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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,4.5.3 4.5.6 4.5.7 4.7.3 4.7.5 4.7.8 4.7.11 4.7.13 and
4.7.16
Problem 4.5.3
X is a Gaussian random variable with zero mean but unknown variance. We do know, however,
that
01 ¡
¢ ¥ FX 10 © FX 10 £ 10 10
σX 2Φ ¦
© ¦
¥ £ X 1 0 55. Using Table 4.1 for the Gaussian CDF, we ﬁnd that 10 σX
¨ ¦
This implies Φ 10 σX
or σX 66 6. 10
¢© P function. 0 15
¦ ¦
§¥ We can ﬁnd the variance Var X by expanding the above probability in terms of the Φ ¨ £
¤¡ 10 ¨ PX © ¨ ¦ Problem 4.5.6
We are given that there are 100 000 000 men in the United States and 23 000 of them are at least
7 feet tall, and the heights of U.S men are independent Gaussian random variables with mean 5 10 .
(a) Let H denote the height in inches of a U.S male. To ﬁnd σX , we look at the fact that the probability that P H 84 is the number of men who are at least 7 feet tall divided by the total
number of men (the frequency interpretation of probability). Since we measure H in inches,
we have ¥ Qx,
¨ ¦
! © ¦
!
© © 3 5 or σX
¦ ¦
! ¨ ¦ From Table 4.2, this implies 14 σX 2 3 10
© Q 14 σX 4
" ¦
§¥ Φx ¦ 1 0 00023
¦ x 70 84
σX Φ ¨ Since Φ 23 000
100 000 000 84 PH 4. (b) The probability that a randomly chosen man is at least 8 feet tall is
Q 65
© ¦ 70
4 ¨ 96 Q 96 ¦
§¥ PH Unfortunately, Table 4.2 doesn’t include Q 6 5 , although it should be apparent that the probability is very small. In fact, Q 6 5
4 0 10 11 .
" ¨ © # 1 ¨ ¦
¨ © (c) First we need to ﬁnd the probability that a man is at least 7’6”.
β
2 87 10 7 .
¨ ¦
© ¦ ¦
¥ Although Table 4.2 stops at Q 4 99 , if you’re curious, the exact value is Q 5 7 " 3 10 $
! Q5 70
4 ¦ 90 Q 90 " PH © ¨ © Now we can begin to ﬁnd the probability that no man is at least 7’6”. This can be modeled as
100,000,000 repetitions of a Bernoulli trial with parameter 1 β. The probability that no man
is at least 7’6” is
100 000 000 94 10
# β 14
" % ¨ ¦ % 1
© (d) The expected value of N is just the number of trials multiplied by the probability that a man is
at least 7’6”.
100 000 000 β 30
¦ EN ¦
&¥ Problem 4.5.7
N µ σ2 distribution, the integral we wish to evaluate is ¥ ' dw dw σ and
¦ © e ' ∞ x2 2
" dx " '
(¦
"
∞ 3 ¦ 1
2π 2σ2 ) µ σ, we have dx
) ©
4¦ I 2 " w 0
1" (a) Using the substitution x ∞ wµ
" 2πσ2 e 2 ∞ ∞ 1 fW w dw 3 ∞ I ¦ (b) When we write I2 as the product of integrals, we use y to denote the other variable of integration
so that x2 y2 2
32 5 0
1" e dx 1
2π ∞
∞ e y2 2 dy " " '
"
' ∞ 3 dx dy " 0 2 e dθ
¦ 2π 2π 0 r2 2 r2 2
" '
' 0 e
' 0 ∞ 3 2π " ' 1
2π
1
2π
1
2π r2 and dx dy
3 6 ¦ "
¦ ¦ ¦ I2 y2
¦ ¦ (c) By changing to polar coordinates, x2 1 r dr d θ so that r dr d θ ∞
0 ¦ )
' ∞ x2 2 ) ∞ ∞
∞ e ' 1
2π ∞ " 1
2π 3 I2 dθ Problem 4.7.3
X2 0. We can conclude that FY y
© PX
¦ dx 1
¦ 9x 0,
" 9e 8
9' © ¦
! 0 y 9y e
" ¦
! ¦ y FY y 0, 8 £ From the exponential PDF of X , we see that for y 0. For y ¥ y £ P X2 ) FY y 0 for y ¦
! 0 we see that Y 7 (a) Since X © Hence, the CDF of Y is
0
0 8 e y
y 9y
" 0
1 7 ¦
! FY y
© By taking the derivative of the CDF, we obtain the PDF
9y
"
8 92e
0 ) © ¦
! fY y yy 0
otherwise © (b) Finding the moments of Y is easiest using the moments of X . Speciﬁcally, we know from Probn! 9n . Hence, E Y
E X2
2 81.
lem 4.4.10 that E X n ¦
§¥ 24 94 4 94
2 ¦ ¦
@¥ EY 20 6561
¦ ¦ ©
U.
