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EE 5375/7375 Random Processes September 30, 2003
Homework #4 solutions
Problem 1. textbook problem 6. A discrete-time random process is defined by Xn = sn^ for n ≥ 0 where s is selected at random from the interval (0,1). (a) Sketch a sample path of the process. (b) Find the PDF of Xn. (c) Find the joint PDF of Xn and Xn+1. (d) Find the mean and autocovariance functions of Xn.
(a) The sample path will decrease geometrically by X 0 = 1, X 1 = s, X 2 = x^2 , and so on, as shown in Fig. 1. (b) The PDF of Xn at a specific n is
P (Xn ≤ x) = P (sn^ ≤ x) = P (s ≤ x^1 /n) = x^1 /n
The last step follows because s is uniformly distributed in (0,1). (c) By definition, the joint PDF is
P (Xn ≤ x, Xn+1 ≤ y) = P (sn^ ≤ x, sn+1^ ≤ y) = P (s ≤ x n^1 , x ≤ y n^1 + ) = P (s ≤ min(x
(^1) n , y n+1^1 )) = min(x n^1 , y n^1 + )
We use the minimum of x n^1 and y n^1 + because the joint event that s is less than both of them means that s must be less than the smaller of them. Again, the last step follows because s is uniformly distributed. (d) The mean of Xn depends on knowledge of the pdf of s which is fs(s) = 1 over the interval (0,1):
E(Xn) = E(sn) =
0
snfs(s)ds =
0
snds =
n + 1
The autocovariance function for lag k is defined as
CX (n, n + k) = E(XnXn+k) − E(Xn)E(Xn+k)
= E(snsn+k) −
n + 1
n + k + 1 = E(s^2 n+k) −
(n + 1)(n + k + 1)
=
0
s^2 n+kds −
(n + 1)(n + k + 1)
=
2 n + k + 1
(n + 1)(n + k + 1)
Problem 2. textbook problem 6. Find an expression for E(|Xt 2 − Xt 1 |^2 ) in terms of autocorrelation function.
The absolute value does not matter because it is squared.
E(|Xt 2 − Xt 1 |^2 ) = E((Xt 2 − Xt 1 )^2 ) = E(X t^22 − 2 Xt 1 Xt 2 + X t^21 ) = E(X t^22 ) − 2 E(Xt 1 Xt 2 ) + E(X t^21 ) = RX (t 2 , t 2 ) − 2 RX (t 1 , t 2 ) + RX (t 1 , t 1 )
1
where RX (t, s) is the autocorrelation function.
Problem 3. textbook problem 6.
Let Yn be the process that results when individual 1’s in a Bernoulli process are erased with probability α. Find the PMF of S n′, the counting process for Yn. Does Yn have independent and stationary increments?
Let In be a Bernoulli process where a 1 occurs with probability p and a 0 occurs with probability 1 − p. Whenever a 1 occurs, it is not erased with probability (1 − α) in the derived process Yn. The probability of a 1 in Yn is therefore (1 − α)p. The counting process S′ n gives the number of ones in the first n Bernoulli trials where each trial has probability (1 − α)p of success. Therefore S n′ is binomial with parameters (n, (1 − α)p) and has PMF
P (S′ n = x) =
n x
((1 − α)p)x(1 − (1 − α)p)n−x
Yn has independent and stationary increments because it is a Bernoulii process (sequence of independent Bernoulli random variables).
Problem 4. textbook problem 6.
Consider the following autoregressive processes:
Wn = 2Wn− 1 + Xn , W 0 = 0
Zn =
Zn− 1 + Xn , Z 0 = 0
(a) Flip a coin 10 times to obtain a realization of the Bernoulli process. Find the resulting realizations for Wn and Zn. What trends do the processes exhibit? Is the sample mean meaningful for either of these proceses? (b) Express Wn and Zn in terms of Xn, Xn− 1 ,... , X 1 and then find E(Wn) and E(Zn). Do these results agree with the trends observed in part (a)? (c) Does Wn or Zn have independent increments? Stationary increments?
