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ECE 534 RANDOM PROCESSES FALL 2008
SOLUTIONS TO PROBLEM SET 7
1 Some linear transformations of some random processes (a) Yes, X is the result of passing the stationary process U through the linear time-invariant system with impulse response function h(k) = akI{k≥ 0 }. Thus X is stationary. Since μU = μ and
CU (n) = σ^2 I{n=0}, we find E[Xk] = μ
k∈Z h(k) =^
μ 1 −a , and^ CX^ (n) =^ h^ ∗^ ˜h ∗ CU (n) = σ^2 h ∗ ˜h(n) =
σ^2
k∈Z h(k)h
∗(k − n) = σ^2 a|n| 1 −a^2.^ (Think of the case^ n^ ≥^ 0 and^ n <^ 0 separately.) (b) Yes. Fix a time k and view it as the present time. The past of X is determined by (Uj : j ≤ k), and the future of X is determined by Xk and (Uj : j ≥ k + 1), through the update equations: Xn+1 = aXn + Un+1 for n ≥ k. Since (Uj : j ≤ k) and (Uj : j ≥ k + 1) are independent, the Markov property follows. (c) Yes, X is mean ergodic in the m.s. sense, because CX (n) → 0 as n → ∞. (d) Yes. This is similar to part (a), but now h is random: h(k) = AkI{k≥ 0 }. Y is stationary for the same reason X is stationary. The mean is μY = μE[ (^1) −^1 A ]. To find CY we first find RY as follows:
RY (n) = E
[
RX (n)
a=A
]
= μ^2 E
[
(1 − A)^2
]
+ E
[
σ^2 A|n| 1 − A^2
]
Then, CY (n) = RY (n) − μ^2 Y = μ^2 Var
1 1 −A
+ E
[
σ^2 A|n| 1 −A^2
]
(e) No. Fix a time k and view it as the present time. The past of X would yield a good estimate of the value of A, which cannot be inferred from Xk alone. Thus, the future of X, which depends on both Xk and A, is not conditionally independent of the past given Xk. (f) No. 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. Another justification is to note that as n → ∞, CX (n) → μ^2 Var( (^1) −^1 A ) 6 = 0.
2 Causal prediction of a Poisson process (a) The random variable S, which is equal to the time of the third count, T 3 , is the sum of three independent Exp(λ) random variables, and so has the Gamma(3, λ) density. Thus, fS (t) = λ^3 t^2 e−λtI{t≥ 0 }
(b) Observe that N 20 = N 10 + (N 20 − N 10 ). The term N 10 is a linear function of (Ns : 0 ≤ s ≤ 10) and the increment N 20 − N 10 has mean 10λ and is independent of (Ns : 0 ≤ s ≤ 10). Thus, Ê [N 20 |Ns, 0 ≤ s ≤ 10] = N 10 + 10λ.
3 On an M/D/infinity system (a) Since N (t − 1 , t] is a Poisson random variable with mean λ, E[Xt] = λ. If |s − t| ≥ 1 then Xs is independent of Xt, because the intervals (s− 1 , s] and (t− 1 , t] don’t overlap. If |s−t| < 1 then these two intervals have an overlap of length 1 − |s − t|. For example, if 0 ≤ t − s < 1, then (s − 1 , s] = (s − 1 , t − 1] ∪ (t − 1 , s] and (t − 1 , t] = (t − 1 , s] ∪ (s, t]. Therefore, Xs = N (s − 1 , t − 1] + N (t − 1 , s], and Xt = N (t − 1 , s] + N (s, t]. Since the three random variables N (s − 1 , t − 1], N (t − 1 , s] and
N (s, t] are independent,
CX (s, t) = Cov(N (s − 1 , t − 1] + N (t − 1 , s], N (t − 1 , s] + N (s, t]) = Var(N (t − 1 , s]) = λ(s − t + 1) = λ(1 − |t − s|)
A similar computation shows CX (s, t) = λ(1 − |t − s|) if 0 ≤ s − t < 1 as well. Thus, X is wide sense stationary, and CX (τ ) = λ(1 − |τ |)+ for all τ. (b) Yes and yes. One proof that X is stationary is that it can be viewed as the output of a time- invariant system, where the input is the stationary process N ′, as explained in the statement of the next problem. Here is another proof. Let n ≥ 1 and fix t 1 ,... , tn. Let u 0 < u 1 < · · · < um, be such that ti − 1 and ti are both in {u 0 ,... , um} for 1 ≤ i ≤ n. Then there is an m × n matrix G, with each entry equal 0 or 1, such that
(Xt 1 +s,... , Xtn+s) = (N (u 0 + s, u 1 + s],... , N (um− 1 + s, um + s]) G (1)
