Random Processes for Engineers 1st Edition By Bruce Hajek INSTRUCTORS SOLUTION MANUAL

INSTRUTOR'S SOLUTION MANUAL

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Solutions to Odd Numbered Problems Random Processes for Engineers

1.1 Simple events (a) Ω = {0, 1}8 , or Ω = {x1 x2 x3 x4 x5 x6 x7 x8 : xi ∈ {0, 1} for each i}. It is natural to let F be the set of all subsets of Ω. Finally, let |A| , where |A| denotes the cardinality of a set |A|.

P (A) = 256 1 2 = 128 . (b) E1 = {01010101, 10101010} and P (E1 ) = 256 E2 = {00110011, 01100110, 11001100, 10011001} and P (E 2 ) = 4/256 = 1/64.

8 E3 = {x ∈ Ω : x1 + · · · + x8 = 4} and P (E2 ) = 4 /256 = 70/256 = 35/128.

E4 = {11111111, 11111110, 11111101, 10111111, 01111111, 00111111, 01111110, 11111100} and P (E4 ) = 8/256 = 1/32. (c) E1 ⊂ E3 , so P (E1 |E3 ) = |E1 |/|E3 | = 2/70 = 1/35.

E2 ⊂ E3 , so P (E2 |E3 ) = |E2 |/|E3 | = 4/70 = 2/35.

1.3 Ordering of three random variables P {X < u < Y } = P {X < u}P {u < Y } = (1 − e−λu )e−λu = e−λu − e−2λu . Averaging over the choices of u using the pdf of U yields,

P {X < U < Y } =

Z 1 0

e−λu − e−2λu du =

0.5 − e−λ + 0.5e−2λ . λ

1.5 Congestion at output ports (a) One possibility is Ω = {1, 2, . . . , 8}4 = {(d1 , d2 , d3 , d4 ) : 1 ≤ di ≤ 8 for 1 ≤ i ≤ 4}, where the packets are assumed to be numbered one through four, and di is the output port of packet i. Let F be all the subsets of Ω, and for any event A, let P (A) = |A| 84 . (b) P {X1 = k1 , . . . , X8 = k8 } =

1 4 84 k1 k2 · · · k8

where k1 k24···k8 = k1 !k24!!···k8 ! is the multinomial coefficient.

P4 (c) One way to do this problem is to note that Xj = i=1 Xij , where Xij = 1 if packet i is routed to output port j, and Xij = 0 otherwise. Suppose j 6= j 0 . Then 2 1 Xij Xij 0 ≡ 0, and so also, E[Xij Xij 0 ] = 0. Thus, Cov(Xij , Xij 0 ) = 0 − 81 = − 64 .

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Also, Cov(Xij , Xi0 j 0 ) = 0 if i 6= i0 . Thus, 4 4 X X Cov(Xj , X ) = Cov( Xij , Xi0 j 0 ) j0

i=1

=

4 X 4 X

i0 =1

Cov(Xij , Xi0 j 0 )

i=1 i0 =1

=

4 X i=1

Cov(Xij , Xij 0 ) = 4(−

1 1 )=− .

64 16

(d) Consider the packets one at a time in order. The first packet is routed to a random output port. The second is routed to a different output port with probability 78 . Given the first two packets are routed to different output ports, the third packet is routed to yet another output port with probability 86 . Similarly, given the first three packets are routed to distinct output ports, the fourth packet is routed to yet another output port with probability 58 . The answer is thus 8·7·6·5 = 105 84 256 ≈ 0.410. (e) The event is not true if and only if there are either exactly 3 packets assigned to one output port or all four packets assigned to one output port. There are 4 · 8 · 7 possibilities for exactly three packets to be assigned to one output port, since there are four choices for which packet is not with the other three, eight choices of output port for the group of three, and given that, seven choices of output port for the fourth packet. There are 8 possibilities for all four packets to be routed to the same output port. Thus, some output port has three or more 29 = 4·7+1 = 512 ≈ 0.0566. Thus, packets assigned to it with probability 4·8·7+8 84 83 29 P {Xi ≤ 2 for all i} = 1 − 512 ≈ 0.9434.

1.7 Conditional probability of failed device given failed attempts (a) P (first attempt fails)=0.2+(0.8)(0.1)=0.28 (b) P (server is working | first attempt fails ) = P (server working, first attempt fails)/P (first attempt fails) =(0.8)(0.1)/0.28≈ 0.286 (c) P (second attempt fails | first attempt fails ) =P (first two attempts fail)/P (first attempt fails) = [0.2 + (0.8)(0.1)2 ]/0.28 ≈0.783 (d) P (server is working | first and second attempts fail ) =P (server is working and first two attempts fail)/P (first two attempts fail) = (0.8)(0.1)2 /[0.2 + (0.8)(0.1)2 ] ≈0.0385 1.9 Conditional lifetimes; memoryless property of the geometric distribution (a) P {X > 3} = 1 − p(3) = 0.8, P (X > 8|X > 5) = P ({X>8}∩{X>5}) = P {X>5} P {X>8} 0 P {X>5} = 0.40 = 0.

(So a five year old working battery is not equivalent to a new one!) (b) P {Y > 3} = P (miss first three shots) = (1 − p)3 . On the other hand, P (Y > 8|Y > 5) =

P ({Y > 8} ∩ {Y > 5}) P {Y > 8} (1 − p)8 = = = (1 − p)3 .

P {Y > 5} P {Y > 5} (1 − p)5

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(A player that has missed five shots is equivalent to a player just starting to take shots.) (c) Y has a geometric distribution. (Part (b) illustrates the fact that the geometric distribution is the memoryless lifetime distribution on the positive integers.

The exponential distribution is the continuous type distribution with the same property.)

1.11 Distribution of the flow capacity of a network One way to solve this problem is to compute X for each of the 32 outcomes for the links. Another is to use divide and conquer by conditioning on the state of a key link, such as link 4.

P {X = 0} = P (((F1 F3 ) ∪ (F2 F5 ))F4c ) + P ((F1 ∪ F2 )(F3 ∪ F5 )F4 )

= ((0.2)2 + (0.2)2 − (0.2)4 )(0.8) + (0.2 + 0.2 − (0.2)2 )2 (0.2) = 0.08864.

P {X = 10} = P (F1c F3c (F2 F5 )c F4c ) + P (F1c F2c F3c F5c F4 )

= (0.8)3 (1 − (0.2)2 ) + (0.8)4 (0.2) = 0.57344.

P {X = 5} = 1 − P {X = 0} − P {X = 10} = 0.33792.

1.13 A CDF of mixed type (a) FX (0.8) = 0.5. (b) There is a half unit of probability mass R 2 at zero and a density of value 0.5 between 1 and 2. Thus, E[X] = 0 × 0.5 + 1 x(0.5)dx = 3/4 and, R2 (c) E[X 2 ] = 02 × 0.5 + 1 x2 (0.5)dx = 7/6. So Var(X) = 7/6 − (3/4)2 = 29/48 .

1.15 Poisson and geometric random variables with conditioning i P∞ −µ P∞ P∞ −µ i = e−µp (a) P {Y < Z} = i=0 j=i+1 e i! µ p(1 − p)j−1 = i=0 e [µ(1−p)] i!

Pi−1 e−µ µj (b) P (Y < Z|Z = i) = P (Y < i|Z = i) = P {Y < i} = j=0 j!

−µ i (c) P (Y = i|Y < Z) = P {Y = i < Z}/P {Y < Z} = e i! µ (1 − p)i /e−µp = e−µ(1−p) [µ(1−p)]i , which is the Poisson distribution with mean µ(1 − p) i!

(d) µ(1 − p) 1.17 Transformation of a random variable (a) Observe that Y takes values in the interval [1, +∞).

P {X ≤ ln c} = 1 − exp(−λ ln c) = 1 − c−λ c ≥ 1 FY (c) = P {exp(X) ≤ c} = 0 c<1 Differentiate to obtain fY (c) =

λc−(1+λ) 0

c≥1 c<1

(b) Observe that Z takes values in the interval [0, 3].  0 c<0  FZ (c) = P {min{X, 3} ≤ c} = P {X ≤ c} = 1 − exp(−λc) 0 ≤ c < 3  1 c≥3

The random variable Z is neither discrete nor continuous type. Rather it is a

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mixture, having a density over the interval [0, 3) and a discrete mass at the point 3.

1.19 Moments and densities of functions of a random variable The pdf E[C] = 2E[L] + 2E[W ] = 2 Var(C) = 4Var(L) + 4Var(W ) = 32 of C is the convolution of the pdf of 2L with the pdf of 2W . But 2L and 2W are each uniformly distributed over the interval [0, 2], so their pdfs are rectangular pulse functions. The convolution of such a function with itself is a triangular pulse function. The base of the triangle, equal to the support of fC , is the interval [0, 4]. The peak of the triangle is at the midpoint, and must have  height 1/2 in or x/4 0 ≤ x ≤ 2 4−x der that the area of the triangle be one. Therefore, fC (x) = 2≤x≤4  4 0 else E[A] = E[L]E[W ] = ( 21 )2 = 14 7 .

Var(A) = 91 − ( 41 )2 = 144

E[A2 ] = E[L2 ]E[W 2 ] = ( 13 )2 = 19

so

R1 For 0 ≤ c ≤ 1, P {A ≤ c} =area of {(x, y) ∈ [0, 1]2 : xy ≤ c} = c + c xc dx =

− ln(c) 0 ≤ c ≤ 1 c(1 − ln c), so fA (c) = 0 else 1.21 Using the Gaussian Q function (a) P {X ≥ 16} = P { X−10 > 16−10 3 3 } = 16−10 Q( 3 ) = Q(2). −4−10 (b) P {X 2 ≥ 16} = P {X ≥ 4} + P {X ≤ −4} = Q( 4−10 ) = 3 ) + 1 − Q( 3 14 14 Q(−2) + 1 − Q(− 3 ) = 1 − Q(2) + Q( 3 ). (c) Z is N (0, 5) so P {|Z| > 1} = P {Z > 1} + P {Z < −1} = Q( √15 ) + 1 − Q(− √15 ) = 2Q( √15 ).

1.23 Correlation of histogram values (a) X1 is Bernoulli( 16 ), so E[X1 ] = 16 5 and Var(X1 ) = 61 (1 − 16 ) = 36 . n (b) E[X] = nE[X1 ] = 6 and Var(X) = nVar(X1 ) = 5n 36 . (c) We begin by computing Cov(X1 , Y1 ). Since X1 Y1 = 0 with probability one, E[X1 Y1 ] = 0. Therefore Cov(X1 , Y1 ) = E[X1 Y1 ] − E[X1 ]E[Y1 ] = 0 − 16 61 = −1 36 . −1 So Cov(Xi , Yi ) = 36 for any i. On the other hand, if i 6= j then Xi is independent of Xj . So −1 if i = j 36 Cov(Xi , Yj ) = 0 if i 6= j (d) Cov(X, Y ) =

XX i

j

Cov(Xi , Yj ) =

X

Cov(Xi , Yi ) = nCov(X1 , Y1 ) =

i

−n .

36

and Cov(X, Y ) −1 ρ(X, Y ) = p = .

5 Var(X)Var(Y )

(e) Given that x of the dice show a 1, each of the remaining dice is equally likely

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to show 2,3,4,5, or 6. Thus, each of the remaining n − x dice shows a 2 with conditional probability 51 . Therefore E[Y |X = x] = n−x 5 .

1.25 A function of jointly distributed random variables The square has unit area so that the joint density is unit valued within the square. The range of X is the interval [0, 1], so fix c in [0, 1] and consider the event {U V ≤ c}. The probability of this event is the area of the square minus the upper right region above the curve v = c/u. This area is one minus the area of the region inside the square above the curve v = c/u. Therefore,   0 c≤0  R1 c FX (c) = 1 − c (1 − u )du = c − c ln c 0 ≤ c ≤ 1   1 c≥1 Differentiating yields

fX (c) =

− ln c 0 < c ≤ 1 0 else

1.27 Working with a two dimensional density (a) The parallelogram has base and height one, and thus area  one, so that the density is one on the region.

0.5x 0≤x≤1    0.5 1≤x≤2 (b) By inspection, fX (x) =  0.5(3 − x) 2 ≤x≤3   0 else (c) Since the density of X is symmetric about 1.5 and the mean exists, E[X] = R1 R2 R3 8 1.5. E[X 2 ] = 0.5[ 0 x3 dx + 1 x2 dx + 2 x2 (3 − x)dx] = 0.5[ 41 + 37 + 11 4 ] = 3 , so 8 3 2 5 Var(X) = 3 − ( 2 ) = 12 . A slicker way to find the variance is to observe the X has the same distribution as U1 + 2U2 , where U1 and U2 are independent and 5 uniformly distributed over [0, 1], so Var(X) = Var(U1 ) + 4Var(U2 ) = 12 . (d) If 0 ≤ x ≤ 1, the conditional density of Y given X = x is the uniform density 2 0 ≤ y ≤ x2 x over the interval [0, x2 ]. That is, for 0 < x ≤ 1: fY |X (y|x) = 0 else (e) By inspection, if 1 ≤ x ≤ 2, the conditional density of Y given X = x is the x uniform density over the interval [ x−1 2 , 2 ]. That is, for 1 < x ≤ 2: fY |X (y|x) =

x−1 x 2 2 ≤y ≤ 2 0 else (f) E[Y |X = x] is well defined over the support of fX , namely, over the interval [0, 3]. For each X in this interval, the conditional density of Y give X = x is a uniform density, so the  conditional mean is the midpoint of the interval. x/4 0≤x≤1    (x − 0.5)/2 1 ≤ x ≤ 2 Therefore. E[Y |X = x] =  (x + 1)/4 2 ≤ x ≤ 3   undefined x 6∈ [0, 3] 1.29 Uniform density over a union of two square regions (a) Region has area 2 so the density function is 1/2 in the region and zero outside.

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if |x| ≤ 1 else

1 if 0 ≤ y ≤ 1 (c) If 0 < a ≤ 1, fY |X (y|a) = 0 else

1 if − 1 ≤ y ≤ 0 (d) If −1 ≤ a < 0, fY |X (y|a) = 0 else

−0.5 if − 1 ≤ a < 0 (e) E[Y |X = a] = 0.5 if 0 < a < 1 2 (f) E[X] = E[Y ] = 0, Var(X) = E[X ] = 1/3, Var(Y R R R R R 1 R 1) = 1/3, 1 1 0 0 1 1 E[XY ] = 2 0 0 xydxdy + 2 −1 −1 xydxdy = 0 0 xydxdy = 1/4. So ρXY = √ 1/4 = 34 .

(b) fX (x) =

0.5 0

1/3×1/3

(g) No, because fXY (x, y) doesn’t factor into the product of a function of x and a function of y.  if − 0 ≤ |z| ≤ 1  |z|/2 (h) The range of Z is [−2, 2]. fZ (z) = (Shape is 1 − |z|/2 if 1 ≤ |z| ≤ 2  0 else two triangles.)

R1R1 R1R1 1.31 Transformation of densities (a) 0 0 (u − v)2 dudv = 0 0 (u2 − 2uv + v 2 )dudv = 61 , so c = 6. (b) The map from the u, v plane to the x, y plane given by x = u2 and y = u2 v 2 maps the unit square [0, 1] × [0, 1] into the triangular region 0 ≤ y ≤ x ≤ 1 in one-to-one fashion. The inverse mapping is given by u = v 1/2 and v = (y/x)1/2 .

2u 0 ∂(x,y) Also, ∂(u,v) = 4u3 v = 4xy 1/2 . Therefore, = 2 2uv 2u2 v −1

∂(x, y) fXY (x, y) = fU V (u, v) ∂(u, v) ( 6(x1/2 − (y/x)1/2 )2 4xy11/2 = 0

if 0 ≤ y ≤ x ≤ 1 else

X 1.33 Transformation of joint densities To be definite, assume takes valY

W ues in the positive quadrant of the u − v plane and takes values in the Z

W X α − β plane. We have Z = g Y where the transformation g is given by α = u − v and β = u2 + u − v. The transformation is invertible. In fact, we see √ √ that u = β − α and v = β − α − α, for (α, β) in the range of g, which is the set {(α, β) : β > α+(max{0, α})2 }. (To understand the geometry of the function better, note as u varies over u > 0 with v = 0 the function g(u, v) traces out the curve β = α2 + α for α > 0. Then for any u fixed with u > 0, the function g(u, v) traces out a half line of slope one as v ranges over v > 0.) The determinant of the Jacobian of g is given by

p 1 −1 det(J) = det = u = β − α.

2u −1

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Therefore, fX,Y (u.v) fW,Z (α, β) = = det J

(

√ exp(−λ(2 β−α−α)) √ β−α

0

β > α + (max{0, α})2 else

1.35 Conditional densities and expectations (a) Z 1Z u uv(4u2 )dvdu E[XY ] = 0 0

Z u Z 1 3 vdv du = 4u 0

0

=

Z 1

2u5 du =

0

(b) fY (v) =

Z 1

4u2 du =

v

(c)

fX|Y (u|v) =

   0,          

1 .

3

4 3 3 (1 − v ),

0,

0≤v≤1 elsewhere

0 < v < 1, 0 < u < v 4u2

4 3 3 (1−v )

2

3u = 1−v 3,

undefined,

0 < v < 1, v < u < 1 v < 0 or v > 1

R1 3u2 3 1−v 5 (d) For 0 < v < 1, E[X 2 |Y = v] = v u2 1−v 3 du = 5 1−v 3 2.1 Limits and infinite sums for deterministic sequences (a) Before beginning the proof we observe that | cos(θ)| ≤ 1, so |θ(1 + cos(θ))| ≤ 2|θ|. Now, for the proof. Given an arbitrary with > 0, let δ = /2. For any θ with |θ − 0| ≤ δ, the following holds: |θ(1 + cos(θ)) − 0| ≤ 2|θ| ≤ 2δ = . Since was arbitrary the convergence is proved. (b) Before beginning the proof we observe that if 0 < θ < π/2, then cos(θ) ≥ 0 and 1+cos(θ) ≥ 1/θ. Now, for the proof. Given an arbitrary positive number K, θ 1 let δ = min{ π2 , K }. For any θ with 0 < θ < δ, the following holds: 1+cos(θ) ≥ θ 1/θ ≥ 1/δ ≥ K. Since K was arbitrary the convergence is proved. √ PN n (c) The sum is by definition equal to limN →∞ sN where sN = n=1 1+ 1+n2 . The sequence SN is increasing in N . Note that the n = 1 term of the sum is 1 and for any n ≥ 1 the nth term of the sum can be bounded as follows: √ √ 1+ n 2 n ≤ 2 = 2n−3/2 .

1 + n2 n Therefore, comparing the partial sum with an integral, yields Z N N X sN ≤ 1 + 2n−3/2 ≤ 1 + 2x−3/2 dx = 5 − 4N −1/2 ≤ 5. n=2

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In summary, the sequence (SN : N ≥ 1) is an increasing, bounded sequence, and it thus has a finite limit.

2.3 The reciprocal of the limit is the limit of the reciprocal Let > 0. Let x2 0 = min{ |x2∞ | , 2∞ }. By the hypothesis, there exists a value of no so large that for all n ≥ no , |xn −x∞ | ≤ 0 . This condition implies that |xn | ≥ |x∞ |/2, because of the choice of 0 . Therefore, for all n ≥ no ,

1 |xn − x∞ | 2 0 1 = ≤ 2 ≤ , − xn x∞ |xn ||x∞ | x∞

which, by definition, shows that (1/xn ) converges to 1/x∞ .

2.5 On convergence of deterministic sequences and functions (a) Note that 1 1 xn − 83 = 3n . Thus, given any > 0, let n = d 3 e. Then for any n ≥ n , 8 1 1 |Xn − 3 | ≤ 3n ≤ 3n ≤ . Thus, by definition, limn→∞ xn = 38 . (b) Let = 1/3 and let xn = (2/3)1/n for n ≥ 1. Note that xn ∈ [0, 1) and fn (xn ) = 32 . Thus, there is no positive integer n such that |fn (x) − 0| ≤ for all x ∈ [0, 1). So it is impossible to select n with the property required for uniform convergence. Therefore fn does not converge uniformly to zero. (c) Let c < supD f . Then there is an x ∈ D so that c ≤ f (x). Therefore, c ≤ f (x) − g(x) + g(x) ≤ supD |f − g| + supD g. Thus, c < supD f implies c < supD |f − g| + supD g. Equivalently, supD f ≤ supD |f − g| + supD g, or supD f − supD g ≤ supD |f − g|. Exchanging the roles of f and g yields supD g − supD f ≤ supD |f − g|. Combining yields the desired inequality, | supD f − supD g| ≤ supD |f −g|. As an application, suppose fn → f uniformly on D. Then given any > 0, there exists an n so large, that supD |fn − f | ≤ , whenever n ≥ n . But then by the inequality proved, | supD fn − supD f | ≤ supD |fn − f | ≤ , whenever n ≥ n . Thus, by definition, supD fn → supD f as n → ∞.

2.7 On the Dirichlet criterion for convergence of a series Pn (a) Let Rn = k=0 ak . By assumption, the sequence (Rn ) has a finite limit, so it is a Cauchy sequence, i.e. limm,n→∞ |Rm − Rn | = 0. Now for n < m, Pm Pm Pm |Sm − Sn | = | k=n+1 dk | ≤ k=n+1 |dk | ≤ k=n+1 Lak = L|Rm − Rn |. Therefore, limm,n→∞ |Sm − Sn | = 0. That is, (Sn ) is also a Cauchy sequence, and hence also has a finite limit.

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(b) Sn =

n X

k=0

= = =

n X

Ak B k − Ak B k −

k=0 n X

n X

Ak Bk−1

k=1 n−1 X

Ak+1 Bk

k=0

(Ak − Ak+1 )Bk

k=0 n X

k=0

ak Bk

!

since B−1 = 0

!

− An+1 Bn

− An+1 Bn .

Pn (c) Since |ak Bk | ≤ Lak for all k, the sequence of sums k=0 ak Bk is convergent by the result of part (a). Also, |An+1 Bn | ≤ LAn+1 → 0 as n → ∞. Thus, by part (b), Sn has a finite limit.

2.9 Convergence of a random sequence (a) The sequence Xn (ω) is monotone nondecreasing in n for each ω. Also, by induction on n, Xn (ω) ≤ 1 for all n and ω. Since bounded monotone sequences have finite limits, limn→∞ Xn exists in the a.s. sense and the limit is less than or equal to one with probability one. (b) Since a.s. convergence of bounded sequences implies m..s. convergence, limn→∞ Xn also exists in the m.s. sense. (c) Since (Xn ) converges a.s., it also converges in probability to the same random variable, so Z = limn→∞ Xn a.s. It can be shown that P {Z = 1} = 1.

Here is one of several proofs. Let 0 < < 1. Let a0 = 0 and ak = ak−12+1− for k ≥ 1. By induction, ak = (1 − )(1 − 2−k ). Consider the sequence of events: {Ui ≥ 1 − } for i ≥ 1. These events are independent and each has probability .

So with probability one, for any k ≥ 1, the probability that at least k of these events happens is one. If at least k of these events happen, then Z ≥ ak . So, P {(1 − )(1 − 2−k ) ≤ Z ≤ 1} = 1. Since can be arbitrarily close to zero and k can be arbitrarily large, it follows that P {Z = 1} = 1.

2 .

ANOTHER APPROACH is to calculate that E[Xn |Xn−1 = v] = v + (1−v) 2 2

2

Thus, E[Xn ] = E[Xn−1 ]+ E[(1−X2 n−1 ) ] ≥ E[Xn−1 ]+ (1−E[X2 n−1 ]) . Since E[Xn ] → 2

E[Z], it follows that E[Z] ≥ E[Z] + (1−E[Z]) . So E[Z] = 1. In view of the fact 2 P {Z ≤ 1} = 1, it follows that P {Z = 1} = 1.

