Probability and random processes for
electrical and computer engineers
second edition solution
Problem
2.1
(a)
(b)
(c)
(d)
(e)
(f)
(g)
2.2
(a) I) {a1 ,a 2 , a3 , a 4 ,a 6 }
II) {a2 }
III) {a2 ,a 6 }
(b) I) A1 ( A 2 + A3 ) ={ a1 , a2 } , A 1 A 2+ A 2 + A3 ={ a1 , a2 } ,
∴ A 1 ( A 2 + A 3 )= A 1 A 2 + A 2 + A 3
II) A1 + A2 A 3= { a1 , a2 , a 3 , a 4 } , ( A 1+ A 2 ) ( A 1 + A 3 )= { a1 , a2 , a3 , a 4 } ,
∴ A 1+ A 2 A 3=( A1 + A2 ) ( A1 + A3 )
III) A1 ( A 2 + A3 ) ={ a5 } , A1 A 2+ A 1 A 3={ a 5 } ,
∴ A 1 ( A 2+ A 3 )= A 1 A2 + A1 A 3
2.3
I) area always ≥ 0
II) normalize the area of the Venn diagram. So Pr[S] = 1
III)
The area of the union of two events is the sum of the areas
(no intersection)
IV) cannot be shown by Venn diagram
Corollaries (table 2.4)
Pr [ A c ] =1−Pr ⁡[ A ]
area of A always > 0
0 ≤ Pr [ A ] ≤ 1
A ⊆ S ∴area ≤ that of S
If A1 ⊆ A 2, then Pr [ A 1 ] ≤ Pr [ A 2 ]
A1 A 2=∅ ⇒ Pr [ A 1 A2 ] =0 Null set
Pr [ A 1+ A 2 ] =Pr [ A1 ] + Pr [ A 2 ]−Pr [ A1 A 2 ]
Intersection is included twice if we sum the areas
has area size 0
2.4 This solution does not contain the process.
2.5
I)
1
6
II)
5
6
III)
2
6
000
2.6
001
010
011
100
101
110
111
a) S={0 12 3 4 56 7 }
b) Pr [ A 1 ] =
1
4
c) Pr [ A 2 ] =
1
2
2.7
a)
SX
1
5
SY
1
b) Pr [ y <1 ] =
c) Pr
Pr
[
[
1.35
1
5
1
1
≤ x≤1 =
2
10
]
1
≤ y ≤1 =0.1819
2
]
2.8
a)sample spaces (not to scale)
Sx
-1
0
1
2
Sy
0.65
79.25
b) Pr [ y ≤16 ] =
2
3
Pr [ 2 ≤ y ≤ 20 ] =0.4797
2.9
a) Pr [ A+ B ] =
7
12
b) Pr [ A+ B ] =
1
2
c) this is not possible
2.10
a) Pr [ A ]=
1
3
b) Pr [ B ] =
4
9
c) Pr [ AB ] =
1
9
2.11
Pr [ { I =−1 } ∪ { I =5 } ]=1 /6
2.12
a) Pr [ AB c + A c B ] =0.3
[
c
]
b) Pr ( A+ B ) =0.6
2. 13
a) probability = 2.787∗10−6
b) Pr [ S∈10 th poition ] =00038
2.14
a) Pr [ one∨more good ]=
7
8
b) Pr[one or more good] = 0.947
2.15
a) Pr[all good] = 0.512
b) Pr[all bad] =0.008
c) Pr[one bad] = 0.384
d) Pr [ good> bad ] =0.896
2.16
a) total number = 65,536
b) possible commands = 43,680
2. 17
a) different ways two errors occur = 28
2
b) probability = p (1− p )
6
2
c) Pr [ 2 bit errors ]=28∗p ( 1− p )
6
d) Notice to the log scale
p
Pr
0.0001
2.79832041994400e-07
0.001
2.78324194404198e-05
0.01
0.00263614441832280
0.1
0.148803480000000
2.18
a) Pr [ Retrans∈3rd packet ] =0.9801
b) Pr [ retrans∈10th packet ] =9.135∗10
−3
c) Pr[retrains within the first 5 packet] = 0.049
d) N > 10
2. 19
a) 9 Q ( 1−Q )
8
b) Pr [ erros∧not detected ] =
∑ ( 9k )Qk ( 1−Q )9−k
k even
2.20
F = C ( A+ B)(D + E)
2.21
F = A(D + CE) + B(E + CD) or (A + B(C+E))(D+E(B+C))
2.22
Pr [ F ] =00.00361
2.23
Pr [ F ] = p+2 p 2−3 p 3+ p 4
2.24
a) F = AB+ AC
b) F = A + BC
st
C) P1=0.019 , P 2=0.109 1 net has lower probability.
2.25
a) F = A + BCD + E
b) Pr [ F ] =0.9800990199
2.26
a) communicate= ( A ∩ ( C ∪ D ) ) ∪ B
b) Pr [ C ∪ D ]=0.75
c) Pr[communicate] = 0.6875
d) b failed (explain why)
2.27
a) p < 0.001
b) p <0.046
2.28
a) X = (C+D)(A+B+E)
b) Pr [ Y ] =1−( 2 p− p2 ) ( 3 p−3 p2+ p 3 )
c) P[Y] = 0.9994
2.29
a) Pr
B
1
=
A 4
[ ]
b) Pr [ A|B ]=1
c) Pr[1] = 0.5
d) Pr[0] = 0.5
2.30
a) Not independent
b) Pr ⁡[memory failure∨harddisk failure]= 0.07
2.31
a) Independent
b) Pr ⁡[memory failure∨harddisk failure] = 0.02
2.32
a) Yes
b) Pr [ M |D ] =0.015
2.33
Pr [ single bit errors ] =p=0.01
2.34
Pr[hardware failure|software failure] = 0.004
2.35
a)
11
111
110
1
10
101
100
011
11
010
0
001
10
000
1111
1110
1101
1100
1011
1010
1001
1000
0111
0110
0101
0100
0011
0010
0001
0000
b) Pr [ more zeros than ones| first bit zero ¿=
1
2
c) Pr [ more zeros than ones| first two bits are 10 ¿=
d) Pr [ first bit zero|more zeros than ones ¿=
2.36
4
5
1
4
Pr [ Late|breaks down ] =0.48
2.37
G
a)
GB
GBF Probability:0.81 Success
Probability:0.09 Success
Probability:0.025Success
Probability:0.025Fail
Probability:0.025Fail
