Quantum Field Theory for the Gifted Amateur (2016) – Solutions Manual – Lancaster

ALL 10 CHAPTERS COVERED




SOLUTIONS MANUAL

,1 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|
P (A) = 256 , where |A| denotes the cardinality of a set |A|.
2 1
(b) E1 = {01010101, 10101010} and P (E1 ) = 256 = 128 .
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,

Z 1
0.5 − e−λ + 0.5e−2λ
P {X < U < Y } = e−λu − e−2λu du = .
0 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)
 
1 4
P {X1 = k1 , . . . , X8 = k8 } =
84 k 1 k 2 · · · k 8



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 ′ . Then
2 1
Xij Xij ′ ≡ 0, and so also, E[Xij Xij ′ ] = 0. Thus, Cov(Xij , Xij ′ ) = 0 − 81 = − 64 .

,2 Solutions to Odd Numbered Problems Random Processes for Engineers



Also, Cov(Xij , Xi′ j ′ ) = 0 if i 6= i′ . Thus,
4
X 4
X
Cov(Xj , X ) = Cov(
j′ Xij , X i′ j ′ )
i=1 i′ =1
4 X
X 4
= Cov(Xij , Xi′ j ′ )
i=1 i′ =1
4
X 1 1
= Cov(Xij , Xij ′ ) = 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
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 68 . Similarly,
given the first three packets are routed to distinct output ports, the fourth packet
is routed to yet another output port with probability 85 . The answer is thus
8·7·6·5
84 = 105
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
packets assigned to it with probability 4·8·7+884 = 4·7+1
83
29
= 512 ≈ 0.0566. Thus,
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 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; 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} (1 − p)8
P (Y > 8|Y > 5) = = = = (1 − p)3 .
P {Y > 5} P {Y > 5} (1 − p)5

, Solutions to Odd Numbered ProblemsRandom Processes for Engineers 3




(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 geomet-
ric 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 prob-
lem 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
P ∞ P∞ −µ i P∞ −µ i
(a) P {Y < Z} = i=0 j=i+1 e i! µ p(1 − p)j−1 = i=0 e [µ(1−p)] i! = e−µp
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
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 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

λc−(1+λ) c≥1
fY (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 type. Rather it is a