Homework Solutions

Document Actions

Utah State University
ECE 6010
Stochastic Processes
Homework # 7 Solutions

Suppose $\{ X_{t}, t \geq 0 \}$ is a homogeneous Poisson process with parameter $\lambda$ . Define a random variable $\tau$ as the time of the first occurrence of an event. Find the p.d.f. and the mean of $\tau$ .

p.d.f. :
We have,

$\displaystyle P(X_{t} = k) = \frac{e^{- \lambda t} (\lambda t)^k}{k!}$

So, $P(\tau > t) = P(X_t = 0) = e^{-\lambda t}$ t) = P(X_t = 0) = e^{-\lambda t}$" align="middle" border="0" height="39" width="228" />
and $P(\tau \leq t) = 1 - e^{-\lambda t}$ , i.e. $F_{\tau}(t) = 1 - e^{-\lambda t}$ .
Therefore,

$\displaystyle f_{\tau}(t) = \lambda e^{-\lambda t}$

Mean :

$\displaystyle E[\tau] = \int_{0}^{\infty} \tau \lambda e^{-\lambda \tau} d \t... ...tau} \right\vert _{0}^{\infty} = -0-0+0+ \frac{1}{\lambda} = \frac{1}{\lambda}$
Suppose $\{ X_{t}, t \in \mathbb{R} \}$ is a w.s.s. random process with autocorrelation function $R_{X}(\tau)$ . Show that if $R_{X}$ is continuous at $\tau = 0$ then it is continuous for all $\tau \in \mathbb{R}$ .

$\displaystyle [R_X (\tau + \delta) - R_X(\tau)]^{2}$ $\displaystyle =$ $\displaystyle (E[X_{\tau + \delta} X_{0}] - E[X_{\tau } X_{0}])^2$

$\displaystyle =$ $\displaystyle (E[X_{0}(X_{\tau + \delta}- X_{\tau })])^2$

$\displaystyle \leq$ $\displaystyle E[X_{0}^2]E[(X_{\tau + \delta}- X_{\tau })^2]$ (Cauchy-Schwartz)

$\displaystyle =$ $\displaystyle R_{X}(0) [ R_{X}(0) - 2 R_{X}(\delta) + R_{X}(0)]$

$\displaystyle =$ $\displaystyle 2R_{X}(0) [ R_{X}(0) - R_{X}(\delta)]$

$\displaystyle =$ $\displaystyle \leq 2 R_{X}(0) \varepsilon$

So,

$\displaystyle \Rightarrow \vert R_X (\tau + \delta) - R_X(\tau)\vert \leq \sqrt{2 R_{X}(0) \varepsilon}$

$\displaystyle \Rightarrow \vert R_X (\tau + \delta) - R_X(\tau)\vert < \varepsilon '$ < \varepsilon ' $" align="middle" border="0" height="37" width="221" />

So continuous at all $\tau$ .
Under the conditions of problem 2, show that for 0$" align="middle" border="0" height="31" width="45" /> ,

$\displaystyle P(\vert X_{t+\tau} - X_{t}\vert \geq a) \leq \frac{2(R_X(0) - R_X(\tau))}{a^2}.$

Let . is non-negative, non-decreasing on $[0,\infty)$ and symmetric about 0. Then,

$\displaystyle P(\vert X\vert \geq b) \leq \frac{E[g(x)]}{b^2}.$

Now, our example corresponds to:

$\displaystyle P(\vert X_{t+\tau} - X_{t}\vert \geq a)$ $\displaystyle \leq$ $\displaystyle \frac{E[\vert X_{t+\tau} - X_{t}\vert^2]}{a^2}$

$\displaystyle =$ $\displaystyle \frac{R_{X}(0) - 2R_{X}(\tau) + R_{X}(0)}{a^2}$

$\displaystyle =$ $\displaystyle \frac{2(R_X(0) - R_X(\tau))}{a^2}$
Suppose and are random variables with $E[A^2] < \infty$ < \infty$" align="middle" border="0" height="38" width="88" /> and $E[B^2] < \infty$ < \infty$" align="middle" border="0" height="38" width="89" /> . Define the random processes $\{ X_{t}, t \in \mathbb{R} \}$ and $\{ Y_{t}, t \in \mathbb{R} \}$ by

$\displaystyle X_{t} = A+Bt \quad \quad Y_{t} = B+At, \quad t \in \mathbb{R}.$

Find the mean, autocorrelation, and cross correlations of these random processes in terms of the moments of and .

