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Solutions to Final Exam
Problem 1 (9 points) Explain why each of the following is NOT a valid autocorrelation function for
a continuous-time, real-valued WSS random process:
(a) RX (τ ) =
e−τ^ τ ≥ 0
e^2 τ^ τ < 0
(b) RX (τ ) = e−|τ^ |I{|τ |≤ 1 } (c) RX (τ ) = 1+τ^
2
(a) Not symmetric: RX (τ ) 6 = RX (−τ ).
(b) Not continuous, even though it is continuous at zero. Violates equivalence of (i′) and (iii′) in Proposition 7.1.9.
(c) Violates Schwarz inequality, because, for example, RX (0) = 1 < RX (
0 .5) = 1. 2 , or RX (τ ) = (1 + τ 2 )(1 − τ 4 + τ 8 +
· · · ) = 1 + τ 2 + O(τ 4 ) > R(0) for all sufficiently small, nonzero values of τ. Alternatively, we could note that if R were
the autocorrelation function of a random process X, then X would be mean square differentiable (because RX is twice
continuously differentiable) with 0 ≤ RX′ (0) = −R′′ X (0) = − 2 , which is impossible. So, by argument by contradiction,
RX is not a valid autocorrelation function.
Problem 2 (8 points) (a) Suppose Z is a N (μ, σ^2 ) random variable. Express E[Z^3 ] in terms of μ and
σ^2.
(b) Suppose
X
Y
is a N
random vector, where |ρ| < 1. Express E[X^3 |Y ] in
terms of ρ and Y.
(a) Z has the same distribution as W + μ, where W is a N (0, σ^2 ) random variable. Since E[W ] = 0 and E[W 3 ] = 0,
E[Z^3 ] = E[W 3 + 3W 2 u + 3W μ^2 + μ^3 ] = 3σ^2 μ + μ^3. An alternative approach is to use the characteristic function of Z.
(b) The conditional distribution of X given Y = y is N (ρy, 1 − ρ^2 ). Therefore, the answer to this part is obtained by
replacing μ by ρY and σ^2 by 1 − ρ^2 in the answer to part (a). That is, E[X^3 |Y ] = 3ρ(1 − ρ^2 )Y + ρ^3 Y 3.
Problem 3 (12 points) (This problem uses the notation (f, g) =
−∞ f^ (t)g
∗(t)dt, and ||f || = √(f, f ).
Recall that a real-valued Gaussian white noise process Y = (Yt : t ∈ R) with parameter σ^2 = 1 is a
generalized random process with mean zero, RY (τ ) = δ(τ ), and SY (ω) = 1 for all ω. The interpretation
is that all random variables of the form (f, Y ) are jointly Gaussian, mean zero, and E[(f, Y )(g, Y )∗] =
(f, g). The focus of this problem is an ordinary random process X which is approximately a white noise
process.)
Let X be a real-valued stationary Gaussian process with mean zero and RX (τ ) = α 2 e−α|τ^ |. Note that
for α large, RX is an approximation to a delta function because it is narrow, supported near zero, and
integrates to one.
(a) (2 pts) Give SX and verify that for any fixed ω, SX (ω) → 1 as α → ∞.
(b) (5 pts) For real-valued functions f and g, express E[(X, f )(X, g)] in terms of SX , f ,̂ and ̂g.
(c) (5 pts) Suppose f and g are real-valued baseband functions, so that f̂ (ω) = 0 and ̂g(ω) = 0 for
|ω| ≥ ωo. Give a sufficient condition involving α and ωo to ensure that
|E[(f, X)(g, X)] − (f, g)| ≤ (0.01)||f || · ||g||.