2.11 Convergence of some sequences of random variables (a)For each fixed ω., V (ω) → 0 so Xn (ω) → 1. Thus, Xn → 1 in the a.s sense, and hence also n in the p. and d. senses. Since the random variables Xn are uniformly bounded (specifically, |Xn | ≤ 1 for all n), the convergence in p. sense implies convergence in m.s. sense as well. So Xn → 1 in all four senses. (b)To begin we note that P {V ≥ 0} = 1 with P {V > 1} = e−3 > 0. For any ω such that V (ω) < 1, Yn (ω) → 0, and for any ω such that V (ω) > 1, Yn (ω) → +∞, so (Yn ) does not converge in the a.s. sense to a finite random variable.

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Let us show Yn does not converge in d. sense. For any c > 0 limn→∞ Fn (c) = limn→∞ P {Yn ≤ c} = P {V < 1} = 1 − e−3 . The limit exists but the limit function F satisfies F (c) = e−1 for all c > 0, so the limit is not a valid CDF. Thus, (Yn ) does not converge in the d. sense (to a finite limit random variable), and hence does not converge in any of the four senses to a finite limit random variable. (c)For each ω fixed, Zn (ω) → eV (ω) . So Zn → eV in the a.s. sense, and hence also V in the p. and d. senses. Using the inequality 1+u ≤ eu shows that R ∞Zn2u≤ e−3ufor all V V 2 2V n so that |Zn | ≤ e for all n. Note that E[(e ) ] = E[e ] = 0 e 3e du = 3 < ∞. Therefore, the sequence (Zn ) is dominated by a single random variable with finite second moment (namely, eV ), so the convergence of (Zn ) in the p. sense to eV implies that (Zn ) converges to eV in the m.s. sense as well. So Zn → eV in all four senses.

2.13 On the maximum of a random walk with negative drift (a) By the strong law of large numbers, P {Sn /n → −1} = 1. Therefore, with probability one, Sn /n ≤ 0 for all sufficiently large n. That is, with probability one, Sn > 0 only finitely many times. The random variable Z, with probability one, is thus the maximum of only finitely many nonnegative numbers. So Z is finite with probability one. (b) Suppose P {X1 = c − 1} = P {X1 = −c − 1} = 0.5 for a constant c > 0.

Then X1 has mean -1 as required. Following the hint, for c ≥ 1, we have E[Z] ≥ E[max{0, X1 }] = (c − 1)/2. Observe that E[Z] can be made arbitrarily large by taking c arbitrarily large. So the answer to the question is no. (Note: More can be said about E[Z] if the variance of X1 is known. A celebrated bound of J.F.C. Kingman is that E[Z] ≤ Var(X1 ) .) −2E[X1 ]

2.15 Convergence in distribution to a nonrandom limit Suppose P {X = c} = 1 and limn→∞ Xn = X d. Let > 0. It suffices to prove that P {Xn − X| ≤ } → 1 as n → ∞. Note that P {|Xn − X| ≤ } ≥ P {c − < Xn ≤ c + } = Fn (c + ) − Fn (c − ). Since c − is a continuity point of FX and FX (c − ) = 0, it follows that Fn (c − ) → 0. Similarly, Fn (c + ) → 1. Thus Fn (c + ) − Fn (c − ) → 1, so that P {|Xn − X| ≤ } → 1. Therefore convergence in probability holds.

Note: A slightly different approach would be to prove that for any > 0, there is an n so large that P {|Xn − c| ≤ } ≥ 1 − .

2.17 Convergence of a product (a) Examine Sn = ln Xn . The sequence Sn , n ≥ 1 is the sequence of partial sums of the independent R 2 and identically distributed random variables ln Uk . Observe that E[ln Uk ] = 0 ln(u) 21 du = 12 (x ln x − x)|20 = ln 2 − 1 ≈ −0.306. Therefore, by the strong law of large numbers, limn→∞ Snn = ln 2 − 1 a.s. This means that, given an > 0, there is an a.s. finite random variable N so large that | Snn − (ln 2 − 1)| ≤ for all n ≥ N . Equivalently,

n

n 2(1 − ) 2(1 + ) ≤ Xn ≤ for n ≥ N . e e Conclude that limn→∞ Xn = 0 a.s., which implies that also limn→∞ Xn = 0 p.

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and limn→∞ Xn = 0 d. It remains to check for convergence in the m.s. sense. If Xn were to converge in the m.s. sense, it would have to converge to the same random variable in probability. But Xn does not converge to zero in the mean square. One way to see that Xn does not converge to zero in the m.s. sense is to note that E[Xn ] = 1 for all n, so limn→∞ E[Xn ] 6= 0. Another way to see that Xn does not converge to zero in the m.s. sense is to observe n 4 6→ 0.

E[|Xn − 0|2 ] = E[U12 . . . Un2 ] = E[U12 ] · · · E[Un2 ] = 3 In summary, Xn converges to zero in the a.s., p., and d. senses, but does not converge in the m.s. sense. (b) As noted in part (a), limn→∞ Snn = ln 2 − 1 a.s., so if θ = −1, then Yn = Snn converges in distribution to the constant ln 2 − 1.

2.19 Sums of i.i.d. random variables, I Let Xi denote the gamblers net gain for the ith play. Then X1 , X2 , . . . , X100 are iid with  0.5 x = −1    0.1 x = 0 pXi (x) =  0.4 x = 1   0 otherwise Thus E[Xi ] = −0.1, Var(Xi ) = E[Xi2 ] − E[Xi ]2 = 0.9 − (0.1)2 = .89. Let S = X1 + · · · + X100 . Therefore ES = −10, Var(S) = (100)(0.89) = 89. By Chebychev’s inequality P (S ≥ 0) = P (S + 10 ≥ 10) ≤ P (|S + 10| ≥ 10) ≤

Var(S) = .89.

(10)2

By the central limit theorem,

S + 10 10 √ P (S ≥ 0) = P ≥√ ≈ 1 − Φ(1.06) = 0.1446.

89 89 θ To calculate the + 0.4e−θ ) and √ Chernoff bound, we find M (θ) = log(0.5e + 0.1√ `(0) = exp(.4 5 + 0.1), yielding the upper bound P (S ≥ 0) ≤ (.4 5 + 0.1)100 = 0.57187.

It is not difficult to calculate P (S ≥ 0) numerically. The result is P (S ≥ 0) = 0.1572.... Thus, in this example, the approximation based on the central limit theorem is fairly accurate, the Chernoff bound is somewhat loose, and the Chebychev inequality is very loose.

2.21 Sums of i.i.d. random variables, III (a) ΦXi,n (u) = E ejuXi,n = 1 +

n λ ju − 1) so ΦYn (u) = 1 + nλ (eju − 1) . n (e α ju (b) Since limn→∞ 1 + α = eα , it follows that limn→∞ ΦYn (u) = eλ(e −1) . n This limit as a function of u is the characteristic function of a random variable Y with the Poisson distribution with mean λ. (c) Thus, (Yk ) converges in distribution, and the limiting distribution is the

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Poisson distribution with mean λ. There is not enough information given in the problem to determine whether Yn converges in any of the stronger senses (p., m.s., or a.s.), because the given information only describes the distribution of Yn for each n but gives nothing about the joint distribution of the Yn ’s. Note that Yn has a binomial distribution for each n.

2.23 On the growth of the maximum of n independent exponentials (a) Let n ≥ 2. Clearly FZn (c) = 0 for c ≤ 0. For c > 0, FZn (c) = P {max{X1 , . . . , Xn } ≤ c ln n}

= P {X1 ≤ c ln n, X2 ≤ c ln n, . . . , Xn ≤ c ln n}

= P {X1 ≤ c ln n}P {X2 ≤ c ln n} · · · P {Xn ≤ c ln n} = (1 − e−c ln n )n = (1 − n−c )n .

(b) Or, FXn (c) = (1 + xnn )n , where xn = −n1−c . Observe that as n → ∞,   −∞ c < 1 xn → −1 c = 1  0 c > 1,

so by Lemma 2.3.1 (and the monotonicity of the function ex to extend to the case x = −∞),  c<1  0 FZn (c) → e−1 c = 1  1 c > 1.

Therefore, if Z∞ is the random variable that is equal to one with probability one, then FZn (c) → FZ∞ (c) at the continuity points (i.e. at c 6= 1) of FZ∞ . So the sequence (Zn ) converges to one in distribution.

2.25 Limit behavior of a stochastic dynamical system Due to the persistent noise, just as for the example following Theorem 2.1.5 in the notes, the sequence does not converge to an ordinary random variables in the a.s., p., or m.s. senses.

To gain some insight, imagine (or simulate on a computer) a typical sample path of the process. A typical sample sequence hovers around zero for a while, but eventually, since the Gaussian variables can be arbitrarily large, some value of Xn will cross above any fixed threshold with probability one. After that, Xn would probably converge to infinity quickly. For example, if Xn = 3 for some n, and if the noise were ignored from that time forward, then X would go through the sequence 9, 81, 6561, 43046721, 1853020188851841, 2.43 × 1030 , . . ., and one suspects the noise terms would not stop the growth. This suggests that Xn → +∞ in the a.s. sense (and hence in the p. and d. senses as well.

(Convergence to +∞ in the m.s. sense is not well defined.) Of course, then, Xn does not converge in any sense to an ordinary random variable.

We shall follow the above intuition to prove that Xn → ∞ a.s. If Wn−1 ≥ 3 for some n, then Xn ≥ 3. Thus, the sequence Xn will eventually cross above the threshold 3. We say that X diverges nicely from time n if the event En =

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{Xn+k ≥ 3 · 2k for all k ≥ 0} is true. Note that if Xn+k ≥ 3 · 2k and Wn+k ≥ −3 · 2k , then Xn+k+1 ≥ (3 · 2k )2 − 3 · 2k = 3 · 2k (3 · 2k − 1) ≥ 3 · 2k+1 . Therefore, En ⊃ {Xn ≥ 3 and Wn+k ≥ −3 · 2k for all k ≥ 0}. Thus, using a union bound and the bound Q(u) ≤ 12 exp(−u2 /2) for u ≥ 0: P (En |Xn ≥ 3) ≥ P {Wn+k ≥ −3 · 2k for all k ≥ 0}

k = 1 − P ∪∞ k=0 {Wn+k ≤ −3 · 2 } ∞ ∞ X X √ ≥1− P {Wn+k ≤ −3 · 2k } = 1 − Q(3 · 2k · 2) ≥1− =1−

k=0 ∞ X

1 2

k=0

exp(−(3 · 2k )2 ) ≥ 1 −

k=0 −9

e ≥ 0.9999.

2(1 − e−9 )

∞ 1X

2

(e−9 )k+1

k=0

The pieces are put together as follows. Let N1 be the smallest time such that XN1 ≥ 3. Then N1 is finite with probability one, as explained above. Then X diverges nicely from time N1 with probability at least 0.9999. However, if X does not diverge nicely from time N1 , then there is some first time of the form N1 + k such that XN1 +k < 3 · 2k . Note that the future of the process beyond that time has the same evolution as the original process. Let N2 be the first time after that such that XN2 ≥ 3. Then X again has chance at least 0.9999 to diverge nicely to infinity. And so on. Thus, X will have arbitrarily many chances to diverge nicely to infinity, with each chance having probability at least 0.9999. The number of chances needed until success is a.s. finite (in fact it has the geometric distribution), so that X diverges nicely to infinity from some time, with probability one.

2.27 Convergence analysis of successive averaging (b) The means µn of Xn for all n are determined by the recursion µ0 = 0, µ1 = 1, and, for n ≥ 1, µn+1 = (µn + µn−1 )/2. This second order recursion has a solution of the form µn = Aθ1n +Bθ2n , where θ1 and θ2 are the solutions to the equation θ2 = (1+θ)/2.

This yields µn = 23 (1 − (− 12 )n ). (c) It is first proved that limn→∞ Dn = 0 a.s.. Note R 1 that Dn = U1 · · · Un−1 . Since ln Dn = ln(U1 ) + · · · ln(Un−1 ) and E[ln Ui ] = 0 ln(u)du = (x ln x − x)|10 = −1, Dn the strong law of large numbers implies that limn→∞ lnn−1 = −1 a.s., which in turn implies limn→∞ ln Dn = −∞ a.s., or equivalently, limn→∞ Dn = 0 a.s., which was to be proved. By the hint, for each ω such that Dn (ω) converges to zero, the sequence Xn (ω) is a Cauchy sequence of numbers, and hence has a limit. The set of such ω has probability one, so Xn converges a.s.

2.29 Mean square convergence of a random series Let Yn = X1 + · · · + Xn .

We are interested in determining whether limn→∞ Yn exists in the m.s. sense. By Proposition 2.11, the m.s. limit exists if and only if the limit limm,n→∞ E[Ym Yn ] Pn∧m P∞ exists and is finite. But E[Ym Yn ] = k=1 σk2 which converges to k=1 σk2 as P∞ 2 n, m → ∞. Thus, (Yn ) converges in the m.s. sense if and only if k=1 σk < ∞.

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Since E[X12 ] = 2 > 1, Cramér’s theorem implies that R∞ R ∞ − x12 p 2 b = `(2), which we compute. For a > 0, −∞ e−ax dx = −∞ e 2( 2a ) dx = πa , so Z ∞ 2 2 1 1 1 √ e−x ( 2 −θ) dx = − ln(1 − 2θ).

M (θ) = ln E[eθx ] = ln 2 2π −∞ 2.31 A large deviation

1 `(a) = max θa + ln(1 − 2θ) θ 2

1 1 = 1− 2 a 1 1 θ∗ = (1 − ) 2 a 1 b = `(2) = (1 − ln 2) = 0.1534 2 e−100b = 2.18 × 10−7 .

2.33 Sums of independent Cauchy random variables (a) ΦSn /nθ (u) = E[exp(j nuθ X1 )]n = Φ( nuθ )n = (exp(−| nuθ |)n = exp(−|u|n1−θ ) (b) Taking θ = 1 we see that Snn has the same distribution as X1 for all n.

Thus, Snn converges in distribution, and the limiting distribution is the standard Cauchy distribution. (This answer suggests the answers to parts (c) and (d).) (c) Taking θ = 2 yields that the characteristic function of Snn2 is exp(−|u|/n) which converges to 1 for all u. But 1 is the characteristic function of a random variable that is zero with probability one. Thus. Snn2 → 0 d. as n → ∞. A different approach to part (c) is to use part (b) to find that for any > 0, S n P | n2 | ≤ = P {|X1 | ≤ n } → 1 as n → ∞. n (d) Taking θ = 1/2 yields that the characteristic function of nS1/2 is exp(−|u|n1/2 ) which converges to I{u=0} for all u. But I{u=0} is not a valid characteristic n does not converge function (since it is not continuous, for example) so that nS1/2 in distribution. A different approach to part (d) is to use part (b) to find that for any constant c, n P { nS1/2 ≤ c} = P {X1 ≤ cn−1/2 } → P {X1 ≤ 0} → 1/2. But the function equal to the constant 1/2 is not a valid distribution function (it doesn’t converge to 0 at −∞, for example) so the same conclusion follows.

2.35 Chernoff bound for Gaussian and Poisson random variables 2 2 (a−µ)2 θ2 (a) MX (θ) = µθ+ θ 2σ and l(a) = maxθ aθ−(µθ+ 2σ 2) = 2σ 2 . Taking a = µ+c c2 in the optimized Chernoff inequality yields P {X ≥ E[X] + c} ≤ exp(− 2σ 2 ) for c ≥ 0. (b) The log moment generating function of Y is given by P∞ θk k −λ θ MY (θ) = ln k=0 e λk!e = ln(eλ(e −1) ) = λ(eθ − 1). Therefore, a l(a) = max aθ − eλ (θ − 1) = a ln( ) + λ − a. θ λ

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Setting a = λ + c in the optimized Chernoff inequality yields

λ+c +c .

P {Y ≥ E[Y ] + c} ≤ exp −(λ + c) ln λ 2

c (c) Using the definition of ψ yields 2λ ψ( λc ) = λg(1 + λc ) = (λ + c) ln λ+c λ − c as desired. For more information see Shorack and Wellner, Empirical Processess, 1986.

2.37 Large deviation exponent for a mixture distribution (a) e l(a) = maxθ {θa − MX (θ)} where

MX (θ) = log E[exp(θX)] = log{f E[exp(θY )] + (1 − f )E[exp(θZ)]} = log{f exp(MY (θ)) + (1 − f ) exp(MZ (θ))}.

(b) View f exp(MY (θ)) + (1 − f ) exp(MZ (θ)) as an average of exp(MY (θ)) and exp(MZ (θ)). The definition of concavity (or Jensen’s inequality) applied to the concave function log u implies that log(average) ≥ average(log), so that log{f exp(MY (θ)) + (1 − f ) exp(MZ (θ)) ≥ f MY (θ) + (1 − f )MZ (θ), where we also used the fact that log exp MY (θ) = MY (θ). Therefore, e l(a) ≤ l(a) for all a. en S Remark: This means that n is more likely to have large deviations than Snn . e

That is reasonable, because Snn has randomness due not only to FY and FZ , but also due to the random coin flips. This point is particularly clear in case the Y ’s and Z’s are constant, or nearly constant, random variables.

2.39 Bernstein’s inequality in various asymptotic regimes (a) The bound becomes !

1 2

√ 2θ .

P Sn ≥ θ n ≤ exp − 2 √ σ + 3θL n

(b) The bound becomes

P {Sn ≥ cn} ≤ exp

1 2 c n − 22 cL σ + 3

!

.

(b) The bound becomes P {Sn ≥ α} ≤ exp

1 2 α − 2 αL γ+ 3

!

.

3.1 Rotation of a joint normal distribution yielding independence

1 −1 (a) |Cov(X)| = 1 and Cov(X)−1 = so −1 2

T

!

1 1 x1 − 10 1 −1 x1 − 10 fX (x) = exp − x2 − 5 −1 2 x2 − 5 2π 2

1 1 2 2 = exp − {(x1 − 10) − 2(x1 − 10)(x2 − 5) + 2(x2 − 5) } .

2π 2

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√ 1 = λ2 − 3λ + 1, so the eigenvalues are 3±2 5 . The 1−λ

√ √ 2 1 v1 v1 3± 5 eigenvectors satisfy = 2 or v2 = −1±2 5 v1 .

1 1 v2 v2 ! √ 1√ 2 and U can be Unit length eigenvectors are thus given by √ √ −1± 5 (b)

det

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2−λ 1

5∓ 5

2

taken to be the matrix with columns given by the two eigenvectors: q  q  2√

U =  q 5− 5 2√ 5+ 5

2√

−

q 5+ 5  .

2√ 5− 5

The other possible choices for U can be obtained by swapping the two columns or

cos(θ) − sin(θ) multiplying either column by −1. Note that U has the form . sin(θ) cos(θ) 3.3 Calculation of some minimum mean square error estimators (a) Note that Cov(X, Y ) = Var(X) and Var(Y ) = Var(X) + Var(N ). By the formulas for linear MMSE estimation, (X,Y ) 1/λ2 1 1 1 1 b E[X|Y ] = E[X] + Cov Var(Y ) (Y − E[Y ]) = λ + 1/λ2 +σ2 (Y − λ ) = λ + 1+λ2 σ2 (Y − 1 Y +λσ 2 λ ) = 1+λ2 σ 2 2

2

σ E[e ] = Var(X) − Cov(X, Y )2 /Var(Y ) = λ12 − (1/λ4 )/( λ12 + σ 2 ) = 1+λ 2 σ2 b (b) No. Observe that X is nonnegative, but the estimator E[X|Y ] can be negb = max{0, E[X|Y b ative. An estimator with smaller MSE is X ]}, because (X − 2 2 b ≤ (X − E[X|Y b X) ]) with probability one, and the inequality is strict whenever b E[X|Y ] < 0.

3.5 Conditional probabilities with joint Gaussians I 2

N (ρY, 1 − ρ ). Thus, P (X ≤ 1|Y ) = P 2

(a) Given Y , X is

X−ρY 1−ρY √ ≤√ 2 2 1−ρ

1−ρ

Y

=Φ

1−ρY √ .

2 1−ρ

(b) Given Y = y, X is N (ρy, 1 − ρ ). Therefore, given Y = y, X − y is N ((ρ − 1)y, 1 − ρ2 ). Thus, E[(X − Y )2 |Y = y] = E[(X − y)2 |Y = y] = (ρ − 1)2 y 2 + 1 − ρ2 .

3.7 An estimation error bound (a) There’s not enough information to comb pute E[e2 ]. However, the LMMSE estimator E[X|Y ] has error b E[(X − E[X|Y ])2 ] = Var(X) − Cov(X, Y )2 /Var(Y ) = 8 −

7 9 = .

2 2

This is an upper bound on E[(X − E[X|Y ])2 ], since the MMSE estimate has MSE less than or equal to that of the best linear estimator. (b) It is sufficient (and necessary) that E[X|Y ] be linear, for then and only then b is it true that E[X|Y ] = E[X|Y ], and that the error for the MMSE estimator E[X|Y ] achieves its worst case value: E[e2 ] = 7/2. There is more than one b distribution satisfying the given moment conditions such that E[X|Y ] = E[X|Y ].

One possibility is to take X and Y to be jointly Gaussian. For another possibility,

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X be discrete with the joint probability mass function given in the Y following table: √ √ X =2+ 8 X =2− 8 √ 1 1 .

Y = −2 + √2 4 +α 4 −α 1 1 Y = −2 − 2 4 −α 4 +α let

3 .

This joint distribution has the given mean vector and covariance matrix if α = 16 Since Y has only two possible values, any function of Y has the form aY + b, so b E[X|Y ] = E[X|Y ].

3.9 Comparison of MMSE estimators for an example (a) The best estimator 1 of X of the form g(U ) is given by g(u) = 1+u , because this estimator has MSE equal to zero. That is, E[X|U ] = X. (b) We need to calculate some moments.

E[U ] =

Z 1 0

E[U 2 ] =

Z 1 0

udu =

1 2

u2 du =

1 3

Var(U ) = E[U 2 ] − E[U ]2 = E[X] = E[X 2 ] =

Z 1

1 − 3

2 1 1 = 2 12

0

1 du = ln(1 + u)|10 = ln 2 ≈ 0.6931 1+u

0

1 1 1 du = − | = 0.5 (1 + u)2 1+u 0

Z 1

Var(X) = E[X 2 ] − E[X]2 = 0.5 − (0.6931)2 = 0.0195 Z 1 Z 1 u 1 E[XU ] = du = 1 − du = 1 − ln 2 ≈ 0.3069 1 + u 1 + u 0 0 Cov(X, U ) = E[XU ] − E[X]E[U ] = 0.3069 − (0.5)(0.6931) = −0.0397.

(X,U ) 0.0397 1 b So E[X|U ] = E[X] + Cov Var(U ) (U − E[U ]) = 0.6931 − 1 (U − 2 ) = (0.9313) − 12

(.4764)U and the mean square error is (0.0397)2 (X,U )2 Var(X) − Cov = 0.0195 − = 0.00059.

1 Var(U ) 12 3.11 Diagonalizing a two-dimensional Gaussian distribution

1 −1 1 If U = √2 , and Y = U T X, then X = U Y and Cov(Y ) = U T KU = 1 1

λ1 0 , where the variances are λ1 = 1 + ρ and λ2 = 1 − ρ. The columns 0 λ2 of U could be swapped, in which case the eigenvalues would be swapped, and either column of U could be multiplied by minus one. Otherwise the choice of U is unique. One way to find U is to first find the eigenvalues by solving the characteristic equation det(Iλ − K) = 0, find the eigenvectors by solving the

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eigen-equation Kv = λv, and let the columns of U be the normalized eigenvectors.

3.13 Projections onto nested linear subspaces (a) By the Orthogonality Principle, since Z1∗ ∈ V1 , it must only be shown that (Z0∗ − Z1∗ ) ⊥ Z for all Z ∈ V1 .