Probability:0.025Fail
b) I) 0.075
II) 0.025
c) I) 0
II)
2
3
III)
2
3
d) Pr[mission succeeded | beer cooler failed] = 0
2.38
a) Pr [ browser fail ¿access ] =
b) Pr [ IE|fails¿access ] =
1
(ϵ +δ)
2
ϵ
ϵ+δ
2.39
a) Pr[at least 1 IC is good] = 0.975
b) Pr[IC is good] = 0.8125
2NB3
a) Pr[defective component select] = 0.1167
b) Pr[Manufacturer A |defective component] = 0.2857
2.40
a) Pr [ 1 R ] =
4
9
b) Pr [ error ]=
c) Pr [ 1 s|1 R ] =
1
6
7
8
2.41
a) Pr [ 1 R|0 S ] =0.4
b) Pr [ 1 R ] =0.6333
c) Pr [ error ]=0.3
d) Pr [ 0 s|1R ] 0.211
2.42
a) p=0.001
b) p=0.001
c) Pr [ 0 s|0 R ] =0.999
2.43
a)
b)
S
0
R
0
0
1
1
0
1
1
Probability
1
(1−ϵ 0 )
2
1
ϵ
2 0
1
ϵ
2 1
1
(1−ϵ 1 )
2
c) Pr [ 1 S|error ] =
ϵ1
ϵ 0+ ϵ 1
1
Pr [ error ]= (ϵ 0 + ϵ 1)
2
d) Pr [ 1 S|1 R ¿=
1−ϵ 1
1−ϵ 1 +ϵ 0
2.44
a) Pr [ A R ] =0.25
b) Pr [ error ]=0.1
c) I) Pr [ A S|D R ] =0.05
II) Pr [ BS| D R ] =0
III) Pr [ C S|D R ] =0.05
IV) Pr [ D S|D R ] =0.9
2.45
a) Pr [ A R ] =0.1944
b) Pr [ error ]=0.04
c) I) Pr [ A S| D R ¿=0
II) Pr [ BS| D R ¿=0
III) Pr [ C S| D R ¿=0.0386
IV) Pr [ A S| D R ¿=0.9619
2.46
a) Pr [ error|0S ] =0.20
b) Pr [ error|1 S ] =0.20
c) Pr [ 1 S|1R ] =0.821
d) Pr [ error|1 R ] =0.179
2.47
Pr[under 3 error] = 0.9777
2.48
a)
1
0.1
0.9
0.95
0.05
1
b) Pr [ 1 R ] =0.568
c) Pr [ 0 R ∨0 D ] =0.9383
2.49
a) Pr [ α ] =0.52
b) for α received, choose A, for β received, choose B
c) for α received, choose A, for β received, choose A, Pr [ error ]=0.4
2.50
a) Pr [ SPACE ] =
Pr [ M ] =
2
1
2
1
2
1
, Pr [ A ]= , Pr [ C ] = , Pr [ E ] = , Pr [ F ] = , Pr [ L ] = ,
19
19
19
19
19
19
1
2
2
3
2
, Pr [ O ] = , Pr [ R ] = , Pr [ S ] = , Pr [ U ]=
19
19
19
19
19
b) H = 3.3661
bit
letter
2.51
a) Pr [ 0 S|0 R ] =0.66 , Pr [ 0S|1R ] =0.34 , Pr [ 1S|1 R ] =0.83 Pr [ 1S|0 R ] =0.17
b) Pr[error] = 0.255
c) No (explain why)
2.52
a) Pr[A] = Pr[T] =
b) H = 2.8050
c)
LETTER
M
SPACE
A
T
E
N
C
I
5
1
4
6
, Pr[C] = Pr[I] =
, Pr[E] = Pr[N] =
, Pr[M] = Pr[SPACE] =
32
32
32
32
bit
letter
CODE
10
11
001
010
011
0000
00010
00011
d)
bit
´
2.875
Length=¿
letter
3.1
a)
10
9
8
7
6
5
4
3
2
1
b)
0
0
1
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
3.2
0
a)
0
1
9
8
7
6
5
4
3
2
1
0
b)
-1
0
1
2
3
4
5
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-1
0
1
2
3
4
5
3.3
a) k -> 3, 0, 2, 3, 2, 1, 3, 1, 3, 0, 0 ,3, 3, 2, 3, 0, 2, 2, 0, 0
(b) i
(b) k
8
8
6
6
4
4
2
2
0
-1
0
1
i
(c) i
2
3
4
0
-1
0
1
k
(c) k
2
3
4
-1
0
1
k
2
3
4
1
0.4
0.3
0.5
0.2
0.1
0
-1
0
1
i
2
3
4
0
3.4
a) S I = {0,1,2,3,4,5,6,7,8 } , Pr [ 1 ] =
1
1
, Pr [ 0 ] =
2
2
b) Pr[I = 0] = Pr[I = 8] =
1
256
Pr [ I =1 ] =Pr [ I =7 ] =
8
256
Pr [ I =2 ] =Pr [ I =6 ] =
Pr [ I =3 ]=Pr [ I =5 ] =
28
256
56
256
70
256
Pr [ I =4 ] =
70
60
50
40
30
20
10
0
-1
0
1
2
3
4
5
6
7
8
3.5
F I [ i ]=
a)
(3i ) ( 0.56) ( 0.44 )
i
3−i
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-1
0
1
2
3
4
b) Fr [ { I =0 } ∪ { I =3 } ] =0.2608
3.6
a)
4
0.2
∑ f k ( k ) =0.4
0.18
0.16
0.14
k =1
0.12
0.1
0.08
0.06
0.04
0.02
0
-1
0
1
2
3
4
5
6
7
8
9
10
11
9
b)
4
1
∑ f k ( k ) =0.1
0.9
0.8
k =1
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-1
c)
0
1
2
3
4
5
4
0.4
X=1
Y = 0.387
0.35
0.3
∑ f k ( k ) =0.6497
X=0
Y = 0.349
k =1
0.25
X=2
Y = 0.194
0.2
0.15
0.1
X=3
Y = 0.0574
0.05
X=4
Y = 0.0112
0
0
1
2
5
X3= 5 4
Y = 0.00149
6
7
8
9
10
4
d)
0.12
∑ f k ( k ) =0.3439
0.1
k =1
0.08
0.06
0.04
0.02
0
0
1
2
3
4
5
6
7
8
9
3.7
Pr[less than 5 errors] = 0.000968
3.8
Pr [ 1 st error with ∈5 bytes ] =0.4451
3.9
∞
a)
∑ f k ( k ) =1
k =1
b) 0< α <1
3.10
a)
geometric
0.2
0.1
0
0
1
2
3
4
5
6
7
8
9
5
6
7
8
9
5
6
7
8
9
uniform
0.2
0.1
0
0
1
2
3
4
Poisson
0.4
0.2
0
0
1
2
3
4
b)
type
Geometri
c
Uniform
Poisson
Probability
Pr [ I >5 ] =0.5129
1
4
Pr [ I >5 ] =0.01719
Pr [ I >5 ] =
c)
type
Geometri
c
Uniform
Poisson
Probability
p ≈ 0.1294
N = 12
No solution