Mean :

$\displaystyle \mu_{X}(t) = E[ A+Bt] = E[A] +tE[B]$

$\displaystyle \mu_{Y}(t) = E[ B+At] = E[B] +tE[A]$

Autocorrelation :

$\displaystyle R_{X}(t,s)$ $\displaystyle =$ $\displaystyle E[X_{t}X_{s}]$

$\displaystyle =$ $\displaystyle E[(A+Bt)(A+Bs)]$

$\displaystyle =$ $\displaystyle E[A^2 + ABs + ABt + B^2 ts]$

$\displaystyle =$ $\displaystyle E[A^2] + E[AB](s+t) + E[B^2]ts$

Similarly,

$\displaystyle R_{Y}(t,s)$ $\displaystyle =$ $\displaystyle E[B^2] + E[AB](s+t) + E[A^2]st$

Cross correlation :

$\displaystyle R_{XY}(t,s)$ $\displaystyle =$ $\displaystyle E[X_{t}Y_{s}]$

$\displaystyle =$ $\displaystyle E[(A+Bt)(B+As)]$

$\displaystyle =$ $\displaystyle E[AB + A^2 s + B^2 t +AB ts]$

$\displaystyle =$ $\displaystyle E[AB](1+st) + E[A^2]s + E[B^2]t$
Homogeneous Poisson
Simply take and substitute it back into the differential equation and show that it works.

$\displaystyle \frac{\partial}{\partial t} p_{k}(t,s)=\frac{\partial}{\partial t} \frac{e^{-\lambda(t-s)} (\lambda(t-s))^{k}}{k!}$

$\displaystyle = \frac{-\lambda e^{-\lambda(t-s)}(\lambda(t-s))^{k}+e^{-\lambda(t-s)} k \lambda(\lambda(t-s))^{k-1}}{k!}$

$\displaystyle =\lambda \left[\frac{-e^{-\lambda(t-s)}(\lambda(t-s))^{k}}{k!}+\frac{e^{-\lambda(t-s)}(\lambda(t-s))^{k-1}}{(k-1)!} \right]$

$\displaystyle =\lambda [p_{k-1}(t,s)-p_{k}(t,s)]$
Inhomogeneous Poisson
Simply take and substitute it back into the differential equation and show that it works.

$\displaystyle \frac{\partial}{\partial t} p_{k}(t,s)= \frac{-\lambda_{t}e^{-\int_s^t\lambda_x dx}+\lambda_t k(\int_s^t\lambda_x dx)^{k-1}}{k!}$

$\displaystyle =\lambda_t \left[\frac{-e^{-\int_s^t \lambda_x dx}}{k!}+ \frac{(\int_s^t \lambda_x dx)^{k-1}}{(k-1)!} \right]$

$\displaystyle = \lambda_t[p_{k-1}(t,s)-p_k(t,s)$

Mean: $\int_s^t \lambda_x dx$
Covariance: Assume that s$" align="middle" border="0" height="31" width="41" /> :

$\displaystyle \begin{aligned} E[X_t X_s] &= E[(X_t-X_s)X_s] + E[X_s^2] = E[X_t... ...&= \left[\int_0^t \lambda_x dx +1\right] \int_0^s \lambda_x dx \end{aligned}$

Similarly when < s$" align="middle" border="0" height="31" width="41" /> . Then

$\displaystyle R_X(t,s) = \left[\int_0^s \lambda_x dx +1\right] \int_0^t \lambda_x dx$
Suppose $\{ X_{t}, t \in \mathbb{R} \}$ is a random process with power spectral density

$\displaystyle S_{X} (\omega) = \frac{1}{(1+ \omega^{2})^{2}}.$

Find the autocorrelation function of $X_{t}$ .

