# 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.

N/A
N/A
Protected

Share "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."

Copied!
13
0
0

Bezig met laden.... (Bekijk nu de volledige tekst)

Hele tekst

(1)

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,

Referenties

GERELATEERDE DOCUMENTEN

Copyright and moral rights for the publications made accessible in the public portal are retained by the authors and/or other copyright owners and it is a condition of

In the paper, we draw inspiration from Blommaert (2010) and Blommaert and Omoniyi (2006) and their analyses of fraudulent scam emails, texts that show high levels of competence

Assuming this is not a case of association, but of a grave of younger date (Iron Age) discovered next to some flint implements from the Michelsberg Culture, the flint could be

Note 3: 47 patients in drug-free remission at year 10 achieved this (i.e. achieved and maintained remission allowing to taper to drug-free) on the following treatment

In beide jaarrekeningen 2017 is echter de volgende tekst opgenomen: “Er is echter sprake van condities die duiden op het bestaan van een onze- kerheid van materieel belang op

This is an open access article distributed under the terms of the Creative Commons Attribution License (CC-BY-NC-ND 4.0), which permits unrestricted use, distribution,

H5: The more motivated a firm’s management is, the more likely a firm will analyse the internal and external business environment for business opportunities.. 5.3 Capability

I envisioned the wizened members of an austere Academy twice putting forward my name, twice extolling my virtues, twice casting their votes, and twice electing me with