# Advanced Signals and Systems - Stochastic Processes II

### 5. Bivariate probability densitiy functions.

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: ?

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

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.

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.

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.

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*}