But this follows from the three facts: Z0∗ − Z1∗ = (Y − Z1∗ ) − (Y − Z0∗ ), ∗ (Y − Z1 ) ⊥ Z for all Z ∈ V1 , (Y − Z0∗ ) ⊥ Z for all Z ∈ V0 , so for all Z ∈ V1 (b) V1 V0 (i) {a + bY1 + cY2 : a, b, c ∈ R} {a + bY1 : a, b ∈ R} (ii) {g(Y1 , Y2 ) : g so E[g(Y1 , Y2 )2 ] < ∞} {g(Y1 ) : g so E[g(Y1 )2 ] < ∞} R (iii) {a + bY1 : a, b ∈ R} 3.15 Some identities for estimators, version 2 (a) True. The estimator b E[X|Y ] yields a smaller MSE than any other function of Y , including E[X|Y, Y 2 ]. (b) True. Equality holds because the unconstrained estimator with the smallest b mean square error, E[X|Y ], is linear, and the MSE for E[X|Y, Y 2 ] is less than or equal to the MSE of any linear estimator. (c) False. For example if X is a nonconstant random variable, Y = X, and Z ≡ 0, then, on one hand, E[X|Z] = E[X], so E[E[X|Z]|Y ] = E[X], and thus E[(X − E[E[X|Z]|Y ])2 ] = Var(X). On the other hand, E[X|Y ] = X so that E[(X − E[X|Y ])2 ] = 0. (d) FALSE. For example, suppose X = Y W , where W is a random variable independent of Y with mean zero and variance one. Then given Y , the conditional distribution of X has mean zero and variance Y 2 . In particular, E[X|Y ] = 0, so that E[(X − E[X|Y ])2 ] = Var(X), but X and Y are not independent. The intuition here is that many different distributions can have the same mean, so even if the conditional mean of X given Y is the same for all Y , it doesn’t mean that the conditional distribution of X given Y is the same for all Y. 3.17 The square root of a positive semidefinite matrix (a) Yes. Suppose B is an m × m matrix over the reals. Since (BB T )T = BB T , the matrix BB T is symmetric. If α is an m dimensional vector, and β = B T α, then αT BB T α = Pm β T β = i=1 βi2 ≥ 0, so that BB T is positive semidefinite. (b) Yes. If Λ is a positive semidefinite matrix with diagonal elements λi , then λi ≥ 0. Thus, Λ1/2 can be defined as the diagonal matrix with diagonal ele1/2 ments λi . Given a symmetric positive semidefinite matrix K, it can be expressed as K = U T ΛU for some orthonormal matrix U and positive semidefinite diagonal matrix Λ. Let S = U T Λ1/2 U . Then S is positive semidefinite, and S 2 = U T Λ1/2 U U T Λ1/2 U = U T Λ1/2 Λ1/2 U = U T ΛU = K. 3.19 A quadratic estimator (a) By the Orthogonality Principle, we want (|Y | − a − bY − cY 2 ) ⊥ Y i for i = 0, 1, 2. Equivalently, E[|Y |] − a − bE[Y ] − cE[Y 2 ] = 0

E[|Y |Y ] − aE[Y ] − bE[Y 2 ] − cE[Y 3 ] = 0

E[|Y |Y 2 ] − aE[Y 2 ] − bE[Y 3 ] − cE[Y 4 ] = 0.

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or equivalently, 0.8 − a − c = 0

−b = 0

1.6 − a − 3c = 0. b = 0.4 + 0.4Y 2 . MSE=E[X 2 ] − E[X̂ 2 ] = yielding a = c = 0.4 and b = 0, so X 2 2 2 4 E[Y ] − (0.4) (1 + 2E[Y ] + E[Y ]) = 1 − 0.16(1 + 2 + 3) = 0.04.

3.21 Estimation for an additive Gaussian noise model The vectors x and y are jointly Gaussian since they are obtained from x and n via linear operations.

Thus, the conditional distribution of x given y is a Gaussian distribution. For a given value of y, that distribution has mean x + Σxy Σ−1 yy (y − y) and covariance −1 matrix equal to Σee = Σxx − Σxy Σyy Σyx where x = Ex, y = Ey, y = E[x + n] = x + n, Σxy = Cov(x, x + n) = Σx + Σxn = Σx , and Σyy = Cov(x + n, x + n) = Σx + Σn .

The proof of the equivalent formulas is left to the reader.

3.23 Steady state gains for one-dimensional Kalman filter (a) Let bk = σk2 .

The solution to Problem 3.7 shows that the sequence bk satisfies the recursion bf 2 bk+1 = F (bk ), where F (b) = 1+b +1, with the initial condition b0 = σ 2 ≥ 0 given.

The function F is positive, strictly increasing, and bounded. Thus if bk ≤ bk+1 for some k, then bk+1 = F (bk ) ≤ F (bk+1 ) = bk+2 . Therefore, if b0 ≤ b1 , then the sequence (bk : k ≥ 0) is monotone nondecreasing and bounded. Similarly, if b0 ≥ b1 then the sequence (bk : k ≥ 0) is monotone nonincreasing and bounded.

Since bounded monotone sequences have finite limits, the sequence (bk : k ≥ 0) converges.

2 (b) Denote the limit by b∞ (so b∞ = σ∞ ). Since F is a continuous function, letting k → ∞ in the equation bk+1 = √ F (bk ) yields b∞ = F (b∞ ), which has the f 2+

f 4 +4

. unique nonnegative solution b∞ = 2 (c) If f = 0, the states xk are uncorrelated with variance one. The observations y0 , . . . , yk−1 are therefore orthogonal to xk , and the variance of the error, σk2 , is just the variance of xk , equal to one for all k. The limiting variance of error is thus also one.

3.25 Estimation with jointly Gaussian random variables (a) E[W ] = E[X + 2Y + 3] = E[X] + 2E[Y ] + 3 = 13.

We use the fact Cov(X, Y ) = ρσX σY = (0.2)(3)(5) = 3 to get Var(W ) = Var(X + 2Y + 3) = Var(X) + 4 · Var(Y ) + 2 · 2 · Cov(X, Y )

= 9 + 100 + 12 = 121.

(b) Since W is a linear combination of jointly Gaussian random variables, W is

19

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Gaussian. So P {W ≥ 20} = P

W − 13 20 − 13 ≥ 11 11

Random Processes for Engineers

=Q

20 − 13 11

= Q(0.6364) = 0.2623.

(c) Since W and Y are linear combinations of the jointly Gaussian random variables X and Y , the variables W and Y are jointly Gaussian. Therefore, the best unconstrained estimator of Y given W is the best linear estimator of Y given W. So b |W ] = E[Y ] + Cov(Y, W ) (W − E[W ]). g ∗ (W ) = E[Y Var(W ) Using

Cov(Y, W ) = Cov(Y, X + 2Y + 3) = Cov(Y, X) + 2Var(Y ) = 3 + 50 = 53, we find

and

b |W ] = 4 + g ∗ (W ) = E[Y MSE = Var(Y ) −

53 (W − 13).

121

(Cov(Y, W ))2 (53)2 = 25 − = 1.7851.

Var(W ) 121

3.27 Innovations and orthogonal polynomials for the normal distribution n 2 ] (a) The coefficient of un in E[euX ] is E[X and the coefficient of un in eu /2 is n!

1 if n is even and zero if n is odd. Equating these expressions for each n (n/2)!2n/2 yields the desired identity. (b) The formula recursive formula for the linear innovations sequence yields: Ye1 = X, Ye2 = X 2 − 1, Ye3 = X 3 − 3X, and 2 4 ] E[X 4 (X 2 −1)] 15−3 2 4 Ye4 = X 4 − E[X (X − 1) = E[1] · 1 − E[(X 2 −1)2 ] (X − 1) = X − 3 − 2 4 2 X − 6X + 3. Note: This sequence of orthogonal (with respect to the N (0, 1) distribution) polynomials 1, Y , Y 2 − 1, Y 3 − 3Y , Y 4 − 6Y + 3 are known as the Hermite polynomials. The nth polynomial is n! times the coefficient of un in the power series expansion of exp(uY − u2 /2).)

3.29 Representation of three random variables with equal cross covariances (a) A necessary and sufficient condition for K to be a covariance matrix is that it be symmetric and positive semidefinite. Since it is symmetric for any value of a, the only requirement is to make K positive semidefinite. Equivalently, a must be such that the determinant of every matrix obtained by deleting a set of rows and the corresponding set of columns from K, is nonnegative. Only three distinct matrices arise in such a way:

1 a (1) K itself. a 1 Thus, K is a valid covariance matrix if and only if 1 − a2 ≥ 0 and det(K) = 1 + 2a3 − 3a2 = (1 − a)2 (1 + 2a) ≥ 0. Equivalently, it is necessary and sufficient

Solutions to Odd Numbered ProblemsRandom Processes for Engineers

that − 12 ≤ a ≤ 1.

(Another way to solve part (a) would have been to observe that K is symmetric, so that it has real-valued eigenvalues, and then K is positive semidefinite if and only the eigenvalues, found in part (b) below, are nonnegative.) (b) The eigenvalues of K are zeros of the characteristic polynomial in the variable λ, det(Iλ−K) = λ3 −3λ2 +3(1+a2 )λ−1−2a3 −3a2 . By the hint, one of the factors of the polynomial must be λ − 1 − 2a. Taking out this factor leaves a quadratic polynomial, which can be factored further. The result is that det(Iλ − K) = (λ − (1 − a))2 (λ − 1 − 2a). Thus, the eigenvalues are λ1 = λ2 = 1 − a and λ3 = 1 + 2a. We selected this ordering of the eigenvalues because the given third column of U is the eigenvector for eigenvalue λ3 = 1 + 2a. The first two columns of U can be any two vectors of length one that are orthogonal to each other and each have sum of entries equal to zero. The diagonal entries of Λ are the eigenvalues, λ1 , λ2 , λ3 . For example, one possible choice of U gives a representation of a random vector X with the covariance matrix K: 

 X=

√1 2 − √12

0

√1 6 √1 6 − √26

√1 3 √1 3 √1 3



√  ( 1 − a)Z1 √  ( 1 − a)Z2  .  √ ( 1 + 2a)Z3

(Note: It is interesting to see what this representation reduces to in the two cases that K is singular, namely when a = 1 or a = − 21 .)

3.31 Kalman filter for a rotating state (a) Given a nonzero vector xk , the vector F xk is obtained by rotating the vector xk one tenth revolution counterclockwise about the origin, and then shrinking the vector towards zero by one percent. Thus, successive iterates F k xo spiral in towards zero, with one-tenth revolution per time unit, by about ten percent per revolution. and shrinking

a c (b) Suppose Σk|k−1 = . Note that HkT Σk|k−1 Hk is simply the scaler c b a. The equations for Σk+1|k and Kk can thus be written as "

Σk+1|k = F Σk|k−1 − Kk = F

a 1+a c 1+a

!

a2 1+a ac 1+a

ac 1+a c2 1+a

!#

FT + Q

.

4.1 Event probabilities for a simple random process (a) There are four sample functions corresponding to the four possible values of (A, B) as shown in Figure 1.1.

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(1,1) 2 (-1,1) -2

2 -2

(1,-1)

(-1,-1)

Figure 1.1 The four sample paths.

(b) Let Eij be the event that A = i and B = j. Then  P (E−1,−1 ∪ E1,−1 ) = 1/2 if     1 − P (E−1,1 ) = 3/4 if  P {Xt ≥ 0} = P (E1,−1 ∪ E1,1 ) = 1/2 if    1 − P (E−1,−1 ) = 3/4 if   P (E1,1 ∪ E−1,1 ) = 1/2 if

t < −2 t = −2 −2 < t < 2 t=2 t>2

(c) The set {ω : Xt (ω) ≥ 0 all t} is empty, because for each ω, Xt (ω) < 0 for some t. Thus P {Xt ≥ 0 all t} = 0.

4.3 A sinusoidal random process Since Xt = A cos(2πV t) cos(Θ) − A sin(2πV t) sin(Θ) and E cos(Θ) = E sin(Θ) = 0, it follows that µX (t) ≡ 0. The autocorrelation function of X is given by RX (s, t) = E[A2 ]E[cos(2πV s + Θ) cos(2πV t + Θ)] E[A2 ] = E[cos(2πV (s − t)) + cos(2πV (s + t) + 2Θ)] 2 Z 5

E[A2 ] 1 = cos(2πv(s − t))dv + 0 = 4sinc(10(s − t)).

2 0 5 where we used the fact that E[A2 ] = (EA)2 + Var(A) = 8 and the definition sinc(u) = sin(πu) if u 6= 0 and sinc(0) = 1. πu 4.5 A random line (a) Three sample paths are shown in Figure 1.2. The graph of each sample path is a line with positive intercept at t = 0 and negative slope. (b) No, X is not Markov. For example, given X0 , the conditional distribution of X1 has a pdf, namely fX1 |X0 (x1 |x0 ) = f (x0 − x1 ), where f is the Rayleigh pdf with parameter σ 2 . But given X0 and X−1 , X1 is completely determined: X1 = X0 + (X0 − X−1 ). Thus, the conditional distribution of X1 given X0 is not equal to the conditional distribution of X1 given X0 and X−1 . So X is not Markov. (c) No, X does not have independent increments. For example, the increments X1 − X0 and X2 − X1 are both random, and are equal to each other with

Solutions to Odd Numbered ProblemsRandom Processes for Engineers

R

slope !S t

Figure 1.2 Three sample paths

probability one. So they are not independent, even though the intervals (0,1) and (1,2) are disjoint.

R2 (d) The triangle has height R and base of width R S , so A = 2S . Since R and S are independent, E[A] = E[R2 ]E[ S1 ]/2. Since R2 has the same distribution as the 2 2 . ) random variables, E[R2 ] = 2σR sum of the squares of two independent N (0, σR By the law of the unconscious statistician, Z ∞

1 s2 1 s E exp − 2 ds = S s σS2 2σS 0 √ Z ∞ r

s2 1 2π 1 π p = exp − 2 ds = .

2 σS 0 2σS σS 2 2πσS σ2 p Combining the above yields E[A] = σRS π2 . Note: As a partial check of this answer, we check the dimensional units. If t is time in seconds and Xt is location in meters, then σR is in meters, σS is in meters/second, and E[A] is in (meters)×(seconds).

4.7 Some probabilities for a Brownian motion The event in question is the same as {2W3 ≥ W2 + W4 + 2} or {(W3 − W2 ) − (W4 − W3 ) ≥ 2}. Since W3 − W2 and W4 − W3 are mutually independent N (0, 1) random variables, their difference√is a N (0, 2) random variable.

Thus, the probability in question is Q( √22 ) or Q( 2).

ALTERNATIVELY, the event in question is the same as {Z ≥ 2} where Z = −W2 + 2W3 − W4 . Since W is a Gaussian process, Z is a Gaussian random variable. Since the W ’s have mean zero, E[Z] = 0. Finally, Var(Z) = Cov(Z, Z) = Var(W2 ) + 4Var(W3 ) + Var(W4 ) − 4Cov(W2 , W3 ) − 4Cov(W3 , W4 ) + 2Cov(W2 , W4 ) = 2 + 12 + 4 √ − 8 − 12 + 4 = 2. Thus, the probability in question is P {Z ≥ 2} = Q( √22 ) or Q( 2).

W2

t √ t has the N (0, 1) distribution so for any t, (b) For any t, W t has the same t distribution as Z 2 where Z is a N (0, 1) random variable. The pdf is found as fol√ √ lows. P {Z 2 ≥ c} = 2P {Z ≥ c} = 2Q( c). Taking the derivative with respect 2 to c and using the chain rule and definition of Q yields fZ 2 (c) = − dP {Zdc ≥c} = 1 √1 1 2 2√ e−c/2 = √2πc e−c/2 for c ≥ 0. (Z 2 has the chi-square distribution with c 2π

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one degree of freedom, or, equivalently, the gamma distribution with parameter 1/2.)

4.9 Brownian bridge (a) W t t 1

Bt

t 1

Figure 1.3 Corresponding sample paths of a Brownian motion and a Brownian bridge

(b) For s, t ∈ [0, 1], if s ≤ t: RB (s, t) = E[(Ws − sW1 )(Wt − tW1 )] = s − st − st + st = s − st = s(1 − t). It follows by switching s and t that if s ≥ t: RB (s, t) = (1 − s)t. Therefore,

s(1 − t) for 0 ≤ s ≤ t ≤ 1 RB (s, t) = t(1 − s) for 0 ≤ t ≤ s ≤ 1 Equivalently, RB (s, t) = (s ∧ t)(1 − s ∨ t) for s, t ∈ [0, 1]. (c) Letting ρ(s, t) denote the correlation coefficient between Xs and Xt , we find that for s < t: r r s 1−t Cov(Bs , Bt ) s(1 − t) . ρ(s, t) = p =p = 1 − s t Var(Bs )Var(Bt ) s(1 − s) · t(1 − t)

Therefore, for 0 ≤ r < s < t ≤ 1, ρ(r, s)ρ(s, t) = ρ(r, t), so that by Example 4.8.3 of the notes, the Gaussian process B is a Markov process. (d) It is enough to show that Cov(Bt , W1 ) = 0 for 0 ≤ t ≤ 1, because uncorrelated jointly Gaussian vectors are independent. But Cov(Bt , W1 ) = Cov(Wt − tW1 , W1 ) = Cov(Wt , W1 ) − tCov(W1 , W1 ) = t − t = 0. s t (e) For 0 ≤ s ≤ t < 1, 1−s ≤ 1−t . Therefore, s , (1 − t)W t ) RX (s, t) = Cov((1 − s)W 1−s 1−t s , W t ) = (1 − s)(1 − t) = (1 − s)(1 − t)Cov(W 1−s 1−t

s = s(1 − t).

1−s

This expression holds for 0 ≤ s ≤ t = 1 as well. Similarly, if 0 ≤ t ≤ s ≤ 1, RX (s, t) = (1 − s)t. That is, X has the same autocorrelation function as the process B. Since both X and B are mean zero Gaussian random processes with identical autocorrelation functions, they have the same finite-dimensional distributions. Also, both have continuous sample paths. So X is also a Brownian bridge process.

4.11 Some Poisson process calculations (a) (Method 1) Given N2 = 2, the distribution of the locations of the first two jump points are as if they are independently and uniformly distributed over the interval [0, 2]. Thus, each falls in the second half of the interval with probability 1/2, and they both fall into the second interval with probability ( 12 )2 = 41 . Thus, at least one of them falls in the

Solutions to Odd Numbered ProblemsRandom Processes for Engineers

first half of the interval with probability 1 − 41 = 34 .

(Method 2) Since N1 and N2 − N1 are independent, Poi(λ) random variables, P {N1 ≥ 1, N2 = 2} = P {N1 = 1, N2 = 2} + P {N1 = 2, N2 = 2}

= P {N1 = 1, N2 − N1 = 1} + P {N1 = 2, N2 − N1 = 0} 2 −λ 3λ2 e−2λ λ e e−λ = = (λe−λ )(λe−λ ) + 2 2

2 −2λ

2 −2λ

2 −2λ

and P {N2 } = (2λ) 2e . Thus, P (N1 ≥ 1|N2 = 2) = 3λ e2 /( (2λ) 2e ) = 34 .

2 −2λ (b) P {N1 ≥ 1} = 1 − e−λ . Therefore, P (N2 = 2|N1 ≥ 1) = 3λ e2 /(1 − e−λ ). (c) Yes. The process N is Markov (because N0 is constant and N has independent increments), and since Xt and Nt are functions of each other for each t, the process X is also Markov. The state space for X is S = {0, 1, 4, 9, . . .}. For i, j ∈ Z and t, τ ≥ 0, ( (λτ )j−i e−λτ 0≤i≤j 2 2 (j−i)! pi2 ,j 2 (τ ) = P (Xt+τ = j |Xt = i ) = 0 else 4.13 A simple discrete-time random process (a) The thick lines in the following sketch indicate a sample path of U , and the corresponding sample path of X is given by thick and thin lines taken together.

(b) Yes, X is stationary, since it is obtained from the stationary process U be applying a time-invariant function to U . (c) No, it is not Markov. For example, P (Xk+1 ≥ 0.5|Xk ≥ 0.5) < 1, but P (Xk+1 ≥ 0.5|Xk ≥ 0.5, Xk−1 < 0.5) = 1. (d) For any n, the range of Xn is the interval [0, 1]. For 0 ≤ c ≤ 1, P {Xn ≤ c} = P {Un−1 ≤ c and Un ≤ c} = c2 . Thus,   0 c≤0 FX,1 (c, n) = c2 0 ≤ c ≤ 1  1 c≥1 Equivalently,

fX,1 (c, n) =

2c 0 ≤ c ≤ 1 0 else

(e) If |n − m| ≥ 2 then Xn and Xm are independent, and their joint CDF is simply the product of their individual CDFs. So consider the joint distribution

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of (Xn , Xn+1 ) for a fixed n. This takes values in the unit square [0, 1]2 , so let (c1 , c2 ) ∈ [0, 1]2 . Then P {Xn ≤ c1 , Xn+1 ≤ c2 } = P {Un−1 ≤ c1 , Un ≤ min{c1 , c2 }, Un+1 ≤ c2 } = c1 min{c1 , c2 }c2 . Thus, for (c1 , c2 ) ∈ [0, 1]2 ,  2 2 n=m  min{c1 , c2 } FX,2 (c1 , n; c2 , m) = c1 c2 min{c1 , c2 } |n − m| = 1  c21 c22 |n − m| ≥ 2

4.15 Sliding function of an i.i.d. Poisson sequence (a) It suffices to show that the sum of two independent Poisson random variables is again a Poisson random variable. One proof is to use z-transforms (characteristic functions could be used instead). If Ai is a Poisson random variable with parameter λi for i = 1, 2, then Ai has z transform exp(λi (z − 1)). If A1 is independent of A2 then the z-transform of the sum A1 + A2 is the product of the z-transforms, exp(λ1 (z − 1)) exp(λ2 (z − 1)) = exp((λ1 + λ2 )(z − 1)), which is the z-transform of a Poisson random variable with mean λ1 + λ2 . Thus, A1 + A2 is a Poisson random variable with parameter λ1 + λ2 . (b) For any distinct integers t1 , . . . , tn and any integer s, the vector (X(t1 + s), . . . , X(tn + s)) is a vector of n independent, Poi(λ) distributed random variables. This distribution of the vector is the same for all s, so, by definition, X is a stationary random process. (c) Yes, Y is a stationary random process. The vector (Y (t1 + s), . . . , Y (tn + s)) can be written as G(X(t1 +s), X(t1 +1+s), X(t2 +s), X(t2 +1+s) . . . , X(tn +s), X(tn +1+s)) where G is the function defined by G(x1 , . . . , x2n ) = (x1 + x2 , x3 + x4 , . . . , x2n−1 + xn ).

The function G does not depend on s. Since the distribution of the vector that G is applied to does not depend on s, the distribution of the result, (Y (t1 + s), . . . , Y (tn + s)), also does not depend on s, so Y is a stationary random process.

4.17 Invariance of properties under transformations (a) FALSE. For example, let P {X0 = i} = 31 for i ∈ {−1, 0, 1} and let the sample paths all have the form . . . , −1, 0, 1, −1, 0, 1, −1, 0, 1, . . .. Then X is a Markov process on {−1, 0, 1}   0 1 0 with one-step probability matrix P =  0 0 1 . But Y has sample paths of 1 0 0 the form . . . , 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, . . .. Thus, P (Yn+1 = 1|Yn = 1) = 0.5 and P (Yn+1 = 1|Yn = 1, Yn−1 = 0) = 1. Since these probabilities are not equal, Y is not Markov. (The problem is that Yn has less information than Xn because squaring is not a one-to-one function.) (b) TRUE, because Xn and Zn have the same information for each n. (c) FALSE. For example, let X have the three possible sample paths looking like . . . , −1, 1, 1, −1, 1, 1 − 1, 1, 1, . . ., each with probability one third. Then X is not Markov (similar to Y in part (a) ), but Yn = 1 for all n, so that Y is a Markov

Solutions to Odd Numbered ProblemsRandom Processes for Engineers

27

process.

The problem is that Yn has less information than Xn . (d) TRUE. If the distribution of (Xs+t1 , . . . , Xs+tn ) is the same for all s, then 2 2 the distribution of (Xs+t , . . . , Xs+t ) is the same for all s.

