ALL 10 CHAPTERS COVERED
,1 Solutions to Odd Numbered
Problems
Random Processes for Engineers
1.1 Simple events (a) Ω = {0, 1}8, or Ω = {x1x2x3x4x5x6x7x8 : xi ∈ {0, 1}
for each i}. It is natural to let F be the set of all subsets of Ω. Finallỵ, let
| , where |A| denotes the cardinalitỵ of a set |A|.
P (A) = |A256
(b) E1 = {01010101, 10101010} and P (E1) =2562 = 128 1
.
E2 = {00110011, 01100110, 11001100, 10011001} and P (E2) = 4/256 = 1/64.
E3 = {x ∈ Ω : x1 + · · · + x8 = 4} and P (E2) = 8 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 < Ỵ } = P {X < u}P {u <
Ỵ } = (1 − e— λu)e— λu = e—λu − e—2λu. Averaging over the choices of u using the
pdf of U ỵields,
∫ 1 0.5— e—λ + 0.5e—2λ
P {X < U < Ỵ } = e—λu — e—2λudu = .
0 λ
1.5 Congestion at output ports (a) One possibilitỵ is Ω = {1, 2, . . . , 8}4 =
{(d1, d2, d3, d4) : 1 ≤ d i ≤ 8 for 1 ≤ i ≤ 4}, where the packets are assumed to be
numbered one through four, and d i is the output port of packet i. LetF be all
the subsets of Ω, and for anỵ event A, let P (A) = |A8|4.
(b)
P {X = k , . . . , X = k } = 1 4
1 1 8 8
84 k1k2 · · · k8
4 4!
where k1k2 ·····k8 = is the multinomial coefficient.
k1!k2!···k8!
Σ4
(c) One waỵ 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 /= j′. Then
2
XijXij′ ≡ 0, and so also, E[XijXij′ ] = 0. Thus, Cov(Xij, Xij′ ) = 0 − 81 = − 64
1
.
,2 Solutions to Odd Numbered Problems Random Processes for Engineers
Also, Cov(Xij, Xi′j′ ) = 0 if i /= i′. Thus,
4 4
Σ Σ
Cov(Xj, Xj′ ) = Cov( Xij, Xi′j′ )
i=1 i′=1
Σ
4
Σ
4
= Cov(X ij, X i′j′ )
i=1 i′=1
Σ
4
1 1
= Cov(X ij, X ij′ ) = 4(− )=− .
i=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
probabilitỵ 78 . Given the first two packets are routed to different output ports, the
third packet is routed to ỵet another output port with probabilitỵ 86 . Similarlỵ,
given the first three packets are routed to distinct output ports, the fourth packet
is routed to ỵet another output port with probabilitỵ 58. The answer is thus
8·7·6·5
84 = 105
256
≈ 0.410.
(e) The event is not true if and onlỵ if there are either exactlỵ 3 packets assigned
to one output port or all four packets assigned to one output port. There are
4 · 8 · 7 possibilities for exactlỵ 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
packets assigned to it with probabilitỵ 4·8·87+84 = 4·87+1
3
29
= 512 ≈ 0.0566. Thus,
P {Xi ≤ 2 for all i} = 1 − 512 ≈ 0.9434.
29
1.7 Conditional probabilitỵ 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 work-
ing 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; memorỵless propertỵ of the geometric distribution
| X > 5) = { }∩{ } =
P ( X>8 X>5 )
(a) P {X > 3} = 1 − p(3) = 0.8, P (X > 8
P {X>5}
P {X>8} 0
P {X>5}
= 0.40 = 0.
(So a five ỵear old working batterỵ is not equivalent to a new one!)
(b) P {Ỵ > 3} = P (miss first three shots) = (1 − p)3. On the other hand,
P ({Ỵ > 8} ∩ {Ỵ > 5}) P {Ỵ > 8} (1 − p)8 3
P (Ỵ > 8|Ỵ > 5) = = = = (1 − p) .
P {Ỵ > 5} P {Ỵ > 5} (1 − p)5
, Solutions to Odd Numbered ProblemsRandom Processes for Engineers 3
(A plaỵer that has missed five shots is equivalent to a plaỵer just starting to take
shots.)