$\displaystyle R_{X}(\tau)$ $\displaystyle =$ $\displaystyle \mathcal{F}^{-1} \{S_{X}(\omega) \} = \mathcal{F}^{-1} \left\{ \f... ...ega^{2})} \right\} * \mathcal{F}^{-1} \left\{ \frac{1}{(1+\omega^{2})} \right\}$

$\displaystyle =$ $\displaystyle \frac{1}{2} e^{-\vert\tau\vert} * \frac{1}{2} e^{-\vert\tau\vert}$

$\displaystyle =$ $\displaystyle \frac{1}{4} \int_{-\infty}^{\infty} e^{-\vert t\vert} e^{-\vert\tau - t\vert} dt$

$\displaystyle =$ $\displaystyle \frac{1}{4} \int_{-\infty}^{0} e^{t} e^{t-\tau} dt + \frac{1}{4... ...{t-\tau} dt + \frac{1}{4} \int_{\tau}^{\infty} e^{-t} e^{\tau -t} dt$ $\displaystyle \mbox{ (for $\tau \geq 0$)}$

$\displaystyle =$ $\displaystyle \frac{1}{4} \left[ \int_{-\infty}^{0} e^{2t - \tau} dt + \int_{0}^{\tau} e^{-\tau} dt + \int_{\tau}^{\infty} e^{t - 2\tau} dt \right]$

$\displaystyle =$ $\displaystyle \frac{1}{4} \left[ \frac{1}{2} e^{-\tau} + \tau e^{-\tau} + \frac{1}{2} e^{-\tau} \right]$

$\displaystyle =$ $\displaystyle \frac{1}{4} \left( \tau e^{-\tau} + e^{-\tau} \right) \$ $\displaystyle \mbox{ (for $\tau \geq 0$)}$

Similarly,

$\displaystyle R_{X}(\tau) = \frac{1}{4} \left( - \tau e^{\tau} + e^{\tau} \right)$ $\displaystyle \mbox{ (for $\tau < 0$)}$ < 0$)}$" align="middle" border="0" height="35" width="91" /> $\displaystyle$

Therefore,

$\displaystyle R_{X}(\tau) = \frac{1}{4} e^{-\vert\tau\vert} ( \vert\tau \vert +1)$
Suppose that $\omega$ is a random variable with p.d.f. $f_{\omega}$ and $\theta$ is a random variable independent of $\omega$ uniformly distributed in $(-\pi,\pi)$ . Define a random process by $X_{t} = a \cos(\omega t + \theta), t\in \mathbb{R}$ where is a constant. Find the power spectral density of $\{X_{t} \}$ .

$\displaystyle E[X_{t_{1}} X_{t_{2}} ]$ $\displaystyle =$ $\displaystyle E \{ a^{2} \cos(\omega t_{1} + \theta) \cos(\omega t_{2} + \theta)$

$\displaystyle =$ $\displaystyle \frac{1}{2} a^{2} E \{ \cos(\omega t_{1}-\omega t_{2}) - \cos(\omega t_{1} + \omega t_{2} + 2\theta) \}$

$\displaystyle =$ $\displaystyle \frac{1}{2} a^{2} \left [ E \{ \cos(\omega t_{1}-\omega t_{2}) \} - \underbrace{E \{ \cos(\omega t_{1} + \omega t_{2} + 2\theta) \}}_{0} \right]$

$\displaystyle =$ $\displaystyle \frac{1}{2} a^{2} E \{ \cos(\omega t_{1}-\omega t_{2}) \}$

$\displaystyle =$ $\displaystyle \frac{1}{2} a^{2} \int_{-\infty}^{\infty} \cos(\tau \omega) f_{\o... ...frac{ e^{j \omega \tau} + e^{-j \omega \tau} } {2} f_{\omega}(\omega) d\omega$

$\displaystyle =$ $\displaystyle \frac{1}{4} a^{2} 2\pi [ \mathcal{F}^{-1} \{ f_{\omega}(\omega) \} + \mathcal{F}^{-1} \{ f_{\omega}(-\omega) \}]$

$\displaystyle =$ $\displaystyle \frac{\pi a^{2}}{2} [ \mathcal{F}^{-1} \{ f_{\omega}(\omega) \} + \mathcal{F}^{-1} \{ f_{\omega}(-\omega) \}]$

Therefore,

$\displaystyle S_{X}(\omega) = \frac{\pi a^{2}}{2} [f_{\omega}(\omega) + f_{\omega}(-\omega)]$
Suppose that $\{X_t, t \in \mathbb{R} \}$ is a w.s.s., zero-mean, Gaussian random process with auto-correlation function $R_{X}(\tau), \tau \in \mathbb{R}$ and power spectral density $S_X(\omega), \omega \in \mathbb{R}$ . Define the random process $\{ Y_{t}, t \in \mathbb{R} \}$ by $Y_{t} = (X_{t})^{2}, t \in \mathbb{R}$ . find the mean, autocorrelation, and powerspectral density of $\{ Y_{t}, t \in \mathbb{R} \}$ .

