Probability and Stochastic Processes: A Friendly Introduction for Electrical and Computer Engineers Roy D. Yates and David J. Goodman Problem Solutions : Yates and Goodman,3.6.7 3.6.8 3.7.1 3.7.7 4.1.1 and 4.2.1 Problem 3.6.7 M and N . Note that The The key key to solv solvin ing g this this prob proble lem m is to find find the the join jointt PMF PMF of M that N the joint event M m N n has probability m 1 calls
P M
m N
n
P dd
n
M . For n
m,
m 1 calls
d v dd
1
p
m 1
1
p
n 2 2
dv
p 1
p
n m 1
p
p
A complete expression for the joint PMF of M and N is P M N m n
For n
2 3
1 0
p
n 2 2
m 1 2 otherwise
p
1; n
n
1 m
m
2
, the marginal PMF of N satisfies n 1
∑
P N n
1
p
n 2 2
p
1 1
n
p
n 2 2
p
m 1
Similarly, for m
12
, the marginal PMF of M satisfies ∞
∑
P M m
1
p
n 2 2
p
n m 1
p2 1
1
p
p
m 1
1
p
m
m 1
p
The complete expressions for the marginal PMF’s are m 1
P M m
1 0
p
P N n
n 0
1 1
p p
m 1 2 otherwise n 2 2
n 2 3 otherwise
p
Not surprisingly, if we view each voice call as a successful Bernoulli trial, M has a geometric PMF since it is the number of trials up to and including the first success. Also, N has a Pascal PMF since it is the number of trials required to see 2 successes. The conditional PMF’s are now easy to find. P N M n m
P M N m n P M m
1 0
p
1
n m 1
p
n m 1 m otherwise
2
The interpretation of the conditional PMF of N given M is that given M m, N has a geometric PMF with mean 1 p. The conditional PMF of M given N is P M N m n
P M N m n
1 n 0
P N n
1
m 1 n otherwise
m
N where N
1
Given that call N n was the second voice call, the first voice call is equally likely to occur in any of the previous n 1 calls. Problem 3.6.8
(a) The number number of buses, buses, N , must be greater greater than zero. Also, Also, the number number of minutes minutes that pass pass cannot be less than the number of buses. Thus, P N n T t 0 for integers n t satisfying 1 n t . N and T by carefully considering (b) First, we find the the joint PMF PMF of N considering the possible sample sample paths. In particular, P N T n t P ABC P A P B P C where the events A, B and C are
1 buses arrive in the first t
1 minutes
A
n
B
none of the first n
C
at time t a bus arrives and is boarded
1 buses are boarded
These events are independent since each trial to board a bus is independent of when the buses arrive. These events have probabilities PA
1 pn 1
t n
PB
1
P C
pq
q
1
1
p
t 1
n 1
n 1
Consequently, the joint PMF of N and T is P N T n t
t 1 n 1
pn
1
1
p
t n
1
q
n 1
pq
0
n 1 t n otherwise
(c) It It is possib possible le to find the margin marginal al PMF’ PMF’s by summ summing ing the the joint joint PMF PMF. Howev However er,, it is much much easier easier to obtain obtain the margina marginall PMFs by considera consideration tion of the experiment. experiment. Specifical Specifically ly,, when a bus arrive arrives, s, it is boarde boarded d with with probab probabili ility ty q. Moreo Moreover ver,, the experi experime ment nt ends ends when when a bus is board boarded. ed. By viewing whether each arriving bus is boarded as an independent trial, N is the number of trials until the first success. Thus, N has the geometric PMF P N n
1 0
q
n 1
q
n 1 2 otherwise
T , suppos To find find the the PMF PMF of T supposee we regard regard each each minut minutee as an indep independ endenttrial enttrial in which which a succes successs occurs if a bus arrives and that bus is boarded. boarded. In this case, the success probability probability is pq and T is the the numb number er of minu minute tess up to and and incl includ udin ing g the the first first succ succes ess. s. The PMF PMF of T is also also geomet geometric ric.. PT t
1 0
pq
2
t 1
pq t 1 2 otherwise
(d) Once we have the marginal PMFs, the conditional conditional PMFs are easy to find. P N T n t
p 1 q 1 pq
t 1 n 1
P N T n t PT t
n 1
1 p 1 pq
t 1
n 1
0
1 2
n
t
otherwise
t That That is, is, give given n you you depa departat rtat time time T t , thenumber thenumber of buses buses that that arrivedurin arriveduring g minut minutes es 1 1 has has a bino binomi mial al PMF PMF sinc sincee in each each minu minute te a bus bus arri arrive vess with with prob probab abil ilit ity y p. Similarly, Similarly, the conditional PMF of T given N is t 1 n 1
P N T n t
PT N t n
pn 1
p
t n
t n n 1 otherwise
0
P N n
This result can be explained. Given that you board bus N n, the time T when you leave is the time for n buses to arrive. If we view each bus arrival as a success of an independent trial, the time for n buses to arrive has the above Pascal PMF. Problem 3.7.1 Flip a fair coin 100 times and let X be the number of heads in the first 75 flips and Y be the number of heads in the last 25 flips. We know that X and Y are independent and can find their PMFs easily. P X x
75 x
1 2
0
75
x 0 1 otherwise
75
PY y
25 y
0
1 2 25 y 0 1 otherwise
25
The joint PMF of X and N can be expressed as the product of the marginal PMFs because we know that X and Y are independent. P X Y x y
75 x
25 y
1 2
100
01 otherwise
x
0
75 y
01
25
Problem 3.7.7 The key to this problem is understanding that “short order” and “long order” are synonyms for N 1 and N 2. Similarly, Similarly, “vanilla”, “chocolate”, and “strawberry” correspond to the events D 20, D 100 and D 300.