3.11
a) Pr [ 2 ≤ I ≤ 4 ]=0.3123
b) l =2
3.12
a) C =
6
5
b) Pr [ 0.75< X ≤ 1.5 ] =0.6813
c) α =2.4082 , 1.5918 choose 1.5918
0.6
3. 13
0.5
0.4
0.3
a) C = 4
0.2
0.1
0
-1
0
1
2
3
4
5
b) Pr [ 0.35< X ≤ 6 ] =0.39694
3.14
a)
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-0.1
-0.2
-1
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
b) I)
1
9
II)
5
9
III)
1
3
c) F X ( x ) =
X2
9
1.2
1
0.8
0.6
0.4
0.2
0
-0.2
-1
d) b) 와 같음
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
3.15
a) α =
1
6
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-1
b) I) F X ( x ) =
0
−x 2 x 1
+ −
24 2 2
II) F X ( x ) =1
3.16
a) C =
−2
25
b) Pr [ X > 3 ] =0.16
Pr [ 1 ≤ X ≤ 4 ] =0.6
c)
F X ( x ) =0 for x <0
1
2
3
4
5
6
7
2
x
for 0 ≤ x ≤ 5
F X ( x ) =0.4 x−
25
F X ( x ) =1 for x >5
3.17
a) C =
1
8
b) Pr [ 1< X ≤ 8 ] =
3
4
c) M = 5
3.18
증명 문제 생략
3.19
증명 문제 생략
3.20
NOT valid because f X ( x )< 0 ,∈1.5 ≤ x ≤ 2
3.21
a) It is invalid because f X ( x )< 0 , f X ( x ) ≥ 0 by forcing B=A
b) It is valid. Subject is A =
c) It is valid, if A=
3.22
1
2
2
5
(
a) f x ( x ) =
(
x
for 0 ≤ x ≤ A
A
0 otherwise
)
0.4 1−
A=5
b) F X ( x ) =0.4 x−0.04 x
2
3.23
a) A=
2
π
2
π
b)
f x ( x)=
for x ≥ 0
2
1+ X
3.24
a) c =
π
2
0.5
0.4
0.3
0.2
0.1
0
-1
-0.5
b)
0
0.5
1
1.5
2
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-0.5
{
0 for x> 0
0
0.5
1
1.5
1
π
π
F X ( x ) = 1 − 1 cos ( x ) for 0 ≤ x<1 0.7298 for x =1¿1− cos ( x ) for 1< x ≤ ¿1 for < x ¿¿
2
2
2
2 2
¿
3.25
a)
2
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
b) 0< α ≤ ∞
α
π
π
cos ( αx ) for−
≤x ≤
f
(
x
)
=
c) x
2
2α
2α
0 otherwise
{
0.5
0.4
0.3
0.2
0.1
0
-2
d) I)
-1.5
-1
-0.5
0
0.5
1
1.5
1
2
II) Pr [ X ≤−1 ] =¿
III) Pr [| X|≤0.001 ] =0.001 α
3.26
0 for x <0
{
2
x
for 0 ≤ x ≤1
2
(
)
F
x
=
a) X
x2
2 x−1+ for 1 ≤ x ≤2
2
1 for 2 ≤ x
b) Pr
[
1
3
≤ x ≤ =0.75
2
2
]
Pr [ 1 ≤ x ≤5 ] =0.5
2
Pr [ x <0.5 ] =
1
8
Pr [ 0.75 ≤ x ≤ 0.7501 ] =7.5005∗10−6
3.27
a) C =
1
4
b)
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-1
c)
-0.5
0
0.5
1
1.5
2
2.5
3
f x ( x) =
x
2
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-1
-0.5
0
0.5
1
1.5
2
2.5
3
d)
Pr [ 0.5 ≤ X ≤1.5 ] =
3.28
a)
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-0.5
1.5
0
0.5
1
b)
1
=1
1
0.5
3.29
0
a) C
-0.5
0
0.5
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-0.05
0
1
2
3
4
5
6
7
8
9
10
1
2
b) I) Pr [ X ≤ 1 ] =0.264
II) Pr [ X > 2 ] =0.406
c) I) Pr
[
1
1
< X ≤ + 0.001 =3.033∗10− 4
2
2
]
−4
II) Pr [ 2< X ≤ 2+0.001 ] =2.707∗10
Largest probability occurs for X 0=1
d) F X ( x ) =1− ( 1+ x ) e
−x
0< x< ∞
3. 30
a) I)
II)
0.7
0.8
0.6
0.7
0.6
0.5
0.5
0.4
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
III)
0
-5
0
1
2
3
4
5
6
-4
-3
-2
-1
0
IV)
7
0.6
1.5
0.5
0.4
1
0.3
0.5
0.2
0.1
b)
0
0
-1
-8
-7
-6
-5
-4
-3
-2
-1
0
1
2
-0.8
-0.6
-0.4
-0.2
0
0.2
0.4
0.6
0.8
1
xaxis
yaxis
I)
II)
III)
IV)
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-0.05
-4
-3
-2
-1
0
1
2
3
4
xaxis
yaxis
V)
VI)
VII)
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-0.05
-4
-3
-2
-1
0
1
2
3
4
2
xaxis
yaxis
I)
II)
iii)
iV)
1.8
1.6
1.4
5
10-11
xaxis
yaxis
iii)
4.5
4
3.5
1.2
3
1
2.5
0.8
2
0.6
1.5
0.4
1
0.2
0.5
0
c)
0
1
2
3
4
5
6
7
8
9
10
0
0
1
2
3
4
5
6
7
8
9
v) No, if α is negative f x ( x ) <0 ,
3.31
a) Pr ⁡[T >
2
]≈ 0.135
λ
[
1
≈ 0.393
2λ
b) Pr T <
]
c) the same. The probabilities is not depends on the value of λ
10
3.32
a) M =
a+b
2
b) M =
1
ln 2
λ
c) M = m
3.33
a) Pr [ X ≤ 0 ] =0.7602
b) Pr [ X >1 ] =0.0786
c) Pr [−2< X ≤ 2 ] =0.7433
3.34
a) I) Pr [ X >1 ] =
1
2
II) Pr [ X > √ 3 ] ≈ 0.34
II) Pr [| X−1|< 6 ] =0.99943
b) I) Pr [| X−1|> √ 12 ] =0.0456
c) I) Pr [ X > 1+ √ 3 ] =0.136
II) Pr [ X > 1+ 3 √ 3 ] =0.00135
3.35
1.5
Fi
Fk
1
0.5
a)
0
0
0.5
1
1.5
2
b) Pr [ 1 ≤i ≤2 ]=0.4
c) Pr [ 1 ≤ k ≤ 2 ] =0.35
3.36
2.5
3
3.5
4
