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Probability And Stochastic Processes : A Friendly Introduction For Electrical And Computer Engineers, 3rd Edition Solution Manual

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    Probability and Stochastic Processes A Friendly Introduction for Electrical and Computer Engineers Third Edition ’S SOLUTION MANUAL Roy D. Yates, David J. Goodman, David Famolari September 8, 2014 Comments on this Solutions Manual • This solution manual is mostly complete. Please send error reports, suggestions, and comments to ryates@winlab.rutgers.edu . • To make solution sets for your class, use the Solution Set Constructor at the instruc- tors site www.winlab.rutgers.edu/probsolns . • Send email to ryates@winlab.rutgers.edu for access to the instructors site. • 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. 1

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    Problem Solutions – Chapter 1 Problem 1.1.1 Solution Based on the Venn diagram on the right, the complete Gerlandas pizza menu is • Regular without toppings • Regular with mushrooms • Regular with onions • Regular with mushrooms and onions • Tuscan without toppings • Tuscan with mushrooms M O T Problem 1.1.2 Solution Based on the Venn diagram on the right, the answers are mostly fairly straightforward. The only trickiness is that a pizza is either Tuscan ( T ) or Neapolitan ( N ) so { N, T } is a partition but they are not depicted as a partition. Specifically, the event N is the region of the Venn diagram outside of the “square block” of event T . If this is clear, the questions are easy. M O T (a) Since N = T c , N ∩ M 6 = φ . Thus N and M are not mutually exclusive. (b) Every pizza is either Neapolitan ( N ), or Tuscan ( T ). Hence N ∪ T = S so that N and T are collectively exhaustive. Thus its also (trivially) true that N ∪ T ∪ M = S . That is, R , T and M are also collectively exhaustive. (c) From the Venn diagram, T and O are mutually exclusive. In words, this means that Tuscan pizzas never have onions or pizzas with onions are never Tuscan. As an aside, “Tuscan” is a fake pizza designation; one shouldn’t conclude that people from Tuscany actually dislike onions. (d) From the Venn diagram, M ∩ T and O are mutually exclusive. Thus Ger- landa’s doesn’t make Tuscan pizza with mushrooms and onions. (e) Yes. In terms of the Venn diagram, these pizzas are in the set ( T ∪ M ∪ O ) c . 2

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    Problem 1.1.3 Solution At Ricardo’s, the pizza crust is either Roman ( R ) or Neapolitan ( N ). To draw the Venn diagram on the right, we make the fol- lowing observations: R N M O W • 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 mouse click is A 2 = { hc, mc, lc } . (d) The event that a connection is either high speed or low speed is A 3 = { ht, hc, lt, lc } . 3

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    (e) Since A 1 ∩ A 2 = { mc } 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 Z F = { aaf, af f, f af, f f f } . (2) The event that the circuit from X is acceptable is X A = { aaa, aaf, af a, af f } . (3) (c) Since Z F ∩ X A = { aaf, af f } 6 = φ , Z F and X A are not mutually exclusive. (d) Since Z F ∪ X A = { aaa, aaf, af a, af f, f af, f f f } 6 = S , Z F and X A are not collectively exhaustive. (e) The event that more than one circuit is acceptable is C = { aaa, aaf, af a, f aa } . (4) The event that at least two circuits fail is D = { f f a, f af, af f, f f f } . (5) (f) Inspection shows that C ∩ D = φ so C and D are mutually exclusive. (g) Since C ∪ D = S , C and D are collectively exhaustive. 4

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    Problem 1.2.3 Solution The sample space is S = { A ♣ , . . . , K ♣ , A ♦ , . . . , K ♦ , A ♥ , . . . , K ♥ , A ♠ , . . . , K ♠} . (1) The event H that the first card is a heart is the set H = { A ♥ , . . . , K ♥} . (2) The event H has 13 outcomes, corresponding to the 13 hearts in a deck. Problem 1.2.4 Solution The sample space is S =    1 / 1 . . . 1 / 31 , 2 / 1 . . . 2 / 29 , 3 / 1 . . . 3 / 31 , 4 / 1 . . . 4 / 30 , 5 / 1 . . . 5 / 31 , 6 / 1 . . . 6 / 30 , 7 / 1 . . . 7 / 31 , 8 / 1 . . . 8 / 31 , 9 / 1 . . . 9 / 31 , 10 / 1 . . . 10 / 31 , 11 / 1 . . . 11 / 30 , 12 / 1 . . . 12 / 31    . (1) The event H defined by the event of a July birthday is given by the following et with 31 sample outcomes: H = { 7 / 1 , 7 / 2 , . . . , 7 / 31 } . (2) Problem 1.2.5 Solution Of course, there are many answers to this problem. Here are four partitions. 1. 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. 2. To separate students by GPA, let B i denote the subset of students 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 4.0 GPAs. Of course, other schools use different scales for GPA. 3. We can also divide the students by age. Let C i 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. 5