1 n (e) FALSE. For example, let X be the deterministic process such that X0 = −1 and Xn = 1 for n 6= 1. Then X is not stationary, because X does not have the same distribution as X shifted in time. But Yn = 1 for all n, so that Y is trivially stationary. (f) FALSE. For example if the Xn are independent, mean zero, variance one random variables, but they don’t all have the same fourth moments, then X is WSS and Y is not. (g) FALSE. For example, let Xn = 0 for n ≤ 0 and Xn = W1 +· · ·+Wn for n ≥ 0, where the Wn are independent random variables with P {Wn = 0} = P {Wn = 1} = 0.5. Then X has independent increments. But P (Y2 − Y1 = 3|Y1 − Y0 = 1) = 0.5 and P (Y2 − Y1 = 3|Y1 − Y0 = 0) = 0, so that the increment Y1 − Y0 is not independent of the increment Y2 − Y1 . (h) FALSE. For example, let Xn = 0 for n ≤ 0 and Xn = B for n ≥ 1, where P {B = −1] = 32 and P (B = 2) = 13 . Then X is a martingale, because its only jump, of size B, has mean zero. But Z is not a martingale because E[Z1 |Zn , n ≤ 0] = E[B 3 ] = 2 6= Z0 .

4.19 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 0 , as explained in the statement of Problem 8.14. Here is another proof. Let n ≥ 1 and fix t1 , . . . , tn . Let u0 < u1 < · · · < um , be such that ti − 1 and ti are both in {u0 , . . . , um } for 1 ≤ i ≤ n. Then there is an m × n matrix G, with each entry equal 0 or 1, such that (Xt1 +s , . . . , Xtn +s ) = (N (u0 + s, u1 + s], . . . , N (um−1 + s, um + s]) G.

(1.1)

for all s. The random vector in the right-hand side of (1.1) consists of independent Poisson random variables, with respective means u1 − u0 , · · · , um − um−1 . Note that such distribution doesn’t depend on s. Therefore, the distribution of the

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left-hand side of (1.1) 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 X1 = 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 rigorous answer would be to give an explicit example violating the conditional independence property. Here is one, based on the above intuition. Clearly, P (X0.5 = 0|X0 = 1, X−0.25 = 0) = 0, whereas P (X0.5 = 0|X0 = 1) > 0. Since these are not equal, X is not a Markov process. For completeness, we can calculate P (X0.5 = 0|X0 = 1) as follows.

Given X0 = 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 X0 = 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 X0 . So, P (X0.5 = 0|X0 = 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(−2λ). (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, R 1 R 1−t with parameter λ. Then P {Xt > 0 for t ∈ [0, 1]} = P {S + T ≤ 1} = 0 0 λ2 e−λ(s+t) dsdt = 1 − (1 + λ)e−λ .

4.21 Hitting the corners of a triangle Let hi be the mean time to hit {3, 4, 5} from initial state i. We are interested in h1 = E[τB ]. By symmetry, h2 = h6 . By conditioning on the first step, we find: h1 = 1 + h2

1 h2 = 1 + h1 .

2

Solving yields (h1 , h2 ) = (4, 3). Thus, E[τB ] = 4. (b) Redefine hi to be the mean time to hit state 3 from initial state i. By symmetry, h1 = h5 and h2 = h4 . By conditioning on the first step, we find h1 = 1 + (h2 + h6 )/2

1 h2 = 1 + h1 h6 = 1 + h1 .

2

Solving yields (h1 , h2 , h6 ) = (8, 5, 9). Thus, E[τ3 ] = 8. (c) The mean time to reach {3, 4, 5} is four. When {3, 4, 5} is first reached, the state is either 3 or 5. If 3 is reached first, then the process continues until state 5 is reached. If 5 is reached first, then the process continues until state 3 is reached. Once either state 3 or 5 is reached, by symmetry, the mean amount of additional time needed to reach the other is the same as E[τ3 ] in part (b). So

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E[τC ] = 4 + 8 = 12. (d) The difference τR − τC is the time needed to reach state 1 starting from either state 3 or 5, and by symmetry, the mean of this difference is E[τ3 ]. Thus, E[τR ] = 4 + 8 + 8 = 20.

4.23 Time elapsed since Bernoulli renewals (a) We take S to be the set of positive integers (i.e. the natural numbers). Given Xn = i, either Xn+1 = i + 1, which happens if there is no renewal at time n, or Xn+1 = 1, which happens if there is a renewal at time n. Thus,   1 − p if j = i + 1 pij = p if j = 1  0 else. (b) For k ∈ Z and i ≥ 1,

P {Xk = i} = P {Uk−1 = 0, Uk−2 = 0, . . . , Uk−i+1 = 0, Uk−i = 1}

= P {Uk−1 = 0}P {Uk−2 = 0} · · · P {Uk−i+1 = 0}P {Uk−i = 1}

= (1 − p)i−1 p.

So Xk has the geometric distribution with parameter p, Geo(p). (c) Yes, X is a stationary random process. A short justification is that X is obtained by applying a sliding function to a stationary random process. To elaborate, for any n ≥ 1 and integers s, t1 , · · · , tn , the random vector (Xt1 , · · · , Xtn ) is a function of (Uk : k ∈ Z). The vector (Xt1 +s , · · · , Xtn +s ) is obtained by applying that same function to (Uk+s : k ∈ Z). That is, time shifting X is equivalent to time-shifting U . The process U , being comprised of independent and identically distributed random variables, is stationary, so that X is also stationary. (d) For i ≥ 1 and k ≥ 1 and n ∈ Z, given Xn = i, the set of possible values of Xn+k are Xn + k, which happens if there are no renewals at the k times {n, . . . , n + k − 1}, or a number j with 1 ≤ j ≤ k, which happens if Un+k−j = 1, Un+k−j+1 = 0, . . . , Un+k−1 = 0. Thus,  k if j = i + k  (1 − p) j−1 pij (k) = p(1 − p) if 1 ≤ j ≤ k  0 else.

4.25 Reinforcing samples (a) Yes. Given the current state is Xk = j, the future actions are conditionally independent of how state j was arrived at. (b) Yes. Given Xk = j, the probability the new ball added at time k + 1 is blue is kj . Thus, E[Xk+1 |Xk = j] = j + kj = j(1+k) . Conditioning on {Xk = j} k is the same as conditioning on {Mk = kj }. Dividing through by k + 1 yields E[Mk+1 |Mk = kj ] = kj for all j, or E[Mk+1 |Mk ] = Mk . (c) E[Vk+1 |Vk = j(k−j) k2 ] = (k+1−(j+1))(j+1) j (k+1−j)(j) k−j (k+2)(k) (k−j)(j) k(k+2) + = .

Thus, E[V |V ] = k+1 k (k+1)2 k (k+1)2 k (k+1)2 k2 (k+1)2 Vk .

1 It follows that E[Vk+1 ] = k(k+2) (k+1)2 E[Vk ]. With the initial condition E[V2 ] = V2 = 4 , 1 1 2 we find E[Vk ] = k+1 6k → 6 . Therefore, 6 = E[M∞ (1−M∞ )] = E[M∞ ]−E[M∞ ] =

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1 1 1 1 2 1 2 2 2 − E[M∞ ]. Thus, E[M∞ ] = 3 so that Var(M∞ ) = 3 − ( 2 ) = 12 .

(d) It will be shown that Mk is uniformly distributed over the set of k − 1 points: { k1 , . . . , k−1 k }. This is proved by induction on k. The statement is trivially true for k = 2, establishing the base case of the induction proof. If the statement is i } = P {Mk = ki } k−i true for k ≥ 2, then for 1 ≤ i ≤ k, P {Mk+1 = k+1 k +P {Mk = i−1 i−1 1 k−i i−1 1 } = ( + ) = .

The statement is thus true for k + 1, and hence k k k−1 k k k for all k by induction. Therefore, the CDF of Mk converges pointwise to the CDF of a random variable uniformly distributed on the interval [0, 1]. Since (Mk ) converges to M∞ in the a.s. sense, it also converges in distribution. So the limit M∞ is uniformly distributed on the interval [0, 1]. This result gives another proof of 1 . the fact, found in part (c), that Var(M∞ ) = 12

4.27 A space-time transformation of Brownian motion (a) If s ≤ t then ρt = E[(Xs + Xt − Xs )2 ] = E[Xs2 ] + E[(Xt − Xs )2 ] ≥ E[Xs2 ] = ρs . (b) If s ≤ t then E[Xs Xt ] = E[Xs2 ]+E[Xs (Xt −Xs )] = ρs . Generally, RX (s, t) = min{ρs , ρt } = ρmin{s,t} . (c) Increments of Y over disjoint intervals are also increments of W over disjoint intervals, and so are independent. Replacing u by ρt in the identity E[W (u)2 ] = u yields E[Yt2 ] = ρt . st (d) Note that for s, t > 0, RZ (s, t) = E[stW ( 1s )W ( 1t )] = st min{ 1s , 1t } = min{ st s, t }= min{s, t}. This expression continues to hold for s, t ≥ 0. Thus, Z is a Gaussian process with the same autocorrelation function as a standard Brownian motion.

Thus it has independent increments. (Note: It can also be shown that Z is sample path continuous, so Z is a Brownian motion.)

4.29 Identification of special properties of two discrete-time processes (a) Let k ≥ 0. Think of k is the present time. The future Xk+1 , Xk+2 , · · · is determined

by Xk and Uk , Uk+1 , · · · . But (Uk , Uk+1 · · · ) is independent of (X0 , · · · , Xk ). So X is Markovian. Since E[Xk+1 |X0 , · · · , Xk ] = 21 (1 + Xk ), which is not equal to Xk , the process X is not a martingale. To see if X is an independent increment process, we begin by asking, “Is X1 − X0 independent of X2 − X1 ?” or equivalently, “Is 2U0 − 1 independent of U1 + 2U0 U1 − 2U0 ?” Given 2U0 − 1 = 2u − 1, the conditional distribution of U1 + 2U0 U1 − 2U0 is uniform over the interval [−2u, 1]. This distribution depends on u. So these two increments of X are not independent. Thus X is not an independent increment process. (b) As a first test of whether Y is Markovian, let us see if the conditional independence condition Yk − Yk+1 − Yk+2 holds for k ≥ 2. Equivalently, we check to see if Cov(Yk , Yk+2 ) = Cov(Yk , Yk+1 )Var(Yk+1 )−1 Cov(Yk+1 , Yk+2 ).

But this condition boils down to 1 = 2(3)−1 2, which is false. Thus, X is not Markovian. Note that for k fixed, (Y0 , · · · , Yk ) and (V0 · · · , Vk ) are functions of each other. Thus for k ≥ 1, E[Yk+1 |Y0 , · · · , Yk ] = E[Vk+1 + Vk + Vk−1 |Vk , Vk−1 , · · · , V0 ] = Vk + Vk−1 6= Yk ,

Solutions to Odd Numbered ProblemsRandom Processes for Engineers

so Y is not a martingale. Since Cov(Y1 −Y0 , Y4 −Y3 ) = Cov(V1 , V4 −V1 ) = −1 6= 0, Y does not have independent increments.

4.31 Identification of special properties of two continuous-time processes (a) Since for each t, Wt and Zt are functions of each other, they have equivalent information. Since W is Markov, so is Z. Let t0 < t1 < · · · < tn+1 . Conditioning on (Zt0 , Zt1 , · · · , Ztn ) is equivalent to conditioning on (Wt0 , Wt1 −Wt0 , · · · , Wtn − Wtn−1 , Ztn ), which is independent of Wtn+1 − Wtn . Therefore,

σ 2 (tn+1 − tn ) |Zt , · · · Ztn E[Ztn+1 |Zt1 , · · · , Ztn ] = E Ztn exp Wtn+1 − Wtn − 2

σ 2 (tn+1 − tn ) = Ztn . = Ztn E exp Wtn+1 − Wtn − 2 So Z is a martingale. The process Z does not have independent increments. One way to prove it is to compute the covariance between two increments. Here is a less tedious way. Let A = Z1 − Z0 and B = Z2 − Z1 . Note that B = (1 + A)R, 2 where R = exp((W2 − W1 ) − σ2 ) − 1, and that A is independent of R. So E[B 2 |A] = E[R2 ](1 + A)2 , so that E[B 2 |A] depends on A. Thus A and B are not independent, so Z does not have independent increments. (b) R has independent increments because N has independent increments and the D’s are iid. In addition, R0 = 0, so R is a Markov process. Note that for t≥0 ER = =

∞ X

n=0 ∞ X n=o

E[R|Nt = n]P {Nt = n} E[D1 + · · · + Dn ] P {Nt = n} = 0. {z } | 0

So R is a mean zero independent increment process, implying that R is also a martingale.

4.33 A branching process (a) Yes. Let n ≥ 1 be the present time. Given Y0 = y0 , . . . , Yn = yn , the statistics of the future depend only on the fact that there are yn individuals in generation n, because all individuals have the same distribution for the number of offspring, and they are independent. (b) E[Yk+1 |Yk ] = E[Yk+1 |Y0 , . . . , Yk ] = mYk , because each of the Yk individuYk als has an average of m offspring. So Mk = m k is a martingale. That is, take k ck = m .

P∞ (c) am+1 = P {Ym+1 = 0} = j=0 P (Ym+1 = 0|Y1 = j)pj . But given Y1 = j, there are no individuals in the (m + 1)st generation if and only if the j subpopulations from j individuals in generation one are extinct within m generations after they begin. Thus, P (Ym+1 = 0|Y1 = j) = (am )j . Therefore, am+1 = Ψ(am ), P∞ where Ψ is the z-transform of p: Ψ(z) = k=0 z k pk . The iterations are indicated in the figure for a case with m < 1 and for a case with m > 1.

31

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1 m<1

m>1

0 0

a

1

(d) The function Ψ is convex and increasing on the interval [0, 1], with Ψ(0) = p0 and Ψ(1) = 1. Since a1 > a0 = 0, it follows by induction that ak is increasing in k, so that the limit a∞ exists. Since Ψ is continuous, taking the limit in the equation am+1 = Ψ(am ) shows that a∞ = Ψ(a∞ ). If amin is the smallest positive solution to a = Ψ(a), it also follows by induction on m that am ≤ amin for all m, so that a∞ ≤ amin . Therefore, a∞ = amin . Since Ψ0 (1) = m, we see that a∞ = 1 if and only if m ≤ 1 (except a∞ = 1 in the degenerate case that p1 = 1). That is, the population eventually becomes extinct almost surely if and only if the mean number of offspring per individual is less than or equal to one (except for the degenerate case that p1 = 1).

P∞ 1−θ . The equation (e) For the specific choice of p, Ψ(z) = k=1 z k θk (1 − θ) = 1−zθ 2 Ψ(a) = a is equivalent to θa − a + 1 − θ = 0, which is a quadratic equation 1−θ 1 with solutions 1 and 1−θ θ . Thus, a∞ = min{1, θ } = min{1, m }. Note that, consistent with part (d), a∞ < 1 if and only if m > 0.

4.35 Mean hitting time for a discrete-time Markov process  

.6 .4 0 4 2 1 , , .

P =  .8 0 .2  π = 7 7 7 0 .4 .6 The equations for the mean hitting times are a1 = 1 + (.6)a1 + (.4)a2 and a2 = 1 + (.8)a1 + (.2)a3 . Using the fact a3 = 0 and solving yields a1 = 17.5 and a2 = 15.

4.37 Poisson merger Suppose Xi is a Poisson random variable with mean λi for i ∈ {1, 2} such that X1 and X2 are independent. Let X = X1 + X2 , λ = λ1 + λ2 , and pi = λi /λ. Then for k ≥ 0, Pi e−λ1 λi1 e−λ2 λk−j Pi 2 P {X = k} = = j=0 P {X1 = j}P {X2 = k − j} = j=0 i! (k−j)!

−λ k Pn −λ k k j k−j e λ e λ ] = k! , where we use the fact that the sum in square j=0 j p1 p2 k! [ brackets is the sum over the binomial distribution with parameters k and p1 .

Thus, the sum of two independent Poisson random variables is a Poisson random variable. By induction this implies that the sum of any number of independent Poisson random variables is also Poisson.

Now suppose that Nj = (Nj (t) : t ≥ 0) is a Poisson process with rate λj for

Solutions to Odd Numbered ProblemsRandom Processes for Engineers

33

1 ≤ j ≤ K, and that N1 , . . . , NK are mutually independent. Given t0 < t1 < · · · < tp , consider the following array of random variables: N1 (t1 ) − N1 (t0 ) N2 (t1 ) − N2 (t0 ) .. .

N1 (t2 ) − N1 (t1 ) N2 (t2 ) − N2 (t1 ) .. .

··· ··· ··· ···

NK (t1 ) − NK (t0 ) NK (t2 ) − NK (t1 )

N1 (tp ) − N1 (tp−1 ) N2 (tp ) − N2 (tp−1 ) .. .

NK (tp ) − NK (tp−1 ) (1.2) The rows of the array are independent by assumption, and the variables within a row of the array are independent and Poisson distributed by Proposition 1.5.1, characterization (iii). Thus, the elements of the array are mutually independent and each is Poisson distributed. Let N (t) = N1 (t) + · · · + NK (t) for all t ≥ 1.

Then the vector of random variables N (t1 ) − N (t0 ) N (t2 ) − N (t1 )

···

N (tp ) − N (tp−1 )

(1.3)

is obtained by summing the rows of the array. Therefore, the variables of the row are independent, and for any i the random variable N (ti+1 ) − N (ti ) has the P oi(λ(λ(ti+1 − ti )) distribution, where λ = λ1 + · · · + λK . Thus, N is a rate λ Poisson process by Proposition 1.5.1, characterization (iii).

4.39 Poisson method for coupon collector’s problem (a) In view of the previous problem, the number of coupons of a given type i that arrive by time t has the Poisson distribution with mean kt , and the numbers of arrivals of different types are independent. Thus, at least one type i coupon arrives with probability 1 − e−t/K , and p(k, t) = (1 − e−t/k )k . −c (b) Therefore, p(k, k ln k + kc) = (1 − e k )k , which, by the hint, converges to −c e−e as k → ∞. (c) The increments of A over intervals of the form [k, k + 1] are independent, Poi(1) random variables. Thus, the central limit theorem can be applied to A(t) √ for large t, yielding that for any constant D, limt→∞ P {A(t) ≥ t+D t} = Q(D), where Q is the complementary normal CDF. The problem has to do√with deviations of size (c−c0 )k, which for any fixed D, grow to be larger than D k ln k + kc.

Thus, with = |c − c0 |, P {|A(k ln k + kc) − (k ln k + kc)| ≥ k} ≤ Q(D) for k large enough, for any D. Thus, limk→∞ P {|A(k ln k + kc) − (k ln k + kc)| ≥ k} = 0, which is equivalent to the required fact. (d) Let Ct be the event that the collection is complete at time t. Condition on the value of A(t). p(k, t) = P (Ct ) = P (Ct |A(t) ≥ n)P (A(t) ≥ n) + P (Ct |A(t) < n)P (A(t) < n).

(1.4) First we bound from below the right side of (1.4). Since getting more coupons can only help complete a collection, P (Ct |A(t) ≥ n) ≥ P (Ct |A(t) = n) = d(k, n), and other terms on the righthand side of (1.4) are nonnegative. This yields the desired lower bound.

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Similarly, to bound above the right side of (1.4) we use P (Ct |A(t) ≥ n) ≤ 1 and P (Ct |A(t) < n) ≤ d(k, n). (e) Fix c. The first part of the inequality of part (c) yields that d(k, n) ≤ p(k, t)/P {A(t) ≥ n}. Suppose n = k ln k + kc. Let c0 > c and take t = k ln k + kc0 −c0 by part (b) and P {A(t) ≥ n} → 1 Thus, as k → ∞. Then p(k, t) → e−e 0 −c −e−c 0 lim sup d(k, n) ≤ e . Since c is arbitrary with c0 > c, lim sup d(k, n) ≤ e−e . −c −c A argument shows that lim inf d(k, n) ≥ e−e . Therefore, lim d(k, n) = e−e .

4.41 A state space reduction preserving the Markov property (a) 0.6 2 0.3

0.8 0.1 0.8 0.2 1

3

0.2

−π1 + 0.1π2 + 0.2π3 = 0 Solve πP = π, which can be written as 0.8π1 − 0.4π2 + 0.8π3 = 0 Adding four 0.2π1 + 0.3π2 − π3 = 0 times the first equation to the second yields −3.2π1 + 1.6π3 = 0, or π3 = 2π1 .

Substituting that into the first equation yields −0.6π1 + 0.1π2 = 0, or π2 = 6π1 .

Thus, π = (π1 , 6π1 , , 2π1 ). Since the probabilities must sum to one, π = ( 19 , 23 , 29 ). (b) Since p1,2 = p3,2 = 0.8, the probability the next state is 2 given that the current state is either 1 or 3 is 0.8. That is, if states 1 and 3 are grouped together, then given the process is in that group, the probability of jumping to state 2 is 0.8. Thus, we take f (1) = f (3) = a and f (2) = b for distinct values a and b. The 0.2 0.8 one-step transition probability matrix of Y is P = .

0.4 0.6 5.1 Estimation of a Poisson parameter (a)We need find θ > 0 to maximize −θ k pY (k|θ)pΘ (k) = e k!θ λe−λθ Equivalently, dropping the constants k! and λ and taking the logarithm, θ is to be selected to maximize −θ(1 + λ) + k ln θ. The k derivative is −(1 + λ) + k/θ, which is positive for θ < 1+λ and negative for k k b M AP (k) = θ> . Hence, Θ .

1+λ

1+λ

b M AP (k) = θbM L (k). This should be expected, because as λ → 0, the (b) limλ→0 Θ Exp(λ) density becomes ever more uniformly spread over R+ .

5.3 ML estimation of covariance matrix (a) The scaler XiT K −1 Xi is trivially equal to Tr(XiT K −1 Xi ) or, equivalently, Tr(Xi XiT K −1 ). So ln f (X1 , . . . , Xn |K) =

n Y

X T K −1 Xi 1 − i 2 e (2π)p/2 (det K)1/2 i=1

= (2π)−np/2 (det K)−n/2 e− 2 Tr(SK n

−1

)

,

and taking the logarithm on both sides yields the desired expression.

Solutions to Odd Numbered ProblemsRandom Processes for Engineers

35

(b) Since S is symmetric semidefinite it can be expressed as S = U ΛU T , where 1 U is orthonormal and Λ is diagonal with positive diagonal entries. Let S 2 = 1 1 T U Λ 2 U , where Λ 2 is obtained from Λ by taking the square root of each entry.

Note that det(S 1/2 ) = det(S)1/2 .

1 e 12 , so det K = det(S 21 ) det K e det(S 12 ) = det S det K. e (c) Taking the hint, K = S 2 KS −1 1/2 −1 1/2 −1 e Also, Tr(SK ) = Tr(S K S ) = Tr(K ). So

e + Tr(K e −1 ) . e = − n p ln(2π) + ln det(S) + ln det(K) ln f (X1 , . . . , Xn |K) 2 Now

p X ei + 1 , e + Tr(K e −1 ) = ln λ ln det(K) ei λ i=1

ei are the eigenvalues of K. e Simple calculus shows that ln λ + 1 where the λ λ e has all is minimized by λ = 1, so the maximum likelihood estimator for K e that maximizes the likelihood eigenvalues equal to one. That is, the value of K is the p × p identity matrix. Therefore, the ML estimator of K is given by K = S 1/2 IS 1/2 = S, as was to be proved.

5.5 Convergence of the EM algorithm for an example (a) Multiplying through by (θ + σ 2 )2 and rearranging, the fixed-point equation F (θ) = θ becomes θ2 (θ − (y − σ 2 )) = 0, from which the claim follows.

2 2 (b) By the hint, F (θ) = θσy4 (1 − σ2θ2 + o(θ2 )) + θ(1 − σθ2 + ( σθ2 )2 − . . .), yielding the expression given. (c) By the above, if y 2 > σ 2 , then F (θ) > θ for all sufficiently small positive θ and F (θ) < θ for all sufficiently large θ (because F is bounded). So if θ(0) > 0 then the sequence θ(k) converges monotonically to θbM L . If y 2 ≤ σ 2 then F (θ) < θ for all θ > 0, and the sequence θ(k) is monotonically decreasing. By continuity, the limit is a fixed point of F , and since 0 is the only nonnegative fixed point, the limit is zero.