(a) SX (ω) = α
2 ω^2 +α^2 →^ 1 as^ α^ → ∞. (b)
E[(f, X)(g, X)] =
Z ∞
−∞
Z ∞
−∞
f (s)RX (s − t)g(t)dsdt = ( fe ∗ RX ∗ g)(0)
Z ∞
−∞
( fb (ω))∗SX (ω)bg(ω)
dω 2 π
(by inverse transform)
Or, a slightly different approach is:
E[(f, X)(g, X)] =
Z ∞
−∞
Z ∞
−∞
f (s)RX (s − t)g(t)dsdt = (f, RX ∗ g)
= ( f ,b R̂X ∗ g)/ 2 π (Parseval’s relation)
= ( f , Sb X bg)/ 2 π =
Z ∞
−∞
f^ b (ω)(bg(ω))∗SX (ω) dω 2 π
The above two expressions for E[(f, X)(g, X)] are complex conjugates of each other, but since E[(f, X)(g, X)] is real valued, the expressions are equal. (c) ˛˛ ˛˛E[(f, X)(g, X)] − (f, g)
Z ∞
−∞
f^ b (ω)(bg(ω))∗(SX (ω) − 1) dω 2 π
Z (^) ωo
−ωo
f^ b (ω)(bg(ω))∗(SX (ω) − 1) dω 2 π
Z (^) ωo
−ωo
| fb (ω)(bg(ω))∗||SX (ω) − 1 | dω 2 π
For |ω| < ωo, |SX (ω) − 1 | = ω
2 α^2 +ω^2 ≤^
ω o^2 α^2 +ω o^2 ≤^0 .01 if^ α
(^2) ≥ 99 ω 2 o , or^ α^ ≥
99 ωo ≈ 10 ωo. Under this condition, ˛ ˛˛ ˛E[(f, X)(g, X)]^ −^ (f, g)
˛ ≤^ (0.01)
Z (^) ωo
−ωo
| fb (ω)(bg(ω))∗|
dω 2 π
s„Z ωo
−ωo
| fb (ω)|^2
dω 2 π
« „Z (^) ωo
−ωo
|bg(ω)|^2
dω 2 π
(Schwarz inequality)
= (0.01)||f || · ||g|| (Parseval’s relation)
Problem 4 (12 points) Consider a continuous-time Markov process X = (Xt : t ≥ 0) with the state
space S = { 0 , 1 , 2 } × { 0 , 1 , 2 } and the transition rate diagram shown. The transition rates for all arrows
in the diagram are equal to one.
2,
0,0 0,1 (^) 0,
1,0 1,1 1,
2,0 2,
(a) Is the equilibrium distribution for this process equal to the uniform distribution, ( 19 ,... , 19 )? If so,
justify. If not, find the equilibrium distribution.
(b) Let Ut and Vt denote the two coordinates of Xt for each t fixed. That is, Xt = (Ut, Vt) for each
t, where Ut and Vt both take values in { 0 , 1 , 2 }. Let π(t) = (π 0 , 0 (t), π 0 , 1 (t),... , π 2 , 2 (t)) denote the
distribution of X at time t for t ≥ 0. Under what conditions on the intital distribution, π(0), are the
processes U = (Ut : t ≥ 0) and V = (Vt : t ≥ 0) independent of each other? Explain.
(c) Find π(t) for all t ≥ 0 for the initial distribution P {X(0) = (1, 1)} = 1. (Hint: Due to the particular
structure of this problem, it can be solved with little computation.)