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
a)
0
1
2
3
4
5
6
7
8
9
0~1
1~2
2~3
3~4
4~5
5~6
6~7
7~8
8~
1
256
9
256
37
256
93
256
163
256
219
256
247
256
255
256
256
256
3.37
1.5
1
2
3
0.5
0
a)C =
b) F X ( x ) =
0
1
2
3
4
2
( 1−e− x ) for 0< x <2
3
2
1
¿ ( 1−e−x ) + for 2≤ x ≤ ∞
3
3
c) Pr [ 15< X ≤ 2.5 ] =0.4274
5
6
3.38
1.5
1
C=
1
9
0.5
0
-0.5
a)
0
0.5
1
1.5
2
0 for x< 0
{
2
x
for 0 ≤ x<1
18
b) Fx [ x ] =
x2
5+ for 1 ≤ x ≤3
18
1 for 3< x
c) I) Pr [ 0.5 ≤ X ≤2 ]=0.7083
II) Pr [ 2< x ] =0.2778
3.39
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
a)
-0.5
0
0.5
1
1.5
2
2.5
3
3.5
2.5
3
3.5
1.5
0 for x< 0
1
1
1
1+ δ x− for 0< x ≤
f x ( x) =
2
2
2
1
0 for < x
2
{
b)
( )
3.40
a) C = 5.5
0.2
0.15
0.1
0.05
0
-0.05
b)
-3
-2
-1
0
1
2
3
4
c) Pr [−2< X <0 ] =0.2
Pr [ 0< X ≤2.5 ]=0.4
3.41
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
-1
0
1
2
3
4
5
5
6
1
0.5
0
-0.5
0
0.5
1
1.5
{
−e
2
1
1
e 2 + δ ( e ) for 0 ≤ e
f E ( e )= √ 2 π
2
0 fo r e<0
3.42
Y
20
18
fx(x)
0.7
16
14
0.6
12
0.5
10
0.4
8
0.3
6
0.2
4
2
0.1
0
0
a)
-2.5
-2
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
-1
2.5
0
1
fy(y)
0.2
2
3
4
Fy(y)
1
0.9
0.15
0.8
0.7
0.1
0.6
0.5
0.4
0.05
0.3
0.2
0
0.1
0
b)
-0.05
-1
0
1
2
3
4
5
1
y−2
y−2
f y( y)= f x
, F y ( y )=F x
3
3
3
( )
( )
3.43
−α
|μ− ( b+aμ )|
α
f y ( y )=
e |α|
|α|∗2
3.44
[
−3 d
3 d 1 −2 d
=Pr Y =
= e
2
2
2
Pr Y =
−d
d 1
=Pr Y = = 1−e−2 d )
2
2 2
a) Pr Y =
[
] [ ]
] [ ] (
b) Pr [ x >2 d ] =
1 −4 d
e
2
1
Pr [ x <−2 d ] = e−4 d
2
-1
0
1
2
3
4
5
6
5
3.45
x
Y
-x –c
0
x-c
−∞ ¿−c
−c ¿ c
c ¿∞
3.46
|1y|f (ln ( y ) ) , y >0
f y ( y )=
x
1 1
f y( y)=
e
y √2 π
||
−ln y
4
2
y>0
3. 47
fy(y)
1
0.9
0.8
0.7
6
5
2
Pr [ Y =−5 ] = , Pr [ Y =0.5 ] = , Pr [ Y =4 ]=
13
13
13
0.6
0.5
0.4
0.3
0.2
0.1
0
-5
a)
Fy(y)
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
b)
-5
-4
-3
-2
-1
0
1
2
3
4
5
-4
-3
-2
-1
0
1
2
3
4
3.48
a)
x <3
3≤x ≤8
x >8
y=0
Linear increase from 0 to 10
Y = 10
b)
x <3
[
−∞
]
1 y+ 6
f y( y)= f x
2
2
3≤x ≤8
x >8
[
∞
]
f y ( y ) = ∫ f x ( x ) dx δ ( y−10 )
fy(y)
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
6
3.49
1
Pr [ x ≤0 ] = , for y=0
2
a) F y ( y )=
y −z
1
1
Pr [ x ≤ y ] = +
∫ e 8 dz for y > 0
2 √8 π 0
{
b) f y ( y ) =
3.50
3
f y ( y ) = ∫ f x ( x ) dx δ ( y )
1
δ ( y ) +f x ( y ) for 0 ≤ y < ∞
2
2
8
0 for y <0
−2 c
a) F y ( y )= 1−e
for y=0
−2( y+c )
1−e
for y> 0
{
−2 c
b) f y ( y ) =( 1−e
) δ ( y ) +2 e−2( y+c ) , y ≥ 0
3.51
f y ( y )=
1
y
fx
0≤ y < 4
2
2
()
{[ ]
∞
,
∫ f x ( x ) dx δ ( y−4 )
2
fy(y)
0.5
X: 4
Y: 0.3679
0.4
0.3
0.2
0.1
0
0
1
2
3
4
3.52
10
9
8
7
6
5
4
3
2
1
a)
0
0
b) f Y ( y )=
1
2
3
4
5
1
y +4
fx
, y> 0
2
2
( )
6
5
6
0.2
0.15
0.1
0.05
0
-0.05
0
1
2
3
4
-3
-2
-1
-3
-2
5
6
7
8
9
3.53
2.5
2
1.5
1
0.5
0
-0.5
-1
-1.5
-2
-2.5
a)
0
1
2
3
0.25
0.2
0.15
0.1
0.05
0
-0.05
b)
-1
0
1
2
3
3.54
4
3.5
3
2.5
2
1.5
1
0.5
0
a)
-2
-1
0
1
2
3
0.35
0.3
0.25
0.2
f y ( y )=
0.15
0.1
0.05
{
1
for y=0
4
1
for 0< y ≤ 6
8
0
-0.05
b)
-2
-1
0
1
2
3
4
5
6
7
3.55
2
1.5
f x∨ A ( x| A )=
x 0 ≤ x ≤1
{02otherwise
1
0.5
0
a)
-1.5
{
-1
-0.5
9 1
x ≤ x≤1
8 3
b) f x∨ A ( x| A )= 9
8
( 2−x ) 1≤ x ≤
0
0.5
1
1.5
2
2.5
1.2
5
3
1
0.8
0 otherwise
0.6
0.4
0.2
0
-0.5
72
π
x0≤ x≤
2
6
c) f x∨ A ( x| A )= π
0 otherwise
{
0
0.5
1
1.5
2
4
3.5
3
2.5
2
1.5
1
0.5
0
-0.5
3.56
0
0.5
1
1.5
2.5
f x∨ A ( x| A )=λ e−λ (x−2 ) x> 2
1.2
1
0.8
0.6
0.4
0.2
0
3.57
-0.5
a)
1
Pr ¿ −2≤ k < 0
k∨k <0 2
b)
1
Pr ¿ 0 ≤ k <2
k∨k <0 3
0
0.5
1
1.5
2
2.5
3.58
1.2
1
0.8
f X ( x )=f w ( x−z )
0.6
0.4
0.2
0
-0.5
a)
b) Pr [ X > 2 ] =
3.59
1
2
0
0.5
1
1.5
2
2.5
3
3.5
4
4.5
3
3.5
4
4.5
fx(x|H1)
fx(x|H0)
1.2
1.2
1
1
0.8
0.8
0.6
0.6
0.4
0.4
0.2
0.2
0
0
-2
a)
-1
0
1
2
-2
3
b) pd =0.875 , p fa =0
3.60
0.06
xaxis
yaxis
f(x|H0)
f(x|H1)
0.05
0.04
0.03
0.02
0.01
0
a)
-10
-5
0
5
b) Pr [ false alarm ] =
c) Pr [ detection ] =
1