5.7 Using the EM algorithm for estimation of a signal variance −1 (a) SbM M SE = µS + ΣSY Σ−1 Y = θ(θI + ΣN )−1 Y, Y (Y − µY ) = (θI)(θI + ΣN ) and 2 −1 Σe = Cov(S − SbM M SE ) = ΣS − ΣSY Σ−1 = θ(θI + Y ΣY S = θI − θ (θI + ΣN ) −1 ΣN ) ΣN . (b) Since f (y|θ) is the normal density with mean zero and covariance matrix θI + ΣN , T

−1

log |θI + ΣN | y (θI + ΣN ) − θbM L (y) = arg max − θ 2 2

y

.

Even if ΣN is diagonal, there is apparently no analytical closed-form solution for θbM L (Y ). (c) Take (S, N ) as the compete data for the EM algorithm. Then d X d s2i d 1 1 log fcd (s, n|θ) = − log(2πθ) − − log(2π) − log |ΣN | − nT Σ−1 N n.

2 2θ 2 2 2 i=1

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so for the expectation step we obtain d 1 X d E[Si2 |θb(k) , y] + terms not involving θ.

Q(θ|θ(k) ) = − log(2πθ) − 2 2θ i=1

Maximizing with respect to θ we obtain

d 1X θb(k+1) = E[Si2 |θb(k) , y] d i=1 o 1n b = ||SM M SE ||22 + tr{Σe } , d

where SbM M SE and Σe were found in part (a). (d) Suppose d = 2 and that ΣN is the 2 × 2 identity matrix. The ML estimator y12 +y22 found in part (b) reduces to: θbM L (Y ) = arg maxθ − 21 log((1 + θ)2 ) − 2(1+θ) = y 2 +y 2

( 1 2 2 − 1)+ The EM recursion found in part (c) simplifies to: ( ) 2 2

2 θ y + y θ 1 2 (k+1) θb (y) = + | b(k) , 1+θ 2 1 + θ θ=θ

which is very similar to the recursion for d = 1 found in Example 5.8.

5.9 State estimation for an HMM with conditionally Gaussian observations (a) p(z1 , z2 , z3 , z4 ) has the form c2−i for any possible sequence z1 , z2 , z3 , z4 , so it is sufficient to track the exponent i. Smaller values of i correspond to larger probabilities. Give a transition label j if the transition has conditional probably 2−j Since c is common to all paths it is omitted.

0

3

2 3

2 1

2 1

1

0

1

1 0

1

5

2 2

5

1 3

2

t=1

2 2 1

2

1

4

2 2

2

2

2

4

2 1

2

2 2

2

3

2

t=2

winner!

2

2

3

1

t=3

t=4

To show how the Viterbi algorithm progresses, we use bold edges to denote the last edge of the minimum weight path into each node, and inside the circle for each node we denote the minimum weight of paths up to that node. Tracing back from the winner, we see (1,0,2,2) is the most likely path. (b)The joint likelihood p(z, y) is given by p(z, y) = p(z)p(y|z) −i

2 −2

= c2 (2πσ ) P4

= c0 2−(i+

exp −

2 t=1 (yt −zt )

)

4 X (yt − zt )2 t=1

2σ 2

!

Solutions to Odd Numbered ProblemsRandom Processes for Engineers

where i depends on the sequence z as in part (a), and c0 = c(2πσ 2 )−2 . So for y bM AP (y) is the path z minimizing i + P4 (yt − zt )2 . given, the MAP estimate Z t=1 The number i is the sum of edge weights for the trellis below, including the weights on the ingoing edges, which correspond to the prior distribution. A weight (yt −z)2 is placed on node (t, z) for 1 ≤ t ≤ 4 and z ∈ {0, 1, 2}. The MAP estimate is then the path that minimizes the sum of all edge weights and node weights along the path.

4

0

3

0 2

7

2

1 2

2

2

1 1

1

0

1

1

1

2

2

5

5

2 1

t=1

2

winner

9

1

1

1

0 1

4

2 2

4

2 0 2

4 2

5

4 2 7

t=2

2 2

1

9 15

2

1 2 4

16

2

21

1

t=3

t=4

To show how the Viterbi algorithm progresses, we use bold edges to denote the last edge of the minimum weight path into each node, and inside the circle for each node we denote the minimum weight of paths up to that node, including the weight associated with that node. Tracing back from the winner, we see bM AP (y) = (1, 0, 1.0).

Z 5.11 Estimation of a critical transition time of hidden state in HMM (a) Since F is an absorbing set, for 1 ≤ t ≤ T, the event {τF ≤ t} is the same as the event {Zt ∈ F }. So, for 1 ≤ t ≤ T, X P (τF ≤ t|Y ) = P (Zt ∈ F |Y ) = γi (t), i∈F

and the γ’s can be computed by the forward-backward algorithm. (b) This problem is more complicated than part (b) because the state process Z might enter F and then exit it again. One way to handle this is to replace each state in F c by two states, with one of them indicating that the state process has not yet entered F and the other indicating that the process has entered F. To be specific, consider a new state space Se defined by Se = {(i, 0) : i ∈ F c } ∪ {(i, 1) : i ∈ S}.

e (i, 0) and (i, 1), while Note that for each state i ∈ F c , there are two states in S, e for each state i ∈ F, there is only one state, (i, 1), in S. Let Ze be the new state process with statespace Se defined by

(Zi , 0) if i < τF e Zi = (Zi , 1) if i ≥ τF and let F̃ = {(i, 1) : i ∈ S}. Since Ze is a function of Z and the distribution

37

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e Y ) is well defined. In fact, of (Z, Y ) is already specified, the distribution of (Z, e Y ) is an HMM with parameter θe = (e e B) e with (Z, π , A,   πi πi π e(i,4) =  0

if 4 = 0 if i ∈ F and 4 = 1 if i ∈ F c and 4 = 1

  ai,j e(i,4),(j,40 ) = ai,j A  0

if 4 = 40 if 4 = 0, 40 = 1, j ∈ F else

e Since Fe is an and eb(i,4),k = bi,k . Note that τF for Z is equal to τFe for Z. e the conditional distribution of τF given Y can therefore be absorbing set for Z, computed by the method of part (a).

5.13 An underconstrained estimation problem We take Z to deterministically move through the states. To be specific, let S = {1, . . . , T } and P {Zt = t} = 1 for 1 ≤ t ≤ T . So π = (1, 0, . . . , 0) and A satisfies at,t+1 = 1 for 1 ≤ t ≤ T − 1.

Since A is a one-step probability matrix of size Ns × Ns , this determines the first Ns − 1 rows of A. The last row can be any length Ns probability vector. Finally, let bt,l = I{l=yt } for 1 ≤ t ≤ T and any l in the set of possible observation values. So with probability one, the state Zt at time t is t, and given that the state is t, the observation is yt with probability one. Thus, the output sequence is y = (y1 , . . . , yT ) with probability (i.e. likelihood) one.

5.15 Bayes estimation for a simple product form distribution a a ai,2 1 =1,Y2 =2,Y3 =2) so = P 0 ai,10 i,2 P (X = i|Y1 = 1, Y2 = 2, Y3 = 2) = P (X=i,Y P (Y1 =1,Y2 =2,Y3 =2) i ,1 ai0 ,2 ai0 ,2 i b XM AP = arg maxi ai,1 ai,2 ai,2 = 2. Here we use the fact that ai,1 ai,2 ai,2 is 2 for i = 1, 4 for i = 2, and 1 for i = 3. (b) By the calculations in part (a),  2  7     4 P (X = i|Y1 = 1, Y2 = 2, Y3 = 2) = 7      2 7

i=1 i=2 i=3

5.17 Free energy and the Boltzmann distribution (a) The slope of Q(i) log Q(i) converges to −∞ as Q(i) converges down to zero, so the constraint Q(i) ≥ 0 will not be binding at the optimal solution. Using the Lagrange multiplier λ for the P equality constraint: i Q(i) = 1 yields the Lagrangian L(Q, λ) = F (Q) − λ(1 − P ∂L i Q(i)). Solving ∂Q(i) = V (i) − λ + T (log(Q(i)) + 1) = 0 shows that Q(i) is proportional to exp(−V (i)/T ), yielding the desired conclusion. (b) Since limT →∞ exp(−V (i)/T ) = 1 for all i, the distribution BT converges to the uniform distribution over the state space as T → ∞.

T (j) (c) Note that B BT (i) = exp((V (i) − V (j))/T ), which is equal to one for all T if V (j) = V (i), and converges to zero as T → 0 if V (j) > V (i). Thus, as T → 0, the disturbution BT converges to the uniform distribution on the set of minimizers of V as T → 0. (d) Straightforward rearrangement yields F (Q) = T D(Q||BT ) − T log(ZT ).

Solutions to Odd Numbered ProblemsRandom Processes for Engineers

5.19 Inference for a mixture model p(z = i|y, θ) =

(a) Since

p(z = i, y|θ) exp(−||y − θi ||2 /2) =P , 2 p(y|θ) j exp(−||y − θj || /2)

bM AP is the index i that minimizes the distance, ||y − θi ||, from y to θi .

Z (b) Suppose θ is uniformly distributed over a cube in Rd with sides of length L for L >> 1. Then, for θ in that cube,

n Y 1 1 2 ||y − θ || p(θ, z (1) , y (1) , · · · , z (n) , y (n) ) = d n exp − t zt 2 L J (2π)nd/2 t=1 ! n 1 1X 2 = d n exp − ||yt − θzt || 2 t=1 L J (2π)nd/2 (c) Equivalently, for y = (y (1) , . . . , y (n) ) fixed, the alternating algorithm seeks to Pn minimize t=1 ||y (t) − θz(t) ||2 with respect to z and θ. For θ fixed, the algorithm selects z (t) for each t to minimize the distance from y (t) to θz(t) . For z fixed, the algorithm chooses θi to be the average of {y (t) : z (t) = i). (d) Up to an additive constant not depending on θ, Q(θ||θ(k) ) = − =−

n

J

J

n

1 X X (t) (k) (y − θi )2 p(z (t) = i|yt , θi ) 2 t=1 i=1

1 X X (t) (k) (y − θi )2 p(z (t) = i|yt , θi ), 2 i=1 t=1

where for the second equality we simply switched the order of summation, and the function p(z = i|y, θ) is the same as appearing in part (a). Maximizing Q(θ||θ(k) ) with respect to θi yields the EM algorithm: Pn (k) (t) (t) = i|yt , θi ) (k+1) t=1 y p(z θi = P . n (t) = i|y , θ (k) ) t i t=1 p(z This is a soft version of the ICM algorithm. Interpret p(z (t) = i|yt , θ(k) ) as a (k+1) fractional indicator of whether yt was generated using θi . The value θi is the average of the y (t) ’s using the fractional indicators.

5.21 MAP estimation of parameters of a Markov process [ Z(2) M AP = arg max P (Z(2) = i|Z(1) = 0, Z(4) = 1) i

= arg max P (Z(2) = i, Z(4) = 1|Z(1) = 0). i

Now P (Z(2) = 0, Z(4) = 1|Z(1) = 0) =

X

P (Z(2) = 0, Z(3) = j, Z(4) = 1|Z(1) = 0)

j

= (2/3)(2/3)(1/3) + (2/3)(1/3)(2/3) = 8/27.

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and P (Z(2) = 1, Z(4) = 1|Z(1) = 0) =

X

P (Z(2) = 1, Z(3) = j, Z(4) = 1|Z(1) = 0)

j

= (1/3)(1/3)(1/3) + (1/3)(2/3)(2/3) = 5/27. [ Hence, Z(2) M AP = 0. (b) Since one out of three initial values are 1, π b = (1/3, 2/3). Since the transition 0 → 0 occurs twice and the transition 0 → 1 occurs twice, the first row b is (2/4, 2/4). Since the transition 1 → 0 occurs twice and the transition of A b 1 → 1 occurs thrice, the second row of A is (2/5, 3/5). Therefore, θM L = 2/4 2/4 (1/3, 2/3), . (In detail, the likelihood of the observations is 2/5 3/5 π0 p0,0 p0,0 p0,1 π1 p1,0 p0,1 p1,1 π1 p1,1 p1,1 p1,0 , or equivalently, π0 π12 p20,0 p20,1 p21,0 p31,1 , which is to be maximized over valid choices of θ.

6.1 Mean hitting time for a simple Markov process (a) 1 0

a 1

1!a

0.5 2

0.5

3 1

1−a 1 2a a Solve π = πP and πe = 1 to get π = ( 2(1+a) , 2(1+a) , 2(1+a) , 2(1+a) ) for all a ∈ [0, 1] (unique). (b) A general way to solve this is to let hi = E[min{n ≥ 0|X(n) = 3}|X(0) = i], for 0 ≤ i ≤ 3. Our goal is to find h0 . Trivially, h3 = 0. Derive equations for the P other values by conditioning on the first step of the process: hi = 1+ j pij hj for     4 h0 = 1 + h1 h0 a +1 i 6= 3. Or h2 = 1 + (1 − a)h0 + ah2 yielding  h1  =  a4 . Thus, 2 h2 = 1 + (0.5)hi h2 a +1 h0 = a4 + 1 is the required answer.

6.3 Equilibrium distribution of the jump chain The choice ( of B insures that 0 if i = j (−Bqii πi )i∈S is a probability distribution. Since pJij = , we qij if i 6= j −qii have for fixed j,

X X qij (−Bqii πi )pJij = −Bqii πi −qii i∈S i:i6=j X =B πi qij = −Bπj qjj . i:i6=j

Thus, −Bπi qii is the equilibrium vector for P J . In other words, −Bπi qii = πiJ .

Another justification of the same result goes as follows. For the original Markov −1 process, −qii is the mean holding time for each visit to state i. Since πiJ represents the proportion of jumps of the process which land in state i, we must have πi ∝ πiJ (−qii )−1 , or equivalently, πiJ ∝ (−qii )πiJ .

Solutions to Odd Numbered ProblemsRandom Processes for Engineers

6.5 A simple question of periods (a) The period of state 4 is GCD{4, 6, 8, 10, 12, 14, . . .} = GCD{4, 6} = 2. (b) The process is irreducible, so all states have the same period. So state 6 must also have period 2.

6.7 A simple nonzero entries:  birth-death Markov process (a) Showing −1 1   1 −3 2     1 −5 4     1 −5 4   .  Q=  8 −12 4     8 −12 4    .. .. ..   . . .  (b) S1 = 1 + 1 + 2 + 8(1 + 0.5 + 0.52 + · · · ) = 1 + 1 + 2 + 16 = 20.

So π = (1/20, 1/20, 2/20, 8/20, (8/20)/2, (8/20)/22 , (8/20)/23 , · · · ).

1 (c) M1 = π1 (−q = 20, and M1 is the sum of the mean holding time of state 1,1 ) 1, which is 1, and the mean time to return to state 1 from state 2 (because state 2 is always the next state visited after state 1). So the mean time to return back to state 1 from state 2 is 19.

6.9 Generating a random spanning tree (a) Yes. Let T and Tf be any two distinct spanning trees. Suppose |T ∩ Tf | = k, that is, suppose the two trees have k edges in common. It suffices to show that there is a spanning tree T 0 with pT,T 0 > 0 such that |T 0 ∩ Tf | = k + 1. Beginning with tree T , there is positive n−1−k probability ( m−n+1 , to be exact) that the edge e selected from E − T is an edge in Tf . Given that e ∈ Tf , since Tf does not have any cycles, at least one edge in the cycle in T ∪ {e} is not in Tf , and the conditional probability that that edge is deleted is positive. Deletion of that edge would result in a tree T 0 satisfying the given requirements, and pT,T 0 > 0. (b) Yes, because pT,T > 0 for each state T ; each state has period one. (c) Consider any pair of distinct states, T and T 0 . Then pT,T 0 > 0 if and only if |T ∪ T 0 | = n, which is true if and only if pT,T 0 > 0. Furthermore, if pT,T 0 > 0, then there is a unique cycle C in T ∪ T 0 , and pT,T 0 and pT 0 ,T are both equal to 1 (m−n+1)|C| . (d) If P is a symmetric one-step transition probability matrix on a state space P P 1 S and πi = |S| for all states i ∈ S, then for any state j, i πi pi,j = S1 i pi,j = P 1 1 i pj.i = S = πj That is, π = πP, so π is the equilibrium distribution. (More S generally, if P is a one-step transition probability matrix and π is a probability distribution satisfying the detailed balance condition: πi pi,j = πj pj,i for all i, j, then π is an equilibrium distribution and the process is reversible.)

6.11 A birth-death process with periodic rates (a) Let θ = µλaa λµbb . Then S1 = P∞ n λa n=0 θ (1 + µa ), which is finite if and only if θ < 1. Thus, the process is positive recurrent if and only if θ < 1. (In case θ < 1, π2n = θn /S1 and π2n+1 = θn Sλ1 µa a .)

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(b) λa = λb . In general, rk = πk λk /λ, so that r ≡ π if and only if the arrival rates are all equal (corresponding to Poisson arrivals, and then PASTA holds.)

6.13 An unusual birth-death process P∞ P∞ ...µn n = n=0 ( 1−p (a) S2 = n=0 µλ11 ...λ p ) < +∞ (because p > 0.5) so X is transient. n (b) aQ = 0 is equivalent to λk ak = µk ak+1 for k ≥ 0, which is easily checked. (c) pJ01 = 1 and, for i ≥ 1, pJii−1 = 1 − p and pJii+1 = p. All other transition probabilities are zero. p

p

1 0

1 1!p

.,.

2 1!p

1!p

(d) S2 is the same as in (a), so X J is transient (as is also obvious from part (c).)

Note: X reaches the graveyard state 4 in finite time. Otherwise, the fact that Qa = 0 for a probability distribution a would imply that all states are positive recurrent.

6.15 Limit of a discrete time queueing system (q) The transition rate diagram for the number in the system is shown: "q

"q

0

"q 2

1 !p

1!" q

"q 3

!p

...

!p

!p

1! " q!!p 1! " q!!p 1! " q!!p

ak = P (no arrivals for k − 1 slots, then arrival) = (1 − αq)k−1 αq for k ≥ 1.

Similarly, bk = (1 − βq)k−1 βq for k ≥ 1. Thus, an interarrival time is q times a geometric random variable with parameter αq, and the mean interarrival time is α1 . and a service time is q times a geometric random variable with parameter βq, and the mean service time is β1 . (b) βqpk+1 = αqpk for k ≥ 0, impies that pk = (1 − ρ)ρk , where ρ = α β.

This distribution is the same for all q, so it trivially converges as q → 0 to the same value: pk → (1 − ρ)ρk . The interarrival times and service times become exponentially distributed in the limit. The limit system is a continuous time M/M/1 system.

6.17 Statistical multiplexing (a) Use an M/M/1 model, with λ = 40 pack1 ets/second, and µ = 50 packets per second. Then T = λ−µ = 0.1 seconds. (b) Model each link as an M/M/1 queue, with λ = 20 packets/second, and 1 µ = 25 packets per second. Then T = λ−µ = 0.2 seconds. (c) Model as an M/M/2 queue with λ = 40 packets/second and µ = 25 packets/second. !

0

!

µ

!

2

1 2µ

!

3

2µ

! ...

4 2µ

2µ

Solutions to Odd Numbered ProblemsRandom Processes for Engineers

43

λ 0.5 k Let ρ = 2µ . Then π0 = 0.5+ρ+ρ for k ≥ 1. Thus 2 +··· and πk = 2π0 ρ

T =

P∞ k ρ/(1 − ρ)2 N k=1 kρ = 0.111 seconds. = = 1 2 λ λ(0.5 + ρ + ρ + · · · ) λ(( 1−ρ ) − 0.5)

We can view model (c) as a variation of (b) with statistical multiplexing at the packet level, and (a) as a variation with multiplexing at the subpacket level. Here (c) is a substantial improvement over (b), and (a) is a little better than (c).

6.19 Three queues and an autonomously traveling server (a) The 1 1 1 equilibrium distribution for the server is π = ( αZ , βZ , γZ ), where Z = α1 + β1 + γ1 , and the service capacity of station i is thus πi , for 1 ≤ 1 ≤ 3. (b) The equilibrium distributions for the individual stations can be found one at a time, by the method of phases (even though the stations are not independent).

For a fixed station i, the state space would be {(l, θ) : l ≥ 0, 1 ≤ θ ≤ 3}, where l is the number of customers at station i and θ is the phase of the server.

6.21 Recurrence of mean zero random walks (Note: The boundedness condition is imposed to make the problem easier – it is not required for the results.) (a) Let V (x) = |x|, the absolute value of x. Then for x ≥ M and any k ≥ 0, |x + Bk | = x + Bk with probability one, by the boundedness of Bk . Thus, if x ≥ M , P V (x) − V (x) = E[|x + Bk |] − x = E[x + Bk ] − x = 0. Similarly, if x ≤ −M , P V (x) − V (x) = E[|x + Bk |] − |x| = E[|x| − Bk ] − |x| = −E[Bk ] = 0.

Therefore, the Foster-Lyapunov criteria for recurrence are satisfied by P , V , and the set C = {x : |x| < M }, so that X is recurrent. (b) Let V (y) = y. As in part (a), we see that P V (y) − V (y) = 0 off the finite set C = {y : 0 ≤ y < M }, so that Y is recurrent.

6.23 Routing with two arrival streams We begin by describing the system by stochastic evolution equations. For i ∈ {0, 1}, let Bi (t) be a Bernoulli(ai ) random variable, and let Ui (t) be a Bernoulli(ui ) random variable, and for j ∈ {1, 2, 3} let Dj (t) be a Bernoulli(dj ) random variable. Suppose all these Bernoulli random variables are independent. Then the state process can be described by Xi (t + 1) = Xi (t) + Ai (t) − Di (t) + Li (t)

for i ∈ {1, 2, 3}

(1.5)

where A1 (t) = U1 (t)B1 (t) A2 (t) = (1 − U1 (t))B1 (t) + U2 (t)B2 (t) A3 (t) = (1 − U2 (t))B2 (t).

and Li (t) = (−(Xi (t) + Ai (t) − Di (t)))+ .

Assume that the parameters satisfy the following three conditions: a1 + a2 < d1 + d2 + d3

(1.6)

a1 < d1 + d2

(1.7)

a2 < d2 + d3 .

(1.8)

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These conditions are necessary for any routing strategy to yield positive recurrence. Condition (1.6) states that the total arrival rate to the network is less than the sum of service rates in the network. Condition (1.7) states that the total arrival rate to servers 1 and 2 is less than the sum of the service rates of those servers. Condition (1.8) has a similar interpretation. We will show that conditions (1.6)-(1.8) are also sufficient for positive recurrence for random routing for a particular choice of u. To that end we consider the Lyapunov-Foster criteria.

Letting V = (x21 + x22 + x23 ) and arguing as in Example 1a, and using the fact (Ai (t) − Di (t))2 ≤ 1 for i = 1, 3 and (A2 (t) − D2 (t))2 ≤ 4 yields (writing ui for 1 − ui ): P V (x) − V (x) ≤

3 X i=1

xi E[Ai (t) − Di (t)|X(t) = x]

!

+3

= x1 (a1 u1 − d1 ) + x2 (a1 u1 + a2 u2 − d2 ) + x3 (a2 u2 − d3 ) + 3

(1.9)

≤ −(x1 + x2 + x3 ) min{d1 − a1 u1 , d2 − a1 u1 − a2 u2 , d3 − a2 u2 } + 3.(1.10)

To obtain the best bound possible, we select u = (u1 , u2 ) to maximize the min term in (1.10). For any u, min{d1 − a1 u1 , d2 − a1 u1 − a2 u2 , d3 − a2 u2 }, is the minimum of three numbers, which is less than or equal to the average of the three numbers, is less than or equal to the average of any two of the three numbers, and is less than or equal to any one of the three numbers. Therefore, for any choice of u ∈ [0, 1]2 , min{d1 − a1 u1 , d2 − a1 u1 − a2 u2 , d3 − a2 u2 } ≤

(1.11)

where = min

d1 + d2 + d3 − a1 − a2 d1 + d2 − a1 d2 + d3 − a2 , , , d1 , d2 , d3 .