(b) Therefore, bθM L(y) is the value of θ that minimizes the function (y − θs)T^ K−^1 (y − θs) with respect to θ. This is a quadratic function of θ, so it is minimized at the unique point the derivative is zero. Calculating,
d(y − θs)T^ K−^1 (y − θs) dθ
= −sT^ K−^1 (y − θs) − (y − θs)T^ K−^1 s
= −sT^ K−^1 y + 2θsT^ Ks − yT^ K−^1 s = 2(θsT^ K−^1 s − yT^ K−^1 s),
Setting the derivative equal to zero yields bθM L(y) = y
T (^) K− (^1) s sT^ K−^1 s. (c) The MAP estimator θbM AP (y) is the value of θ that maximizes fY (y|θ)fΘ(θ) with respect to θ, or equivalently, taking negative logarithms, minimizes (y − θs)T^ K−^1 (y − θs) 2
θ^2 2 with respect to θ. This is again quadratic in θ, and by a modification of the calculation above, the derivative is given by θsT^ K−^1 s − yT^ K−^1 s + θ. Setting the derivative equal to zero yields
θ^ bM AP (y) = y
T (^) K− (^1) s 1 + sT^ K−^1 s
An alternative derivation is to note that since Θ and Y are jointly Gaussian and have mean zero (under the Bayesian assumption) bθM AP (y) = θbM L(y) = Eb[Θ|Y = y] = Cov(Θ, Y )Cov(Y)−^1 y = sT^ (ssT^ + K)−^1 y. The two approaches yield
equivalent answers, because s
T (^) K− 1 1+sT^ K−^1 s =^ s
T (^) (ssT (^) + K)− (^1) , as can be verified by multiplying both sides on the right by
ssT^ + K.
Problem 6 (12 points) Suppose (Xn : n ≥ 1) is a discrete-time Markov process with state space { 0 , 1 }, initial distribution
π(1) = (0. 5 , 0 .5), and one-step transition probability matrix P =
1 − p p p 1 − p
, where 0 < p < 1. Let Sn =
X 1 + · · · + Xn. (a) (3 pts) Find E[Sn] for all n ≥ 1. (b) (3 pts) Does limn→∞ S nn exist in the a.s. sense? Justify your answer. (c) (6 pts) Let θ > 0. The purpose of this problem is to compute E[exp(θSn)] for all n ≥ 1. (This could be used in the Chernoff inequality to provide a bound on large deviation probabilities of the form P { S nn ≥ α}.) Let an = E[eθSn^ I{Xn=0}] and bn = E[eθSn^ I{Xn=1}]. Note that an + bn = E[exp(θSn)]. Identify a recursive way to compute (an, bn) for all n ≥ 1. Start by finding the initial condition (i.e. the value of (a 1 , b 1 ).)
(a) The initial probability distribution π(1) is the equilbrium distribution, that is, π(1)P = π(1). Therefore, Xk has probability vector π(1) for all k ≥ 1. Hence, E[Sn] = E[X 1 + · · · + Xn] = E[X 1 ] + · · · + E[Xn] = n 2. (b) Yes. Since X has a finite state space it is positive recurrent, and it is irreducible. Hence, the a.s. limit of the time averages of the X’s exists and is equal to the statistical average, namely E[Xn] = 0. 5. (See Section 6.5.) (c) Since S 1 = X 1 , the initial values are given by (a 1 , b 1 ). Let n ≥ 1 and use the fact Sn+1 = Sn + Xn+1, the law of total probability, and the Markov property, to get
an+1 = E[eθSn+1^ I{Xn+1=0}] = E[eθSn^ I{Xn+1=0}] = E[eθSn^ I{Xn=0,Xn+1=0}] + E[eθSn^ I{Xn=1,Xn+1=0}] = an(1 − p) + bnp
Similarly,
bn+1 = E[eθSn+1^ I{Xn+1=1}] = E[eθSn^ I{Xn+1=1}]eθ
=
E[eθSn^ I{Xn=0,Xn+1=1}] + E[eθSn^ I{Xn=1,Xn+1=1}]
eθ^ = anpeθ^ + bn(1 − p)eθ
Summarizing, (an+1, bn+1) = (an, bn)B, where B =
1 − p peθ p (1 − p)eθ
. Thus, (an, bn) = (0. 5 , (0.5)eθ^ )Bn−^1.
(Note: The recursive method is essentially the same as the forward part of the forward-backward algorithm for HMMs. The powers of B, and therefore E[exp(θSn)], grow like λn 1 , where λ 1 is the largest magnitude eigenvalue of B. It turns out that λ 1 = p + (1 − p)eθ^ , with corresponding right eigenvector (1, 1)T^ .)
