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Probability and Stochastic Processes:
A Friendly Introduction for Electrical and Computer Engineers
Roy D. Yates and David J. Goodman
Problem Solutions : Yates and Goodman,2.2.1 2.2.9 2.3.10 2.3.11 2.3.12 2.4.3 2.5.10 2.5. 2.6.5 2.7.7 2.7.8 2.7.9 2.8.10 and 2.9.
Problem 2.2.
(a) We wish to find the value of c that makes the PMF sum up to one.
PN
n ^ ^
c
1 2 n^ n 0 1 2
0 otherwise
Therefore, ∑^2 n 0 PN
n ^ ^ c^^ c ^2 c ^4 1, implying^ c^^ ^4 ^ 7.
(b) The probability that N 1 is
P N 1 P N 0 P N 1 4 7 2 7 6 7
Problem 2.2.
(a) In the setup of a mobile call, the phone will send the “SETUP” message up to six times. Each time the setup message is sent, we have a Bernoulli trial with success probability p. Of course, the phone stops trying as soon as there is a success. Using r to denote a successful response, and n a non-response, the sample tree is
p ^ r
1 p^ n
p r
1 p^ n
p r
1 p^ n
p r
1 p^ n
p r
1 p^ n
p r
1 p^ n
K 1 K 2 K 3 K 4 K 5 K 6
K 6
(b) We can write the PMF of K , the number of “SETUP” messages sent as
PK
k ^
^1 p ^ k^ ^1 p^ k^^ ^1 ^2 ^5
1 p 5 p
1 p 6
1 p ^5 k 6
0 otherwise
Note that the expression for PK
6 is different because^ K^^ 6 if either there was a success or
a failure on the sixth attempt. In fact, K 6 whenever there were failures on the first five
attempts which is why PK
6 simplifies to^
1 p ^5.
(c) Let B denote the event that a busy signal is given after six failed setup attempts. The probability
of six consecutive failures is P B
^1 p ^6. To be sure that^ P^ B ^0 ^ 02, we need^ p^^ ^1
0 02 1 ^6 0 479.
Problem 2.3.
(a) Since each day is independent of any other day, P W 33 is just the probability that a winning lottery ticket was bought. Similarly for P L 87 and P N 99 become just the probability that a losing ticket was bought and that no ticket was bought on a single day, respectively. Therefore
P W 33 p 2 P L 87
1 p 2 P N 99 1 2
(b) Supose we say a success occurs on the k th trial if on day k we buy a ticket. Otherwise, a failure
occurs. The probability of success is simply 1 2. The random variable K is just the number
of trials until the first success and has the geometric PMF
PK
k^ ^
1 2 k^ ^1
1 2 k^ k 1 2
0 otherwise
(c) The probability that you decide to buy a ticket and it is a losing ticket is
1 p 2, independent
of any other day. If we view buying a losing ticket as a Bernoulli success, R , the number of losing lottery tickets bought in m days, has the binomial PMF
PR
r ^ ^
m r
1 p ^ ^2
r 1 p 2 m r r 0 1 m
0 otherwise
(d) Letting D be the day on which the j -th losing ticket is bought, we can find the probability that
D d by noting that j 1 losing tickets must have been purchased in the d 1 previous days.
Therefore D has the Pascal PMF
PD
d^ ^
j 1
d 1
1 p ^ ^2
d 1 p 2 d j d j j 1
0 otherwise
Problem 2.3.
(a) Let Sn denote the event that the Sixers win the series in n games. Similarly, Cn is the event that the Celtics in in n games. The Sixers win the series in 3 games if they win three straight, which occurs with probability
P S 3
The Sixers win the series in 4 games if they win two out of the first three games and they win the fourth game so that
P S 4
1 2 ^3
The Sixers win the series in five games if they win two out of the first four games and then win game five. Hence,
P S 5
1 2 ^4
Problem 2.4.
(a) Similar to the previous problem, the graph of the CDF is shown below.