10
9
10
d) s = 20
3.61
a) pfa =
1
32
b) pfa =
1
1
−τ + τ 2
2
2
10
15
20
25
30
-1
0
1
2
3
c) Pr [ nodetect ] =
1
32
3.62
증명문제 생략
3.63
a) X T =4.1
b) pfa =0.09
3.64
a) x T =3.1073
b) pd =0.9985
3.65
a) x T =4.6526
b) pd =0.0 .748
3.66
a) x T =1.25
b) Pr [ 1 R|0 s ] =Pr [ 0 R|1 s ] =0.1271 , Pr [ 1R|1 s ] =Pr [ 0 R|0 s ] =0.8729
c) SNR=1.15
4.1
a) Pr [ Cludia ] =
3
1
, Pr [ Ursla ] =
4
4
b) Pr [ Cludia Accepts ] =
3
8
c) E { euros }=100
1
0.9
0.8
0.7
0.6
E { C }=100
0.5
0.4
0.3
0.2
0.1
0
d)
0
20
40
60
80
100
120
140
160
180
4.2
E { k }=7
4.3
m k =2 , E [ ( k −2 )4 ]=6.8
4.4
E [ I ]=1.68
4.5
0.3
0.25
0.2
0.15
0.1
0.05
a)
0
-4
-2
0
2
4
6
200
b) E [ K ] =1
4.6
a)
Ed
1
you
1
2
3
4
1
4
1
Pr ¿
4
1
Pr ¿
4
1
Pr ¿
4
Pr ¿
2
3
4
5
b) E [ w ] =−85 ₡
c) E [ w ] =−2.65 $
4.7
a) E [ k ] =0.5
b) E [ I ]=4.5
c) E [ I + k ] =5
d) E [ kI ] =
101
6
4.8
a) E [ k ] =0
b) E [ X∨ X >0 ]=
4.9
b
2
1
5
1
Pr ¿
20
1
Pr ¿
20
1
Pr ¿
20
1
Pr ¿
20
Pr ¿
2
3
4
5
6
1
5
1
Pr ¿
20
1
Pr ¿
20
1
Pr ¿
20
1
Pr ¿
20
Pr ¿
3
4
5
6
7
1
5
1
Pr ¿
20
1
Pr ¿
20
1
Pr ¿
20
1
Pr ¿
20
Pr ¿
4
5
6
7
8
2
5
1
Pr ¿
10
1
Pr ¿
10
1
Pr ¿
10
1
Pr ¿
10
Pr ¿
E [ k ]=
3
8
4.10
a) E [ I ∨I >0 ] =2
b) E [ I ∨I ≤ 0 ] =
−2
3
4.11
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
a)
0
1
2
3
4
5
'
b) I) Pr [ K =3 , Date|dosn t 2 day ] =
'
6
2
13
II) Pr [ K =5 , Date|dosn t 2 day ] =
10
39
c) claudia say yes to rolf E [ α ] =4
claudia say no to rolf E [ α ] =4.231
cloudia will say ‘no’ to Rolf
d) Rolf is wrong in his thinking
e) after the 3rd full day of the week we have E { α }=3.182
Rolf should ask cloudia as early as possible on the 4 th day.
4.12
a) c = 4
b) E [ 3+2 X ] =
19
3
4.13
1.2
1
0.8
0.6
0.4
0.2
0
a) A = 1.0524
-1
0
b) E [ X ] =0.8428
4.14
E [ k ] =3.5 , var [ k ] =2.917
4.15
σ=
√
2
3
4.16
6
for k=0
16
2
for k=5
a) f k [ k ] =
16
8
for k=8
16
{
b) E [ k ] =4.625
c) σ 2=13.7344
1
2
3
4
5
4.17
1
0.9
0.8
0.7
0.6
0.5
0.4
Value
frequency
0.3
0.2
0.1
0
a)
0
1
2
3
4
5
6
7
8
9
1
5
6
6
9
4
10
b) E { j } =5.1333Var [ j ] =9.9822
4.18
a)
Value
frequency
2
5
5
2
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
0
1
2
3
4
5
b) E [ I 2 ] =33.8125
4.19
mk =0 ,σ 2=2
4.20
6
7
8
9
10
7
7
8
2
1
0.9
0.8
0.7
0.6
1-p
p
0.5
0.4
0.3
0.2
0.1
0
a)
-5
-4
-3
-2
-1
0
1
2
3
4
5
b) E [ k ] =2 p−1
E [ k 2 ]=1
var [ k ] =4 p( 1− p)
1
2
c) var [ k ] is maximized for p= . Max variance is 1
4.21
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
a)
-5
-4
-3
-2
-1
0
1
2
3
4
5
3
3
E [ S ] = E [ S 2 ]= σ 2=0.24
5
5
1
0.9
0.8
0.7
0.6
E [ R ] =0.604 E [ R2 ]=0.604 σ 2=0.2392
0.5
0.4
0.3
0.2
0.1
0
b)
4.22
-5
-4
-3
-2
-1
0
1
2
3
4
5
E { k }=
n+m
1
var [ k ]= [ ( n−m+1 )2−1]
2
12
4.23
a) E [ I ]=
1
5
b) E [ I 2 ] =4.125
c) I) σ 2=1.875 II ¿ σ 2 =1.875
d) Pr [ I >m I ] =0.4375
4.24
a) E [ I ]=
1− p
p
2
b) E [ I ] =
(2− p)(1− p) 2 1−p
σ = 2
p
p
2
c) m I =2 , σ I =6
4.25
a) E [ I ]=
2
5
σ 2=1.04
b) E [ I ]=0
σ 2=
3
2
4.26
a) E [ X ] =
4
7
b) σ 2=2.2925
4.27
E [ y 2 ] =8.4557
4.28
E [ X ] =1.6125 σ 2 =3.8123
4.29
E { k 3 }=0 E { k 4 } =6.8
4.30
0.3
0.25
0.2
¿
1
6
0.15
0.1
0.05
0
a) A
-1
0
1
2
3
4
5
6
7
1
( x−1 ) for 1≤ x <2
6
1
b) F X ( x ) = 0.3667 + ( x−2 ) for 2 ≤ x ≤ 4
6
0.07 for 4 ≤ x <5
1 for x >5
{
c) Pr [ 1.5 ≤ X ≤ 2.5 ]=0.3667 Pr [ 1.5≤ X ≤ 2.5 ] =0.4666
2
d) E [ x ] =3.15 σ =1.8775
4.31
0.5
0.45
0.4
0.35
0.3
0.25
0.2
0.15
0.1
0.05
0
-1
a) c = 0.3012
0
1
2
3
4
5
6
7
1
0.9
0.8
0.7
0.6
1−e−0.4 x for 0 ≤ x <1
FX ( x) =
1.3012−e−0.4 x for 1< x ≤ 3
{
0.5
0.4
0.3
0.2
0.1
0
b)
-1
0
1
c) E[Y] = 3.4338 σ 2=4.4774
4.32
0.25
0.2
0.15
A ≈ 3.3
0.1
0.05
0
a)
-6
-4
-2
0
2
4
6
b) Pr [ X ≤ 0 ] =0.45 Pr [ X ≥ 1 ] =0.3586 Pr [ X ≥ 3 ] =0.0168
2
3
4
5
6
7
1
X: 3.36
Y: 1.001
0.9
0.8
0.7
X: 1
Y: 0.742
0.6
X: 0.99
Y: 0.6402
0.5
X: 0
Y: 0.4505
0.4
0.3
0.2
0.1
0
-6
-4
-2
0
2
4
c)
6
약간의 오차가 있습니다.
