_v \end{equation*}