3 2 2

Note that > 0 under the necessary conditions (1.6)-(1.8). Taking u∗1 ∈ [0, 1] as large as possible subject to d1 − a1 u∗1 ≥ , and taking u∗2 ∈ [0, 1] as large as possible subject to d3 − a2 u∗2 ≥ , yields the choice u∗1 =

d1 − ∧1 a1

u∗2 =

d3 − ∧ 1. a2

It is not hard to check that d2 − a1 u∗1 − a2 u∗2 ≥ , so that equality holds in (1.11) for u = u∗ .

This paragraph shows that the above choice of and u∗ can be better understood by applying the max-flow min-cut theorem to the flow graph shown.

Solutions to Odd Numbered ProblemsRandom Processes for Engineers

q

45

1 d !!

1

a

1

v

1 q

s a

2

v

2

d !!

2

t

2 q

3

d !!

3

In addition to the source node s and sink node t, there are two columns of nodes in the graph. Nodes v1 and v2 correspond to the two arrival streams, and nodes q1 , q2 and q3 correspond to the three queues. There are three stages of links in the graph. The capacity of a link (s,vi ) in the first stage is ai , the capacities of the links in the second graph are very large, and the capacity of a link (qj , t) in the third stage is dj − . The choice of above is the largest possible so that (1) all links have nonnegative capacity, and (2) the capacity of any s − t cut is greater than or equal to a1 + a2 . Thus, the min flow max cut theorem insures that there exists an s − t flow with value a1 + a2 . Then, u∗i can be taken to be the fraction of the flow into vi that continues to qi , for i = 1 and i = 2.

Under (1.6)-(1.8) and using u∗ , we have P V (x) − V (x) ≤ − (x1 + x2 + x3 ) + 3.

Thus, the system is positive recurrent, and X 1 + X 2 + X 3 ≤ 3 .1 (b) Suppose each arrival is routed to the shorter of the two queues it can go to.

For simplicity, let such decisions be based on the queue lengths at the beginning of the slot. As far as simple bounds are concerned, there is no need, for example, to take into account the arrival in the top branch when making the decision on the bottom branch, although presumably this could improve performance. Also, it might make sense to break ties in favor of queues 1 and 3 over queue 2, but that is not very important either. We can denote this route-to-shorter policy by thinking of u in part (a) as a function of the state, u = ϕRS (x). Note that for given x, this choice of u minimizes the right hand side of (1.9). In particular, under the route-to-shorter policy, the right hand side of (1.9) is at least as small as its value for the fixed control u∗ . Therefore, writing P RS for the one-step transition matrix for the route to shorter policy, P RS V (x)−V (x) ≤ − (x1 +x2 +x+ 3) +3, and under the necessary conditions (1.6)-(1.8), the system is positive recurrent, and X 1 + X 2 + X 3 ≤ 3 .

6.25 Allocation of service Let > 0. The largest we can make µ1 (u) is m1 and the largest we can make µ3 (u) is m2 . Thus, we can select u so that λ1 + ≤ µ1 (u)

and λ3 + ≤ µ3 (u)

(1.12)

if and only if m1 ≥ λ1 + and m2 ≥ λ3 + . If we select u so that equality holds in (1.12) then the value of µ2 (u) is maximized subject to (1.12) , and it is given 1

By using the fact that Xi = 0 if the arrival rate to queue i is zero, it can be shown that for u = u∗ , X 1 + X 2 + X 3 ≤ 30 , where 0 = min{ d1 +d2 +d33 −a1 −a2 , d1 +d22 −a1 , d2 +d23 −a2 }.

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by µ2 (u) = m1 + m2 − µ1 (u) − µ3 (u) = m1 + m2 − λ1 + λ3 − 2 .

Therefore, in order that λ2 + ≤ µ2 (u) hold in addition to (1.12), it is necessary and sufficient that ≤ λ1 +λ2 +λ33 −µ1 −µ2 and m1 ≥ λ1 + and m2 ≥ λ2 + . Thus, the value of given in the example is indeed the largest possible.

6.27 Routing to two queues – continuous time model Suppose λ < µ1 + µ2 , which is a necessary condition for positive recurrence. (a) Under this condition, we can find u so that µ1 > λu and µ2 > λu. If each customer is independently routed to queue 1 with probability u, and if V (x) = (x21 + x22 )/2, (6.30) becomes

2QV (x) = (x1 + 1)2 − x21 λu + (x2 + 1)2 − x22 λu

+ (x1 − 1)2+ − x21 µ1 + (x2 − 1)2+ − x22 µ2 .

Since (x1 − 1)2+ ≤ (xi − 1)2 , it follows, with γ = λ + µ1 + µ2 , that

γ .

(1.13) 2 Thus, the combined stability criteria and moment bound applies, yielding that the process is positive recurrent, and X1 (µ1 − λ) + X2 (µ2 − uλ) ≤ γ2 .

In analogy to Example 1a, we let QV (x) ≤ − (x1 (µ1 − uλ) + x2 (µ2 − uλ)) +

= max min{µ1 − λu, µ2 − λu} 0≤u≤1

µ1 + µ2 − λ = min µ1 , µ2 , 2

and the corresponding value u∗ of u is given by  0 if µ1 − µ2 < −λ  µ1 −µ2 ∗ 1 u = + 2λ if |µ1 − µ2 | ≤ λ  2 1 if µ1 − µ2 > λ

γ . The remark at the end of Example For the system with u = u∗ , X 1 + X 2 ≤ 2 1a also carries over, yielding that for splitting probability u = u∗ , X 1 + X 2 ≤ γ µ1 +µ2 −λ

(b) Consider now the case that when a packet arrives, it is routed to the shorter queue. To be definite, in case of a tie, the packet is routed to queue 1. Let QRS denote the transition rate matrix in case the route to short queue policy is used.

Note that for any u, λ(x1 u + x2 u) ≥ λ(x1 ∧ x2 ). Thus, (1.13) continues to hold if Q is replaced by QRS . In particular, if λ < µ1 + µ2 , then the process X under γ the route-to-shorter routing is positive recurrent, and X 1 + X 2 ≤ 2 .

6.29 Stability of a system with two queues and modulated server The equilibrium distribution of the server state is (w(0), w(1), w(2)) = ( 21 , 14 , 14 ).

Thus, over the long run, the server can serve queue 1 at most 43 of the time, it can serve queue 2 at most 34 of the time, and, of course, it can offer at most one service in each slot. Thus, it is necessary that a1 < 43 , a2 < 34 , and a1 +a2 < 1. (It can be

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shown that these conditions are also sufficient for stability by using the Lyapunov x2 +x2 function V (x) = 1 2 2 applied to the time-sampled process Y (n) = X(nk), where k is a constant large enough so that the mean fraction of time the server is in each state over k time slots is close to the statistical average, no matter what the state of the server at the beginning of such interval.)

7.1 Calculus for a simple Gaussian random process (a) Differentiation with respect to t would derivative of X is given by Xt0 = B +2Ct. suggest that the m.s.

2 Verifying: E Xt+hh−Xt − (B + 2Ct) = E[h2 C 2 ] = h2 → 0 as h → 0.

R1 1 1 2 2 (b)n 0 Xs ds = A o + B/2 + C/3, which is a N (0, σ ) with σ = 1 + 4 + 9 . So R1 P 0 Xs ds ≥ 1 = Q(1/σ).

7.3 Properties of a binary valued process (a) Yes. Let n ≥ 1 and t1 < · · · < tn < tn+1 . Note that Ytn+1 = (−1)D Ytn where D = Ntn+1 −Ntn . (We write D for brevity, even though it depends on tn and tn+1 ). The distribution of D is Poisson with mean λτ , where τ = tn+1 − tn . The vector (Yt1 , . . . , Ytn ) is determined by (Nt : 0 ≤ t ≤ tn ), and is thus independent of D by the independent increment property of Poisson processes. To summarize, if tn is viewed as the present time, the future value Ytn+1 is a function of the present value Ytn , and D, which is independent of the past and has a distribution depending on (tn , tn+1 ) only through the difference τ . Thus, Y is a time-homogeneous Markov process.

One way to identify the transition rate matrix Q and the transition probabilities is to directly find the transition probabilities, as follows. For i, j ∈ {1, −1}, P (Ytn+1 = j|Ytn = i, Yt1 = ii , . . . , Ytn−1 = in−1 ) = P ((−1)D Ytn = j|Ytn = i, Yt1 = ii , . . . , Ytn−1 = in−1 ) = P {(−1)D = j/i}.

Since ∞ X

∞

e−λτ X (λτ )k + (−λτ )k e−λτ (λτ )k = k!

2 k! k=0 k=0, k even

e−λτ eλτ + e−λτ 1 + e−2λτ = = , 2 2 ( 1+e−2λτ i=j 2 it follows that, for i, j ∈ {1, −1}, pij (τ ) = The transition 1−e−2λτ i 6= j 2

−λ λ rates are given by qij = p0ij (0), yielding Q = . λ −λ Another way is to identify the transition rates first, by considering tn+1 decreasing to tn , or τ → 0. Note that λτ e−λτ = P {D = 1} ≤ P (D is odd) ≤ 1 − P {D = 0} = 1 − e−λτ , so that limτ →0 τ1 P (D is odd) = λ. Therefore, for i, j ∈ {1, −1} with i 6= j, qij = τ1 P (Y (t + τ ) = j|Y (t) = i) = τ1 P (D is odd) → λ.

Thus, the Q matrix is given as before, and the transition rates can be found be P {(−1)D = 1} =

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solving the Kolmogorov forward equations. (Take α = β = λ in the example at the end of Chapter 4.)

Yet another approach to part (a) is to use the fact that the intercount times of the Poisson process are independent, exponentially distributed with parameter λ, and to use the memoryless property of the exponential distribution. (b) E[|Yt+τ − Yt |2 ] = 4P {Yt+τ 6= Yt } = 2(1 − e−2λτ ) → 0 as τ → 0, so that Y is m.s. continuous. (c) If Y were m.s. differentiable, then E[| Yt+ττ−Yt |2 ] would have a finite limit −2λτ

as τ → 0. But this expectation is equal to 2(1−eτ 2 ) , which converges to +∞ as τ → 0, so that Y is not m.s. continuous.

7.5 Differentiation of the square of a Gaussian random process (a) Since An → A, the corollary in Section 2.2 of the notes implies that E[AAn ] → E[A2 ] and E[A2n ] → E[A2 ]. Since A and the An are jointly Gaussian, the hint and the statement just proved imply that: E[A4 ] = 3E[A2 ]2 E[A4n ] = 3E[A2n ]2 → 3E[A2 ]2

E[A2n A2 ] = 2E[AAn ]2 + E[A2 ]E[A2n ] → 3E[A2 ]2 .

Therefore E[(A2n − A2 )2 ] = E[A4n ] − 2E[A2n A2 ] + E[A4 ] → 0. (b) By the assumptions and part (a), A2n → A2 m.s., Bn2 → B 2 m.s., and (An + Bn )2 → (A + B)2 m.s.. Therefore, (An + Bn )2 − A2n − Bn2 → (A + B)2 − A2 − B 2 m.s., which is equivalent to what is to be proved. (c) Fix any t. Then for h 6= 0, Yt+hh−Yt = Xt+hh−Xt (Xt+h + Xt ) = Ah Bh , where Ah → X 0 (t) m.s. and Bh → 2X(t) m.s. as h → 0. Therefore, by part (b), Yt+h −Yt = Ah Bh → 2Xt0 Xt m.s. That is, if X is mean zero, Gaussian and m.s. h differentiable, then X 2 is also m.s. differentiable, and, just as for deterministic functions, (X 2 )0 = 2XX 0 .

7.7 Mean square differentiability of some random processes (a) Yes, the integral of a m.s. continuous process is (continuously) m.s. differentiable. (b) No, RY is not even continuous, so Y is not even m.s. continuous, so is not m.s. differentiable. (c) Yes. We should expect the answer is yes because if the sum is differentiated term by term, the resulting sum is still m.s. convergent. That is, we expect 4 P∞ Zt0 = Yt = n=0 Vn cos(nt) . To check this out, note that n " 2 # Zt+h − Zt E − Yt h  

!2 ∞ X sin(n(t + h)) − sin(nt) cos(nt)  =E Vn − n2 h n n=1 =

∞ X 1 sin(n(t + h)) − sin(nt)

n=1

n2

nh

2 − cos(nt) .

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For each n ≥ 1, sin(n(t+h))−sin(nt) − cos(nt) → 0 as h → 0 and the terms nh sin(n(t+h))−sin(nt) − cos(nt) are bounded by 2 (use the intermediate value form nh of Taylor’s theorem). Hence by the dominated convergence theorem, 2 P∞ 1 sin(n(t+h))−sin(nt) − cos(nt) → 0 as h → 0.

2 n=1 n nh ALTERNATIVE, note that, since the terms in the series defining Z are orthogP∞ sin(ns) sin(nt) onal random variables, RZ (s, t) = . From this we see that n=1 n4 ∂2 RZ and ∂1 ∂2 RZ exist and are continuous. Therefore, RZ is continually m.s. differentiable.

7.9 A two-state stationary Markov process (a) Since E[Xt ] = 0 for each t and since Xt takes values in {−1, 1}, the distribution of Xt for any t must be πi = 0.5 for i ∈ {−1, 1}. Solving the Kolmogorov forward equations (see Example 4.9.3) yields the distribution of the process at any time t ≥ 0 for any given initial distribution: π(t) = π(0)e−2αt + (0.5, 0.5)(1 − e−2αt ).

So the transition probability functions are given by:

0.5(1 + e−2ατ ) 0.5(1 − e−2ατ )

X

X

pij (τ ) =

if i = j if i 6= j

so that for τ ≥ 0, RX (τ ) = E[X(τ )X(0)] = =

X

X

i∈{−1,1} j∈{−1,1}

P {X(0) = i, X(τ ) = j}ij

πi pi,j (τ )ij

i∈{−1,1} j∈{−1,1}

=

1 (1 + e−2ατ ) + (1 + e−2ατ ) − (1 − e−2ατ ) − (1 − e−2ατ ) = e−2ατ .

4

Hence for all τ, RX (τ ) = e−2α|τ | . (b) For all α, because RX is continuous. (c) For α = 0, because RX is twice continuously differentiable in a neighborhood of zero if and only if α = 0. (d) For α > 0, because limτ →∞ RX (τ ) = 0 for α > 0 and limτ →∞ RX (τ ) 6= 0 if α = 0.

7.11 Fundamental theorem of calculus for m.s. calculus Fix tR > 0. The quess −Yt −Yt 1 tion is whether Yss−t → Xt m.s. as s → t. Now Yss−t −Xt = s−t (Xu −Xt )du, t

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so

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"

2 # Ys − Yt − Xt E s−t

2 # 2 " Z s 1 = (Xu − Xt )du E s−t t

2 Z s Z s 1 (Xv − Xt )dv (Xu − Xt )du = E s−t t t

2 Z t Z t 1 RX (u, v) − RX (t, v) − RX (u, t) + Rx (t, t)dudv. = s−t s s

Let > 0. Since RX is continuous, there exists δ > 0 so |RX (a, b)−RX (t, t)| ≤ /3 whenever |a − t| < E and |b − t| < δ. So if |s − t| < δ and u, v are between s and v) − RX (t, v) − RX (u, t) + RX (t, t)| ≤ . So if |s − t| < then t, then |RX (u, 2 Ys −Yt −Yt E → Xt m.s. . Thus Yss−t s−t − Xt

7.13 An integral Rof white noise R t times an exponential RsRt s (a) RX (s, t) = E[ 0 Zu e−u du 0 Zv e−v dv] = 0 0 σ 2 e−u−v δ(u − v)dudv R s∧t 2 −2v 2 = 0 σ e dv = σ2 (1 − e−2(s∧t) ). (b) No. If the ratio Xt+hh−Xt converged in the m.s. sense as h → 0, its mean square R t+h −2u σ2 value would converge to a finite number. But E[( Xt+hh−Xt )2 ] = 2h e du → 2 t +∞, so the answer is no. (c) Yes. Observe from part (a) that lims,t→∞ RX (s, t) exists and is finite. Thus, the Cauchy criterion for m.s. convergence holds.

7.15 An integrated Poisson process (a) The sample paths of Y are piecewise linear. The initial slope is zero, and the slope of Y increases by one at each jump of N .

Rt Rt (b) E[Yt ] = 0 ENs ds = λ 0 sds = λt2 /2. (c) Using the CN (u, v) = λ(u ∧Rv),R yields Rt R t Rfact 3 t t v Var(Yt ) = 0 0 CN (u, v)dudv = 2λ 0 0 ududv = λ 0 v 2 dv = λt3 . (d) Note that if Nt/2 ≥ 2, then Yt ≥ t. So λt P {Yt < t} ≤ P {Nt/2 ≤ 1} = (1 + λt 2 ) exp(− 2 ) → 0 as t → ∞.

7.17 A random Taylor’s approximation (a) For brevity, write RX (τ ) = f (τ ).

Then E[X00 Xt ] = −f 0 (t) and E[(X00 )2 ] = −f 00 (0). So     X0 f (0) 0 f (t) Cov  X00  =  0 −f 00 (0) −f 0 (t)  .

Xt f (t) −f 0 (t) f (0)

bt = (b) Let K denote the matrix of part (b), and let αT = (1, −t, 1). Then Xt − X   X0 2 αT  X00 , and the MSE is αT Kα = 2{f (0) − f (t) − t2 f 00 (0) + tf 0 (t)}.

Xt (c) Since X0 is orthogonal to X00 and the random variables are mean zero,

Solutions to Odd Numbered ProblemsRandom Processes for Engineers

0

b t |X0 , X 0 ] = E[X b t |X0 ] + E[X b t |X 0 ] = f (t) X0 + −f00(t) X 0 .

E[X 0 0 f (0) −f (0) 0 (d) (This part was not required.) Since f is an even function, f 0 (0) = f 000 (0) = 0.

2 3 t4 0000 So f (t) = f (0)+ t2 f 00 (0)+ 24 f (0)+o(t4 ) and f 0 (t) = tf 00 (0)+ t6 f 0000 (0)+o(t3 ).

Using these relations yields that, as t → 0, the MSE for both the Taylor’s ap4 proximation and the optimal estimator is t4 f 0000 (0) + o(t4 ). In particular, the ratio of the MSE for the Taylor’s approximation to that of the optimal estimator converges to one as t → 0.

7.19 Integral of a Brownian bridge (a) The integral over the finite interval [0, 1] exists in the m.s. sense because B is m.s. continuous, or equivalently, RB is continuous. (b) X and W1 are jointly Gaussian because they are linear R 1 functions of the Gaussian process W = (Wt : t ≥ 0). Also, E[X] = 0 E[Bt ]dt = 0 and R1 R1 R1R1 R1Rt E[X 2 ] = E[ 0 Bs ds 0 Bt dt] = 0 0 RB (s, t)dsdt = 2 0 0 RB (s, t)dsdt = R1Rt R1 2 1 . By a previous problem, the 2 0 0 s(1 − t)dsdt = 0 t (1 − t)dt = 31 − 14 = 12 whole process B is independent of W1 . (Since the variables are jointly Gaussian, this amounts to the fact Cov(Bt , W1 ) = 0 for all t.) Therefore, Cov(X, W1 ) = 0 too, or equivalently (since X and W1 are jointly Gaussian) X is independent of W1 . To summarize, X and W1 are independent random variables, with X being 1 ) and W1 being N (0, 1).

N (0, 12 7.21 A random process which changes at a random time (a)

Y

Z

U

(b) FX,1 (c, t) = P {Xt ≤ c} = P (Xt ≤ c|U ≤ t)P {U ≤ t} + P (Xt ≤ c|U > t)P {U > t} = P {Yt ≤ c}FU (t) + P {Zt ≤ c}(1 − FU (t)) = FY,1 (c, t)FU (t) + FZ,1 (c, t)(1 − FU (t)).

That is, for each t fixed, the distribution of Xt is a mixture of the distributions of Yt and Zt . However, for this problem, Yt and Zt both have the N (0, 1) distribution, and hence, Xt also has the N (0, 1) distribution for any t.

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(c) By part (b), E[Xt ] = 0 for all t. Suppose s ≤ t. Then RX (s, t) = E[Xs Xt ] = E[Xs Xt |U ≤ s]P {U ≤ s} + E[Xs Xt |s < U ≤ t]P {s < U ≤ t} + E[Xs Xt |U > t]P {U > t}

= E[Zs Zt |U ≤ s]FU (s) + E[Ys Zt |s < U ≤ t](FU (t) − FU (s)) + E[Ys Yt |U > t](1 − FU (t))

= E[Zs Zt ]FU (s) + E[Ys Zt ](FU (t) − FU (s)) + E[Ys Yt ](1 − FU (t))

= RZ (s, t)FU (s) + RY Z (s, t)(FU (t) − FU (s)) + RY (s, t)(1 − FU (t))

= e−|s−t| FU (s) + 0 + e−|s−t| (1 − FU (t)) = e−|s−t| (FU (s) + 1 − FU (t)).

In general, RX (s, t) = e−|s−t| (FU (s ∧ t) + 1 − FU (s ∨ t)). (d) X is m.s. continuous if and only if RX is continuous, which, in view of the answer to part (c), is true if and only if FU is continuous. (i.e., there is no fixed number u so that P {U = u} > 0.) (e) If there is a constant uo such that P {U = uo } = 1, or equivalently if FU (u) = I{u≥uo } , then X is clearly a Gaussian random process. because the variables comprising Y and Z are all jointly Gaussian. Otherwise, X is not a Gaussian process, even though Xt is a Gaussian random variable for each t. Here is a proof. If U is not deterministic, there is a choice of s such that 0 < FU (s) < 1. Since FU (t) converges to one as t → ∞ there is a fixed choice of t so that 0 < FU (s) < FU (t) ≤ 1. Let a = FU (s) + 1 − FU (t) and ρ = e−|s−t| . Then a = P {s ≤ U < t} and 0 < a < 1. The joint distribution Xt ) is a mixture of (X s ,

0 1 0 0 1 ρ of the N , distribution and the N , dis0 0 1 0 ρ 1 tribution, with respective weights a and 1 − a. More explicitly, 2

2

2 x −2ρx1 x2 +x2 x2 1−a a 1 +x2 − 1 2(1−ρ 2) e , fX,2 (x1 , s; x2 , t) = √ e− 2 + p 2π 2π(1 − ρ2 )

which is not a Gaussian density.

7.23 First order differential equation driven by Gaussian white noise (a) Since µN ≡ 0 it follows that µX (t) = x0 e−t .The covariance function of X is given by: Z sZ t CX (s, t) = e−(s−u) e−(t−v) σ 2 δ(u − v)dvdu 0 0 Z s Z s −(s−u) −(t−u) 2 2 −s−t = e e σ du = σ e e2u du 0

σ 2 s−t = e − e−s−t .

2

By the symmetry of CX , it is given in general by

σ 2 −|t−s| CX (s, t) = e − e−t−s .

2

0

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(b) Let r < s < t. It must be checked that CX (r, s)CX (s, t) = CX (v, t) CX (s, s) or (er−s − e−r−s )(es−t − e−s−t ) = (er−t − e−r−t )(1 − e−2s ), which is easily done.

2 (c) As t → ∞, µX (t) → 0 and CX (t + τ, t) → σ2 e−|τ | .

7.25 KL expansion of a finite rank process Since K is positive definite, it can be factored as K = U ΛU T , where Λ is a diagonal matrix with nonnegative diagonal entries λ1 , . . . , λN , and U is orthonormal: U −1 = U T . By the given PN PN assumptions, RZ (s, t) = i=1 j=1 Kij ξi (s)ξj (t), or   ξ1 (t)   ..   .

RZ (s, t) = [ξ1 (s)| · · · |ξN (s)]U ΛU T   . So letting   ξN (t) PN [ϕ1 | · · · |ϕN ] = [ξ1 | · · · |ξN ]U , or equivalently, ϕj (t) = i=1 ξi (t)uij , we have   ϕ1 (t)  P  ..   N .

RZ (s, t) = [ϕ1 (s)| · · · |ϕN (s)]Λ   = i=1 λi ϕi (s)ϕi (t). Also,   ϕN (t).

Z T

ϕi (t)ϕj (t)dt =

0

Z TX N X N 0

=

N N X X

a=1 b=1

=

N X

a=1

ξa (t)uai ξb (t)ubj dt

a=1 b=1

I{a=b} uai ubj

uai uaj = I{i=j} .