d) E [ x ] =0.1
σ 2=3.3
4.33
1
X: 3
Y: 1
0.9
0.8
0.7
X: 2
Y: 0.75
0.6
0.5
0.4
0.3
0.2
X: 1
Y: 0.25
0.1
a)
0
-1
b) E { X }=
0
1
3
2
c) var [ X ] =0.5833
4.34
a) A =
1
6
b) E [ X ] =
2
3
2
3
4
5
4.35
E { X }=
( b−a )
b +a
var [ x ]=
2
12
2
4.36
E { X }=2 √ 2 σ =2
4.37
E [ X 2 ]=
7
6
4.38
a) E [ X ] =2.5
b) E[X|1 <= X <= 3.5 ] = 2.25
4.39
a) E [ Y ] =5
b) var[ Y ] =4
c) yes
4.40
1
0.9
0.8
0.7
X: 3
Y: 0.5
0.6
0.5
0.4
0.3
0.2
0.1
0
a)
-1
0
1
b) var [ w ] =
c) var [ x ] =
2
3
4
5
6
1
6
2
3
4.41
x for 0 ≤ x ≤ 1
(
)
a) f X x = 2−x for 1 ≤ x ≤2
0 otherwise
{
b) E [ x ] =1, var [ x ] =
1
6
4.42
2
a) m X =1 , E [ X ] =
2
b) σ X =
5
3
2
3
c) Pr [ m X −σ X ≤ X ≤ m X +σ X ] =0.6498
4.43
2
2
a) m X =μ , E [ X ] =μ +
2 2 2
, σ X= 2
α2
α
b) Pr | X−m X|>2 σ X =0.05911 , Pr | X−m X|<σ X =0.7569
[
]
[
]
4.44
m X=
19
349
313
=4.75 , E [ X 2 ]=
=29.08333 , σ 2X =
=6.521
4
12
48
4.45
skewness=0 , kurtosis=
−6
5
in particular, skewness would be non-zero
4.46
2
E [ X ] =μ , var [ X ]=σ X
4.47
x for 0 ≤ x ≤ 1
{
a) f X ( x )= 2−x for 1 ≤ x ≤2
0 otherwise
2
b) m X ( s )=
1−e s
e s−1
∨
S
S
2
( ) ( )
c) E { x }=1
4.48
a) M
X
( s )=e
1 2 2
sm+ S σ
2
2
2
3
4
b) m X =0 , E [ X ] =σ , E [ X ] =0 , E [ X ] =σ
var [ X ] =σ 2 , skewness=0 , kurtosis =0
c) m X =μ other :not change
4.49
2
E [ X 2 ] =24
4.50
1
1−S 2
a) m X ( s )=
b) E[X] = 0
c) var [ x ] =2
d) f y ( y ) =
e) f y ( y ) =
1 −| y−1|
e
2
1
2 √2
y
2
| |
e √
−
4.51
m X ( s )=
λ eS
λ−s
4.52
Gk ( z )=( 1− p ) z−1+ pz
4.53
E [ I 2 ] =6
4.54
a) G k ( z )=
1
3−2 z
2
b) E [ I ]=2, σ =6
4.55
a) H k =−( 1− p ) log 2 ( 1−p )− plo g2 p
1
0.9
0.8
X: 0.25
Y: 0.8113
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
b)
-1
-0.5
0
0.5
1
1.5
2
4.56
b)
letter
A
pro
2
14
5
14
2
14
1
14
1
14
2
14
1
14
E
H
K
N
S
Y
c) E [ L ] =
32
≈ 2.29
14
d) E [ L ] =
4.57
82
≈ 3.15
26
Codeword
01
Length
2
0
1
0000
4
101
3
10
2
000
3
1011
4
1
H X = ( 1+ ln 2 π ) +lnσ
2
4.58
a) both game hae equal entropy.
b) the second game would have greater entropy.
5.1
a) 0.1357
i
b) f I 1=0.1696 ( 0.875 ) 0 ≤ i 1 ≤ 9
f I =0..1 0 ≤i 1 ≤9
2
b) f I 1∨ I 2 ( i 1|i 2) =
f I ∨ I ( i 2|i 1) =
2
1
0.1696 ( 0.845 )i 0<i1 , i 2 ≤ 9
0 otherwise
{
0.1 0<i 1 , i 2 ≤ 9
0 otherwise
{
5.2
a) c= 0.908
i
b) f I1 [ i1 ] =0.1908 (3−i ) 0 ≤ i 1 ≤ 2
i
f I2 [ i2 ] =0.9542 ( 1−0.8i +1 ) 0 ≤i 2 ≤2
2
5.3
4
4
a) f I [ i1 ] =∑ f I I [ i 2 ,i 1 ] , f I [ i 2 ] =∑ f I I [ i 2 ,i 1 ]
1
i2 =1
b) f I 1∨ I 2 ( i 1|i 2) =
1 2
2
1
i i =1
2
f i ,i [i 1 , i 2 ]
f I [i 2 ]
1
2
2
f I ∨ I ( i 1|i 2=1 )=
1
2
f i , i [i 1 ,1]
f i ,i [i 1 , 2]
, f I ∨I ( i 1|i 2=2 ) =
⋯
f I [1]
f I [2]
1
2
1
1
2
i1
d)
2
2
2
i2
F I [ i1 , i2 ]= ∑ ∑ f I , I ¿ ¿
1
l 1=1 l2=1
1
2
5.4
a) = C =
fk1(k1)
1
1
21
0.9
0.8
0.8
0.7
0.7
0.6
0.6
0.5
0.5
0.4
1
b) f K 1 [ k 1 ] =
( k +1 ) ; 0 ≤k 1 ≤ 5
21 1
fk2(k2)
1
0.9
0.4
0.3
0.3
0.2
0.2
0.1
0.1
0
-1
0
1
2
3
4
5
6
7
0
-1
0
1
2
3
4
5
6
7
f K 2 [ k 2 ]=
1
(6−k 2 )0 ≤ k 2 ≤5
21
Not independent
c) f K 1∨K 2 ( k 1|k 2 )=
f K ∨K ( k 2|k 1 )=
2
1
1
k ≤ k ≤5
6−k 2 2
1
0 ≤ k2 ≤ k
k 1 +1
5.5 f K [ k ] =0.2 ( 0.6 )
K −1
+ 0.3 ( 0.4 ) K−1 k ≥ 1
5.6
증명 생략
5.7
a) f X1 ( X 1 ) =
1
−2 ≤ x 1 ≤2
4
1
f X ( X 2 ) = −1 ≤ x 2 ≤1
2
2
b) i) Pr [ 0 ≤ X 1 ≤1 ,−0.5≤ X 2 ≤ 2 ] =
ii) Pr [ X 1−X 2 <0 ] =
[ 2
2
1
2
]
iii) Pr X 1+ X 2 ≤1 =0.3927
5.8
a) Pr [ −2 ≤ X 1 ≤ 1 , , X 2 ≥ 0.5 ] =
b) Pr [ X 1< 2 X 2 ] =
1
2
3
16
3
16
5.9
a) C=
1
12
b) f X1 ( X 1 ) =
x
1
4− 1 0 ≤ x 1 ≤ 4
12
2
(
)
2
f X ( X 2 ) = (2−x2 ) 0≤ x 2 ≤ 1
3
2
c) not independent
5.10
a) sketch!!!!!