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    4. 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 1. 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) 2. 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) 3. If we need to check whether each resistance doesn’t fall below a minimum 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) 4. If we want to check whether the resistors in parallel are within an acceptable range of 90 to 110 ohms, a partition is D 1 = { (1 /R 1 + 1 /R 2 ) − 1 < 90 } , (5) D 2 = { 90 ≤ (1 /R 1 + 1 /R 2 ) − 1 ≤ 110 } , (6) D 2 = { 110 < (1 /R 1 + 1 /R 2 ) − 1 } . (7) 6

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    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. (a) Thus P[ R 3 W 2 ] = P[ R 3 ] P[ W 2 ] = 1 36 . (b) In fact, each pair of possible rolls R i W j has probability 1 / 36. To find P[ S 5 ], we add up the probabilities of all pairs that sum to 5: P[ S 5 ] = P[ R 1 W 4 ] + P[ R 2 W 3 ] + P[ R 3 W 2 ] + P[ R 4 W 1 ] = 4 / 36 = 1 / 9 . Problem 1.3.3 Solution An outcome is a pair ( i, j ) where i is the value of the first die and j is the value of the second die. The sample space is the set S = { (1 , 1) , (1 , 2) , . . . , (6 , 5) , (6 , 6) } . (1) with 36 outcomes, each with probability 1 / 36 Note that the event that the absolute value of the difference of the two rolls equals 3 is D 3 = { (1 , 4) , (2 , 5) , (3 , 6) , (4 , 1) , (5 , 2) , (6 , 3) } . (2) Since there are 6 outcomes in D 3 , P[ D 3 ] = 6 / 36 = 1 / 6. 7

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    Problem 1.3.4 Solution (a) FALSE. Since P[ A ] = 1 − P[ A c ] = 2 P[ A c ] implies P[ A c ] = 1 / 3. (b) FALSE. Suppose A = B and P[ A ] = 1 / 2. In that case, P [ AB ] = P [ A ] = 1 / 2 > 1 / 4 = P [ A ] P [ B ] . (1) (c) TRUE. Since AB ⊆ A , P[ AB ] ≤ P[ A ], This implies P [ AB ] ≤ P [ A ] < P [ B ] . (2) (d) FALSE: For a counterexample, let A = φ and P[ B ] > 0 so that A = A ∩ B = φ and P[ A ] = P[ A ∩ B ] = 0 but 0 = P[ A ] < P[ B ]. 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) 8

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    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 prob- ability 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 H i denote the event that the first card drawn is the i th heart where the first heart is the ace, 9

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    the second heart is the deuce and so on. In that case, P[ H i ] = 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 =1 P [ H i ] = 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 1.3.8 Solution Let s i denote the outcome that the down face has i dots. The sample space is S = { s 1 , . . . , s 6 } . The probability of each sample outcome is P[ s i ] = 1 / 6. From Theorem 1.1, the probability of the event E that the roll is even is P [ E ] = P [ s 2 ] + P [ s 4 ] + P [ s 6 ] = 3 / 6 . (1) Problem 1.3.9 Solution Let s i equal the outcome of the student’s quiz. The sample space is then composed of all the possible grades that she can receive. S = { 0 , 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 , 9 , 10 } . (1) Since each of the 11 possible outcomes is equally likely, the probability of receiving a grade of i , for each i = 0 , 1 , . . . , 10 is P[ s i ] = 1 / 11. The probability that the student gets an A is the probability that she gets a score of 9 or higher. That is P [Grade of A] = P [9] + P [10] = 1 / 11 + 1 / 11 = 2 / 11 . (2) The probability of failing requires the student to get a grade less than 4. P [Failing] = P [3] + P [2] + P [1] + P [0] = 1 / 11 + 1 / 11 + 1 / 11 + 1 / 11 = 4 / 11 . (3) 10

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    Problem 1.3.10 Solution Each statement is a consequence of part 4 of Theorem 1.4. (a) Since A ⊂ A ∪ B , P[ A ] ≤ P[ A ∪ B ]. (b) Since B ⊂ A ∪ B , P[ B ] ≤ P[ A ∪ B ]. (c) Since A ∩ B ⊂ A , P[ A ∩ B ] ≤ P[ A ]. (d) Since A ∩ B ⊂ B , P[ A ∩ B ] ≤ P[ B ]. 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 , . . . , A n , we define A = A 1 ∪ A 2 ∪ · · · ∪ A n − 1 , B = A n . (4) By our induction hypothesis, P [ A ] = P [ A 1 ∪ A 2 ∪ · · · ∪ A n − 1 ] ≤ P [ A 1 ] + · · · + P [ A n − 1 ] . (5) This permits us to write P [ A 1 ∪ · · · ∪ A n ] = P [ A ∪ B ] ≤ P [ A ] + P [ B ] (by the union bound for n = 2) = P [ A 1 ∪ · · · ∪ A n − 1 ] + P [ A n ] ≤ P [ A 1 ] + · · · P [ A n − 1 ] + P [ A n ] (6) which completes the inductive proof. 11

