Download Final Exam - 6 Problems on Random Processes with Solution | ECE 534 and more Exams Electrical and Electronics Engineering in PDF only on Docsity!
Solutions to Final Exam December, 2005
Problem 1(12 points) Let U = (Uk : k ∈ Z) such that the Uk are independent, and each is uniformly
distributed on the interval [0, 1]. Let X = (Xt : t ∈ R) denote the continuous time random process
obtained by linearly interpolating between the U ’s. Specifically, Xn = Un for any n ∈ Z, and Xt is
affine on each interval of the form [n, n + 1] for n ∈ Z.
(a) Sketch a sample path of U and a corresponding sample path of X.
(b) Let t ∈ R. Find and sketch the first order marginal density, fX, 1 (x, t). (Hint: Let n = btc and
a = t − n, so that t = n + a. Then Xt = (1 − a)Un + aUn+1. It’s helpful to consider the cases and
0. 5 < a < 1 separately. For brevity, you need only consider the case 0 ≤ a ≤ 0 .5.)
(c) Is the random process X WSS? Justify your answer.
(d) Find P {max 0 ≤t≤ 10 Xt ≤ 0. 5 }.
(a)
n n+1 t
X t
(b) Xt is the sum of two random variables, (1 − a)Ut, which is uniformly distributed on the interval [0, 1 − a], and aUn+1, which is uniformly distributed on the interval [0, a]. Thus, the density of Xt is the convolution of the densities of these two variables:
1
1!^1 a^1!^1 a
0 1 !a 0 a a 1!a
(^1) a
0
(c) CX (t, t) = a (^2) +(1−a) 2 12 for^ t^ =^ n^ +^ a. Since this depends on^ t,^ X^ is not WSS. (d) P {max 0 ≤t≤ 10 Xt ≤ 0. 5 } = P {Uk ≤ 0 .5 for 0 ≤ k ≤ 10 } = 0. 511.
Problem 2 (9 points) Let X = (Xt : t ∈ Z) be a real stationary Gaussian process with mean zero and
RX (t) = 1+^1 t 2. Answer the following unrelated questions.
(a) Is X a Markov process? Justify your anwer.
(b) Find E[X 3 |X 0 ] and express P {|X 3 − E[X 3 |X 0 ]| ≥ 10 } in terms of Q, the standard Gaussian com-
plementary cumulative distribution function.
(c) Describe the joint probability density of (X 0 , X 1 , X 2 )T^. You need not write it down in detail.
(a) No. All mean zero stationary, Gaussian Markov processes have autocorrelation functions of the form RX (t) = Aρ|t|, where A ≥ 0 and 0 ≤ ρ ≤ 1 for continuous time (or |ρ| ≤ 1 for discrete time). (b) E[X 3 |X 0 ] = Eb[X 3 |X 0 ] = R RXX^ (3)(0) X 0 = X 100. The error is Gaussian with mean zero and variance MSE = Var(X 3 ) − Var( X 100 ) = 1 − 0 .01 = 0. 99. So P {|X 3 − E[X 3 |X 0 ]| ≥ 10 } = 2Q( √^100. 99 ).
(c) The vector
X 0
X 1
X 2
A (^) has the N
A ,
1 2 1 1 1 2 5 1 2 1
A
A (^) distribution.
Problem 3 (12 points) Let U = (Uk : k ∈ Z) be a random process such that the variables Uk are
independent, identically distributed, with E[Uk] = μ and Var(Uk) = σ^2 , where μ 6 = 0 and σ^2 > 0.
Please keep in mind that μ 6 = 0.
Let X = (Xn : n ∈ Z) be defined by Xn =
k=0 Un−ka
k, for a constant a with 0 < a < 1.
(a) Is X stationary? Find the mean function μX and autocovariance function CX for X.
(b) Is X mean ergodic in the m.s. sense?
Let U be as before, and let Y = (Yn : n ∈ Z) be defined by Yn =
k=0 Un−kA
k, where A is a
random variable distributed on the interval (0, 0 .5) (the exact distribution is not specified), and A is
independent of the random process U.
(c) Is Y stationary? Find the mean function μY and autocovariance function CY for Y.
(d) Is Y mean ergodic in the m.s. sense?