Thus, ϕ1 , . . . , ϕN are orthonormal eigenfunctions for RZ , and λ1 , . . . , λN are the eigenvalues. The j th coordinate in the K-L expansion of Z can be expressed in terms of the X by R T random vector PN R T PN hZ, ϕj i = 0 Zt ϕj (t)dt = j=1 0 Zt ξi (t)uij dt = j=1 Xi uij . Equivalently, the vector of coordinates for the K − L expansion of Z is given by the vector U T X. Intuitively, in the case of a finite rank process, finding the K-L expansion is equivalent to diagonalizing the correlation matrix K of the random vector of coefficient variables.

7.27 An infinitely differentiable process (a) We show the following statement 2 by induction on n: the nth derivative of RX , R(n) (τ ), has the form Hn (τ )e−τ /2 , where Hn is a polynomial of degree n. The base case, for n = 0, is trivially true, 2 (n) with H0 (τ ) = 1. For the induction step, suppose RX (τ ) = Hn (τ )e−τ /2 , where Hn is a polynomial of degree n. Then (n+1)

RX

2

2

2

(τ ) = (Hn (τ )e−τ /2 )0 = (Hn0 (τ ) − τ Hn (τ ))e−τ /2 = Hn+1 (τ )e−τ /2 ,

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where Hn+1 (τ ) = Hn0 (τ ) − τ Hn (τ ), and indeed Hn+1 is a polynomial of degree n + 1. The statement is thus true for all n by induction. Now RX (n) (τ ) = 2 (2n) (−1)n RX = (−1)n H2n (τ )e−τ /2 . By induction on n, X (n) exists for all n. (b) Since limτ →∞ RX (n) (τ ) = 0, X (n) is mean ergodic in the m.s. sense, for any n ≥ 0.

7.29 Periodicity of a random frequency sinusoidal process (a) We need to check that µX does not depend on time, RX (s, t) depends only on s − t, and RX (T ) = RX (0). To begin, µX (t) = E[A]E[exp(2πjΦt)]. By the assumption P {A > 0} = 1, E[A] 6= 0. Therefore, the only way µX (t) can be constant in t is if E[exp(2πjΦt)] is constant in t. But this is the inverse Fourier transform of the distribution of Φ, and it is equal to a constant if and only P {Φ = 0} = 1.

In this case, we have X(t) ≡ A. For this choice of Φ, RX (s, t) ≡ E[A2 ], so RX (τ ) ≡ RX (0). Thus, under the assumption of (a), X is a WSS periodic random process if and only if P {Φ = 0} = 1 (or, equivalently, if and only if Xt ≡ A.) (b) Since Xt ≡ A, the long time averages of Xt trivially converge to A. In order for X to be mean ergodic in the m.s. sense, the limit must be deterministic.

Hence, under the assumption of (a), X is WSS, periodic, and mean ergodic process if and only if Xt ≡ a for some positive constant a. (c) The mean function, given by µX (t) = E[A]E[exp(2πjΦt)], does not depend on t if either E[A] = 0 or if E[exp(2πjΦt)] is constant in t. By the analysis in part (a), it means µX (t) does not depend on t if either E[A] = 0 or X(t) ≡ A. If X(t) ≡ A then X is WSS and periodic. We turn attention to the other case, namely E[A] = 0. RX (s, t) = E[A2 ]E[exp(2πjΦs) exp(−2πjΦs)] = E[A2 ]E[exp(2πjΦτ ) where τ = s − t, so that RX (s, t) is a function of s − t alone with no additional assumptions. Thus, X is WSS if E[A] = 0. Finally, in order for X to be periodic, we need that for some T > 0, RX (0) = RX (T ) or 1 = E[exp(2πjΦτ )], or equivalently, E[cos(2πΦτ )] = 1 and E[sin(2πΦτ )] = 0, which is true if and only if cos(2πΦτ ) ≡ 1. Equivalently, if and only if P {Φ ∈ { Tk : k ≥ 0}} = 1 for some T > 0.

In summary, under the assumption Var(A) > 0, the process X is WSS and periodic if and only if either X ≡ A or (E[A] = 0 and P {Φ ∈ { Tk : k ≥ 0}} = 1 for some T > 0). (d) If X ≡ A, then since it is assumed that Var(A) > 0, the long time averages of X converge to a random quantity, and X is not mean ergodic in the m.s. sense in that case. If E[A] = 0 and P {Φ ∈ { Tk : k ≥ 0}} = 1, then on the event Φ = Tk , theR limit of the long time averages is the average over one time period, given by 1 T A T 0 Xt dt = T I{k=0} . Therefore, if P {Φ = 0} > 0, the limiting time average is random, and X is not mean ergodic in the m.s. sense. If that case is avoided (i.e. if X has no power at zero frequency with probability one) X is mean ergodic in the m.s. sense. Summarizing, under the assumption of part (c), X is WSS, periodic, and mean ergodic if and only if E[A] = 0 and P {Φ ∈ { Tk : k ≥ 1}} = 1 for some constant T > 0.

7.31 Application of the KL expansion to estimation

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55

(a) E

"Z

0

T

#

Xt2 dt

2

= AE[B ]

Z T 0

2

sin

πt T

dt = A

2

Z T 0

1 − cos( 2πt A2 T T ) dt = .

2 2

(b) E[Xt ] = AE[B] sin

πt T

≡0

2

CX (s, t) = A sin

πs T

sin

πt T

0 ≤ s, t ≤ T.

(c) Suppose ϕ is an eigenfunction of CX for some λ 6= 0; Then

Z T πs πt λϕ(s) = A2 sin sin ϕ(t)dt T T 0

πs Z T πt 2 A sin ϕ(t)dt . = sin T T |0 {z } a constant

so ϕ must be proportional to the function sin πs T . Thus, r

2 πt A2 T sin 0 ≤ t ≤ T, λ1 = . ϕ1 (t) = T T 2 and the K − L expansion of X is Xt = (X, ϕ1 )ϕ1 (t) where r r

Z T πt 2 πt T hX, ϕ1 i = AB sin sin dt = AB .

T T T 2 0 q

The coordinates of X relative to ϕ1 , ϕ2 , · · · are given by X ↔ AB T2 , 0, 0, · · · . (d) The coordinates of N are given by N ↔ (N1 , N2 , · · · ) where the Ni are N (0, σ 2 ), and the coordinates of Y are given by r T Y ↔ (N1 + AB , N2 , N3 , · · · ).

2 | {z } Y1

The random variables B and Y1 are mean 0, so q (A T2 )Y1 Cov(B, Y1 )Y1 E[B|Y1 ] = =1− 2 Var(Y1 ) σ 2 + A2T

Z T A πt

Yt dt. = sin A2 T 2 T σ + 2 0

and

2

A T Cov(B, Y1 )2 2 MMSE = Var(B) − =1− 2 Var(Y1 ) σ 2 + A2T

=

σ2 .

2 σ 2 + A2T

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8.1 Baseband limiting (a) The total power in X is RX (0) = 1. The power 2 spectral density of X is the Fourier transform of RX , SX (2πf ) = 1+(2πf )2 . The fo Rf 2 2 −1 = 0.99. desired cutoff frequency fo satisfies 2 0 o 1+(2πf df = tan (2πf ) 2 ) π 0

1 Or fo = 2π tan( 0.99π 2 ) = 10.13. (b) Yes, they are independent. They are jointly Gaussian processes, because they are both obtained from the Gaussian process X by linear operations, and they are uncorrelated because the cross spectral density is given by SX,Y (2πf ) = I{|f |≤fo } I{|f |>fo } SX (2πf ) ≡ 0. Here we use the fact that X − Y can be viewed as the output of a linear filter with input X and transfer function 1 − H(2πf ) = I{|f |>fo } . (c) By (b) we know that the total power of X is the sum of the total power of Y and the total power of X −Y. Since Y takes 99% of the power in X, the remaining 1% is in X −Y. That is, the power of X −Y is 0.01. Thus, for any t, Xt −Yt has the N (0, 0.01) distribution. So P {|Xt − Yt | ≥ 0.1} = 2Q( √0.1 ) = 2Q(1) = 0.3174.

0.01

8.3 On filtering a WSS random process (a) True. Since SY (ω) = |H(ω)|2 SX (ω), it follows that SY (ω) ≤ SX (ω) for all ω.

R ∞Integrating over all frequencies yields the result, because the power of X is −∞ SX (ω) dω 2π and similarly for Y . (b) True. The process Y is WSS because X is WSS and the system is timeinvariant. Since with some period T . Thus, R ∞X is WSS and periodic, RX isRperiodic ∞ RY (τ +T ) = −∞ RX (τ +T −t)(h∗ e h)(t)dt = −∞ RX (τ −t)(h∗ e h)(t)dt = RY (τ ), so that RY is also periodic with period T . Therefore, Y is periodic. (c) False. The output power is zero if (and only if) SX (ω)|H(ω)|2 ≡ 0. This happens, for example, if H(ω) = I{|ω|≤2π} and SX (ω) = I{4π≤|ω|≤6π} .

8.5 Modulating and filtering a stationary process (Note: Since cos(80πt + Θ) = cos(Θ) for any integer time t, parts (a) and (b) of the problem as stated are degenerate. It would make more sense either in continuous time, or if the frequency of the cosine term were non-integer, such as Yt = Xt cos(2πt/40 + Θ). The form of the answer given here would be correct for noninteger frequencies, thinking of SX as a period 2π function.) (a) Since Θ and ((Θ + c) mod 2π) have the same distribution, the process cos(80πt + Θ) is stationary. Two independent stationary processes are jointly stationary, and the product of two jointly stationary processes is also stationary. (b) Using the fact cos(a) cos(b) = (cos(a − b) + cos(a + b))/2, we find RY (s, t) = E[Xs Xt ]E[cos(80π(s−t))+cos(80π(s+t)+2Θ)]/2 = RX (s, t) cos(80π(s− ) RX (τ ). In the transform domain this is SY (ω) = t))/2. So RY (τ ) = cos(80πτ 2 1 [S (ω − 80π) + S (ω + 80π)].

X X 4 P∞ (c) The impulse response function is given by h(τ ) = k=0 a(1 − a)k δ(τ − k).

P∞ a Thus, H(ω) = k=0 a[(1 − a)e−jω ]k = 1−(1−a)e −jω . −jω (d) Note that |1 − (1 − a)e | ≥ 1 − (1 − a) = a, so that |H(ω)| ≤ 1 for all ω.

Thus, SZ (ω) ≤ SY (ω) for all ω. Therefore the power of Z is less than or equal

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to the power of Y , which is given by RY (0) = 21 RX (0) = 1/2. (e) Examining H further shows that |H(ω)| = 1 if and only if ω is a multiple of 2π, and in that case H(ω) = 1. Now, if X is a periodic process with power one and period one (or a smaller period that divides one), then Y will have power 0.5 and also have period one (or a smaller period that divides one) because the frequencies in Y are the frequencies in X plus or minus 40 Hz. Then, Y will pass through the filter H unchanged, and the power of Z will therefore be 0.5.

R∞ 8.7 Slight smoothing (a) RY X (τ ) = h∗RX (τ ) = −∞ h(τ −t)RX (t)dt. Setting τ = 0, and using the symmetry of h and of RX yields RY X (0) = h ∗ RX (0) Z ∞ = h(0 − t)RX (t)dt −∞ Z a/2

2 e−t dt a 0

2 a 1 a 2 2 1 a 3 1−1+ − = (1 − e−a/2 ) = + + ··· a a 2 2 2 3! 2 2 a a a =1− + + · · · = 1 − + o(a).

4 24 4 R∞ (b) RY (τ ) = h ∗ e h ∗ RX (τ ) = −∞ (h ∗ e h)(τ − t)RX (t)dt. First we find that h ∗ e h is the triangle function over the interval [−a, a] with height 1/a. Thus, taking τ = 0 and using the symmetry of h ∗ e h and of RX yields Z ∞ RY (0) = (h ∗ e h)(0 − t)RX (t)dt −∞ Z a 2 (a − t)e−t dt = 2 a 0

2 2 a2 a3 a4 = 2 (a − 1 + e−a ) = 2 − + ··· a a 2 3!

4! a a2 a =1− + + · · · = 1 − + o(a).

3 12 3 =

(c) E[|Xt − Yt |2 ] = E[(Xt − Yt )(Xt − Yt )∗ ] = RX (0) − RXY (0) − RY X (0) + RY (0).

Since RX (0) = 1 and RXY (0) = RY∗ X (−0) = RY X (0), it follows that E[|Xt −Yt |2 ] = 1−2RY X (0)+RY (0) = 1−2(1− a4 +o(a))+1− a3 +o(a) = a6 +o(a) 8.9 A stationary two-state Markov process in continuous time πQ = 0 implies π = ( 21 , 12 ) is the equilibrium distribution so P {Xt = 1} = P {Xt = −1} = 12 for all t. Thus µX = 0. For τ ≥ 0 RX (τ ) = P {Xτ = 1, X0 = 1} − P {Xτ = −1, X0 = 1}

− P {Xτ = 1, X0 = −1} + P {Xτ = −1, X0 = −1} 1 + e−2ατ 1 − e−2ατ 1 − e−2ατ 1 + e−2ατ = − − + = e−2ατ .

4 4 4 4

So in general, RX (τ ) = e−2α|τ | . (This is the same as the autocorrelation function

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of a Gaussian Markov process.) The corresponding power spectral density is given by: SX (ω) = 4α24+ω2 .

8.11 Linear time invariant, uncorrelated scattering channel P∞ (a) Yi = u=−∞ E[hu ]xi−u = 0 P P P E[Yi Yj∗ ] = u v xi−u x∗j−u E[hu h∗v ] = u xi−u x∗j−u gu P (b) So E[|Yi |2 ] = u |xi−u |2 gu . The average total energy of Y is P P P P P P 2 2 2 2 i u |xi−u | gu = u i |xi−u | gu = ( i |xi | ) u gu = G||x|| , or G times the energy of the input signal.

P P ∗ ∗ (c) Y has mean zero as before. E[Yi Yj∗ ] = u v E[Xi−u Xj−u ]E[hu hv ] = P P P u v RX (i − j)gu I{u=v} = u RX (i − j)gu = GRX (i − j). Thus Y is WSS with mean zero and RY = GRX .

P P (d) RH (ω1 , ω2 ) = E[H(ω1 )H ∗ (ω2 )] = m n e−jω1 m ejω2 n E[hm h∗n ] P −j(ω1 −ω2 )m = me gm = gb(ω1 − ω2 ). Thus, H is a WSS periodic random process with autocorrelation function gb and power spectral mass function g.

8.13 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) = ak I{k≥0} . Thus X is stationary.

P µ , and Since µU = µ and CU (n) = σ 2 I{n=0} , we find E[Xk ] = µ k∈Z h(k) = 1−a P 2 |n| σ a 2 2 ∗ e e CX (n) = h ∗ h ∗ CU (n) = σ h ∗ h(n) = σ h(k)h (k − n) = 2 . (Think k∈Z

1−a

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) = Ak I{k≥0} . Y is 1 stationary for the same reason X is stationary. The mean is µY = µE[ 1−A ]. To find CY we first find RY as follows:

RY (n) = E RX (n) a=A = µ2 E Then, CY (n) = RY (n) − µ2Y = µ2 Var

1 1−A

2 |n| σ A 1 +E .

2 (1 − A) 1 − A2

+E

h 2 |n| i σ A 1−A2

.

(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 . µ (f) No. For A fixed, the time average of Y is 1−A , by parts (a) and (c). Since this time average is not a constant, Y is not mean ergodic in the m.s. sense. Another 1 justification is to note that as n → ∞, CX (n) → µ2 Var( 1−A ) 6= 0.

1 8.15 A linear system with a feedback loop (a) K = 1 + H + H 2 + · · · = 1−H , where H denotes the loop transfer function, for all ω such that |H(ω)| < 1. That

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is, K(ω) = 1−

1

α 1+jω

=

1 + jω α =1+ .

1 − α + jω (1 − α) + jω

(b) K ↔ k(t) = δ(t) + αI{t≥0} e−(1−α)t .

1+ω 2 −2 (c) Note that |K(ω)|2 = (1−α) , with equality at ω = 0. There2 +ω 2 ≤ (1 − α) fore, for any choice of SX , the ratio R to be maximized satisfies

R=

R∞

SX (ω)|K(ω)|2 dω 2π R∞ ≤ dω S (ω) 2π −∞ X

−∞

R∞

−∞

SX (ω)(1 − α)−2 dω 2π R∞ = (1 − α)−2 . dω S (ω) 2π −∞ X

with equality if SX (ω) = δ(ω), or in other words, if all the power is at frequency zero. If we insist on having an ordinary spectral density, the ratio (1 − α)−2 can be approached arbitrarily closely by densities SX supported on an interval of length around zero, with small.

8.17 Sampling a cubed Gaussian process (a) Yes, Y is WSS, because it is obtained from the stationary process Y by sliding a time-invariant function along the stationary process X. By the hint, RY (τ ) = Cov(Xτ3 , X03 ) = 6RX (τ )3 + 9RX (0)2 RX (τ ). Therefore, SY (ω) = 6(SX ∗ SX ∗ SX )(ω) + 9RX (0)2 SX (ω). That is, a triple convolution is involved. (b) Due to the triple convolution, we can see that Y is a baseband random process with one-sided band limit 3fo . Therefore, Y can be linearly reconstructed from its samples taken at sampling rate T10 = 6fo . (c) If nonlinear processing is allowed, then Y can be recovered from its samples 1/3 taken at sampling frequency 2fo , because XnT = YnT . That is:

Yt =

∞ X

n=−∞

1/3 YnT sinc

t − nT T

!3

.

8.19 Simulating a baseband random process (a) With Bn = Xn for n ∈ Z, the discrete time process B has SB (ω) = SX (ω) = 1 for |ω| ≤ π. Thus, RB (k) = I{k=0} so the (Xn : n ∈ Z) are independent N (0, 1) random variables. (b) The approximation error is given by

(N )

Xt − Xt

=

∞ X

n=N +1

sinc(t − n)Xn +

−∞ X

n=−(N +1)

sinc(t − n)Xn .

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The terms are independent, mean 0. So if t ∈ [−500, 500] and N ≥ 501, (N ) 2

E[(Xt − Xt

) ]=

∞ X

n=N +1

≤

∞ X

n=N +1 Z ∞

sinc(t − n)2 + 1 + 2 π (n − t)2

−∞ X

n=−(N +1) −∞ X

n=−(N +1) Z ∞

sinc(t − n)2 1

π 2 (t − n)2

1 1 du + du 2 (u − t)2 2 (t − u)2 π π N −N 1 1 = 2 + . π (N − t) π 2 (N + t) ≤

The maximum of this bound over t ∈ [−500, 500] occurs at the endpoints, so for any t ∈ [−500, 500], (N ) 2

E[(Xt − Xt

1 1 + π 2 (N − 500) π 2 (N + 500) ≤ 0.01 if N ≥ 511.

) ]≤

8.21 Filtering to R ∞maximize signal to noise ratio The problem is to select H to minimize σ 2 −∞ |H(ω)|2 dω 2π , subject to the constraints (i) |H(ω)| ≤ 1 for all ω, Rω and (ii) −ωo o |ω| |H(ω)|2 dω ≥ (the power of X)/2. First, H should be zero outside 2π of the interval [−ωo , ωo ], for otherwise it would be contributing to the output noise power with no contribution to the output signal power. Furthermore, if |H(ω)|2 > 0 over a small interval of length 2π contained in [ωo , ωo ], then the contribution to the output signal power is |ω||H(ω)|2 , whereas the contribution to the output noise is σ 2 |H(ω)|2 . The ratio of these contributions is |ω|/σ 2 , which we would like to be as large as possible. Thus, the optimal H has the form H(ω) = I{a≤|ω|≤ωo } , where a is selected to meet the signal power constraint with √ e equality: (power of X)= (power of X)/2. This yields a = ωo / 2. In summary, the optimal choice is |H(ω)|2 = I{ωo /√2≤|ω|≤ωo } .

8.23 Sampling a signal or process that is not band limited (a) Evaluating the co at nT , and using the fact that 2ωo T = 2π yields inverse transform of x R ∞ jωnT R R ωo jωnT P o co (ω) dω = ωo ejωnT x co (ω) dω = ∞ x (nT ) = −∞ e x x b(ω+ m=−∞ −ωo e 2π 2π −ωo 2mωo ) dω R ∞ jωnT P∞ 2π R (2m+1)ω x b(ω) dω = m=−∞ (2m−1)ωoo ejωnT x b(ω) dω 2π = −∞ e 2π = x(nT ). (b) The observation follows from the sampling theorem for the narrowband signal xo and part (a). o (c) The fact RX (nT ) = RX (nT ) follows from part (a) with x(t) = RX (t). o (d) Let X be a WSS baseband random process with autocorrelation function o RX . Then by the sampling theorem for baseband random processes, Xto =

P∞ t−nT o . But the discrete time processes (XnT : n ∈ Z) and n=−∞ XnT sinc T o (XnT : n ∈ Z) have the same autocorrelation function by part (c). Thus X o and Y have the same autocorrelation function. Also, Y has mean zero. So Y is WSS o with mean zero and autocorrelation function RX .

Solutions to Odd Numbered ProblemsRandom Processes for Engineers

61

P∞ o (e) For 0 ≤ ω ≤ ω o , SX (ω) = n=−∞ exp(−α|ω + 2nωo |) P∞ P−∞ = n=0 exp(−α(ω + 2nωo )) + n=−1 exp(α(ω + 2nωo )) o )) o ))+exp(α(ω−ω0 )) o −ω)) = exp(−αω)+exp(α(ω−2ω = exp(−α(ω−ω = cosh(α(ω . Thus, 1−exp(−2αωo ) exp(αωo )−exp(−αωo )) sinh(αωo ) for any ω, cosh(α(ωo − |ω|)) o .

SX (ω) = I{|ω|≤ωo } sinh(αωo ) 8.25 Another narrowband Gaussian process (a) Z ∞ µX = µR h(t)dt = µR H(0) = 0 −∞

4

SX (2πf ) = |H(2πf )|2 SR (2πf ) = 10−2 e−|f |/10 I5000≤|f |≤6000 .

(b) Z ∞

2 102

Z 6000

4

e−f /10 df = (200)(e−0.5 − e−0.6 ) = 11.54 −∞ 5000

6 ≈ Q(1.76) ≈ 0.039.

X25 ∼ N (0, 11.54) so P {X25 > 6} = Q √ 11.54

RX (0) =

SX (2πf )df =

(c) For the narrowband representation about fc = 5500 (see Figure 1.4),

SX

SU=SV

For fc= 5500

j SUV

SU=SV

For fc= 5000

SUV

j

Figure 1.4 Spectra of baseband equivalent signals for fc = 5500 and fc = 5000.

h i 4 4 SU (2πf ) = SV (2πf ) = 10−2 e−(f +5500)/10 + e−(−f +5500)/10 I|f |≤500 =

e−.55 cosh(f /104 )I|f |≤500 .

50

h i 4 4 −je−.55 SU V (2πf ) = 10−2 je−(f +5500)/10 − je−(−f +5500)/10 I|f |≤500 = sinh(f /104 )I|f |≤500 .

50 (d) For the narrowband representation about fc = 5000 (see Figure 1.4), 4

SU (2πf ) = SV (2πf ) = 10−2 e0.5 e−|f |/10 I|f |≤1000 .

SU V (2πf ) = jsgn(f )SU (2πf ).

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8.27 Declaring the center frequency for a given random process (a) SU (ω) = g(ω + ωc ) + g(−ω + ωc ) and SU V (ω) = j(g(ω + ωc ) − g(−ω + ωc )). (b) R∞ R∞ R ∞ The integral becomes: 2 dω (g(ω + ωc ) + g(−ω + ωc ))2 dω 2π = 2 −∞ g(ω) 2π + 2 −∞ g(ω + ωc )g(−ω + −∞ 2 ωc ) dω 2π = 2||g|| + g ∗ g(2ωc ) Thus, select ωc to maximize g ∗ g(2ωc ).

