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ECE 534 RANDOM PROCESSES FALL 2008
SOLUTIONS TO PROBLEM SET 6
1 A random process created by interpolation (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 } = (0.5)^11.
2 A stationary Gaussian process (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 ] = Ê [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) RX′^ (τ ) = −R′′ X (τ ) = 2 −^6 τ^ 2 (1+τ 2 )^3.^ In particular, since^ −R
′′ X exists and is continuous,^ X^ is continu- ously differentiable in the m.s. sense. (d) The vector has a joint Gaussian distribution because X is a Gaussian process and differ- entiation is a linear operation. Cov(Xτ , X 0 ′) = RXX′^ (τ ) = −R′ X (τ ) = (^) (1+^2 ττ 2 ) 2. In particular,
Cov(X 0 , X 0 ′) = 0 and Cov(X 1 , X 0 ′) = 24 = 0. 5. Also, Var(X 0 ′) = RX′ (0) = 2. So (X 0 , X 0 ′, X 1 )T^ has
the N
(^) distribution.
3 A state space reduction preserving the Markov property (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 1 , , 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 =
4 Integral of a Brownian bridge (a) The integral over the finite interval [0, 1] exists in the m.s. sense because B is m.s. continuous, or equivalently, RB is continuous. (b) X and W 1 are jointly Gaussian because they are linear functions of the Gaussian process W = (Wt : t ≥ 0). Also, E[X] =
0 E[Bt]dt^ = 0 and^ E[X
2 ] = E[∫^1
0 Bsds^
0 Btdt] =^
0
0 RB^ (s, t)dsdt^ = 2
0
∫ (^) t 0 RB^ (s, t)dsdt^ = 2^
0
∫ (^) t 0 s(1^ −^ t)dsdt^ =^
0 t
(^2) (1 − t)dt = 1 3 −^
1 4 =^
1
- By a previous problem, the whole process B is independent of W 1. (Since the variables are jointly Gaussian, this amounts to the fact Cov(Bt, W 1 ) = 0 for all t.) Therefore, Cov(X, W 1 ) = 0 too, or equivalently (since X and W 1 are jointly Gaussian) X is independent of W 1. To summarize, X and W 1 are independent random variables, with X being N (0, 121 ) and W 1 being N (0, 1).
5 An infinitely differentiable process (a) We show the following statement by induction on n: the nth^ derivative of RX , R(n)(τ ), has the form Hn(τ )e−τ^2 /^2 , where Hn is a polynomial of degree n. The base case, for n = 0, is trivially true,
with H 0 (τ ) = 1. For the induction step, suppose R( Xn )(τ ) = Hn(τ )e−τ^ (^2) / 2 , where Hn is a polynomial of degree n. Then
R X(n +1)(τ ) = (Hn(τ )e−τ^ (^2) / 2 )′^ = (H n′(τ ) − τ Hn(τ ))e−τ^ (^2) / 2 = Hn+1(τ )e−τ^ (^2) / 2 ,
where Hn+1(τ ) = H′ n(τ )−τ Hn(τ ), and indeed Hn+1 is a polynomial of degree n+1. The statement is thus true for all n by induction. Now RX(n) (τ ) = (−1)nR(2 X n)= (−1)nH 2 n(τ )e−τ^ (^2) / 2
. By induction on n, X(n)^ exists for all n. (b) Since limτ →∞ RX(n) (τ ) = 0, X(n)^ is mean ergodic in the m.s. sense, for any n ≥ 0.
6 KL expansion for derivative process (a) Since φ′ n(t) = (2πjn)φn(t), the derivative of each φn is a constant times φn itself. Therefore, the equation given in the problem statement leads to: X′(t) =
n(X, φn)φ ′ n(t) = ∑ n[(2πjn)(X, φn)]φn(t),