b) C = 1
c) f x 1 ( x1 ) =2 ( 1−x 1 ) 0 ≤ x 1 ≤ 1
f x ( x2 ) =1−
2
x2
0 ≤ x1 ≤ 1
2
d)
{
1
x
f X ∨ X ( x 1| x2 ) = 1− 2
2
1
2
0≤ x 1 ≤ 1 ,0 ≤ x 2 ≤ 2 ( 1−x1 )
0 otherwise
1
0 ≤ x 1 ≤ 1 ,0 ≤ x 2 ≤ 2(1−x 1)
f X ∨ X ( x 2|x1 ) = 2 ( 1−x1 )
0 otherwise
2
{
1
e) No
5.11
a) k =
1
2
b) i) F X 1 X2 ( x 1 , x 2 )=
ii) F X 1 X2 ( x 1 , x 2 )=
1
x x
2 1 2
1
1
x 1 x 2− ( x 1+ x2 2 )2
2
4
c Pr [ x 1 <0.5 , X 2 >0.5 ] =
5
16
5.12
a) c = 2
b) f X1 X2 ( x 1 , x 2 )=
1
x2
5.13
a) F X 1 X2 ( x 1 , x 2 )=
{
2
0 otherwise
1−e− x x ≥ 0
0 otherwise
{
( ) {
b) F X 1 ( x1 ) =
−X
(
) x1 >0 , x 2 >0
(1−e−X
1 ) 1−e
1
1
− x 2 x 2≥ 0
F X x2 = 1−e
0 otherwise
2
−x 1
c) f X1=e
x1 ≥ 0
f X =e−x x 2 ≥ 0
2
2
d) yes
5.14
a) F X 1 X2 ( x 1 , x 2 )=¿
[
b) Pr x1 >
1
1
9
, x 2< =
2
2 32
]
5.15
a) C =
4
27
b) f x 1 ( x1 ) =
f X ( x 2) =
2
x
4
1− 1 0≤ x 1 ≤ 1
3
2
(
)
x
4
3− 2 0≤ x 2 ≤ 3
27
2
(
)
c) f X X ( x 1 , x 2 )=
1
2
{
0 for x1 <0 , x 2< 0
2
2
x x
4
3 x 1 x 2− 1 2 for 0 ≤ x 1 ≤ 1 , 0≤ x 2 ≤ 3
27
4
[
x 21
4
1−
for 0 ≤ x1 ≤1 , x 2> 3
27
4
( )
x 22
4
3 x 2−
for x 1>1 , 0 ≤ x 2 ≤3
27
4
1 for x1 >1 , x 2>3
(
)
5.16
f X X ( x 1| x2 ) =
1
2
3−x 1 x 2 0 ≤ x1 ≤ 2
6−2 x 2 0 ≤ x2 ≤3
5.17
a) f X2 ( x 2|x 1 )=
1
for−1≤ x 2 ≤ 1
2
b) they are independent
5.18
a)
fx2(x2)
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
0.1
0
b)
-2
-1
0
]
1
2
3
4
5
fx1(x1)
1
0.9
0.8
0.7
0.6
0.5
for 0 ≤ x 2 ≤1∨2 ≤ x2 ≤ 3
0.4
0.3
0.2
0.1
0
-2
c)
-1
0
1
2
3
4
5
5.19
a) C=1
−x
b) f x 1 ( x1 ) =e 1 1 x1 >0
f X ( x 2) =
2
1
x >0
2 2
( 1+ x2 )
c)there are not independent
2
− x1 (1+x 2)
d) f X1 ¿ X 2 ( x 1| x 2 ¿=( 1+ x 2) x 1 e 1
x 1> 0 , x 2 >0
−x x
e) f X2 ¿ X 1 ( x 2| x 1 ¿=x1 e 1 1 2 x1 >0 , x 2 >0
5.20
a) C=
23
12
2
b) f X ( x 1) =
1
( 1−x 1 ) x 1 23
+
2
12
( 1−x 1) 0 ≤ x 1 ≤ 1
2
f X ( x 2) =
2
( 1−x 2 ) x 2 23
2
+
12
( 1−x 2 ) 0≤ x 2 ≤ 1
c) thet are not independent
d) f x 1 ( x1|x 2 )=
x1 x2 +
23
12
2
x 2 ( 1−x 2 ) 23
+ ( 1−x 2 )
2
12
0 ≤ x1 ≤1−x2
0≤ x 2 ≤ 1
5.21
a) Pr [ 2< x ≤ 2.01 ] =0.00223
b) f XY ( x , y )=
e−( x− y ) for x ≥ 0 ; 0 ≤ y ≤ x ≤ ∞
0 otherwise
{
1
0.9
0.8
0.7
0.6
0.5
0.4
0.3
0.2
1−e− x for 0 ≤ x ≤1
c) f x ( x ) = ( e−1 ) e− x for 1< x< ∞
0 otherwise
0.1
{
0
-2
5.22
a)
1
0.9
0.8
0.7
0.6
f X ( x 1) =¿
1
0.5
0.4
0.3
0.2
0.1
0
b)
-3
-2
-1
0
1
2
3
-1.5
-1
-0.5
0
0.5
1
1.5
2
2.5
3
1
4
f x ∨ X ( x 1|x 2 ) =
1−¿ x 2∨¿| X 1|≤2−2∨X 2∨¿ ¿
1
c)
2
5.23
a) Pr [ T 1> T 2 ] =
5
=0.4545
11
b) Pr [ T 1+ 0.5>T 2 ] =
c) Pr [ miss ] =0.4691
d) Pr [ miss ] =0.4836
e) Pr [ miss bus ]=
5.24
증명문제 생략
5.25
a) E [ X 1 ] =
1
3
b) E [ X 2 ] =
2
3
(
−1
)
5 6
+
1−e 12 =0.4982
11 11
4
=0.16
25
[ 2 ] 16
c) E X 1 =
[ 2 ] 32
d) E X 2 =
e) E [ X 1 X 2 ] =
2
f) σ X 1=
2
1
6
1
18
g) σ X 2=
2
9
h) ρ X 1 X 2=
−1
2
i) E [ X 1|X 2 ] =
1
( 2−X 2)
4
j) E [ X 1|X 2 ] =1−x 1
5.26
a) f x 1∨ X2 ( x 1|x 2 ) =
b) E [ X 1|X 2 ] =
