Probability and Stochastic Processes
A Friendly Introduction for Electrical and Computer Engineers
Third Edition
INSTRUCTOR’S SOLUTION MANUAL
Roy D. Yates, David J. Goodman, David Famolari
January 5, 2014
1
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Guidelines and tips
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Community
Ask the community
Ask the community for help and clear up your study doubts
University Rankings
Discover the best universities in your country according to Docsity users
Free resources
Our save-the-student-ebooks!
Download our free guides on studying techniques, anxiety management strategies, and thesis advice from Docsity tutors
Probability and Stochastic processes 3rd edition Roy D Yateshw solution manual, Study notes of Electrical Engineering
textbook homework solution manual
Typology: Study notes
2015/2016
1 / 787
This page cannot be seen from the preview
Don't miss anything!
On special offer
Related documents
Partial preview of the text
Download Probability and Stochastic processes 3rd edition Roy D Yateshw solution manual and more Study notes Electrical Engineering in PDF only on Docsity!
Probability and Stochastic Processes
A Friendly Introduction for Electrical and Computer Engineers
Third Edition
INSTRUCTOR’S SOLUTION MANUAL
Roy D. Yates, David J. Goodman, David Famolari
January 5, 2014
Comments on this Solutions Manual
- This solution manual is currently under construction. Some problems may not have solutions, although none so far have been identified
- To make solution sets for your class, try the Solution Set Constructor at the instructors site www.winlab.rutgers.edu/probsolns. For access, send email: ryates@winlab.rutgers.edu.
- Please send error reports, suggestions, comments or access requests to ry- ates@winlab.rutgers.edu.
- Matlab functions written as solutions to homework problems can be found in the archive matsoln3e.zip (available to instructors). Other Matlab functions used in the text or in these homework solutions can be found in the archive matcode3e.zip. The .m files in matcode3e are available for download from the Wiley website. Two other documents of interest are also available for download: - A manual probmatlab3e.pdf describing the matcode3e .m functions - The quiz solutions manual quizsol.pdf.
- This manual uses a page size matched to the screen of an iPad tablet. If you do print on paper and you have good eyesight, you may wish to print two pages per sheet in landscape mode. On the other hand, a “Fit to Paper” printing option will create “Large Print” output.
(e) Yes. In terms of the Venn diagram, these pizzas are in the set (T ∪ M ∪ O)c.
Problem 1.1.3 Solution
At Ricardo’s, the pizza crust is either Roman (R) or Neapoli- tan (N ). To draw the Venn diagram on the right, we make the following observations:
R N M
W O
- The set {R, N } is a partition so we can draw the Venn diagram with this partition.
- Only Roman pizzas can be white. Hence W ⊂ R.
- Only a Neapolitan pizza can have onions. Hence O ⊂ N.
- Both Neapolitan and Roman pizzas can have mushrooms so that event M straddles the {R, N } partition.
- The Neapolitan pizza can have both mushrooms and onions so M ∩ O cannot be empty.
- The problem statement does not preclude putting mushrooms on a white Roman pizza. Hence the intersection W ∩M should not be empty.
Problem 1.2.1 Solution
(a) An outcome specifies whether the connection speed is high (h), medium (m), or low (l) speed, and whether the signal is a mouse click (c) or a tweet (t). The sample space is
S = {ht, hc, mt, mc, lt, lc}. (1)
(b) The event that the wi-fi connection is medium speed is A 1 = {mt, mc}.
(c) The event that a signal is a tweet is A 2 = {ht, mt, lt}.
(d) The event that a connection is either high speed or low speed is A 3 = {ht, hc, lt, lc}.
(e) Since A 1 ∩A 2 = {mt} and is not empty, A 1 , A 2 , and A 3 are not mutually exclusive.
(f) Since
A 1 ∪ A 2 ∪ A 3 = {ht, hc, mt, mc, lt, lc} = S, (2)
the collection A 1 , A 2 , A 3 is collectively exhaustive.
