Download Problem Set 5 Solutions - Random Processes | ECE 534 and more Assignments Electrical and Electronics Engineering in PDF only on Docsity!
ECE 534 RANDOM PROCESSES SPRING 2008
SOLUTIONS TO PROBLEM SET 5
5.1. Calculus for a simple Gaussian random process (a) Differentiation with respect to t would suggest that the m.s. derivative of X is given by X′ t = B + 2Ct. Direct verification: E[( Xt+h h− Xt− (B + 2Ct))^2 ] = E[h^2 C^2 ] = h^2 → 0 as h → 0. (b)
0 Xsds^ =^ A^ +^ B/2 +^ C/3, which is a^ N^ (0, σ
(^2) ) random variable with σ (^2) = 1 + 1 4 +^
1 9.^ So P [
0 Xsds^ ≥^ 1] =^ Q(1/σ).
5.5. Differentiation of the square of a Gaussian random process (a) Since An → A, the corollary in Section 2.2 of the notes implies that E[AAn] → E[A^2 ] and E[A^2 n] → E[A^2 ]. Since A and the An are jointly Gaussian, the hint and the statement just proved imply that:
E[A^4 ] = 3 E[A^2 ]^2 E[A^4 n] = 3 E[A^2 n]^2 → 3 E[A^2 ]^2 E[A^2 nA^2 ] = 2 E[AAn]^2 + E[A^2 ]E[A^2 n] → 3 E[A^2 ]^2
Therefore E[(A^2 n − A^2 )^2 ] = E[A^4 n] − 2 E[A^2 nA^2 ] + E[A^4 ] → 0. (b) By the assumptions and part (a), A^2 n → A^2 m.s., B^2 n → B^2 m.s., and (An + Bn)^2 → (A + B)^2 m.s.. Therefore, (An + Bn)^2 − A^2 n − B^2 n → (A + B)^2 − A^2 − B^2 m.s., which is equivalent to what is to be proved. (c) Fix any t. Then for h 6 = 0, Yt+h h− Yt = Xt+h h− Xt(Xt+h + Xt) = AhBh, where Ah → X′(t) m.s. and Bh → 2 X(t) m.s. as h → 0. Therefore, by part (b), Yt+h h− Yt= AhBh → 2 X t′Xt m.s. That is, if X is mean zero, Gaussian and m.s. differentiable, then X^2 is also m.s. differentiable, and, just as for deterministic functions, (X^2 )′^ = 2XX′.
5.7. Fundamental theorem of calculus for m.s. calculus Fix t > 0. The question is whether Ys s−−Yt t→ Xt m.s. as s → t. Now Ys s−−Yt t−Xt = (^) s^1 −t
∫ (^) s t (Xu^ −Xt)du, so
E
[(
Ys − Yt s − t − Xt
) 2 ]
s − t
E
[(∫
s
t
(Xu − Xt)du
) 2 ]
s − t
E
[∫ (^) s
t
(Xu − Xt)du
∫ (^) s
t
(Xv − Xt)dv
]
s − t
) 2 ∫ (^) t
s
∫ (^) t
s
RX (u, v) − RX (t, v) − RX (u, t) + RX (t, t)dudv
Let > 0. Since RX is continuous, there exists δ > 0 so that |RX (a, b) − RX (t, t)| ≤ /3 whenever |a − t| < δ and |b − t| < δ. So if |s − t| < δ and u, v are between s and t, then
|RX (u, v) − RX (t, v) − RX (u, t) + RX (t, t)| ≤ .
So if |s − t| < then
E
[(
Ys − Yt s − t
− Xt
) 2 ]
Thus Ys s−−Yt t→ Xt m.s.