FX
x ^
0 x " 3
0 4 3 x 5
0 8 5 x 7
1 x 7 −3 0 5 7
0
1
x
FX
(x)
(b) The corresponding PMF of X is
PX
x^
0 4 x # 3
0 4 x 5
0 2 x 7
0 otherwise
Problem 2.5. By the definition of the expected value,
E Xn
n
x 1
x n x px^
1 p n^ x
np
n
x 1
n 1 !
x 1 !
n 1
x 1 !
px^ ^1
1 p n ^1 ^ %$^ x ^1 &
With the substitution x ' x 1, we have
E Xn np
n 1
x () 0
n 1
x '
px (^
1 p n^ x (
1
np
n 1
x () 0
PXn * 1
x^ ^ np
The above sum is 1 because it is he sum of a binomial random variable for n 1 trials over all possible
values.
Problem 2.5. We write the sum as a double sum in the following way: ∞
i 0
P X + i
∞
i 0
∞
j i 1
PX
j
At this point, the key step is to reverse the order of summation. You may need to make a sketch of the feasible values for i and j to see how this reversal occurs. In this case,
∞
i 0
P X + i
∞
j 1
j 1
i 0
PX
j ^ ^
∞
j 1
jPX
j^ ^ E^ X
Problem 2.6.
(a) The source continues to transmit packets until one is received correctly. Hence, the total num-
ber of times that a packet is transmitted is X x if the first x 1 transmissions were in error.
Therefore the PMF of X is
PX
x^ ^
qx ^1
1 q x 1 2
0 otherwise
(b) The time required to send a packet is a millisecond and the time required to send an acknowl- edgment back to the source takes another millisecond. Thus, if X transmissions of a packet
are needed to send the packet correctly, then the packet is correctly received after T 2 X 1
milliseconds. Therefore, for an odd integer t + 0, T t iff X
t 1 2. Thus,
PT
t^ ^ PX
t 1 2 q
$ t 1 & , 2 1 q t 1 3 5
0 otherwise
Problem 2.7. Let W denote the event that a circuit works. The circuit works and generates revenue of k dollars if all of its 10 constituent devices work. For each implementation, standard or ultra-reliable, let R denote the profit on a device. We can express the expected profit as
E R P W E R - W P Wc^ E R - Wc
Let’s first consider the case when only standard devices are used. In this case, a circuit works with
probability P W
1 q 10. The profit made on a working device is k 10 dollars while a non-
working circuit has a profit of -10 dollars. That is, E R - W . k 10 and E R - Wc %/ 10. Of course,
a negative profit is actually a loss. Using Rs to denote the profit using standard circuits, the expected profit is
E Rs 0
1 q 10
k 10
1 q 10
0 9 ^10 k 10
And for the ultra-reliable case, the circuit works with probability P W %
1 q 2 10. The profit per
working circuit is E R - W k 30 dollars while the profit for a nonworking circuit is E R - Wc 1 30
dollars. The expected profit is
E Ru 0
1 q 2 ^10
k 30
1 q 2 ^10
0 95 ^10 k 30
Now we wish to determine which implementation will generate the most profit. Realizing that both profit functions are linear functions of k , we can plot them versus k and find for which values of k
each plan is preferable. The two lines intersect at a value of k 80 21 dollars. So for values of
k $80 21 using all standard devices results in greater revenue, and for values of k + $80 21 more
revenue will be generated by implementing all ultra-reliable devices. So we can see that when the price commanded for each working circuit is sufficiently high it is worthwhile to spend the extra money to ensure that more working circuits can be produced.