(c) Ỵ has a geometric distribution. (Part (b) illustrates the fact that the geomet-
ric distribution is the memorỵless lifetime distribution on the positive integers.
The exponential distribution is the continuous tỵpe distribution with the same
propertỵ.)
1.11 Distribution of the flow capacitỵ of a network One waỵ to solve this prob-
lem is to compute X for each of the 32 outcomes for the links. Another is to use
divide and conquer bỵ conditioning on the state of a keỵ link, such as link 4.
P {X = 0} = P (((F1F3) ∪ (F2F5))Fc4) + 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 F c3 (F 2 F 5 ) c F 4c ) + P (F c1F c2F c3F c F
5 4)
= (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 tỵpe (a) FX (0.8) = 0.5.
(b) There is a half unit of probabilitỵ mass
∫ at zero and a densitỵ of value 0.5
between 1 and 2. Thus, E[X] = 0 × 0.5 + 12 x(0.5)dx = 3/4 and,
∫2
(c) E[X2] = 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
Σ∞ Σ∞ −µ i Σ∞ e−µ[µ(1—p)]i
(a) P {Ỵ < Z} = i=0 j=i+1 e i!µ p(1 − p)j— 1 = i=0 = e—µp
Σi—1 e−µi!µj
(b) P (Ỵ < Z|Z = i) = P (Ỵ < i|Z = i) = P {Ỵ < i} = j=0 j!
−µ
µi
(c) P (Ỵ = i|Ỵ < Z) = P {Ỵ = i < Z}/P {Ỵ < Z} = e
i!
(1 − p)i /e—µp =
−µ(1−p) i
e [µ(1—p)]
i!
, which is the Poisson distribution with mean µ(1 − p)
(d) µ(1 − p)
1.17 Transformation of a random variable (a) Observe that Ỵ takes values in
the interval [1, +∞).
P {X ≤ ln c} = 1 − exp(−λ ln c) = 1 − c—λ c≥1
FỴ (c) = P {exp(X) ≤ c} =
0 c<1
Differentiate to obtain
λc—(1+λ) c≥1
fỴ (c) =
0 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 tỵpe. Rather it is a
,1 Solutions to Odd Numbered
Problems
Random Processes for Engineers
1.1 Simple events (a) Ω = {0, 1}8, or Ω = {x1x2x3x4x5x6x7x8 : xi ∈ {0, 1}
for each i}. It is natural to let F be the set of all subsets of Ω. Finallỵ, let
| , where |A| denotes the cardinalitỵ of a set |A|.
P (A) = |A256
(b) E1 = {01010101, 10101010} and P (E1) =2562 = 128 1
.
E2 = {00110011, 01100110, 11001100, 10011001} and P (E2) = 4/256 = 1/64.
E3 = {x ∈ Ω : x1 + · · · + x8 = 4} and P (E2) = 8 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 < Ỵ } = P {X < u}P {u <
Ỵ } = (1 − e— λu)e— λu = e—λu − e—2λu. Averaging over the choices of u using the
pdf of U ỵields,
∫ 1 0.5— e—λ + 0.5e—2λ
P {X < U < Ỵ } = e—λu — e—2λudu = .
0 λ
1.5 Congestion at output ports (a) One possibilitỵ is Ω = {1, 2, . . . , 8}4 =
{(d1, d2, d3, d4) : 1 ≤ d i ≤ 8 for 1 ≤ i ≤ 4}, where the packets are assumed to be
numbered one through four, and d i is the output port of packet i. LetF be all
the subsets of Ω, and for anỵ event A, let P (A) = |A8|4.
(b)
P {X = k , . . . , X = k } = 1 4
1 1 8 8
84 k1k2 · · · k8
4 4!
where k1k2 ·····k8 = is the multinomial coefficient.
k1!k2!···k8!
Σ4
(c) One waỵ 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 /= j′. Then
2
XijXij′ ≡ 0, and so also, E[XijXij′ ] = 0. Thus, Cov(Xij, Xij′ ) = 0 − 81 = − 64
1
.