Problem 1.2.2 Solution
(a) The sample space of the experiment is
S = {aaa, aaf, af a, f aa, f f a, f af, af f, f f f }. (1)
(b) The event that the circuit from Z fails is
ZF = {aaf, af f, f af, f f f }. (2)
The event that the circuit from X is acceptable is
XA = {aaa, aaf, af a, af f }. (3)
(c) Since ZF ∩ XA = {aaf, af f } 6 = φ, ZF and XA are not mutually exclu- sive.
(d) Since ZF ∪ XA = {aaa, aaf, af a, af f, f af, f f f } 6 = S, ZF and XA are not collectively exhaustive.
Problem 1.2.5 Solution
Of course, there are many answers to this problem. Here are four partitions.
- We can divide students into engineers or non-engineers. Let A 1 equal the set of engineering students and A 2 the non-engineers. The pair {A 1 , A 2 } is a partition.
- We can also separate students by GPA. Let Bi denote the subset of stu- dents with GPAs G satisfying i− 1 ≤ G < i. At Rutgers, {B 1 , B 2 ,... , B 5 } is a partition. Note that B 5 is the set of all students with perfect 4. GPAs. Of course, other schools use different scales for GPA.
- We can also divide the students by age. Let Ci denote the subset of students of age i in years. At most universities, {C 10 , C 11 ,... , C 100 } would be an event space. Since a university may have prodigies either under 10 or over 100, we note that {C 0 , C 1 ,.. .} is always a partition.
- Lastly, we can categorize students by attendance. Let D 0 denote the number of students who have missed zero lectures and let D 1 denote all other students. Although it is likely that D 0 is an empty set, {D 0 , D 1 } is a well defined partition.
Problem 1.2.6 Solution
Let R 1 and R 2 denote the measured resistances. The pair (R 1 , R 2 ) is an outcome of the experiment. Some partitions include
- If we need to check that neither resistance is too high, a partition is
A 1 = {R 1 < 100 , R 2 < 100 } , A 2 = {R 1 ≥ 100 } ∪ {R 2 ≥ 100 }. (1)
- If we need to check whether the first resistance exceeds the second resistance, a partition is
B 1 = {R 1 > R 2 } B 2 = {R 1 ≤ R 2 }. (2)
- If we need to check whether each resistance doesn’t fall below a mini- mum value (in this case 50 ohms for R 1 and 100 ohms for R 2 ), a partition is C 1 , C 2 , C 3 , C 4 where
C 1 = {R 1 < 50 , R 2 < 100 } , C 2 = {R 1 < 50 , R 2 ≥ 100 } , (3) C 3 = {R 1 ≥ 50 , R 2 < 100 } , C 4 = {R 1 ≥ 50 , R 2 ≥ 100 }. (4)
- If we want to check whether the resistors in parallel are within an ac- ceptable range of 90 to 110 ohms, a partition is
D 1 =
(1/R 1 + 1/R 2 )−^1 < 90
D 2 =
90 ≤ (1/R 1 + 1/R 2 )−^1 ≤ 110
D 2 =
110 < (1/R 1 + 1/R 2 )−^1
Problem 1.3.1 Solution
(a) A and B mutually exclusive and collectively exhaustive imply P[A] + P[B] = 1. Since P[A] = 3 P[B], we have P[B] = 1/4.
(b) Since P[A ∪ B] = P[A], we see that B ⊆ A. This implies P[A ∩ B] = P[B]. Since P[A ∩ B] = 0, then P[B] = 0.
(c) Since it’s always true that P[A ∪ B] = P[A] + P[B] − P[AB], we have that P[A] + P[B] − P[AB] = P[A] − P[B]. (1) This implies 2 P[B] = P[AB]. However, since AB ⊂ B, we can conclude that 2 P[B] = P[AB] ≤ P[B]. This implies P[B] = 0.
Problem 1.3.2 Solution
The roll of the red and white dice can be assumed to be independent. For each die, all rolls in { 1 , 2 ,... , 6 } have probability 1/6.