for all s. The random vector in the right-hand side of (1) consists of independent Poisson random variables, with respective means u 1 − u 0 , · · · , um − um− 1. Note that such distribution doesn’t de- pend on s. Therefore, the distribution of the left-hand side of (1) also doesn’t depend on s. So X is stationary. (c) No, X is not Markov. For example, think of t = 1 as the present time. Given X 1 = 1, we know there is one customer in the system at time 1. But if we also knew the past of X, we would know when that customer arrived, so that we would know exactly when that customer will depart. Therefore, the past tells us more about the future than the present alone. A more rigor- ous answer would be to give an explicit example violating the conditional independence property. Here is one, based on the above intuition. Clearly, P [X 0. 5 = 0|X 0 = 1, X− 0. 25 = 0] = 0, whereas P [X 0. 5 = 0|X 0 = 1] > 0. Since these are not equal, X is not a Markov process. For complete- ness, we can calculate P [X 0. 5 = 0|X 0 = 1] as follows. Given X 0 = 1, or equivalently, given that there is exactly one arrival during (− 1 , 0], the distribution of the time of that arrival is uniformly distributed over (− 1 , 0]. Thus, given X 0 = 1, the departure time of the customer in the system at time zero is uniformly distributed over the interval (0, 1]. Hence, it departs by time 0.5 with probability 0.5. Also, the probability there are no arrivals during (0, 0 .5] is exp(−λ/2), and that event is independent of X 0. So, P [X 0. 5 = 0|X 0 = 1] = (0.5) exp(−λ/2). (d) The system is empty during [0, 1] if and only if there are no arrivals during (− 1 , 1]. So P {Xt = 0 for t ∈ [0, 1]} = P {N (− 1 , 1] = 0} = exp(− 2 λ). (e) The event is equivalent to there being at least one customer in the queue at time zero, and a new customer arriving after time zero before the departure of the last customer arriving before time zero. Let T be the time of the first arrival after time zero, and let −S be the time of the last arrival before time zero. Then S and T are independent, exponentially distributed random variables, with param- eter λ. Then P {Xt > 0 for t ∈ [0, 1]} = P {S + T ≤ 1 } =
0
∫ (^1) −t 0 λ
(^2) e−λ(s+t)dsdt = 1 − (1 + λ)e−λ.
4 Filtering Poisson white noise (a) Since μN ′^ = λ, μX = λ
−∞ h(t)dt.^ Also,^ CX^ =^ h^ ∗^ ˜h ∗ CN ′ (^) = λh ∗ ˜h. (In particular, if
h(t) = I{ 0 ≤t< 1 }, then CX (τ ) = λ(1 − |τ |)+, as already found in Problem 3.) (b) In the special case, in between arrival times of N , X decreases exponentially, following the equation X′^ = −X. At each arrival time of N , X has an upward jump of size one. Formally, we can write, X′^ = −X + N ′. For a fixed time to, which we think of as the present time, the process after
fall within linear regions (happens when 0 ≤ U ≤ 0 .5). We can try reconstructing X using both cases. With probability one, only one of the cases will yield a reconstruction with change points having spacing one. That must be the correct reconstruction of X. The algorithm is illustrated in Figure 1. Figure 1(a) shows a sample path of B and a corresponding sample path of X, for
1
0 1 2 3 4 5 6 (b)
(a)
(c)
(d)
0 2 3 4 5 6
1
0 2 3 4 5 6
1
0 2 3 4 5 6
Figure 1: Nonlinear reconstruction of a signal from samples
U = 0. 75. Thus, the breakpoints of X are at times of the form n + 0.75 for integers n. Figure 1(b) shows the corresponding samples, taken at integer multiples of T = 0. 5. Figure 1(c) shows the result of connecting pairs of the form (Xn, Xn+0. 5 ), and Figure 1(d) shows the result of connecting pairs of the form (Xn+0. 5 , Xn+1). Of these two, only Figure 1(c) yields breakpoints with unit spacing. Thus, the dashed lines in Figure 1(c) are connected to reconstruct X.
7 Estimation of a process with raised cosine spectrum (a)-(b) The situation is a special case of Example 9.1.1 of the notes.
H(ω) =
SX (ω) SX (ω) + N 2 o
1 + cos( πωωo ) 1 + cos( πωωo ) + No
I{|ω|≤ωo}
σ e^2 =
−∞
SX (ω)SN (ω) SX (ω) + SN (ω)
dω 2 π
∫ (^) ωo
−ωo
No 2
1 + cos( πωωo ) 1 + No + cos( πωωo )
dω 2 π
The graph of H is similar to the graph of SX itself, although 0 ≤ H(ω) < 1 for all ω. (c) As No decreases to zero, H converges up to the rectangle function I{|ω|≤ωo} and σ^2 e → 0. One way to see this is that the term in brackets in the second integral expression for σ^2 e is less than one, so in general, σ^2 e ≤ Noωo. Basically, as No → 0 , the noise becomes negligible so that H becomes the all-pass filter in the band [−ωo, ωo]. (d) As No increases to infinity, H decreases to zero, and σ^2 e increases to
−∞ SX^ (ω)^
dω 2 π =^
ωo 2 π ,^ be- cause the integrand in the expressions for σ^2 e converge up to SX (ω). In this limit, the observations become useless so the optimal estimator converges to zero, and the MSE converges to E[|Xt|^2 ].
