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Midterm 2 Solution
ECE534 Spring 2009 There are a total of five problems. Apr. 16, 7:00-8:30 pm Only one sheet of handwritten note is allowed. Each problem is worth 20 points
- Let {Wk} be a sequence of i.i.d. random variables which take the values +1 or −1 with equal probability. Define the discrete-time Markov chain X as follows:
Xk+1 =
Xk + Wk, if − 2 ≤ Xk + Wk ≤ 2 2 , if Xk + Wk > 2 − 2 , if Xk + Wk < − 2 ,
with X 0 = 0. Thus, X increases or decreases by 1 at each time instant but is not allowed to go above 2 or below −2.
(a) Write down the one-step transition probability matrix of X.
(b) What is the steady-state distribution of this Markov chain?
Solution (a) Let the state space be S = {− 2 , − 1 , 0 , 1 , 2 }. Then the one-step transition probability matrix is
P =
(b) The steady-state distribution ∑ π = (0.2 0.2 0.2 0.2 0.2) can be calculated using πP = π and 5 i=1 π(i) = 1.
- Let N be a Poisson process of rate 1. Write down an expression for P (N 1 = 2, N 2 = 5), simpli- fying it as much as possible.
Solution P (N 1 = 2, N 2 = 5) = P (N 1 − N 0 = 2, N 2 − N 1 = 3) = P (N 1 − N 0 = 2)P (N 2 − N 1 = 3) = e 12 −^2.
- Consider a bag which contains one white ball and three black balls. At each time instant, the following operations are performed:
- A ball is picked at random from the bag.
- If the randomly picked ball is white, then it is removed along with a black ball from the bag.
- If the randomly picked ball is black, then it is put back in the bag along with an extra black ball and a white ball.
The process is said to terminate when the number of white balls in the bag is either zero or three. Find the expected time for the process to terminate. Hint: Let Ti be the expected time for the process to terminate starting with i white balls.
Solution Note that no matter which ball is picked, the number of black balls in the bag will always be 2 more than the number of white balls. Let i denote the number of white balls in the bag, then the probability of picking a white (or black) ball is (^) i+(ii+2) = (^) 2(ii+1) (or (^) 2(i+2i+1) ). Let T denote the number of time steps for the process to terminate, then Ti = E[T |X 0 = i] by the hint. Obviously, T 0 = T 3 = 0. By conditioning on all possible states at X 1 , we can get
T 1 = E[T |X 0 = 1, X 1 = 0]P (X 1 = 0|X 0 = 1)
- E[T |X 0 = 1, X 1 = 2]P (X 1 = 2|X 0 = 1)
=
4 (1 +^ T^0 ) +
4 (1 +^ T^2 ) = 1 +
4 T^2.
Similarly, we can get T 2 = 1 + 13 T 1 using T 3 = 0. After solving these equations, we get T 1 = 73 and T 2 = 169. T 1 is the expected time that is asked to calculate.
- Let W be a Brownian motion with σ^2 = 1. Compute P (W 5 ≤ 3 |(W 2 )^3 = − 1 , W 1 = 1). Express your answer in terms of the Q-function where Q(x) is the probability that a Gaussian random variable with mean 0 and variance 1 is greater than x.
Solution Being given (W 2 )^3 = −1 and W 1 = 1 is equivalent to being given W 2 = −1 and W 1 = 1 (and is also equivalent to being given W 2 − W 1 = −2 and W 1 − W 0 = 1). As the increment W 5 − W 2 is Gaussian N (0, 3) and is independent of W 2 − W 1 and W 1 − W 0 ,
P (W 5 ≤ 3 |(W 2 )^3 = − 1 , W 1 = 1) = P (W 5 ≤ 3 |W 2 − W 1 = − 2 , W 1 − W 0 = 1) = P (W 5 − W 2 ≤ 4 |W 2 − W 1 = − 2 , W 1 − W 0 = 1)
= P (W 5 − W 2 ≤ 4) = Φ( √^4 3
) = 1 − Q( √^4
- Let X be a continuous-time WSS random process with mean 2 and autocorrelation function RX (τ ) = 3 + e−|τ^ |. The process Y is defined by the stochastic differential equation
Y (^) t′ = −Yt + Xt; Y 0 = 1.
(a) Find the mean of Yt.
(b) Find the autocorrelation function RY.
Solution It can be verified that Yt = Y 0 e−t^ +
∫ (^) t 0 e
−(t−v)Xvdv is the solution to the differential equation.
Therefore, (a) μY (t) = e−t^ +
∫ (^) t 0 e
−(t−v)E[Xv ]dv = 2 − e−t.