Problem 1.3.5 Solution
The sample space of the experiment is
S = {LF, BF, LW, BW }. (1)
From the problem statement, we know that P[LF ] = 0.5, P[BF ] = 0.2 and P[BW ] = 0.2. This implies P[LW ] = 1 − 0. 5 − 0. 2 − 0 .2 = 0.1. The questions can be answered using Theorem 1.5.
(a) The probability that a program is slow is
P [W ] = P [LW ] + P [BW ] = 0.1 + 0.2 = 0. 3. (2)
(b) The probability that a program is big is
P [B] = P [BF ] + P [BW ] = 0.2 + 0.2 = 0. 4. (3)
(c) The probability that a program is slow or big is
P [W ∪ B] = P [W ] + P [B] − P [BW ] = 0.3 + 0. 4 − 0 .2 = 0. 5. (4)
Problem 1.3.6 Solution
A sample outcome indicates whether the cell phone is handheld (H) or mobile (M ) and whether the speed is fast (F ) or slow (W ). The sample space is
S = {HF, HW, M F, M W }. (1)
The problem statement tells us that P[HF ] = 0.2, P[M W ] = 0.1 and P[F ] = 0 .5. We can use these facts to find the probabilities of the other outcomes. In particular,
P [F ] = P [HF ] + P [M F ]. (2)
This implies
P [M F ] = P [F ] − P [HF ] = 0. 5 − 0 .2 = 0. 3. (3)
Also, since the probabilities must sum to 1,
P [HW ] = 1 − P [HF ] − P [M F ] − P [M W ] = 1 − 0. 2 − 0. 3 − 0 .1 = 0. 4. (4)
Now that we have found the probabilities of the outcomes, finding any other probability is easy.
(a) The probability a cell phone is slow is
P [W ] = P [HW ] + P [M W ] = 0.4 + 0.1 = 0. 5. (5)
(b) The probability that a cell hpone is mobile and fast is P[M F ] = 0.3.
(c) The probability that a cell phone is handheld is
P [H] = P [HF ] + P [HW ] = 0.2 + 0.4 = 0. 6. (6)
Problem 1.3.7 Solution
A reasonable probability model that is consistent with the notion of a shuffled deck is that each card in the deck is equally likely to be the first card. Let Hi denote the event that the first card drawn is the ith heart where the first heart is the ace, the second heart is the deuce and so on. In that case, P[Hi] = 1/ 52 for 1 ≤ i ≤ 13. The event H that the first card is a heart can be written as the mutually exclusive union
H = H 1 ∪ H 2 ∪ · · · ∪ H 13. (1)
Using Theorem 1.1, we have
P [H] =
∑^13
i=
P [Hi] = 13/ 52. (2)
This is the answer you would expect since 13 out of 52 cards are hearts. The point to keep in mind is that this is not just the common sense answer but is the result of a probability model for a shuffled deck and the axioms of probability.
Problem 1.3.11 Solution
Specifically, we will use Theorem 1.4(c) which states that for any events A and B,
P [A ∪ B] = P [A] + P [B] − P [A ∩ B]. (1)
To prove the union bound by induction, we first prove the theorem for the case of n = 2 events. In this case, by Theorem 1.4(c),
P [A 1 ∪ A 2 ] = P [A 1 ] + P [A 2 ] − P [A 1 ∩ A 2 ]. (2)
By the first axiom of probability, P[A 1 ∩ A 2 ] ≥ 0. Thus,
P [A 1 ∪ A 2 ] ≤ P [A 1 ] + P [A 2 ]. (3)
which proves the union bound for the case n = 2.Now we make our induction hypothesis that the union-bound holds for any collection of n − 1 subsets. In this case, given subsets A 1 ,... , An, we define
A = A 1 ∪ A 2 ∪ · · · ∪ An− 1 , B = An. (4)
By our induction hypothesis,
P [A] = P [A 1 ∪ A 2 ∪ · · · ∪ An− 1 ] ≤ P [A 1 ] + · · · + P [An− 1 ]. (5)
This permits us to write
P [A 1 ∪ · · · ∪ An] = P [A ∪ B] ≤ P [A] + P [B] (by the union bound for n = 2) = P [A 1 ∪ · · · ∪ An− 1 ] + P [An] ≤ P [A 1 ] + · · · P [An− 1 ] + P [An] (6)
which completes the inductive proof.