5.11. An integrated Poisson process (a) The sample paths of Y are piecewise linear. The initial slope is zero, and the slope of Y increases by one at each jump of N. (b) EYt =
∫ (^) t 0 ENsds^ =^ λ^
∫ (^) t 0 sds^ =^ λt
(c) Using the fact CN (u, v) = λ(u ∧ v), yields Var(Yt) =
∫ (^) t 0
∫ (^) t 0 CN^ (u, v)dudv^ = 2λ^
∫ (^) t 0
∫ (^) v 0 ududv^ = λ
∫ (^) t 0 v
(^2) dv = λt^3
(d) Note that if Nt/ 2 ≥ 2, then Yt ≥ t. So P [Yt < t] ≤ P [Nt/ 2 ≤ 1] = (1 + λt 2 ) exp(− λt 2 ) → 0 as t → ∞.
5.15. A random process which changes at a random time (a)
Figure 1: Sample path for X
(b)
FX, 1 (c, t) = P [Xt ≤ c] = P [Xt ≤ c|U ≤ t]P [U ≤ t] + P [Xt ≤ c|U > t]P [U > t] = P [Yt ≤ c]FU (t) + P [Zy ≤ c](1 − FU (t)) = FY, 1 (c, t)FU (t) + FZ, 1 (c, t)(1 − FU (t)).
That is, for each t fixed, the distribution of Xt is a mixture of the distributions of Yt and Zt. However, for this problem, Yt and Zt both have the N (0, 1) distribution, and hence, Xt also has the N (0, 1) distribution for any t. (c) By part (b), E[Xt] = 0 for all t. Suppose s ≤ t. Then
RX (s, t) = E[XsXt]
= E[XsXt|U ≤ s]P [U ≤ s] + E[XsXt|s < U ≤ t]P [s < U ≤ t] + E[XsXt|U > t]P [U > t] = E[ZsZt|U ≤ s]FU (s) + E[YsZt|s < U ≤ t](FU (t) − FU (s)) + E[YsYt|U > t](1 − FU (t)) = E[ZsZt]FU (s) + E[YsZt](FU (t) − FU (s)) + E[YsYt](1 − FU (t)) = RZ (s, t)FU (s) + RY Z (s, t)(FU (t) − FU (s)) + RY (s, t)(1 − FU (t)) = e−|s−t|FU (s) + 0 + e−|s−t|(1 − FU (t)) = e−|s−t|(FU (s) + 1 − FU (t))
In general, RX (s, t) = e−|s−t|(FU (s ∧ t) + 1 − FU (s ∨ t)). (d) X is m.s. continuous if and only if RX is continuous, which, in view of the answer to part (c),
(c) As t → ∞, μX (t) → 0 and CX (t + τ, t) = σ 2 2 (e
−|τ | (^) − e− 2 t−τ (^) ) → σ^2 2 e
−|τ |.
- From Exam 2, Spring 2006
(a) If a random process {Xt} is such that ∂ (^2) RX ∂s∂t (s, t) exists and is continuous, then it is m.s. differentiable. Now ∂^2 RX ∂s∂t
(s, t) = 2 + 4st exists and is continuous
Thus X is m.s. differentiable.
(b) Using the formulas derived in class:
μY = μ′ X = 0 and RY (t, s) =
∂^2 RX
∂s∂t
(s, t) = 2 + 4st
The derivative is obtained by taking limit of the expression Xt+ − Xt as → 0. This is a linear combination of jointly Gaussian random variables, which is Gaussian for every . The limit Yt is therefore Gaussian. Furthermore, in a similar manner we can conclude that all finite joint distributions of the process {Yt} are Gaussian. Therefore the process {Yt} is Gaussian and is hence characterized by its mean and autocorrelation function.
- From Exam 2, Spring 2006 (a) The mean is obviously 0. The covariance is:
RY (k) = CY (k) = CY (k − 0) = E[(Xk + A)(X 0 + A)] = E[A^2 ] + α|k|^ = 1 + α|k|
Since the process is WSS and the limit exists but does not equal 0, property (b) from proposition 5.4.1, p.115 of the notes can be used to conclude that {Xk} is not mean ergodic.
(b) Again the mean is 0, since A and X are independent. And
RZ (k) = CZ (k) = E(A^2 )RX (k) = α|k|^ → 0 , for k → ∞
Therefore {Zk} is mean ergodic.
