# 1 Week The sample space is: and the event H is The sample space is: a) We have: b) No. c) We have: d) We have: e) Yes.

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De oplossingen van de opgaven zijn natuurlijk onder voorbehoud. Er kunnen altijd fouten in staan. Het melden van deze fouten wordt zeer op prijs gesteld.

### 1 Week 1

1.2.3 The sample space is:

{2♣, 3♣, . . . , K ♣, A♣, 2♦, 3♦, . . . , K ♦, A♦, 2♥, 3♥, . . . , K ♥, A♥,

2♠, 3♠, . . . , K ♠, A♠}

and the event H is

{2♥, 3♥, . . . , K ♥, A♥}.

1.3.2 The sample space is:

S = {H F, H W, M F, MW}

a) P[W ] = 0.5, b) P[M F ] = 0.3, c) P[H ] = 0.6.

1.5.2 a) P[R3|G1] = 1

5, b) P[R6|G3] = 1

3, c) P[G3|E ] = 23, d) P[E |G3] = 2

3. 1.6.4 a) We have:

P[ A ∩ B] =0, P[B] = 14, P[A ∩ Bc] = 3

8, P[A ∪ Bc] = 3

4. b) No.

c) We have:

P[D] = 23, P[C ∩ Dc] = 1

6, P[Cc∩Dc] = 1

6, P[C|D] = 12. d) We have:

P[C ∪ D] = 56, P[C ∪ Dc] = 2

3. e) Yes.

1.7.7 We have

P[H1H 2] = 6 32

and H1and H2are not independent.

1.8.2 We have 64 three-letter words. We have 24 four-letter words if each letter appears only once.

(2)

1.10.1 We get:

6

5

4

1

### W W

3

2

We have:

P[W ] = [1 − q(1 − (1 − q)3)][1 − q2].

### 2 Week 2

2.2.3 a) c = 301,

b) P[V ∈ {u2|u =1, 2, 3, . . .}] = 1730, c) P[V even] = 23,

d) P[V > 2] = 56. 2.3.2 a) We have

PK(k) = ( n

k pk(1 − p)n−k k =0, 1, . . . , n

0 otherwise

b) The minimal value for n equals 2.

2.3.5 a) We have:

PN(n) =((1 − p)n−1p n =1, 2, . . .

0 otherwise

b) The minimal value for p equals 1 − 3

√ 0.05.

2.4.3 We get:

−5 0 5 10

−0.2 0 0.2 0.4 0.6 0.8 1

x FX(x)

(3)

and

PX(x) =









0.4 x = −3 0.4 x = 5 0.2 x = 7 0 otherwise 2.5.1 a) Xmod = {1, 2, . . . , 100}.

b) Xmed= {x |50< x < 51}.

2.5.5 E [X ] = 2.2.

2.6.5 a) We have:

Px(x) =

(qx −1(1 − q) x = 1, 2, . . .

0 otherwise

b) We have:

PT(t) = PX

 t + 1 2



=

(q(t−1)/2(1 − q) t = 1, 3, 5, . . .

0 otherwise

2.8.1 a) E [N ] = 0.9, b) E [N2] =1.1, c) Var[N ] = 0.29, d) σN =

√0.29.

2.8.6 a) σX =

5 2 ,

b) P[µX−σX ≤ X ≤µXX] = 5

8. 2.9.3 We have:

E [X |B] = 17 3 , Var[X |B] = 8

9.

### 3 Week 3

2.9.7 a) P[M> 0] = 1 − q and PM(m) =

(

q(1 − q)m m =0, 1, 2, . . .

0 otherwise

b) r =(1 − q)26 c) We get:

PJ( j) = ( 365

j rj(1 − r)365− j j =0, 1, . . . , 365

0 otherwise

(4)

d) We get

PK | A(k) =((1 − q)kq k =0, 1, 2, . . .

