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ECE 534 Spring 2008
April 21, 2008
Solutions to Exam 2
1. (7 pts) Consider the WSS process {Xt : t ∈ R} with autocorrelation function:
RX (τ ) = e−τ^
2
(a) (1 pt) Determine whether or not {Xt} is m.s. continuous.
Ans: RX (τ ) is continuous at τ = 0 and so {Xt} is m.s. continuous.
(b) (3 pts) Determine whether or not {Xt} m.s. differentiable. If you conclude that it is m.s.
differentiable, find the mean and autocorrelation function of the derivative process.
Ans: RX (τ ) is twice differentiable for all τ ∈ R, with
R′′ X (τ ) =
d dτ
[
− 2 τ e−τ^
2 ]
= − 2 e−τ^
2 [
1 − 2 τ 2
]
which is continuous. Thus {Xt} is m.s. differentiable, with
μX′ = 0 RX′ (τ ) = −R′′ X (τ ) = 2e−τ^
2 [
1 − 2 τ 2
]
(c) (2 pts) Find the autocovariance function CX (τ ).
Ans: Since RX (τ ) → 0 as τ → ∞, μX = 0, and therefore CX (τ ) = RX (τ )
(d) (1 pts) Determine whether or not {Xt} is mean ergodic in the m.s. sense.
Ans: Since CX (τ ) → 0 as τ → ∞, {Xt} is mean ergodic in the m.s. sense.
2. (10 pts) Let {Nt : t ≥ 0 } be a Poisson process with parameter λ = 1.
(a) (5 pts) Find P(N 2 ≥ 2 |N 3 = 3).
Ans: Conditioned on N 3 = 3, the times of the 3 jumps are independent and uniformly distributed on [0, 3]. The event that N 2 ≥ 2 is equivalent to at least two of the three jumps being in the interval [0, 2]. Thus
P(N 2 ≥ 2 |N 3 = 3) = P{all three jumps are in [0,2]} + P{exactly two of the three jumps are in [0,2]}
=
An alternative (longer) way to compute this is to to use the independent increment property as follows:
P(N 2 ≥ 2 , N 3 = 3) = P(N 2 = 2, N 3 = 3) + P(N 2 = 3, N 3 = 3) = P(N 2 = 2, N 3 − N 2 = 1) + P(N 2 = 3, N 3 − N 2 = 0)
= 22 e−^2 2!
1 e−^1 1!
23 e−^2 3!
1 e−^1 0!
= (2 + 4/3)e−^3 = 10 e−^3 3 and P(N 3 = 3) =
33 e−^3 3!
9 e−^3 2 Thus P(N 2 ≥ 2 |N 3 = 3) =
P(N 2 ≥ 2 , N 3 = 3)
P(N 3 = 3)
(b) (3 pts) Find P(N 3 = 3|N 2 ≥ 2).
Ans: P(N 2 ≥ 2) = 1 − P(N 2 ≤ 1) = 1 − (2e−^2 + 1e−^2 ) = 1 − 3 e−^2 Using Bayes rule
P(N 3 = 3|N 2 ≥ 2) =
P(N 2 ≥ 2 |N 3 = 3)P(N 3 = 3)
P(N 2 ≥ 2)
9 e−^3 2
1 − 3 e−^2
10 e−^3 3(1 − 3 e−^2 )
(c) (2 pts) Let Yt =
Nt. Determine whether or not {Yt} is a Markov process?
Ans: We know that {Nt} is Markov. For each t, Nt and Yt are functions of each other. Thus {Yt} inherits the Markov property from {Nt}.
3. (12 pts) Let {Wt : t ≥ 0 } be a Brownian motion with parameter σ^2 = 1.
(a) (2 pts) Find E[W 4 |W 23 = 27]
Ans: Since the cube function is one-one over the real line, E[W 4 |W 23 = 27] = E[W 4 |W 2 = 3]. Now by the Martingale property of {Wt}, E[W 4 |W 2 ] = W 2. Thus
E[W 4 |W 23 = 27] = E[W 4 |W 2 = 3] = 3
You could also have used the independent increment property of {Wt} to arrive at this result.
(b) (4 pts) Find E[W 23 |W 1 = 2].
Ans: Let (W 2 − W 1 ) = X. Then X ∼ N (0, 1) and X is independent of W 1. Thus
E[W 23 |W 1 = 2] = E[(X + W 1 )^3 |W 1 = 2] = E[(X + 2)^3 ] = E[X^3 + 6X^2 + 12X + 8] = 0 + 6 + 0 + 8 = 14
(c) (6 pts) Now suppose we define the random variable Y via the m.s. integral
Y =
0
t Wtdt
Find P{Y > 1 }.
Ans: Since RW (s, t) = min(s, t) is continuous on [0, 1]^2 , the m.s. integral exists. Furthermore, Y is Gaussian with zero mean and variance
E[Y 2 ] =
0
0
ts min(s, t)dsdt = 2
t=
∫ (^) t
s=
ts^2 dsdt (by symmetry in (s, t))
0
t^4 3
dt =
and P{Y > 1 } = Q
= Q
4. (11 pts) Let {Xk : k = 1, 2 ,.. .} be a discrete-time random process with components that are i.i.d.
Gaussian random variables with mean 1 and variance 1, i.e., Xk ∼ N (1, 1). Now define a new
discrete-time random process {Yk : k = 1, 2 ,.. .} by
Yk =
∏^ k
i=
Xi, k = 1, 2 ,...