(a) A realization of Xn, Wn, and Zn are shown in Fig. 2. Wn is approximately exponentially increasing by a factor of 2 in each step. Therefore, the sample mean does not have meaning for Wn. Zn is exponentially decreasing by a factor of 1/2 except when Xn is 1. The sample mean for Zn is roughly twice the sample mean of Xn.
(b) Consider Wn first.
Wn = 2Wn− 1 + Xn = 2(2Wn− 2 + Xn− 1 ) + Xn = 4(2Wn− 3 + Xn− 2 ) + 2Xn− 1 + Xn = 8(2Wn− 4 + Xn− 3 ) + 4Xn− 2 + 2Xn− 1 + Xn = 2n−^1 X 1 + · · · + 4Xn− 2 + 2Xn− 1 + Xn
E(Wn) = E(2n−^1 X 1 ) + · · · + E(4Xn− 2 ) + E(2Xn− 1 ) + E(Xn) = (2n−^1 + · · · + 4 + 2 + 1)E(X)
=
1 − 2 n 1 − 2
E(X) = (2n^ − 1)E(X)
Let Xn consist of an iid sequence of Poisson random variables with mean α. (a) Find the PMF of the sum process Sn. (Hint: use the probability generating function.) (b) Find the joint PMF of Sn and Sn+k.
(a) For a sum of random variables, the PGF of the sum is the product of individual PGFs. In this case,
GSn (z) = E(zSn^ ) = E(zX^1 +···+Xn^ ) = (eα(z−1))n^ = enα(z−1)
This PGF may be recognized as corresponding to a Poisson distribution with mean nα, i.e.,
P (Sn = k) =
(nα)k k!
e−nα
(b) We know that Sn is Poisson with mean nα. Also, the increment (Sn+k − Sn) consists of the sum of k iid Poisson random variables, which has the same distribution as Sk.
P (Sn = i, Sn+k = j) = P (Sn = i)P (Sn+k = j|Sn = i) = P (Sn = i)P (Sn+k − Sn = j − i) = P (Sn = i)P (Sk = j − i)
= (nα)i i!
e−nα^ (kα)j−i (j − i)!
e−kα
Problem 7. textbook problem 6. Let Xn be an iid sequence of zero-mean, unit-variance Gaussian random variables. (a) Find the PDF of Mn defined in Problem 5. (b) Find the joint PDF of Mn and Mn+k. (Hint: use the independent increments property of Sn.)
(a) Let us find the PDF using the characteristic function because the characteristic function of a sum of independent random variables is the product of individual characteristic functions.
φMn (ω) = E(ejω(X^1 +···+Xn)/n) = (E(ejωX/n))n
=
φX
( (^) ω n
))n
e−(^
ωn )^2 / 2 )n
= e−ω
(^2) / 2 n
This characteristic function implies that Mn is Gaussian with zero mean and variance 1/n, ie, the pdf is
fMn (x) =
n 2 π e−nx
(^2) / 2
(b) We note that Mn and Mn+k are related to Sn and Sn+k by
Mn =
n
Sn, Mn+k =
n + k
Sn+k
Thus the joint PDF of Mn and Mn+k can be expressed as
FMn,Mn+k (x, y) = P (Mn ≤ x, Mn+k ≤ y) = P (Sn ≤ nx, Sn+k ≤ (n + k)y) = FSn,Sn+k (nx, (n + k)y)
4
The joint pdf is found by differentiating with respect to x and y:
fMn,Mn+k (x, y) = n(n + k)fSn,Sn+k (nx, (n + k)y)
We note that Sn has independent increments and the increment (Sn+k − Sn) is the sum of k iid N (0, 1) random variables, so it is N (0, k). Hence the joint pdf can be rewritten as
fMn,Mn+k (x, y) = n(n + k)fSn,Sn+k (nx, (n + k)y) = n(n + k)fSn (nx)fSn+k |Sn ((n + k)y|nx) = n(n + k)fSn (nx)fSn+k −Sn ((n + k)y − nx)
= n(n + k)
2 πn
e−(nx)
(^2) / 2 n 1 √ 2 πk
e−((n+k)y−nx)
(^2) / 2 k
Problem 8. textbook problem 6. Consider the following moving average process:
Yn =
(Xn + Xn− 1 ) , X 0 = 0
(a) Is Yn a stationary random process if Xn is an iid process? (b) Is Yn a stationary random process if Xn is a stationary process?