Mean :

$\displaystyle \mu_{y}(t) = E[X_{t}^{2}] = R_{X}(0) = \sigma^{2}$

Autocorrelation :

$\displaystyle R_{Y}(t,s)$ $\displaystyle =$ $\displaystyle E[Y_{t}Y_{s}] = E[X_{t}^{2} X_{s}^{2}]$

$\displaystyle =$ $\displaystyle \left. \frac{\partial ^{4}}{\partial u^{2} \partial v^{2}} \Phi_{X_{t}X_{s}}(u,v) \right\vert _{u=v=0} \frac{1}{i^{4}}$

$\displaystyle \vdots$

$\displaystyle =$ $\displaystyle R_{X}^{2}(0) + 2 R_{X}^{2} (\tau) \quad \quad (\tau = t-s)$

PSD :

$\displaystyle S_{Y}(\omega) = \mathcal{F} \{R_{Y}(\tau) \} = 2 \pi R_{X}^{2}(0) \delta(\omega) + 2 (S_{X}(\omega) *S_{X}(\omega))$
Suppose and are independent random variables with and $\mathrm{var}(U) = \mathrm{var}(V) = 1$ . Define random processes by

$\displaystyle X_t = U \cos t + V \sin t \quad Y_{t} = U \sin t + V \cos t, \quad t \in \mathbb{R}.$

Find the autocorrelation and cross-correlation functions of $\{ X_{t}, t \in \mathbb{R} \}$ and $\{ Y_{t}, t \in \mathbb{R} \}$ . Are $\{X_{t} \}$ and $\{Y_{t} \}$ jointly wide sense stationary? Are they individually wide sense stationary?

$\displaystyle R_{X}(t,s)$ $\displaystyle =$ $\displaystyle E[X_{t}X_{s}] = E[(U \cos t + V \sin t)(U \cos s + V \sin s) ]$

$\displaystyle =$ $\displaystyle E[U^{2} \cos t \cos s + UV (\cos t \sin s + \sin t \cos s) + V^{2} \sin t \sin s ]$

$\displaystyle =$ $\displaystyle \cos t \cos s E[U^{2}] + E[U] E[V] (\cos t \sin s + \sin t \cos s) + E[V^{2}] \sin t \sin s$

$\displaystyle =$ $\displaystyle \cos t \cos s + \sin t \sin s$

$\displaystyle =$ $\displaystyle \cos(t-s)$

Similarly,

$\displaystyle R_{Y}(t,s) = cos(t-s)$

$\displaystyle R_{XY}(t,s)$ $\displaystyle =$ $\displaystyle E[X_{t}Y_{s}] = E[(U \cos t + V \sin t)(U \sin s + V \cos s) ]$

$\displaystyle =$ $\displaystyle E[ U^{2} \cos t \sin s + UV ( \cos t \cos s + \sin t \sin s) + V^{2} \sin t \cos s]$

$\displaystyle =$ $\displaystyle E[ U^{2}] \cos t \sin s + E[UV] ( \cos t \cos s + \sin t \sin s) E[ V^{2}] \sin t \cos s$

$\displaystyle =$ $\displaystyle \cos t \sin s + \sin t \cos s$

$\displaystyle =$ $\displaystyle \sin (t+s)$

$\displaystyle \mu_{X}(t) = 0 \quad \quad \mu_{Y}(t)$

So, $\{X_{t} \}$ and $\{Y_{t} \}$ are individually WSS, but not jointly WSS.

Copyright 2008, Todd Moon. Cite/attribute Resource. admin. (2006, June 13). Homework Solutions. Retrieved November 23, 2009, from Free Online Course Materials — USU OpenCourseWare Web site: http://ocw.usu.edu/Electrical_and_Computer_Engineering/Stochastic_Processes/hw7sol.html. This work is licensed under a Creative Commons License.

Sections

Personal tools

Homework Solutions

Document Actions

Utah State University
ECE 6010
Stochastic Processes
Homework # 7 Solutions

Sections

Personal tools

Homework Solutions

Document Actions

Utah State University ECE 6010 Stochastic Processes Homework # 7 Solutions

Utah State University
ECE 6010
Stochastic Processes
Homework # 7 Solutions