0 otherwise

3.1.2 a) c = 1441

b) P[V > 4] = 14463 c) P[−3< V ≤ 0] = 14421 d) a = 4

√ 3 − 5 3.2.1 a) c = 12

b) P[0 ≤ X ≤ 1] = 14 c) P[−12 ≤ X ≤12] = 1

16

d) We get:

FX(x) =





0 x < 0

x2

4 0 ≤ x ≤ 2 1 x > 2 3.3.2 a) E [X ] = 5 and Var[X ] = 163

b) h(E[X]) = 15, E [h(X)] = 12 c) E [Y ] = 12. Var[Y ] = ln 981

4

3.4.7 a) P[1 ≤ X ≤ 2] = e−1/2−e−1 b) We get:

FX(x) =

(0 x < 0 1 − e−x/2 x ≥0 c) E [X ] = 2

d) Var[X ] = 4 3.5.5 We get:

P[T > 32] = 8(−2215) ≈ 0.071 P[T < 0] = 8(−23) ≈ 0.252 P[T > 60] = 8(−103) ≈ 0.000434 3.6.1 We get:

(5)

−2 −1.5 −1 −0.5 0 0.5 1 1.5 2

−0.2 0 0.2 0.4 0.6 0.8 1

x FX(x)

a) P[X < −1] = 0, P[X ≤ −1] = 0 b) P[X < 0] = 13, P[X ≤ 0] = 23

c) P[0< X ≤ 1] = 13, P[0 ≤ X ≤ 1] = 23 3.7.4 a) We get:

FY(y) =





0 y < 0

1

3 0 ≤ y < 100 1 y ≥100 b) fY(y) = 13δ(y) +23δ(y − 100) c) E [Y ] = 2003

3.8.4 a) We get:

fW |C(w) =

(ew2/32 2

2π w > 0 0 otherwise b) E [W |C] = 8

2π

c) Var[W |C] = 16 −32π

### 4 Week 4

4.1.4 Yes. It is easy to verify the conditions from Theorem 4.1 but that is not sufficient. To verify the results we need additionally that we will get that P[x1 ≤ X ≤ x2, y1 ≤ Y ≤ y2] ≥0 when x2≥ x1and y2≥ y1.

4.2.2 We get:

a) c = 141

b) P[Y < X] = 12 c) P[Y > X] = 12

(6)

d) P[Y = X ] = 0 e) P[X < 1] = 148 4.2.8 We get

PK,X(k, x) =

( n−x −1

k−1  pn−k(1 − p)k x + k ≤ n, x ≥ 0, k ≥ 0

0 otherwise

4.3.2 We get:

a) We obtain:

PX(x) =





6

14 x = −2, 2

2

14 x =0 0 otherwise

PY(y) =





5

14 y = −1, 1

4

14 y =0 0 otherwise b) E [X ] = 0 and E [Y ] = 0.

c) σX =√

4.4.2 We get a) c = 6

b) P[X > Y ] = 25 and P[Y < X2] = 1

4. c) P[min(X, Y ) ≤ 12] = 11

32

d) P[max(X, Y ) ≤ 34] = 3

4

5

4.5.3 We get

a) We obtain:

fX(x) = (2

r2−x2

πr2 −r ≤ x ≤ r

0 otherwise

b) We obtain:

fY(y) = (2

r2−y2

πr2 −r ≤ y ≤ r

0 otherwise

4.6.2 We get

a) We obtain

PW(w) =













3

14 W = −4

3

14 W = −2

1

7 W =0

3

14 W =2

3

14 W =4

(7)

b) E [W ] = 0.

c) P[W > 0] = 37. 4.7.2 We get

a) E [W ] = 6128. b) E [X Y ] = 47. c) Cov[X, Y ] = 47. d) ρX,Y = 2

30. e) Var[X + Y ] = 377 4.8.6 We get:

a) P[ A] = 125 b) We obtain:

fX,Y |A(x, y) = (8

5(2x + y) 0 ≤ x ≤ 1, 0 ≤ y ≤ 12

0 otherwise

fX | A(x) = (1

5(8x + 1) 0 ≤ x ≤ 1

0 otherwise

fY | A(y) = (1

5(8y + 8) 0 ≤ y ≤ 12

0 otherwise

### 5 Week 5

5.1.1 We get:

a) This yields:

PN1,N2,N3,N4(n1, n2, n3, n4) =

 4

n1, n2, n3, n4



pn11p2n2p3n3pn44

b) The probability equals 168756272. c) The probability equals 168758656. 5.2.2 We get c = 2

Pn i =1ai. 5.4.4 Yes

5.5.4 We get

a) The probability equals 1 − [8(2)]10 ≈0.2056.

b) The probability equals 1 − [8(3)]10 ≈0.0134.

c) The probability equals 1 − [8(7)]10 ≈1.28 × 10−11. Unfortunately the latter approx- imation cannot be found in the table of the book.

(8)

5.6.6 We have:

E [K ] = 1p

 1 2 3

,

CK = 1− p

p2

1 1 1 1 2 2 1 2 3

,

RK = 1

p2

2 − p 3 − p 4 − p 3 − p 6 − 2 p 8 − 2 p 4 − p 8 − 2 p 12 − 3 p

.

5.7.1 We get:

a) We have:

RX=

20 30 25 30 68 46 25 46 40

.

b) We have:

fX1,X2(x1, x2) = 1 4π√

3e(x21+x1x2−16x1−20x2+x22+112)/6. c) We get 1 −8(2) ≈ 0.0228.

5.7.6 We get

a) We have Cov[Y1, Y2] =(σ12−σ22) sin θ cos θ.

b) Forθ = kπ2 where k is any integer.

### 6 Week 6

6.1.3 We get

a) We have:

PN1(n) = ( 3

4

n−1 1

4 n =1, 2, . . .

0 otherwise

b) E [N1] =4.

c) We have:

PN4(n4) = ( n−1

3

 3

4

n−4 1 4

4

n =1, 2, . . .

0 otherwise

d) E [N4] =16.

(9)

6.2.3 We have forλ 6= µ:

fW(w) = ( λµ

λ−µ eµw−eλw

w ≥ 0

0 otherwise

while forλ = µ we have:

fW(w) =(λ2weλw w ≥ 0

0 otherwise

6.4.3 We have:

a) φK(s) = 1 − p + pes. b) φM(s) = 1 − p + pesn

.

c) E [M] = np, Var[M] = np(1 − p).

6.5.3 We have:

PY(y) =

(1 y =100 0 otherwise 6.6.2 We have:

a) E [K100] =20.

b) σK100 =4.

c) P[K100≥18] ≈ 0.6915.

d) P[16 ≤ K100 ≤24] ≈ 0.6826.

6.7.1 We get:

P[Wn=n] n =1 n =4 n =25 n =64 exact 0.3679 0.1954 0.0795 0.0498 approxmate 0.3829 0.1974 0.0796 0.0498 6.8.1 We get:

c =1 c =2 c =3 c =4 c=5

Chernoff bound 0.606 0.135 0.011 3.35 × 10−4 3.73 × 10−6 Q(c) 0.1587 0.0228 0.0013 3.17 × 10−5 2.87 × 10−7

### 7 Week 7

10.2.2 The sample space is S = {s0, s1, s2, s3}. The ensemble of sample functions is { x(t, si) | i =0, 1, 2, 3 } where

x(t, si) = cos(2π f0t + π4 +iπ2)

for i = 0, 1, 2, 3. The ensemble is shown below:

(10)

0 0.2T 0.4T 0.6T 0.8T T

−1

−0.5 0 0.5 1

x(t,s0)

0 0.2T 0.4T 0.6T 0.8T T

−1

−0.5 0 0.5 1

x(t,s1)

0 0.2T 0.4T 0.6T 0.8T T

−1

−0.5 0 0.5 1

x(t,s2)

0 0.2T 0.4T 0.6T 0.8T T

−1

−0.5 0 0.5 1

x(t,s3)

t

10.3.1 We get:

FX(t)(x) = (

e(t−x) x < t

1 x ≥ t

10.3.2 We have:

a) p = 0.05.

b) E [T1] = 1

p.

c) PT1(20) = (0.95)19(0.05).

d) E [T5] = 5

p. 10.4.2 No.