Our expected return on a winning ticket is
E Wn nE
Kn 1
1 q 2 n
2 q
Note that when nq 4 1, we can use the approximation that
1 q 2 n
1 2 nq to show that
E Wn
1 2 nq
2 q
n
nq 4 1
However, in the limit as the value of the prize n approaches infinity, we have
n lim 5 ∞ E^ Wn ^
2 q 2
That is, as the pot grows to infinity, the expected return on a winning ticket doesn’t approach infinity because there is a corresponding increase in the number of other winning tickets. If it’s not clear how large n must be for this effect to be seen, consider the following table:
n 106 107 108
E Wn 9 00 3 105 4 13 3 106 4 68 3 106
When the pot is $1 million, our expected return is $900,000. However, we see that when the
pot reaches $100 million, our expected return is very close to 1
2 q , less than $5 million!
Problem 2.7.
(a) There are 466 equally likely winning combinations so that
q
46 6
(b) Assuming each ticket is chosen randomly, each of the 2 n 1 other tickets is independently a
winner with probability q. The number of other winning tickets Kn has the binomial PMF
PKn
k^ ^
2 n 1
k q
k 1 q 2 n 1 k k 0 1 2 n 1
0 otherwise
Since the pot has n r dollars, the expected amount that you win on your ticket is
E V 0
1 q qE
n r
Kn 1
q
n r E
Kn 1
Note that E 1 Kn 1 was also evaluated in Problem 2.7.8. For completeness, we repeat those
steps here.
E
Kn 1
2 n 1
k 0
k 1
2 n 1 !
k!
2 n 1 k !
qk^
1 q ^2 n ^1 ^ k
2 n
2 n 1
k 0
2 n !
k 1 !
2 n
k 1 !
qk^
1 q ^2 n ^ .$^ k^ ^1 &
By factoring out 1 q , we obtain
E
Kn 1
2 nq
2 n 1
k 0
2 n
k 1
qk ^1
1 q 2 n ^ .$^ k ^1 &
2 nq
2 n
j 1
2 n j q j^
1 q 2 n^ j
A We observe that the above sum labeled A is the sum of a binomial PMF for 2 n trials and success
probability q over all possible values except j 0. Thus A 1 ^20 n q^0
1 q ^2 n^ ^0 , which
implies
E
Kn 1
A
2 nq
1 q 2 n
2 nq The expected value of your ticket is
E V
q
n r 6 1
1 q 2 n
2 nq
r n
1 q 2 n
Each ticket tends to be more valuable when the carryover pot r is large and the number of new tickets sold, 2 n , is small. For any fixed number n , corresponding to 2 n tickets sold, a
sufficiently large pot r will guarantee that E V 7+ 1. For example if n 107 , (20 million tickets
sold) then
E V 8 0 44 1
r 107
If the carryover pot r is 30 million dollars, then E V 1 76. This suggests that buying a one
dollar ticket is a good idea. This is an unusual situation because normally a carryover pot of 30 million dollars will result in far more than 20 million tickets being sold.
(c) So that we can use the results of the previous part, suppose there were 2 n 1 tickets sold before
you must make your decision. If you buy one of each possible ticket, you are guaranteed to
have one winning ticket. From the other 2 n 1 tickets, there will be Kn winners. The total
number of winning tickets will be Kn 1. In the previous part we found that
E
Kn 1
1 q 2 n
2 nq Let R denote the expected return from buying one of each possible ticket. The pot had r dollars
beforehand. The 2 n 1 other tickets are sold add n 1 2 dollars to the pot. Furthermore, you
must buy 1 q tickets, adding 1
2 q dollars to the pot. Since the cost of the tickets is 1 q
dollars, your expected profit
E R E
r n 1 2 1
2 q
Kn 1
q
q
2 r 2 n 1 1
2 q
E
Kn 1
q
q
2 r 2 n 1 1
1 q 2 n^
4 nq^2
q
Making the substitution j n 19 yields
E N - B
∞
j 1
j 19
1 p j^ ^1 p 1 p 19
We see that in the above sum, we effectively have the expected value of J 19 where J is
geometric random variable with parameter p. This is not surprising since the N 20 iff we
observed 19 successful tests. After 19 successful tests, the number of additional tests needed
to find the first failure is still a geometric random variable with mean 1 p.
