Advanced Signals and Systems - Stochastic Processes I

 

3. LAPLACE's distribution

Task

Let \(v\) be a random variable with the probability-density function \(f_{v}(V)=\frac{\alpha}{2}e^{-\alpha |V|}\) (density of LAPLACE's distribution).

  1. Determine the probability function \(F_v(V)\).
  2. Determine the mean value and the variance of \(v\).
  3. Determine the characteristic function of LAPLACE's distribution.
  4. Verify the results of (2) with the characteristic function.

Amount and difficulty

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

Solution

  1. In general: \begin{equation*} F_v(V) = \int \limits_{x=-\infty}^V f_v(x) \, dx \end{equation*} For solving the integral a case differentiation for \(V<0\) and \(V\leq0\) is done and the following result can be derived: \begin{equation*} F_v(V) = \frac{1}{2} \cdot e^{-\alpha |V|} + \left[ 1 - e^{-\alpha |V|}\right] \delta_{-1}(V)\end{equation*}
  2. Determining the mean in general: \begin{equation*} \text{E}\left\{ v\right\} = \int \limits_{V=-\infty}^\infty V f_v(V) \, dV = \mu_v \end{equation*} Here we can conclude \begin{equation*} \text{E}\left\{ v\right\} = \int \limits_{V=-\infty}^\infty \underbrace{\underbrace{V}_{\text{odd}} \cdot \underbrace{\frac{\alpha}{2} \cdot e^{-\alpha |V|}}_{\text{even}}}_{\text{odd}} \, dV = 0 \end{equation*} because the integral of an odd function equals to zero.

    Determining the variance in general: \begin{equation*} \sigma_v^2 = \text{E}\{v^2\} - {\text{E}\left\{v\right\}}^2 \end{equation*} Here we can conclude \begin{align*} \sigma_v^2 &= \text{E}\{v^2\} - \underbrace{\text{E}\{v\}^2}_{=0} = \text{E}\{v^2\}\\ &= \int \limits_{-\infty}^\infty V^2 f_v(V) \, dV = \int \limits_{-\infty}^\infty \underbrace{V^2 \cdot \frac{\alpha}{2} e^{-\alpha |V|}}_{\text{even}} \, dV\\ &= \alpha \int \limits_{0}^\infty V^2 e^{-\alpha V} dV \end{align*} After substitution with \(t=\alpha \cdot V\) and applying the gamma function \(\Gamma(x) = \int \limits_0^\infty t^{x-1} e^{-t} dt\) the variance is defined by: \begin{equation*} \sigma_v^2 = \frac{1}{\alpha^2} \cdot \Gamma(3) = \frac{2}{\alpha^2} \end{equation*}
  3. \begin{align*} C(\omega) &= \text{E} \{ e^{j\omega V} \} = \int \limits_{V=-\infty}^\infty e^{j\omega V} f_v(V) \, dV \\ &= \frac{\alpha^2}{\omega^2 + \alpha^2} \end{align*}
  4. Moment theorem: \begin{align*} \text{E} \left\{x^k \right\} &= \left. \frac{1}{j^k} \frac{d^k C(\omega)}{d \omega ^k} \right| _{\omega =0}\\ \Rightarrow \mu_v &= \text{E} \{x \} = \left. \frac{1}{j} \frac{d C(\omega)}{d \omega }\right| _{\omega =0} = \cdots = 0 \\ \Rightarrow \sigma_v^2 &= \text{E} \{x ^2\} = \left. \frac{1}{j^2} \frac{d^2 C(\omega)}{d \omega ^2}\right| _{\omega =0} = \cdots = \frac{2}{\alpha^2} \end{align*}The same results as in (b) can be achieved.

Additional material

Laplace distribution:

Example speech signal:

4. Properties of ACF/CCF.

Task

  1. Show that for the CCF of the complex random signals \(v_1(n)\) and \(v_2(n)\) holds: \begin{equation*}s_{v_1v_2}(\kappa) = s^\ast_{v_2v_1}(-\kappa) \end{equation*} What does this imply for the CCF of real signals \(v_1(n)\) and \(v_2(n)\)?
    How can this relationship be used to determine the ACF of a real signal?
  2. Show that the CAUCHY-SCHWARZ inequality for real random variables holds: \begin{equation*} \left[E\{xy\}\right]^2 \leq E\{x^2\}\cdot E\{y^2\} \end{equation*} What does this imply for the ACF of a real signal?
  3. How can the above results for discrete signals be applied to the CCF and ACF of continuous signals?

Amount and difficulty

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

Solution

  1. Stationarity has to be assumed. According to the lecture, the cross-correlation is defined by \begin{equation*} s_{v_1v_2}(\kappa) = \text{E} \left\{ v_1^*(n) \cdot v_2(n+\kappa) \right\} \end{equation*} Further it can be shown: \begin{align*}s_{v_2v_1}^*(-\kappa) &= \left[ \text{E}\left\{ v_2^*(n) \cdot v_1(n-\kappa) \right\} \right]^*\\ &= \left[ \text{E}\left\{ v_2^*(n+\kappa) \cdot v_1(n) \right\} \right]^*\\ &= \text{E}\left\{ v_2(n+\kappa) \cdot v_1^*(n) \right\}\\ &= s_{v_1v_2}(\kappa)\end{align*} Special case: \(v_1(n), v_2(n) \in \mathbb{R}\) \begin{equation*}s_{v_1v_2}(\kappa) = \text{E} \{ v_1(n) \cdot v_2(n+\kappa)\} = \cdots = s_{v_2v_1}(-\kappa);\end{equation*} Special case ACF: \(v_1(n) = v_2(n) =v(n)\) \begin{align*} v(n) \in \mathbb{C} &\rightarrow s_{vv}(\kappa) = s_{vv}^* (-\kappa)\\ v(n) \in \mathbb{R} &\rightarrow s_{vv}(\kappa) = s_{vv} (-\kappa) \end{align*}
  2. Let \(x,y\) be real random variables and let \(a \in \mathbb{R}\). \begin{align*} \text{E} \{ (ax-y)^2 \} \geq 0 \\ a^2 \, \text{E} \{ x^2 \} +2a\,\text{E} \{ xy \} +\text{E} \{ y^2\}\geq 0 \end{align*} By solving this equation and inserting \(a=\frac{\text{E} \{ xy\}}{\text{E} \{ x^2\}}\) it follows that \begin{equation*}\text{E}^2 \{ xy \} \leq \text{E}\{x^2\} \cdot \text{E}\{y^2\}\end{equation*}
  3. The above results hold true also for continuous signals.

Ergodicity and stationarity.

Additional material

Example 1:

Stationary?

Ergodic?

Example 2:

Stationary?

Ergodic?

Example 3:

Stationary?

Ergodic?