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    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 A n = φ . Since A 1 = ∪ ∞ i =1 A i , we can use Axiom 3 to write P [ A 1 ] = P [ ∪ ∞ i =1 A i ] = P [ A 1 ] + P [ A 2 ] + ∞ ∑ i =3 P [ A i ] . (2) By subtracting P[ A 1 ] from both sides, the fact that A 2 = φ permits us to write P [ φ ] + ∞ ∑ n =3 P [ A i ] = 0 . (3) By Axiom 1, P[ A i ] ≥ 0 for all i . Thus, ∑ ∞ n =3 P[ A i ] ≥ 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 { A i | i = 1 , 2 , . . . } such that i = 1 , . . . , m , A i = B i and for i > m , A i = φ . By construction, ∪ m i =1 B i = ∪ ∞ i =1 A i . Axiom 3 then implies P [ ∪ m i =1 B i ] = P [ ∪ ∞ i =1 A i ] = ∞ ∑ i =1 P [ A i ] . (1) 12

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    For i > m , P[ A i ] = P[ φ ] = 0, yielding the claim P[ ∪ m i =1 B i ] = ∑ m i =1 P[ A i ] = ∑ m i =1 P[ B i ]. Note that the fact that P[ φ ] = 0 follows from Axioms 1 and 2. This problem is more challenging if you just use Axiom 3. We start by observing P [ ∪ m i =1 B i ] = m − 1 ∑ i =1 P [ B i ] + ∞ ∑ i = m P [ A i ] . (2) Now, we use Axiom 3 again on the countably infinite sequence A m , A m +1 , . . . to write ∞ ∑ i = m P [ A i ] = P [ A m ∪ A m +1 ∪ · · · ] = P [ B m ] . (3) Thus, we have used just Axiom 3 to prove Theorem 1.3: P [ ∪ m i =1 B i ] = m ∑ i =1 P [ B i ] . (4) Problem 1.3.14 Solution Each claim in Theorem 1.4 requires a proof from which we can check which axioms are used. However, the problem is somewhat hard because there may still be a simpler proof that uses fewer axioms. Still, the proof of each part will need Theorem 1.3 which we now prove. For the mutually exclusive events B 1 , . . . , B m , let A i = B i for i = 1 , . . . , m and let A i = φ for i > m . In that case, by Axiom 3, P [ B 1 ∪ B 2 ∪ · · · ∪ B m ] = P [ A 1 ∪ A 2 ∪ · · · ] = m − 1 ∑ i =1 P [ A i ] + ∞ ∑ i = m P [ A i ] = m − 1 ∑ i =1 P [ B i ] + ∞ ∑ i = m P [ A i ] . (1) 13

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    Now, we use Axiom 3 again on A m , A m +1 , . . . to write ∞ ∑ i = m P [ A i ] = P [ A m ∪ A m +1 ∪ · · · ] = P [ B m ] . (2) Thus, we have used just Axiom 3 to prove Theorem 1.3: P [ B 1 ∪ B 2 ∪ · · · ∪ B m ] = m ∑ i =1 P [ B i ] . (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 = A c , we have P [ S ] = P [ A ∪ A c ] = P [ A ] + P [ A c ] . (5) Since, Axiom 2 says P[ S ] = 1, P[ A c ] = 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 ) ∪ ( AB c ) , B = ( AB ) ∪ ( A c B ) . (6) Now we apply Theorem 1.3 to write P [ A ] = P [ AB ] + P [ AB c ] , P [ B ] = P [ AB ] + P [ A c B ] . (7) We can rewrite these facts as P[ AB c ] = P[ A ] − P[ AB ] , P[ A c B ] = 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 ) ∪ ( AB c ) ∪ ( A c B ) . (9) 14

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    Once again, using Theorem 1.3, we have P[ A ∪ B ] = P[ AB ] + P[ AB c ] + P[ A c B ] (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 ∪ ( A c B ). By Theorem 1.3 (which uses Axiom 3), P [ B ] = P [ A ] + P [ A c B ] . (12) By Axiom 1, P[ A c B ] ≥ 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 ] = 0 . 4 0 . 6 = 2 3 . (2) (b) The probability of one handoff is P[ H 1 ] = P[ H 1 B ] + P[ H 1 L ] = 0 . 2. The probability that a call with one handoff will be long is P [ L | H 1 ] = P [ H 1 L ] P [ H 1 ] = 0 . 1 0 . 2 = 1 2 . (3) (c) The probability a call is long is P[ L ] = 1 − P[ B ] = 0 . 4. The probability that a long call will have one or more handoffs is P [ H 1 ∪ H 2 | L ] = P [ H 1 L ∪ H 2 L ] P [ L ] = P [ H 1 L ] + P [ H 2 L ] P [ L ] = 0 . 1 + 0 . 2 0 . 4 = 3 4 . (4) 15

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