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Unformatted text preview: Probability and Stochastic Processes:
A Friendly Introduction for Electrical and Computer Engineers
Roy D. Yates and David J. Goodman
Problem Solutions : Yates and Goodman,3.1.2 3.1.4 3.2.3 3.2.5 3.3.3 3.3.5 3.4.1 and 3.4.6
Problem 3.1.2
On the X Y plane, the joint PMF is y £ c ¥ ¨
¨
¨ 3c 2c 2c 3c ¨ ¨ 1 ¡ 1
c ¢ § c ¨ ¦ PX Y x y 2 x c ¨ ¨
¤ (a) To ﬁnd c, we sum the PMF over all possible values of X and Y . We choose c so the sum equals
one. ∑ cx 101 #
$ 1 PX Y 2 ¦ 1 #
$ 1 2 0 2y ¥ ¥ ¥ ¥
©
§ x y 6c ©
∑ xy ¦ ¥ 2c 6c 14c 1 14. © Thus c y x © ∑ ∑ PX Y (b) #
$ 12 © ¥ 7c
&§ © 3c ¥
%§ ¦ 2c ¦ ¥ c PX Y 2 1
%§ ©
"! c PX Y 2 0 ¥ PX Y 0 ¦ X § PY © (c) ©
#¥
¦ 7c 12
#¥
¦ ©
!
©
¦ ' (d) From the sketch of PX Y x y given above, P X PX Y Y 0. ©
"! ©
%§ c
&§ c 20 #¥
¦ 2c PX Y 21
%§ 3c 2 PX Y 0 1 ¥ PX Y ¦ X § PY § © ¥ (e)
21
)§ #¥
¦
%§ 1 PX Y PX Y 0 ¥ 20 ¦ #¥
¦ PX Y #
$
%§ #
$ ©
#¥
¦ 8 14 1 1
%§ 8c 2 PX Y 0 1 ¥ PX Y ¦ 1 § ©
©
(! PX Problem 3.1.4
hh ht th tt and each sample point has probability 1 4. Each samThe sample space is the set S
ple outcome speciﬁes the values of X and Y as given in the following table 2 0
1© outcome
hh
ht
th
tt XY
01
10
11
20 The joint PMF can represented by the table
0y 1
0
14
14
14
14
0
¦ §
¥ ©
©
© Problem 3.2.3
We recognize that the given joint PMF is written as the product of two marginal PMFs PN n
and PK k where ¦ # ©
7§ ©
3§ ¥ ©
3§ ¦ ¦ ¦ 0 100 k § ¦ ∑ PN K n k p k 01
otherwise © ¦ ¥ n0 pk 1 ¦ ©
4§ 100
k nk n 01
otherwise 666
)$$$ 5 ∑ PN K 0 © ¦ PK k k0
∞ 100 100n e
n! ∑ PN K n k 666
$$$ § 100 PN n Problem 3.2.5
N is 1 ∞ 100n 1 e 100
∑ n 1!
100 n k 8 1 § ¦ ¦ k 100
9! ' ©
© w. That is, x ∑
©
"! # 2 x ∞ PX Y x w ¥ w # xY ¦ ∞ © PX x § © © PN § 1∞
∑ PN n
100 n k 666
$$$ ¦ © © ∞ § ©
4§ ∑ 100 100n e 100
∑
n k n 1! ¦
x 100n e
n! ©
4§ ©
7§ ¦ PW w ∞ ¥ x in order for W § ¦ ©
7§ ∞ PK k # Problem 3.3.3
have Y w ¦ , the marginal PMF of K is 01 100n e 100
∑
k 0 n 1! © For k k n PN n 100 § y © xy
0
1
2 © PX Y
x
x
x © Problem 3.3.5 A
GE
"FD The x y pairs with nonzero probability are shown in the
ﬁgure at right. From the ﬁgure, we observe that for w
01
10, © §#
! ' '
7§ wY 2 w 2 1 2w w
10 # ¦ # # 6 ¦ 0 01 21 PW w 2 !§ ' 0 01 10 1 # w §
#§ #
¦
#! © PW 10, 6 © 6 ¦ © ©
7§ PW w 1 ' ¦
' ' To ﬁnd the PMF of W , we observe that for w ! w 666
)$$$ ©
©
(! 0 01 10 w B PX w @ ¦ P min X Y ! 666
)$$$ w C PW © The complete expression for the PMF of W is
w 12
otherwise § 2w © # ¦ 0 01 21
0 666
)$$$ 6 ©
7§ ¦ PW w 10 Problem 3.4.1
In Problem 3.2.1, we found that the joint PMF of X and Y is
y ¦ £ PX Y x y § ¥ 4 3 28 6 28 12 28 1 28 2 28 4 28 ¨H ¨H ¨H 3
2 ¨H ¨H 0
0 1 2 3 4 ¢ ¨H 1 x The expected values and variances were found to be
EX 3 Var X 10 7 EY 52 Var Y 34 ©
"!
©
I! 3 ©(!
©
"! We will need these results in the solution to this problem. Y X has expected value y
PX Y x y
13x 36
2 28 14
4 28 ©
"! ¥ ¥¥ 11
33
12
1 28 1 28 2 28
30 28 15 14 3 12
4 28 ¥ x 1 2 4y ¦ ∑∑ EY X § © © (a) Random variable W © ©
(b) The correlation of X and Y is xyPX Y x y 236
28 PP 212
28 PP 123
28 414
28 PP
PP
PP
¥ ¥¥ 111
28
207 28 ©
"! x 1 2 4y 1 3 4 3 12
28 PP ¥ ∑∑ ¦ E XY § rX Y ©¥ ©
© (c) The covariance of X and Y is
EXEY ©
(! ! #! E XY # σX Y 3
28 ©¥ (d) The correlation coefﬁcient is ¦ © 1 16 ¨¨ H W4
V4 ¦ PX Y x y § 1 12 1 16 ¥ £ W3
V3 W3
V4 18 1 12 1 16 W2
V2 W2
V3 W2
V4 ¨¨ H ¨H ¨¨ H ¨¨H 2 ¨H 3 14 18 1 12 1 16 W1
V1 W1
V2 W1
V3 W1
V4 1 2 3 4 ¨H H¨ 0
0 4 ¢ ¨H 1 ¨H 4 x § ©¥ y © min X Y and V ¦ Q !
!
! Problem 3.4.6
problem is for each possible pair x y, we identify the values of W max X Y . § © Var X Var Y 3
2 210 # Cov X Y ρX Y We observe that W Y and V X . This is a consequence of the statement of the underlying experx are equally likely. Hence Y X
iment in which given X x, each value of Y in the set 1 2
and this implies min X Y
Y and max X Y
X . Using the results from Problem 3.4.4, we have
the following answers. © 2 %$$$ 0 666 R © ¦ ©
7§ © ©
7§ ¦ (a) The expected values are
74 EV EX 52 ©
"! ©
(! EY ©
(! EW ©
(! (b) The variances are
41 48 Var V Var X ©
"! ©
! Var Y ©
(! ©
(! (c) The correlation is
E XY ©
"! ©
(! E WV rX Y 5 ©¥ rW V ©¥ (d) The covariance of W and V is
Cov X Y 10 16 ©
"! Cov W V ©
"! (e) The correlation coefﬁcient is § ¦ § ©¥
¦
5 10 16
41 48 5 4 S ρX Y 0 605 6 ρW V 54 Var W ©¥ ...
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This note was uploaded on 02/11/2012 for the course EEE 352 taught by Professor Ferry during the Spring '08 term at ASU.
 Spring '08
 Ferry

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