(a) X is the result of passing the stationary process U through the linear time-invariant system with impulse response function h(k) = ak^ I{k≥ 0 }. Thus X is stationary. Since μU = μ and CU (n) = σ^2 I{n=0}, we find E[Xk ] = μ P k∈Z h(k) = μ 1 −a , and^ CX^ (n) =^ h^ ∗^ eh^ ∗^ CU^ (n) =^ σ (^2) h ∗ eh(n) = σ 2 P k∈Z h(n)h ∗(n − k) = σ^2 a|n| 1 −a^2.^ (Think of the case^ n^ ≥^ 0 and^ n <^0 separately.) (b) Yes, X is mean ergodic in the m.s. sense, because CX (n) → 0 as n → ∞. (c) This is similar to part (a), but now h is random: h(k) = Ak^ I{k≥ 0 }. Y is stationary for the same reason as X. μY = μE[ (^1) −^1 A ] and RY (n) = E[RX (n)|a=A] = μ^2 E[ (^) (1−^1 A) 2 ] + E[ σ 12 −AA|n 2 | ]. (d) For A fixed, the time average of Y is (^1) −μA , by parts (a) and (c). Since this time average is not a constant, Y is not mean ergodic in the m.s. sense.
Problem 4 (6 points) Consider a time-homogeneous, discrete-time Markov process X = (Xk : k ≥ 0)
with state space S = { 1 , 2 , 3 }, initial state X 0 = 3, and one-step transition probability matrix
P =
(a) Sketch the transition probability diagram and find the equilibrium probability distribution π =
(b) Identify a function f on S so that f (s) = a for two choices of s and f (s) = b for the third choice
of s, such that the process Y = (Yk : k ≥ 0) defined by Yk = f (Xk) is a Markov process, and give the
one-step transition probability matrix of Y. Briefly explain your answer.
(a)
(^1) 0. 2
2
3
0.10.
Solve πP = π, which can be written as
−π 1 + 0. 1 π 2 + 0. 2 π 3 = 0
- 8 π 1 − 0. 4 π 2 + 0. 8 π 3 = 0
- 2 π 1 + 0. 3 π 2 − π 3 = 0
Adding four times the first equation to the second
yields − 3. 2 π 1 + 1. 6 π 3 = 0, or π 3 = 2π 1. Substituting that into the first equation yields − 0. 6 π 1 + 0. 1 π 2 = 0, or π 2 = 6π 1. Thus, π = (π 1 , 6 π 2 , , 2 π 1 ). Since the probabilities must sum to one, π = ( 19 , 23 , 29 ). (b) Since p 1 , 2 = p 3 , 2 = 0.8, the probability the next state is 2 given that the current state is either 1 or 3 is 0.8. That is, if states 1 and 3 are grouped together, then given the process is in that group, the probability of jumping to state 2 is 0.8. Thus, we take f (1) = f (3) = a and f (2) = b for distinct values a and b. The one-step transition probability matrix
of Y is P =
Problem 5 (9 points) Let X be a mean zero, WSS random process with power spectral density
SX (ω) = ω 4 +5^1 ω 2 +.
(a) Find the positive type, minimum phase rational function S X+ such that SX (ω) = |S X+ (ω)|^2.
(b) Let T be a fixed known constant with T ≥ 0. Using the formula H = S^1 +
X
[
S+ X ejωT^
]
+, find
X̂ t+T |t,
the MMSE linear estimator of Xt+T given (Xs : s ≤ t). Be as explicit as possible. (Hint: Convert to
the time domain at the end. Check that your answer is correct in case T = 0 and in case T → ∞).
(c) Find the MSE for the optimal estimator of part (b).
(b) The optimal filter is given by H(ω) = (^) S+^1 X (ω)
S+ X (ω)ejωT^
+.^ The technique of partial fraction expansions yields that S+ X (ω) = (^) jω^1 +1 − (^) jω^1 +2 ↔ I{t≥ 0 }(e−t^ − e−^2 t). Thus S X+ (ω)ejωT^ ↔ I{t≥−T }(e−(t+T^ )^ − e−2(t+T^ )) so
S X+ (ω)ejωT^
I{t≥ 0 }(e−(t+T^ )^ − e−2(t+T^ )), so that
S X+ (ω)ejωT^
+ =^
jωe−+1T −^ e jω−+2^2 T.^ Therefore,^ H(ω) =^ e−T^ (jω^ + 2)^ −^ e−^2 T^ (jω^ + 1). Therefore, Xbt+T |t = Xt(2e−T^ − e−^2 T^ ) + X t′(e−T^ − e−^2 T^ ). (c) The MSE for the optimal predictor is RX (0) −
R ∞
−∞ |^
S X+ (ω)ejωT^
2 dω 2 π = RX (0) − R^ ∞ 0 ((e −(t+T ) (^) − e−2(t+T )) (^2) dt =
RX (0) − e−^2 T^ /2 + 2e−^3 T^ / 3 − e−^4 T^ /4. The constant RX (0) can be found by noting that the MSE is zero for T = 0, so that M M SE = (1 − e−^2 T^ )/ 2 − 2(1 − e−^3 T^ )/3 + (1 − e−^4 T^ )/4.
