Unformatted text preview: ELEC210 Spring 2011 Homework3Solutions 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 = kX < 4) = ( ) )
( ) ( = )
/ = ( )
/ = =1 , = =3 =2 , b) )
( =2 , ( ,
, a) P(X = kX < 4) = =1 =3
, P(X = kX < 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 = 21st drawn k = 1) = ,
9
(X = 21st drawn k = 20) = 0,
P c) ( = 21) = E(Xk = 1) = 2 × + 21 × = , 19911
×+
×=
10 9 10 9 5
, =1 , = 20 ( = 211
( = 211
E(Xk = 20) = 21 1
9
= 1) = 1 = 1) = E(X k = 1) = 4 × + 441 × = E(X k = 20) = 441 , d) E(X) = E(Xk = 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 Poissondistributed 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.
 Spring '11
 Prof.ShenghuiSong

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