Download Solutions to Problem Set 7 in ECE 534: Random Processes, Spring 2010 - Prof. Rayadurgam Sr and more Assignments Electrical and Electronics Engineering in PDF only on Docsity!
ECE 534: Random Processes Spring 2010
Solutions to Problem Set 7
4.1 Event probabilities for a simple random process
Figure 1: The four sample paths.
(a) There are four sample functions corresponding to the four possible values of (A, B) as shown in Figure 1.
(b) Let Eij be the event that A = i and B = j. Then
P [Xt ≥ 0] =
P [E− 1 ,− 1 ∪ E 1 ,− 1 ] = 1/ 2 if t < − 2 1 − P [E− 1 , 1 ] = 3/ 4 if t = − 2 P [E 1 ,− 1 ∪ E 1 , 1 ] = 1/ 2 if − 2 < t < 2 1 − P [E− 1 ,− 1 ] = 3/ 4 if t = 2 P [E 1 , 1 ∪ E− 1 , 1 ] = 1/ 2 if t > 2
(c) The set {ω : Xt(ω) ≥ 0 all t} is empty, because for each ω, Xt(ω) < 0 for some t. Thus P [Xt ≥ 0 all t] = 0.
ECE 534 HW7 Spring 2010
4.3 A sinusoidal random process sinc Since Xt = A cos(2πV t) cos(Θ) − A sin(2πV t) sin(Θ) and E cos(Θ) = E sin(Θ) = 0, it follows that μX (t) ≡ 0. The autocorrelation function of X is given by
RX (s, t) = E[A^2 ]E[cos(2πV s + Θ) cos(2πV t + Θ)]
=
E[A^2 ]
E[cos(2πV (s − t)) + cos(2πV (s + t) + 2Θ)]
E[A^2 ]
0
cos(2πv(s − t))dv + 0
= 4sinc(10(s − t))
where we used the fact that E[A^2 ] = (EA)^2 +Var(A) = 8 and the definition sinc(u) = sin( πuπu) if u 6 = 0 and sinc(0) = 1.
4.9 A random process corresponding to a random parabola
(a) The mean function is μX (t) = 0 + 0t + t^2 = t^2 and the covariance function is given by
CX (s, t) = Cov(A + Bs + s^2 , A + Bt + t^2 ) = Cov(A, A) + stCov(B, B) = 1 + st
Thus, Ê [X 5 |X 1 ] = μX (5)+ C CXX^ (5(1,,1)1) (X 1 −μX (1)) = 25+ 62 (X 1 −1) and MSE = CX (5, 5)− CX (5,1)^2 CX (1,1) = 26^ −^
62 2 = 8.
(b) The variables A and B are jointly Gaussian and X 1 and X 5 are linear combinations of A and B, so X 1 and X 5 are jointly Gaussian. Thus, E[X 5 |X 1 ] = Ê [X 5 |X 1 ] and MSE = 8.
(c) Since X 0 = A and X 1 = A + B + 1, it follows that B = X 1 − X 0 − 1. Thus Xt = X 0 + (X 1 − X 0 − 1)t + t^2. So X 0 + (X 1 − X 0 − 1)t + t^2 is a linear estimator of Xt based on (X 0 , X 1 ) with zero MSE, so it is the LMMSE estimator.
4.12 Poisson process probabilities
(c) The conditional probability should be (^) λ (^2) +4^2 λ+.
4.15 Invariance of properties under transformations
(a) FALSE. For example, let P [X 0 = i] = 13 for i ∈ {− 1 , 0 , 1 } and let the sample paths all have the form... , − 1 , 0 , 1 , − 1 , 0 , 1 , − 1 , 0 , 1 ,.. .. Then X is a Markov process on
{− 1 , 0 , 1 } with one-step probability matrix P =
. But Y has sample
paths of the form... , 1 , 0 , 1 , 1 , 0 , 1 , 1 , 0 , 1 , 1 , 0 ,.. .. Thus, P [Yn+1 = 1|Yn = 1] = 0.5 and P [Yn+1 = 1|Yn = 1, Yn− 1 = 0] = 1. Since these probabilities are not equal, Y is not Markov. (The problem is that Yn has less information than Xn because squaring is not a one-to-one function.)