9.1 A quadratic predictor (a) By the orthogonality principle, the estimator is optimal if and only if bn+1 Xn+1 − X bn+1 Xn+1 − X bn+1 Xn+1 − X

⊥ ⊥ ⊥

1 Xi Xi Xl

for 1 ≤ i ≤ n for 1 ≤ i ≤ l ≤ n

or equivalently, Pn Pj h0 + j=1 k=1 h2 (j, k)E[Xj Xk ] = 0 Pn E[Xn+1 Xi ] − h0 E[Xi ] − k=1 h1 (k)E[Xk Xi ] Pn Pj − j=1 k=1 h2 (j, k)E[Xj Xk Xi ] = 0 for 1 ≤ i ≤ n Pn E[Xn+1 Xi Xl ] − h0 E[Xi Xl ] − k=1 h1 (k)E[Xk Xi Xl ] Pn Pj − j=1 k=1 h2 (j, k)E[Xj Xk Xi Xl ] = 0 for 1 ≤ i ≤ l ≤ n

(b) If X is Gaussian, then the optimal unconstrained estimator of Xn+1 given X1 , . . . , Xn is linear, so the quadratic term h2 is identically zero. Therefore, also, h0 = 0. (This observation can be verified another way. Since third order moments of joint Gaussians, such as E[Xn+1 Xi Xl ] and E[Xk Xi Xl ], are zero, the first and third equations above are true for h0 = 0 and h2 ≡ 0.)

9.3 A simple prediction problem h By the orthogonality principle, it suffices for i RT b XT − XT ⊥ Xu for all u ≤ 0. Or E XT − 0 g(t)Xt dt Xu = 0 for all u ≤ 0.

R0 Which yields the desired equations: RX (T − u) = −∞ g(t)RX (t − u)dt for all u ≤ 0. bT = E[XT |X0 ] = RX (T )X0 = (b) The Markov property of X implies that X RX (0) X0 e−αT . Equivalently, g(t) = δ(t)eT , which we see satisfies the conditions of part (a).

9.5 A simple, noncausal estimation problem (a) Since RX is a sinc function scaled by fo , it’s Fourier transform is a rectangle function of width fo centered at the origin, and RX (0) is the integral of the Fourier transform. So SX (2πf ) = 2

A I{|f |≤fo /2} SX (2πf ) A2 2 fo I{|f |≤fo /2} . Also, SN ≡ σ . Thus, H(2πf ) = SX (2πf )+SN (2πf ) = A2 +σ 2 fo , A2 fo sinc(fo t) which yields h(t) = A2 +σ2 fo .

(b) Since X and N are jointly Gaussian, and Y and D are obtained from them by linear operations, the processes X, N , Y , and D are jointly Gaussian.

All processes have mean zero. The mean square error is given by the formula R∞ 2 2 σ fo SN MSE= −∞ SSXX+S df = AA2 +σ 2 f . Thus, for t fixed, Dt is a N (0, M SE) random o N variable. (c) The conditional distribution is also N (0, M SE), because by the orthogonality principle, Dt is orthogonal, and hence uncorrelated with, the process Y , and

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because uncorrelated jointly Gaussian random variables are independent. Therefore, Dt is independent of the random process Y . (d) First, RDY ≡ 0 by the orthogonality principle. It follows also that RDXb ≡ 0. b + D and R b ≡ 0, it follows that RD = RX − R b . But Since X = X X DX 4 A4 I{|f |≤|fo |/2} S (2πf ) = |H(2πf )|2 S (f ) = . Therefore, R (t) = A sinc(fo t) , b X

Y

2 2 sinc(fo t) . so that RD (t) = A σAf2o+σ 2f o

b X

fo (A2 +σ 2 fo )

A2 +σ 2 fo

9.7 Estimation of a filtered narrowband random process in noise (a) Independent baseband random processes U and V with power spectral density SU (2πf ) = SV (2πf ) = 2I{|f |≤4} can be generated using the sampling theorem as follows.

The cutoff frequency for these baseband processes is 4 Hz, and since the spectrum is flat across the band, the samples in the sampling theorem for sampling frequency 8 Hz form a discrete-time Gaussian white noise sequence. The baseP∞ band processes therefore have the form n=−∞ sinc(8t − n)σWn , where the Wn are independent standard normal random variables, and σ 2 = 16. Then set Xt = cos(2π104 t)Ut − sin(2π104 t)Vt . Of course on a computer the sum would have to be approximated by a finite number of terms, and both the processes and sinc functions would be computed for a finite number of time points, for example, 100 samples per second. The method requires 16 normal variables per unit of simulated time, for long simulation intervals. (For short intervals it is more, due to the fact the sinc terms beyond the interval of interest should be included.) (b) The transfer function H is nearly constant, equal to 1/1.1 ≈ .91 over the narrow band where SX is not zero. Thus, the effect of the filterR is to attenuate the signal by the factor (.91)2 . Since X has power RX (0) = SX = 16, the power out of the filter is approximately 16(.91)2 = 13.2. (c) Note that SZ ≡ 1. The cross spectral density is given SZX (2πf ) = SX (2πf ) = 1 in the narrow band ||f | − 104 | ≤ 4 and is zero outside of that band. The spectral density of Z is 2 over that band. Hence, 4the optimal filter is given by 0.5 ||f | − 10 | ≤ 4 H(2πf ) = SXZ (2πf )/SZ (2πf ) = 0 else R∞ S2 df = 16 × (1 − 0.5) = 8. (d) The mean square error is given by −∞ SX − SXZ Z 9.9 Predicting the future of a simple WSS process (a) First finding that the denominator of SX vanishes for ω 2 = −4 and for ω 2 = −9, we find the factorization: 1 1 SX (ω) = . (jω + 2)(jω + 3) (−jω + 2)(−jω + 3) | {z }| {z } S − (ω)

S + (ω)

+

(b) The optimal filter is given by H(ω) = S +1(ω) SX (ω)ejωT + . The technique X

+ 1 1 − jω+3 ↔ I{t≥0} (e−2t − of partial fraction expansions yields that SX (ω) = jω+2

+ + e−3t ). Thus SX (ω)ejωT ↔ I{t≥−T } (e−2(t+T ) − e−3(t+T ) ) so SX (ω)ejωT + ↔ +

e−2T e−3T I{t≥0} (e−2(t+T ) − e−3(t+T ) ), so that SX (ω)ejωT + = jω+2 + jω+3 . Therefore,

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bt+T |t = Xt (3e−2T − 2e−3T ) + H(ω) = e−2T (jω + 3) − e−3T (jω + 2). Therefore, X 0 −2T −3T Xt (e −e ).

R∞ + (c) The MSE for the optimal predictor is RX (0) − −∞ | SX (ω)ejωT + |2 dω 2π = R ∞ −2(t+T ) −3(t+T ) 2 −4T −5T −6T RX (0) − 0 ((e −e ) dt = RX (0) − e /4 − 2e /5 + e /6.

The constant RX (0) can be found by noting that the MSE is zero for T = 0, so that M M SE = (1 − e−4T )/4 − 2(1 − e−5T )/5 + (1 − e−6T )/6.

9.11 On the MSE for causal estimation Let g be the function with Fourier (ω) transform G(ω) = SSXY , the MSE can be written as − (ω) Y

(MSE) = RX (0) −

Z ∞

2 dω G(ω)ejωT + .

2π −∞

Now G(ω)ejωT ↔ g(t + T ), so that, using the definition of [ ]+ , Parseval’s 2 R∞ R∞ 2 identity, and a change of variables, −∞ G(ω)ejωT + dω 2π = 0 |g(t + T )| dt = R∞ |g(s)|2 ds. Therefore, T Z ∞ dω lim MSE = RX (0) − ||g||2 = RX (0) − |G(ω)|2 T →−∞ 2π −∞ Z ∞ |SXY (ω)|2 dω = RX (0) − SY (ω) 2π −∞ which is the mean square error for noncausal estimation of Xt given all of Y .

This makes sense, because if T is a very large magnitude negative number, the process Y can be observed for long after the time that X is to be estimated.

Also, lim MSE = RX (0),

T →+∞

which is the MSE for no observations. This makes sense, because if T is very large, the observations become nearly useless.

9.13 Filtering a WSS signal plus noise (a) SY = |K1 K2 |2 SX +|K2 |2 SN , where for notational brevity we suppress the argument (ω) for each function.

R (b) SNRoutput =

(c) H = SSXY = Y (d)

∞ |K1 K2 |2 SX dω 2π −∞ R∞ |K2 |2 SN dω 2π −∞ K1∗ K2∗ SX |K! K2 |2 SX +|K2 |2 SN

.

MMSE =

Z ∞

−∞

Z ∞

.

SX − SY |H|2

dω 2π

|K2 |2 SX SN dω 2 |K |2 S + |K |2 S 2π |K | 1 2 X 2 N −∞ Z ∞ SX SN dω . = 2 −∞ |K1 | SX + SN 2π =

(e) Since K2 is invertible, for the purposes of linear estimation of X, using the process Y is the same as using the process Y filtered using a system with transfer

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function K21(ω) . Equivalently, the estimate of X would be the same if the filter K2 were dropped from the original system.

9.15 Properties of a particular Gaussian process (a) Yes, X is mean ergodic because, for example, RX (τ ) → 0 as τ → ∞. (b) No, X is not Markov because RX is not of the form Ae−α|τ | . (c) Yes, because RX is twice continuously differentiable. One way to show this is the following. The right hand derivative of RX at zero is the same as the derivative of (1 + τ )e−τ at zero, which is e−τ − (1 + τ )e−τ |τ =0 = −τ e−τ |τ =0 = 0, and similarly the left-hand derivative is the same as the derivative of (1 − τ )eτ at zero, which is −eτ + (1 − τ )eτ |τ =0 = −τ eτ |τ =0 = 0. Thus, RX is continuously differentiable with  −τ τ >0  −τ e 0 τ RX (τ ) = (1.14) −τ e τ <0  0 τ =0

0 Equivalently, RX (τ ) = |τ |eτ , but since we have to take another derivative, it is 00 is simply more convenient to use the version (1.14). The second derivative RX 0 the first derivative of RX , which is similarly given by

 −τ  (−1 + τ )e 00 RX (τ ) = (−1 − τ )eτ  −1

τ >0 τ <0 τ =0

00 (τ ) = (|τ |−1)e−|τ | . A second approach to finding which can also be written as RX 00 RX at zero is to use the Taylor series expansion of e−|τ | in powers of |τ | to get 2 RX (τ ) = (1 + |τ |)(1 − |τ | + |τ2| − o(|τ |2 ) = 1 − τ 2 + o(|τ |2 ). A third approach to the problem is to work in the frequency domain, noting that (ω)2 SX (ω) is 00 exists and is continuous. integrable, so that RX + − + − 2 2 (d) SX (ω) = SX (ω)SX (ω) where SX (ω) = (1+jω) 2 and SX (ω) = (1−jω)2 . Take + H = SX . (e) If a signal n is the input to a linear system with transfer function H, then the output x satisfies (1 + jω)2 x b(ω) = 2b n(ω), or ((jω)2 + 2jω + 1)b x(ω) = 2b n(ω) 00 0 or x + 2x + x = 2n, so that X satisfies the stochastic differential equation X 00 + 2X 0 + X = 2N .

9.17 A continuous-time Wiener filtering problem (a) The optimal filter without the causality constraint is given by 4

H(ω) =

SXY (ω) SX (ω) 2 = = 4 4+ω No SY (ω) SX (ω) + SN (ω) 4+ω 2 + 2 =

where α =

q

8 No

ω 2 + α2

, −α|t|

4 + N8o , or in the time domain, h(t) = 4eαNo . The corresponding

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MMSE is given by MMSE (noncausal) =

Z ∞ −∞

= =

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Z ∞

SXY (ω)2 SX (ω) − SY (ω)

SX (ω) N2o

dω 2π

dω

No −∞ SX (ω) + 2 2π

Z ∞

dω 1 4 =q 2 + α2 2π ω −∞ 1+ 2

No

As expected, the MMSE increases monotonically from 0 to 1 as N0 ranges from 0 to ∞. (b) We shall use the formula for the optimal filter H(ω). The factorization of SY is given by 8 2 No No (4 + No ) + ω 4 + = SY (ω) = 4 + ω2 2 2 4 + ω2 r r No (jω + α) No (−jω + α) · . = 2 (jω + 2) 2 (−jω + 2) | {z } | {z } + (ω) SY

Thus

where γ =

− (ω) SY

r 4 2 1 SXY (ω) (jω+2)(−jω+2) = = 4 No (jω + 2)(−jω + α) SY− (ω) SY− (ω)

1 1 =γ + . jω + 2 −jω + α

q 4 N2o 2+α

. Therefore SXY (ω) ↔ γe−2t I{t≥0} + γeαt I{t<0} , SY− (ω)

so that jωT

e SXY γe−2(t+T ) I{t ≥ 0} ↔ − γe−2(t+T ) I{t ≥ −T } + γeα(t+T ) I{0 ≤ t ≤ T } SY + Converting back to the transform domain yields  −2T jωT

 γe e SXY jω+2 = γ [eiωT −eαT ] γejωT  SY− + + jω+2

−jω+α

if

T ≥0

if

T <0

if if

T ≥0 T <0

Dividing by SY+ yields the optimal transfer function. For the remainder of this problem solution assume T ≥ 0. The optimal transfer function is thus q γe−2T N2o H(ω) = , jω + α

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so that the optimal impulse response function is given by  −2T 4eq  e−αt t ≥ 0 No (1+ 1+ N2o ) h(t) =  0 t<0

The MMSE is given by

MMSE = RX (0) − =1−

Z ∞

−∞

h(s)RXY (s + T )ds = 1 −

e−4T p .

1 + No + No2 + 2No

Z ∞

h(s)e−2(s+T ) ds

0

As expected, the MMSE converges to one as either T → ∞ or No → ∞, and it converges to 1 − e−4T as No → 0.

9.19 Noiseless prediction of a baseband random process (a) Multiplication by jω in the frequency domain corresponds to applying the differentiation operator ∂ in the time domain. So (jw)n corresponds to differentiating n times. Thus, k k b (n) = Pn (∂ X)|t T . The derivatives can be computed by taking limits X k=0 t+T |t k! from the left, so the derivatives only depend on X up to time t. (b) For |ω| ≤ ωo , |ejωT − H (n) (ω)| = ≤ ≤

∞ X (jωT )k k!

k=n+1 ∞ X

k=n+1 ∞ X k=n+1

(jωT )k k!

(ωo T )k 4 = bn . k!

Thus, 0 ≤ an ≤ bn , and limn→∞ bn = 0 by the convergence of the power series for eωo T . Thus, limn→∞ an = 0. b (n) is the output of a linear system with input (c) The error process Xt+T − X t+T |t X and transfer function ejωT − Hn (ω), so the power spectral density of Xt+T − b (n) is |ejωT − Hn (ω)|2 SX (ω). Therefore, X t+T |t Z ωo dω (n) (MSE using H ) = |ejωT − H (n) (ω)|2 SX (ω) 2π −ω Z ωoo dω ≤ a2n SX (ω) 2π −ωo = a2n RX (0) → 0 as n → ∞.

(d) The scheme would indeed produce Xt+T with zero MSE. But the narrowband filters needed to generate the subprocesses are not causal.

9.21 A singular nonlinear estimation problem (a) We seek the best estimator of X1 in the m.s. closed linear span of (1, (Yu : 0 ≤ u ≤ 0)). Let Yeu = Yu −λu and eu = Nu − λu. Then we seek the best estimator of X e1 in the m.s. closed linear N

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span of (1, (Yeu : 0 ≤ u ≤ 1)). This is just the best linear estimator of X1 in the eu this reduces to m.s. closed linear span of (Yeu : 0 ≤ u ≤ 1). Since Yeu = Xu + N estimation of a signal at a given time from the sum of a signal plus independent noise. The correlation function of the noise, RNe (s, t) = λ(s ∧ t), is proportional to the correlation function of the signal, RX (s, t) = s ∧ t. We can thus apply the problem above on proportional noise, so that the best linear estimator of X1 , based on (1, (Yu : 0 ≤ u ≤ 1)) is b1 = X

Y1 − λ 1 e Y1 = 1+λ 1+λ

λ λ Var(X1 ) = 1+λ . and the MSE is 1+λ (b) The observations (Yt : 0 ≤ t ≤ 1) determine X1 perfectly: X1 is Y1 minus the number of jumps of Y over [0, 1]. So the best nonlinear estimator of X1 given (Yt : 0 ≤ t ≤ 1) is X1 . The mean square error is zero.

9.23 Causal estimation of a channel input process Taking transforms, we have 3 2 SX (ω) = 1+ω 2 , SN (ω) ≡ 1, and K(ω) = jω+3 . Thus,

(3 + ω 2 )(7 + ω 2 ) SY (ω) = SX (ω)|K(ω)|2 + 1 = (1 + ω 2 )(9 + ω 2 ) √ √ √ √ (jω + 3)(jω + 7) (−jω + 3)(−jω + 7) = (jω + 1)(jω + 3) (−jω + 1)(−jω + 3) {z }| {z } | + (ω) SY

− (ω) SY

6 and SXY (ω) = K ∗ (ω)SX (ω) = (−jω+3)(jω+1)(−jω+1) .

(ω) γ1 γ3√ γ2 √ = (jω+1)(−jω+6√3)(−jω+√7) = jω+1 + jω+ . where γ1 = So SSXY + −jω+ − 3 7 (ω) Y

√ 6 √ .

(1+ 3)(1+ 7) i h SXY (ω) γ1 γ1 1 (jω+3) √ = jω+1 , and H(ω) = S +1(ω) jω+1 = (jω+γ√ Thus, S − (ω) = jω+a√3 + 3)(jω+ 7) Y Y + b√ jω+ 7 √ √ √ √ 3+3 √ and b = γ1 −√ 7+3 √ . Or, h(τ ) = I{τ ≥0} (ae− 3τ +be− 7τ ). where a = γ1 −−√3+ 7 − 7+ 3

9.25 Estimation of a process with raised cosine spectrum (a)-(b) The situation is a special case of Example 9.1.1 of the notes. !

1 + cos( πω SX (ω) ωo ) H(ω) = = I{|ω|≤ωo } 1 + cos( πω SX (ω) + N2o ωo ) + No ( ) Z ∞ Z ωo 1 + cos( πω SX (ω)SN (ω) dω No dω ωo ) 2 σe = = . πω 1 + No + cos( ωo ) 2π −∞ SX (ω) + SN (ω) 2π −ωo 2

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 σe2 → 0. One way to see this is that the term in brackets in the second integral expression for σe2 is less than one, so in general, σe2 ≤ No ωo . Basically, as No → 0, the noise becomes negligible so that H becomes the all-pass filter in

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the band [−ωo , ωo ]. (d) to infinity, H decreases to zero, and σe2 increases to R ∞ As No increases ωo dω S (ω) 2π = 2π , because the integrand in the expressions for σe2 converge up −∞ X 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 ].

10.1 Two martingales associated with a simple branching process (a) Let Fn = σ(Y0 , . . . , Yn ). Since each of the Yn individuals in the nth generation has an average of µ offspring, E[Yn+1 |Fn ] = µYn . Thus, E[Gn+1 |Fn ] = µYn µ θ n+1 = θ Gn . Taking θ = µ makes G a martingale. (b) Given Yn , the descendants of each one of the Yn individuals of the nth generation evolve independently. Each of these subpopulations has an extinction probability of α, and the overall population becomes extinct if and only if all Yn subpopulations become extinct. Hence, P (E|Y0 , . . . , Yn ) = αYn . Thus M is a Doob martingale. (Doob martingales converge with probability one. Only the limits 0 and 1 are possible in this case, for otherwise the process Gn keeps moving. Thus, either Yn → 0 or Yn → ∞. ) P∞ (c) M0 = α with probability one, and E[M1 ] = P ∗ (α), where P ∗ (α) = k=0 pk αk and (pk : k ≥ 0) is the probability distribution of the number of offspring of an individual. Hence α satisfies P ∗ (α) = α.

10.3 Doob decomposition Parts (a) and (b) are solved in reverse order. (b) Assume the decomposition exists. We shall show it is unique. First, E[X0 ] = E[M0 + B0 ] = B0 . Therefore, B0 = E[X0 ] and M0 = X0 − B0 . Similarly, for k ≥ 0, Xk+1 −Xk = Bk+1 −Bk +Mk+1 −Mk . Since B is predictable, Bk+1 −Bk is Fk measurable, so that E[Bk+1 − Bk |Fk ] = Bk+1 − Bk . Since M is a martingale, E[Mk+1 − Mk |Fk ] = 0. Therefore, E[Xk+1 − Xk |Fk ] = Bk+1 − Bk . Thus, for n ≥ 0, Bn = E[X0 ] +

n−1 X k=0

E[Xk+1 − Xk |Fk ]

Mn = X0 − E[X0 ] +

n−1 X k=0

and

(Xk+1 − Xk − E[Xk+1 − Xk |Fk ]) .

Thus, M and B are uniquely determined by X and F . (a) The decomposition exits. It is given by the sequences M and B found in part (b).

10.5 A stopped random walk (a) By the fact S0 = 0 and the choice of τ , |Sn | ≤ c for 0 ≤ n ≤ τ − 1. Since the W ’s are bounded by D, it follows that |Sτ | ≤ c + D. Therefore, Sn∧τ is a bounded sequence of random variables, so 0 = E[Sn∧τ ] → E[Sτ ] as n → ∞, by the dominated convergence theorem. (Or we could invoke the first or third form of the optional stopping theorem discussed in class. Note here that we are taking it for granted that T < ∞ with probability one. If we don’t want to make that assumption, we could appeal to part (b) or some other method. )

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h i 2 fn+1 Sen + W f2 − σ (b) Let Fn = σ(W1 , . . . , Wn ). Then Mn+1 − Mn = 2W e so n+1 that fn+1 |Fn ]Sen + E[W f2 − σ E[Mn+1 − Mn |Fn ] = 2E[W e2 ] = 0, so that M is a n+1 2 martingale. Therefore, E[Mτ ∧n ] = 0 for all n, or E[τ ∧n] = E[Sen∧τ ]/e σ 2 . Arguing e as in part a, we see that the sequence |Sn∧τ | ≤ c+a∧b for all n. Thus, E[τ ∧n] ≤ (c+a∧b)2 for all n. But E[τ ∧ n] → E[τ ] as n → ∞ by the monotone convergence σ e2 2

. theorem, so that E[τ ] ≤ (c+a∧b) σ e2 (c) We have that E[|Sn+1 − Sn ||Fn ] = C for all n and E[τ ] < ∞ by part (b), so by the third version of the optional stopping theorem discussed in class, E[Sτ ] = E[S0 ] = 0.

10.7 On the size of a maximum matching in a random bipartite graph (a) Many bounds are possible. Generally the tighter bounds are more complex.

If d = 1, then every vertex in V of degree one or more can be included in a single matching, and such a matching is a maximum matching with expected size n(1 − (1 − n1 )n ) ≥ n(1 − e−1 ). The cardinality of a maximum matching cannot decrease as more edges are added, so a = n(1 − e−1 ) is a lower bound for any d and n. An upper bound is given by the expected number of vertices of V with degree at least one, so b = n(1 − (1 − nd )n ) ≈ n(1 − e−d ) works. (b) The variable Z can be expressed as Z = F (V1 , . . . , Vn ), where Vi is the set of neighbors of ui , and F satisfies the Lipschitz condition with constant 1. We are thus considering a martingale associated with Z by exposing the vertices of U one at a time. The (improved version of the) Azuma-Hoeffding inequality yields √ 2 that P {|Z − E[Z]| ≥ γ n} ≤ 2e−2γ . (c) Process the vertices of U one at a time. Select an unprocessed vertex in U .

If the vertex has positive degree, select one of the edges associated with it and add it to the matching. Remove edges from the graph that become ineligible for the matching, and reduce the degrees of the unprocessed vertices correspondingly. (This algorithm can be improved by selecting the next vertex in U from among those with smallest reduced degree, and also selecting outgoing edges with endpoint in V having the smallest reduced degree in V , or both. If every matched edge has an endpoint with reduced degree one at the time of matching, the resulting matching has maximum cardinality. )