HW3_Spring11_sol

HW3_Spring11_sol - ELEC210 Spring 2011 Homework-3-Solutions...

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Unformatted text preview: ELEC210 Spring 2011 Homework-3-Solutions 1. Let be the maximum of the number of heads obtained when Carlos and Michael each flip a fair coin twice. a) Describe the underlying space of this random experiment and specify the probabilities of its elementary events. b) Show the mapping from to , the range of . c) Find the probabilities for the various values of . d) Compare the pmf of with the pmf of , the number of heads in two tosses of a fair coin. Explain the difference. e) Suppose that Carlos uses a coin with probabilities of heads = 3/4. Find the pmf of . Solutions(20’) a) space Michael , Carlos 1 (p=1/2) 2 (p=1/4) 0 (p=1/4) 00 (p=1/16) 01 (p=1/8) 02 (p=1/16) 1 (p=1/2) 10 (p=1/8) 11 (p=1/4) 12 (p=1/8) 2 (p=1/4) b) 0 (p=1/4) 20 (p=1/16) 21 (p=1/8) 22 (p=1/16) to Maximum 0 1 2 0 0 1 2 1 1 1 2 2 2 2 2 c) Probabilities for the various values of 0 2 1/16 1/2 7/16 0 P( ) 1 1 2 1/4 1/2 1/4 d) pmf of P( ) The mas function shifts the pmf to the larger number of heads. e) Unfair coin Michael , Carlos 0 (p=1/16) 1 (p=6/16) 2 (p=9/16) 0 (p=1/4) 00 (p=1/64) 01 (p=6/64) 02 (p=9/64) 1 (p=1/2) 10 (p=1/32) 11 (p=6/32) 12 (p=9/32) 2 (p=1/4) 20 (p=1/64) 21 (p=6/64) 22 (p=9/64) Probabilities for the various values of 0 1 2 P( ) 1/64 20/64 43/64 2. A modem transmits a +2 voltage signal into a channel. The channel adds to this signal a noise term that is drawn from the set {0, −1, −2, −3} with respective probabilities { , a) b) c) d) , , }. Find the pmf of the output of the channel. What is the probability that the output of the channel is equal to the input of the channel? What is the probability that the output of the channel is positive? Find the expected value and variance of the output signal. Solutions (16’) a) Pmf of the output +2 P( ) +1 0 -1 4 10 3 10 2 10 1 10 b) This event is equally probable to the noise is 0. The probability is c) P(Y = +2 ∪ Y = +1) = d) E(Y) = 2 × 2 E(Y ) = 2 × +1× + +0× . = −1× =1 4 3 2 1 + 12 × + 02 × + (−1)2 × =2 10 10 10 10 D(Y) = E(Y ) − ( ) = 1 3. Consider an information source that produces quaternary information that we designate as = {1,2,3,4}. Find and plot the pmf in the following cases: a) = for all k in . b) = for = 2,3,4. c) = for = 2,3,4. d) Find the conditional pmf for the source in parts a, b, and c given that < 4. Solutions(16’) a) 1=p +p +p +p =p p =p × 1 6 =, 2 25 111 12 1+ + + −→ p = 234 25 p =p × 1 4 =, 3 25 p =p × 1 3 = 4 25 b) 1=p +p +p +p =p 111 8 1+ + + −→ p = 248 15 p =p × 1 4 =, 2 15 p =p × 1 2 =, 4 15 p =p × 1 1 = 8 15 c) 111 64 1+ + + −→ p = 2 8 64 105 1=p +p +p +p =p p =p × 1 32 = , 2 105 p =p × 1 8 = , 8 105 p =p × 1 1 = 64 105 d) , P(X = k|X < 4) = ( ) ) ( ) ( = ) / = ( ) / = =1 , = =3 =2 , b) ) ( =2 , ( , , a) P(X = k|X < 4) = =1 =3 , P(X = k|X < 4) = ( ) ( ) = ( ) / = , =2 , b) =1 =3 4. An urn contains nine $1 bills and one $20 bill. Let the random variable be the total amount that results when two bills are drawn from the urn without replacement. a) Describe the underlying space of this random experiment and specify the probabilities of its elementary events. b) Find the conditional pmf for X given that the first drawn produced dollars. c) Find the conditional expected value corresponding to part b. d) Find [ ] using results from part c. e) Find [ ] and [ ] using the approach in parts c and d. Solutions (20’) a) X= {2,21} X P(X) b) 2 ( = 2) = 21 9 8 72 4 ×= = 10 9 90 5 The probability of the first drawn, P(1st drawn = k) = 8 P(X = 2|1st drawn k = 1) = , 9 (X = 2|1st drawn k = 20) = 0, P c) ( = 21) = E(X|k = 1) = 2 × + 21 × = , 19911 ×+ ×= 10 9 10 9 5 , =1 , = 20 ( = 21|1 ( = 21|1 E(X|k = 20) = 21 1 9 = 1) = 1 = 1) = E(X |k = 1) = 4 × + 441 × = E(X |k = 20) = 441 , d) E(X) = E(X|k = 1) × p(k = 1) + E( |k = 20) × p(k = 20) = e) E(X ) = E(X |k = 1) × p(k = 1) + E(X |k = 20) × p(k = 20) = × + = = × + = D(X) = E(X ) − ( ) = 57.76 5. Let be the binomial random variable. a) Show that ( ) () b) = ℎ (0) = (1 − ) . Show that part a implies that: (1) [ = ] is maximum at x denotes the largest integer that is smaller than or equal to ( + 1) is an integer, then the maximum is achieved at = ( + 1) , where ; and (2) when and − 1. Solutions (10’) (0) = (1 − ) −→ a) (1 − ) P(X = k) = ( ) () ( − 1)( − 2) … ( − + 1) ( − 1)( − 2) … 2 × 1 (1 − ) − +11− ( + 1) = () b) = = ≥ 1 means the probability P(X = k) increases as k increases. = ( + 1) . ( + 1) Therefore, is an integer then ( ) () P(X = k + 1) = P(X = k). Therefore, the maximum is achieved at − 1. = 1, which means and 6. A Christmas fruitcake has Poisson-distributed independent numbers of sultana raisins, iridescent red cherry bits, and radioactive green cherry bits with respective averages 48, 24, and 12 bits per cake. Suppose you politely accept 1/12 of a slice of the cake. a) What is the probability that you get lucky and get no green cherry bits in your slice? b) What is the probability that you get really lucky and get no green cherry bits and two or fewer red cherry bits in your slice? c) What is the probability that you get extremely lucky and get no green or red cherry bits but more than five raisins in your slice? Solutions (18’) Poisson distribution with parameter λ = 48, λ = 24, λ = 12 a) P N = 0 = ( ) ! = = 0.368 b) P N = 0 × P(N ≤ 2) = 0.368 × = 0.368 × ( c) P N = 0, N = 0, N ≥ 5 = + ! +2 ∑ +2 () ! + ! ! ) = 0.249 = ∑ ! = 0.0107 ...
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This note was uploaded on 10/16/2011 for the course ELEC 308,315,10 taught by Professor Prof.shenghuisong during the Spring '11 term at CUHK.

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