(a) Let us look at the joint PDF of three arbitrary points of Yn at n 1 < n 2 < n 3 : P (Yn 1 = y 1 , Yn 2 = y 2 , Yn 3 = y 3 )
= P
(Xn 1 + Xn 1 − 1 ) = y 1 ,
(Xn 2 + Xn 2 − 1 ) = y 2 ,
(Xn 3 + Xn 3 − 1 ) = y 3
= P
(X 2 + X 1 ) = y 1 ,
(Xn 2 −n 1 +2 + Xn 2 −n 1 +1) = y 2 ,
(Xn 3 −n 2 +2 + Xn 3 −n 1 +1) = y 3
The last step is true if Xn is stationary. Next, let us look at the joint PDF at the three points shifted by an arbitrary T : P (Yn 1 +T = y 1 , Yn 2 +T = y 2 , Yn 3 +T = y 3 )
= P
(Xn 1 +T + Xn 1 +T − 1 ) = y 1 ,
(Xn 2 +T + Xn 2 +T − 1 ) = y 2 ,
(Xn 3 +T + Xn 3 +T − 1 ) = y 3
= P
(X 2 + X 1 ) = y 1 ,
(Xn 2 −n 1 +2 + Xn 2 −n 1 +1) = y 2 ,
(Xn 3 −n 2 +2 + Xn 3 −n 1 +1) = y 3
Again, the last step is true if Xn is stationary, which includes the case if Xn is an iid process. Since the time shift T does not effect the joint PDF, Yn is stationary. (b) The steps in part (a) are all true if Xn is stationary, so Yn is stationary if Xn is stationary.
Problem 9. textbook problem 6.62 (optional for EE 5375) A ternary information source produces an iid, equiprobable sequence of symbols from the alphabet {a, b, c}. Suppose that these three symbols are encoded into the respective binary codewords 00, 01, 10. Let Bn be the sequence of binary symbols that result from encoding the ternary symbols. (a) Find the joint PMF of Bn and Bn+1. Is Bn stationary? Cyclostationary? (b) Find the mean and autocorrelation functions of Bn. Is Bn wide-sense stationary? Wide-sense cyclostationary? (c) If Bn is cyclostationary, find the joint PMF, mean, and autocorrelation functions of the randomly phase-shifted version of Bn as defined by equation (6.65): Xs(t) = X(t + θ) , θ uniform in [0, T ]
5
Now the joint PMF for the randomly shifted process is
P (Bns = 0, Bns+1 = 0) = P (Bns = 0, Bsn+1 = 0|n even)
- P (Bsn = 0, Bns+1 = 0|n odd)
Similarly,
P (Bns = 0, Bns+1 = 1) =
P (Bns = 1, Bns+1 = 0) =
P (Bns = 1, Bns+1 = 1) =
The mean is
E(Bns) = 0 ·
The autocorrelation function is
R(n, n + j) =
P (Bns = 1) = 13 if j = 0 P (Bns = 1, Bsn+1 = 1) = 181 if j = 1 1 3 ·^
1 3 =^
1 9 if^ j >^1
Problem 10. Matlab (optional for EE 5375) Start Matlab and use the uniform random number generator rand to generate 1000 random samples by typing:
u = rand(1000,1); Create a sample Bernoulli sequence by typing: w = 0.5 >= u; This sets elements of w equal to 1 if u > 0 .5 or 0 if u < 0 .5. This sequence can be visualized by plotting: stem(w) Scale the Bernoulli sequence to a different range by typing: s = 0.1; w = s(2w - 1.0); Generate a new sequence x by typing: x = cumsum(w); and look at the process by plotting: plot(1:1000,x) What kind of random process does this look like?
It should look like Brownian motion.
- Fig. 1 for Problem
- Fig. 2 for Problem