Problem 6 (12 points) (Note: This problem is about solving the same prediction problem solved
in the previous problem. The two problems will be graded separately, so your solutions should
be independent.) Suppose that W is a Gaussian white noise process with RW (τ ) = δ(τ ). Let
X be the stationary random process solving the following second order linear stochastic differen-
tial equation: X′′^ + 3X′^ + 2X = W , or equivalently, Z defined by Zt =
Xt
X t′
satisfies Z′^ =
Z +
W.
(a) Show that SX (ω) = ω 4 +5^1 ω 2 +.
(b) Find RX (τ ) for τ ∈ R. (Hint: Use a partial fraction expansion of SX .)
(c) The process X is mean square differentiable. Find RX′^ (τ ) and RXX′^ (τ ) for τ ∈ R. (Hint: Check
that RXX′ (0) = 0.)
(d) Explain why Z is a Markov process.
(e) Let T > 0, let t ∈ R, and let X̂ t+T |t be the MMSE estimator of Xt+T given (Xu : u ≤ t), which is
the same as the MMSE estimator of Xt+T given (Zu : u ≤ t), because (Xu : u ≤ t) and (Zu : u ≤ t)
are linearly equivalent. Explain why X̂ t+T |t is a linear combination of Xt and X′ t.
(f) Using parts (c) and (e), find X̂ t+T |t.
(a) In the frequency domain, the linear system is given by ((jω)^2 + 3jω + 2)bx = wb, or H(ω) = (^) (jω) (^2) +3^1 jω+2. Also, SW ≡ 1. Therefore, SX = |H(ω)|^2 = (^) (−ω (^2) +3jω+2)(^1 −ω (^2) − 3 jω+2) = (^) ω (^4) +5^1 ω (^2) +. (b) By the method of partial fraction expansion, SX (ω) = (^) (jω+1)(−jω+1)(^1 jω+2)(−jω+2) = (^) jω^1 /+1^6 + (^) −^1 jω/^6 +1 − (^) jω^1 /+2^12 − (^) −^1 jω/^12 +2 = 16 · (^) ω (^22) +1 − 121 · (^) ω (^24) +. Therefore RX (τ ) = e−| 6 τ^ |− e− 122 | τ^ |. (c) RX′ (τ ) = −R′′ X (τ ) = e−^23 | τ^ |− e−| 6 τ^ |, and RXX′ (τ ) = −R′ X (τ ) = (e −|τ |−e− 2 |τ |)sgn(τ )
(d) Fix a time t and view it as the present time. The process Z up to the present is determined by W up to the present. Given the process at the present time (i.e. given Zt), the future of Z is determined by Zt and the future of W. Since W is a white noise process, the past and future of W are independent. Thus, the past and present of Z are conditinally independent given the present. (e) The past of Z, namely (Zu : u ≤ t), is linearly equivalent to the past of X, namely (Xu : u ≤ t), because X′^ is casually determined by X. Also, Xt+T is a function of Zt+T. Therefore, the Markov property of Z yields that XT is conditionally independent of (Xu : u ≤ t) given Zt. (f) Thus, using the fact that Xt and X t′ are orthogonal and the formula for Eb,
X^ bt+T |t = Eb[Xt+T |
Xt X′ t
] = Cov Var(Xt(+XTt^ ,X) t)Xt + Cov(Xt+T^ ,X t′) Var(X′ t) X t′ =^ RX^ (T^ ) RX (0) Xt^ +^
RXX′ (T ) RX′ (0) X t′
=
e− 6 T − e− 122 T 121 Xt^ +^
(e−T^ − 6 e−^2 T^ ) 16 X t′ = (2e−T^ −^ e−^2 T^ )Xt + (e−T^ −^ e−^2 T^ )X′ t.