Problem 1.3.12 Solution
It is tempting to use the following proof:
Since S and φ are mutually exclusive, and since S = S ∪ φ,
1 = P [S ∪ φ] = P [S] + P [φ].
Since P[S] = 1, we must have P[φ] = 0.
The above “proof” used the property that for mutually exclusive sets A 1 and A 2 ,
P [A 1 ∪ A 2 ] = P [A 1 ] + P [A 2 ]. (1)
The problem is that this property is a consequence of the three axioms, and thus must be proven. For a proof that uses just the three axioms, let A 1 be an arbitrary set and for n = 2, 3 ,.. ., let An = φ. Since A 1 = ∪∞ i=1Ai, we can use Axiom 3 to write
P [A 1 ] = P [∪∞ i=1Ai] = P [A 1 ] + P [A 2 ] +
∑^ ∞
i=
P [Ai]. (2)
By subtracting P[A 1 ] from both sides, the fact that A 2 = φ permits us to write
P [φ] +
∑^ ∞
n=
P [Ai] = 0. (3)
By Axiom 1, P[Ai] ≥ 0 for all i. Thus,
n=3 P[Ai]^ ≥^ 0. This implies P[φ]^ ≤^ 0. Since Axiom 1 requires P[φ] ≥ 0, we must have P[φ] = 0.
Problem 1.3.13 Solution
Following the hint, we define the set of events {Ai|i = 1, 2 ,.. .} such that i = 1 ,... , m, Ai = Bi and for i > m, Ai = φ. By construction, ∪mi=1Bi = ∪∞ i=1Ai. Axiom 3 then implies
P [∪mi=1Bi] = P [∪∞ i=1Ai] =
∑^ ∞
i=
P [Ai]. (1)
Now, we use Axiom 3 again on Am, Am+1,... to write
∑^ ∞
i=m
P [Ai] = P [Am ∪ Am+1 ∪ · · ·] = P [Bm]. (2)
Thus, we have used just Axiom 3 to prove Theorem 1.3:
P [B 1 ∪ B 2 ∪ · · · ∪ Bm] =
∑^ m
i=
P [Bi]. (3)
(a) To show P[φ] = 0, let B 1 = S and let B 2 = φ. Thus by Theorem 1.3,
P [S] = P [B 1 ∪ B 2 ] = P [B 1 ] + P [B 2 ] = P [S] + P [φ]. (4)
Thus, P[φ] = 0. Note that this proof uses only Theorem 1.3 which uses only Axiom 3.
(b) Using Theorem 1.3 with B 1 = A and B 2 = Ac, we have
P [S] = P [A ∪ Ac] = P [A] + P [Ac]. (5)
Since, Axiom 2 says P[S] = 1, P[Ac] = 1 − P[A]. This proof uses Axioms 2 and 3.
(c) By Theorem 1.8, we can write both A and B as unions of mutually exclusive events:
A = (AB) ∪ (ABc), B = (AB) ∪ (AcB). (6)
Now we apply Theorem 1.3 to write
P [A] = P [AB] + P [ABc] , P [B] = P [AB] + P [AcB]. (7)
We can rewrite these facts as
P[ABc] = P[A] − P[AB], P[AcB] = P[B] − P[AB]. (8)
Note that so far we have used only Axiom 3. Finally, we observe that A ∪ B can be written as the union of mutually exclusive events
A ∪ B = (AB) ∪ (ABc) ∪ (AcB). (9)
Once again, using Theorem 1.3, we have
P[A ∪ B] = P[AB] + P[ABc] + P[AcB] (10)
Substituting the results of Equation (8) into Equation (10) yields
P [A ∪ B] = P [AB] + P [A] − P [AB] + P [B] − P [AB] , (11)
which completes the proof. Note that this claim required only Axiom 3.