10.5.3 We get:

PN(t)(n) =

((2t)ne−2t

n! n =0, 1, 2, . . . 0 anders.

10.6.1 We have:

PN(n) = (

100n e−100n! n =0, 1, 2, . . .

0 otherwise

10.7.1 We haveα = 321. 10.8.2 We have:

a) µX(t) = t − 1, b) CX(t, τ) = 1.

10.8.4 We get:

a) E [Cm] =0 and Var[Cm] = 1

4m +64

3



1 − 14m

(11)

b) We have:

CC[m, k] = 1

22m+k + 64 2|k|3



1 − 14min(m,m+k) c) No, the mean is zero.

d) This model is quite reasonable given variance and mean.

10.9.1 Yes.

### 8 Week 8

10.9.5 Yes.

10.10.4 We have:

a) E [X2(t)] = 1,

b) E [cos(2π fct +2)] = 0, c) E [Y(t)] = 0,

d) E [Y2(t)] = 12. 10.11.3 a) RY(t, τ) = RX(τ),

b) RX Y(t, τ) = RX(τ − t0), c) Yes,

d) Yes.

10.12.1 We have E[Y(t)] = 0 and:

RY(t, τ) =





















α(t + τ) t > 0, τ > 0 αt t > 0, −t < τ < 0 0 t > 0, τ < −t

−α(t + τ) t < 0, τ < 0

−αt t < 0, 0 < τ < −t 0 t < 0, τ > −t 11.1.2 E [Y(t)] = −2 × 10−3volts.

11.2.2 πk20 sin(π2k) + sin(π4k)

11.2.3 a) µW =2,

b) RW[n] =













0.5 n = −3, 3 3 n = −2, 2 7.5 n = −1, 1 10 n =0 0 otherwise

,

c) Var[Wn] =6, d) gn =





1/2 n = 0, 2 1 n =1 0 otherwise

(12)

### 9 Week 9

11.3.1 We get:

fX(x) = 4 6π√

2π exp



−2x12 3 −5x22

6 −2x32

3 + 2x1x2

3 + 2x2x3

3



11.3.3 We get:

fY(y) = 16 30π√

6π exp 12y32+11y42+12y52+4y3y4−8y3y5+4y4y5

30



11.4.2 We have

h ≈ 0.6950 − 0.01930 11.5.2 We have:

RY(t, τ) = RX(ατ)

and Y(t) is wide sense stationary. Finally:

SY( f ) = |1α|SX

 f

|α|



11.6.1 We get:

SX(φ) = 2 − 0.2 cos(2πφ) 1.01 − 0.2 cos(2πφ). 11.8.2 We get:

a) RW(τ) = δ(τ), b) SY( f ) =

(1 |f | ≤ B/2 0 otherwise c) E [Y2(t)] = B,

d) E [Y(t)] = 0.

11.8.4 We get:

a) E [X2(t)] = 1, b) SY( f ) =

(1

2eπ f2/4 |f | ≤2

0 anders

c) E [Y2(t)] = 28(√

2π) − 1 ≈ 0.9876.

11.8.5 We get:

a) E [X2(t)] = 0.02, b) SX Y( f ) =

( 10−4

100π+ j2π f |f | ≤100

0 anders

(13)

c) SY X( f ) =

( 10−4

100π− j2π f |f | ≤100

0 anders

d) SY( f ) =

( 10−4

104π2+(2π f )2 |f | ≤100

0 anders

e) E [Y2(t)] = arctan 2

106π2 ≈1.12 × 10−7,

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