(a) The following table table is given in the problem problem statement. statement.
short order long order
vani vanill llaa
choc choc..
stra strawb wber erry ry
0.2
0.2
0 .2
0.1
0.2
0 .1
This table can be translated directly into the joint PMF of N and D. P N D n d n 1 n 2
d
20 d 100 d 300 02 02 02 01 02 01 3
(b) To To find the condit condition ional al PMF P D N d 2 , we first first need need to find find the the prob probab abil ilit ity y of the the cond condit itio ioni ning ng event P N 2
P N D 2 20
P N D 2 100
The conditional PMF of N D given N
2
1 4 d 20 1 2 d 100 1 4 d 300 0 otherwise
P N 2
(c) The conditional conditional expectation expectation of D given N E D N
∑ dP D N
04
2 is
P N D 2 d
P D N d 2
P N D 2 300
d 2
2 is 20 1 4
100 1 2
300 1 4
130
d
(d) To check check independence, we calculate calculate the marginal marginal PMFs of N and D. An easy way to do this this is to find the row and column sums from the table for P N D n d . This yields P N D n d n 1 n 2
d
20 d 100 d 300 02 02 02 01 02 01 03
P D d
04
Some study of the table will show that P N D n d dent.
03 P N n P D d . Hence, N and D are depen-
(e) In terms terms of N and D, the cost (in cents) of a fax is C E C E C
∑ ndP N D
P N n 06 04
ND . The expected value of C is
n d
n d
1 20 0 2
1 100 0 2
2 20 0 3
1 300 0 2
2 100 0 4
2 300 0 3
356
Problem 4.1.1
F X x
0 x 1
1 x
x
1 2
1 x
1
1
Each question can be answered by expressing the requested probability in terms of F X x . (a) P X
1 2
1
P X
1 2
4
1
F X 1 2
1
3 4
1 4
(b) This is a little trickier trickier than it should be. be. Being careful, we can write 1 2
P
3 4
X
1 2
P
3 4
X
1 2
P X
P X
3 4
X is a continuous Since the CDF CDF of X continuous function, function, the probability probability that X takes on any any specific specific value value is zero. This implies P X 3 4 0 and P X 1 2 0. (If this is not clear at this point, it will become clear in Section 4.6.) Thus,
1 2
P
3 4
X
1 2
P
3 4
X
F X 3 4
F X
1 2
5 8
(c) P X
1 2
1 2
P
F X 1 2 Note that P X 1 2 P X 1 2 . Hence P X 1 2 P X
(d) Since F X 1
1 2
P X
1 2
P X
3 4. Since Since the probabil probability ity that P X F X 1 2 1 4. This implies
P X
1, we must have a
1 2
a
1 2
P X
1. For a
P X
Thus a
1 2
X
3 4
1 4
1 2 1 2
0, P X
1 2
1 2
1, we need to satisfy 1
a
F X a
2
08
0 6.
Problem 4.2.1 cx 0
f X x
0 x 2 othe otherw rwis isee
(a) From the above above PDF we can determine determine the value value of c by integrating integrating the PDF and setting setting it equal to 1. 2
cxdx
2c
1
0
Therefore c (b) P 0 (c) P
X
1
1 2
X
1 2. 1 x 0 2
1 2
dx
1 4 1 2 x 0 2
1 16
dx
(d) The CDF CDF of X is found by integrating the PDF from 0 to x. x
F X x
0
f X x
dx
5
0 x 2 x 4 0 1 x
0 x 2
2