(d) Observe that since A ⊂ B, we can write B as the mutually exclusive union B = A ∪ (AcB). By Theorem 1.3 (which uses Axiom 3),
P [B] = P [A] + P [AcB]. (12)
By Axiom 1, P[AcB] ≥ 0, hich implies P[A] ≤ P[B]. This proof uses Axioms 1 and 3.
Problem 1.4.1 Solution
Each question requests a conditional probability.
(a) Note that the probability a call is brief is
P [B] = P [H 0 B] + P [H 1 B] + P [H 2 B] = 0. 6. (1)
The probability a brief call will have no handoffs is
P [H 0 |B] =
P [H 0 B]
P [B]
The conditional probabilities of G 3 given E is
P [G 3 |E] =
P [G 3 E]
P [E]
(d) The conditional probability that the roll is even given that it’s greater than 3 is
P [E|G 3 ] =
P [EG 3 ]
P [G 3 ]
Problem 1.4.3 Solution
Since the 2 of clubs is an even numbered card, C 2 ⊂ E so that P[C 2 E] = P[C 2 ] = 1/3. Since P[E] = 2/3,
P [C 2 |E] =
P [C 2 E]
P [E]
The probability that an even numbered card is picked given that the 2 is picked is
P [E|C 2 ] =
P [C 2 E]
P [C 2 ]
Problem 1.4.4 Solution
Let Ai and Bi denote the events that the ith phone sold is an Apricot or a Banana respectively. Our goal is to find P[B 1 B 2 ], but since it is not clear where to start, we should plan on filling in the table
A 2 B 2 A 1 B 1
This table has four unknowns: P[A 1 A 2 ], P[A 1 B 2 ], P[B 1 A 2 ], and P[B 1 B 2 ]. We start knowing that
P [A 1 A 2 ] + P [A 1 B 2 ] + P [B 1 A 2 ] + P [B 1 B 2 ] = 1. (1)
We still need three more equations to solve for the four unknowns. From “sales of Apricots and Bananas are equally likely,” we know that P[Ai] = P[Bi] = 1/2 for i = 1, 2. This implies
P [A 1 ] = P [A 1 A 2 ] + P [A 1 B 2 ] = 1/ 2 , (2) P [A 2 ] = P [A 1 A 2 ] + P [B 1 A 2 ] = 1/ 2. (3)
The final equation comes from “given that the first phone sold is a Banana, the second phone is twice as likely to be a Banana,” which implies P[B 2 |B 1 ] = 2 P[A 2 |B 1 ]. Using Bayes’ theorem, we have
P [B 1 B 2 ] P [B 1 ]
P [B 1 A 2 ]
P [B 1 ]
=⇒ P [B 1 A 2 ] =
P [B 1 B 2 ]. (4)
Replacing P[B 1 A 2 ] with P[B 1 B 2 ]/2 in the the first three equations yields
P [A 1 A 2 ] + P [A 1 B 2 ] +
P [B 1 B 2 ] = 1, (5)
P [A 1 A 2 ] + P [A 1 B 2 ] = 1/ 2 , (6)
P [A 1 A 2 ] +
P [B 1 B 2 ] = 1/ 2. (7)
Subtracting (6) from (5) yields (3/2) P[B 1 B 2 ] = 1/2, or P[B 1 B 2 ] = 1/3, which is the answer we are looking for.
At this point, if you are curious, we can solve for the rest of the probability table. From (4), we have P[B 1 A 2 ] = 1/6 and from (7) we obtain P[A 1 A 2 ] = 1 /3. It then follows from (6) that P[A 1 B 2 ] = 1/6. The probability table is
A 2 B 2 A 1 1 / 3 1 / 6 B 1 1 / 6 1 / 3