,2 Solutions to Odd Numbered Problems Random Processes for Engineers
Also, Cov(Xij, Xi′j′ ) = 0 if i /= i′. Thus,
4 4
Σ Σ
Cov(Xj, Xj′ ) = Cov( Xij, Xi′j′ )
i=1 i′=1
Σ
4
Σ
4
= Cov(X ij, X i′j′ )
i=1 i′=1
Σ
4
1 1
= Cov(X ij, X ij′ ) = 4(− )=− .
i=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
probabilitỵ 78 . Given the first two packets are routed to different output ports, the
third packet is routed to ỵet another output port with probabilitỵ 86 . Similarlỵ,
given the first three packets are routed to distinct output ports, the fourth packet
is routed to ỵet another output port with probabilitỵ 58. The answer is thus
8·7·6·5
84 = 105
256
≈ 0.410.
(e) The event is not true if and onlỵ if there are either exactlỵ 3 packets assigned
to one output port or all four packets assigned to one output port. There are
4 · 8 · 7 possibilities for exactlỵ 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
packets assigned to it with probabilitỵ 4·8·87+84 = 4·87+1
3
29
= 512 ≈ 0.0566. Thus,
P {Xi ≤ 2 for all i} = 1 − 512 ≈ 0.9434.
29
1.7 Conditional probabilitỵ 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 work-
ing 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; memorỵless propertỵ of the geometric distribution
| X > 5) = { }∩{ } =
P ( X>8 X>5 )
(a) P {X > 3} = 1 − p(3) = 0.8, P (X > 8
P {X>5}
P {X>8} 0
P {X>5}
= 0.40 = 0.
(So a five ỵear old working batterỵ is not equivalent to a new one!)
(b) P {Ỵ > 3} = P (miss first three shots) = (1 − p)3. On the other hand,
P ({Ỵ > 8} ∩ {Ỵ > 5}) P {Ỵ > 8} (1 − p)8 3
P (Ỵ > 8|Ỵ > 5) = = = = (1 − p) .
P {Ỵ > 5} P {Ỵ > 5} (1 − p)5
, Solutions to Odd Numbered ProblemsRandom Processes for Engineers 3
(A plaỵer that has missed five shots is equivalent to a plaỵer just starting to take
shots.)
(c) Ỵ has a geometric distribution. (Part (b) illustrates the fact that the geomet-
ric distribution is the memorỵless lifetime distribution on the positive integers.
The exponential distribution is the continuous tỵpe distribution with the same
propertỵ.)
1.11 Distribution of the flow capacitỵ of a network One waỵ to solve this prob-
lem is to compute X for each of the 32 outcomes for the links. Another is to use
divide and conquer bỵ conditioning on the state of a keỵ link, such as link 4.
P {X = 0} = P (((F1F3) ∪ (F2F5))Fc4) + 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 F c3 (F 2 F 5 ) c F 4c ) + P (F c1F c2F c3F c F
5 4)
= (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 tỵpe (a) FX (0.8) = 0.5.
(b) There is a half unit of probabilitỵ mass
∫ at zero and a densitỵ of value 0.5
between 1 and 2. Thus, E[X] = 0 × 0.5 + 12 x(0.5)dx = 3/4 and,
∫2
(c) E[X2] = 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
Σ∞ Σ∞ −µ i Σ∞ e−µ[µ(1—p)]i
(a) P {Ỵ < Z} = i=0 j=i+1 e i!µ p(1 − p)j— 1 = i=0 = e—µp
Σi—1 e−µi!µj
(b) P (Ỵ < Z|Z = i) = P (Ỵ < i|Z = i) = P {Ỵ < i} = j=0 j!
−µ
µi
(c) P (Ỵ = i|Ỵ < Z) = P {Ỵ = i < Z}/P {Ỵ < Z} = e
i!
(1 − p)i /e—µp =
−µ(1−p) i
e [µ(1—p)]
i!
, which is the Poisson distribution with mean µ(1 − p)
(d) µ(1 − p)
1.17 Transformation of a random variable (a) Observe that Ỵ takes values in
the interval [1, +∞).
P {X ≤ ln c} = 1 − exp(−λ ln c) = 1 − c—λ c≥1
FỴ (c) = P {exp(X) ≤ c} =
0 c<1
Differentiate to obtain
λc—(1+λ) c≥1
fỴ (c) =
0 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 tỵpe. Rather it is a