1 for 0 ≤ x 1 ≤ 2−x 2
2−x2 for 0 ≤ x 2 ≤ 2
2−x 2
0 ≤ x2≤ 2
2
5.27
1
0.9
0.8
0.7
0.6
f XY ( x , y )= 2 y 0 ≤ x ≤ 1 ,0 ≤ y ≤ 1
0 otherwise
{
0.5
0.4
0.3
0.2
0.1
0
-2
a)
b) Pr [ X ≤ Y ] =
2
3
-1.5
-1
-0.5
0
0.5
1
1.5
2
c) var [ Y ] =
1
=0.0555
18
d) ρ XY =0
5.28
In ρ2 ≤1
5.29
2
1.8
1.6
1.4
1.2
2 1−x 2) 0 ≤ x 2 ≤ 1
f X ( x 2) = (
0 otherwise
{
2
1
0.8
0.6
0.4
0.2
0
-2
a)
-1.5
-1
-0.5
0
0.5
1
1.5
2
1
0 ≤ x 1 ≤1−x 2 , 0< x 2 , 1
f X ∨ X ( x 1∨x 2 ) 1−x 2
0 otherwise
1
{
2
b)
c) E [ X 1|X 2 ] =
1−X 2
2
5.30
a) 증명문제 생략
1
− x 2− x 1
2
2∗3
4
(
1
b) f X X [ x 2|x 1 ] =
2
1
2π
√
3
4
e
2
)
gaussian form m X2∨ X 1=
1
3
x , σ2
=
2 1 X ∨X 4
2
1
5.31
a) J(y) = 1
b) X 1 =Y 1 , X 2 =Y 2−a Y 1
( 12 ) y y + y )
(
− y 21 (1+ a+a 2 )−2 a+
f Y Y ( y 1 , y 2 )=
1
2
1
e
2 π √ 1−ρ2
c) if we choose a=
1
2
2
2
2
2 (1− ρ )
−1
2
Y 1 is gaussian with variance σ 2=1 , Y 2 is gaussian with variance σ 2=
3
4
5.32
a=−ρ ¿ make the cross term disammear then
2
1
f Y Y ( y 1 , y 2 )=
e
2 π √ 1−ρ2
1
2
2
−( y 1 (1+ ρ ) + y 2)
2 (1− ρ 2)
2
2
− y1
2
− y1
1
1
=
e 2 ∙
e 2 (1−ρ )
2
√2 π
√ 2 π ( 1−ρ )
2
so Y 1∧Y 2 are independent Gaussian random variables with variance 1∧1− ρ2
5.33
−1
a) f X1 X2 ( x 1 , x 2 )=
( x −1.6 x x + x ) −∞< x <∞
1
1
e 2.88
4∙8 π
−∞¿ x 2< ∞
−1
2
1
2
1
1
2
2
2
2
2
( x +x )
−∞< x 1< ∞
1
2s
b) f X X ( x 1 , x 2 )=
e
2
−∞ ¿ x2 < ∞
2π s
2
1
2
−1
c) f X1 X2 ( x 1 , x 2 )=
5.34
증명 생략
5.35
증명 생략
1 18 ¿ ¿¿
e
18 π
5.36
E [ K ] =7 , va r [ k ]=
35
=5.833
6
5.37
a) Pr
b)
([ x ≤ 14 )+( y ≤ 14 )]=0.393
1 1
4 4
∫∫ e−X e−Y dxdy=0.0489
0 0
[
c) Pr Y −X ≤
−1
1
Y > X =1−e 4 =0.221
4
| ]
d) Pr [ ver <15|Rcv< 15 ] =
0.0489
=0.124
0.393
5.38
a)
Probability=
7
=0.4375
16
b) Probability=
3
8
c) Probability=
7
16
5.39
1
( x+3 y ) for 0≤ x , y ≤ 1
f
(
x
,
y
)
=
a) XY
2
0 otherwise
{
b) f z ( z )=
z0≤ z≤1
{02otherwise
{
1
c) f XY∨Z ( x , y|z )= 2 ( x +3 y )
0 otherwise
d) Pr [ x <3 ] =1
e) E [ Z ] =
2
3
f) NOT independent
5.40
a) 증명생략
b) G J ( z )=e
2 a( z−1)
( 2 α )i −2 α
e for j ≥0
c) f J [ j ] = j!
0 for j<0
{
d) E [ J ] =2 α , var [ J ] =2 α
5.41
증명 생략
5.42
{
y for 0< y <1
a) f Y ( y )= 2− y for 1< y <2
b) M Y ( S )=
0 otherwise
2
1 S
( e −1 )
2
S
c) Pr [ Y ≥1.5 ] =
5.43
1
3
for 0≤ x , y ≤ 1
a) M
Xi
( S )=e
1 2 2
mi S + σ i S
2
n
b)
n
[(∑ ) (∑ ) ]
M ( S )=e
i=1
mi S +
1
2
2
σ S
2 i=1 i
Y
c) m y =
n
n
i=1
i=1
∑ mi σ 2y =∑ σ 2i
5.44
2
a) m x =0 , σ X =
1
6
0 for x ≤−1∧x ≥ 1
(
)
f
x
=
x +1 for−1 ≤ x ≤ 0
b) x
1−x for 0 ≤ x ≤1
{
1
0.9
0.8
0.7
2
c) m x =0 , σ x =
1
6
0.6
0.5
0.4
0.3
0.2
0.1
0
-2
5.45
a) E { X }=
b) f x ( x ) =
5
13
var { X }=
6
36
σ ( e−2 x −e−3 x ) x ≥ 0
0 x <0
{
5.46
a) E[W] = 10.5 S
b) var [ w ] =31.25 S
5.47
2
-1.5
-1
-0.5
0
0.5
1
1.5
2
a) E [ T ]=
n 2 n
σ =
λ T λ2
b) M T ( S )=
c) E [ T ]=
λn
( λ−S )n
n
n
var [ T ] = 2
λ
λ
5.48
(
Pr [ Y >2 ]=e−4 1+ 4+
16 64
+
=0.4335
2 6
)