¦
§¥
ln 1 4! 94 . This implies
¦
§¥ © ¦ Problem 4.7.5
have a sketch of the function X E Y2 ¦ Var Y E X4 (c) The second moment is E Y 2 4
X 2
0
0 0.5
U 1 (a) From the sketch, we observe that X will be nonnegative. Hence FX x
0 for x 0. Since U
has a uniform distribution on 0 1 , for 0 u 1, P U u
u. We use this fact to ﬁnd the
CDF of X . For x 0, £ £ e x
" U PU
¦ P1 £ ¦
¥ © 1 and so
x 1
¦ " e FU 1 e ¦
£ ¦
" © £ © The complete CDF can be written as
¦
B 3 e x
" 0
1 x
x
7 FX x 0
0 x
" FX x £ x 1 U e x
" ¥ x £ e ¦
§¥ 1 © 0, 0 ln 1 ¦
A For x P 7 FX x © (b) By taking the derivative, the PDF is
x e
0 x0
otherwise " ¦
B fX x
© Thus, X has an exponential PDF. In fact, since most computer languages provide uniform 0 1
random numbers, the procedure outlined in this problem provides a way to generate exponential random variables from uniform random variables.
¦
1¥ 1, E X ¥ ¦ (c) Since X is an exponential random variable with parameter a 1. Problem 4.7.8
Let X denote the position of the pointer and Y denote the area within the arc deﬁned by the stopping
position of the pointer.
(a) If the disc has radius r, then the area of the disc is πr2 . Since the circumference of the disc
is 1 and X is measured around the circumference, Y πr2 X . For example, when X 1, the
shaded area is the whole disc and Y πr2 . Similarly, if X 1 2, then Y πr2 2 is half the
area of the disc. Since the disc has circumference 1, r 1 2π and
¦ ¦ ¦ ¦ ¦ ©
C ¦ X
4π
¦ πr2 X ¦ Y £ y PX
¦ X
4π P 4πy
£ £ y ¦
§¥ ¦
§¥ PY ¦
! © Therefore the CDF is
0
y
£ 7 0
y
4πy 0
1
y £ ¦
! FY y 1
4π 1
4π © (c) By taking the derivative of the CDF, the PDF of Y is
4π 0 y 41
π
0 otherwise
£ 1 8π .
¦ 4πy dy ©
C 2 1 4π
0
0 © 3 D
(§¥
¦ Problem 4.7.11
The PDF of U is
£ £ 12
1u1
0
otherwise
¦
! fU u 4 ¦
B (d) The expected value of Y is E Y £ fY y FX 4πy
© FY y (b) The CDF of Y can be expressed as © 0 for w 0 fU u du
© 0 ¦
§¥
' 1 PU 12
¦ 7 © 0 ¦
E ¦
¥ PW 0. Next, we observe that the rectiﬁer output W is a
7 Since W 0, we see that FW w
mixed random variable since
¦ " The above facts imply that 0 12 £ ¦
¥ ¦
! © 1,
w fU u du
© ' 1 w ©
4¦ £ w 12
6 PU
¦
§¥ ¦
7 7 FW w
© " w
120
w
£ 0
w
1 0
w
1
7 6
G £ © © ¦
FW w 1 for w 1. Hence, the complete expression £ £ 1, which implies FW w ¦
F Finally, U 1 implies W
for the CDF is PW
¦ w 0 ¦
¥ Next, we note that for 0 PW FW 0 1 © By taking the derivative of the CDF, we ﬁnd the PDF of W ; however, we must keep in mind that the
discontinuity in the CDF at w 0 yields a corresponding impulse in the PDF.
¦ £ 120 w 1
otherwise
6
δw
0 £ © © ¦
fW w
© From the PDF, we can calculate the expected value
© 6
' 0 w 2 dw
0 ¦ 6
© © 0 1 1 2 dw 14
¦ wδw 1
¦
§¥
' EW Perhaps an easier way to ﬁnd the expected value is to use Theorem 2.10. In this case,
∞ 1 u 1 2 du
© ' 0 14
¦ ¦ © © ' ∞ g u fW w du ¦
§¥ EW " As we expect, both approaches give the same answer.
Problem 4.7.13
shown in the following ﬁgure:
3
Y 2
1
0
0 1 2 3 X 1
© ' ¦
§¥ 5 0 1 fX x dx
¦ 1 ' £ X 0 © £ P0 x 2 dx
12 ¦
F¥ ¦ PY 1. Thus,
¦ X 14
£ 1 2 if and only if 0 £ ¦ (a) Note that Y 12 ¦
© 7 © ¦
B
£ £ 7 12
PY ¦
¥ ¦ 1 14
¦
¥ £ P0 ¦
© £ 2,
© ' £ ¦
¥ y2 4 2. The complete expression of the CDF is
y 12
12 y 1
1y2
y2
0
14
y2 4
1 £ 7 £ 7 7 ¦
! © ¦
! ¦
! £ © FY y fX x dx 0 1 for y y y ¦ 2, FY y PX 7 FY y
Lastly, since Y PY
¦ y 1 2. Also, FY 1 2 FY y
Next, for 1 X 0 for y ¦
§¥ (b) Since Y 1 2, we can conclude that FY y
1 4. Similarly, for 1 2 y 1, © Problem 4.7.16
First, we must verify that F 1 u is a nondecreasing function. To show this, suppose that for u u ,
x F 1 u and x
F 1 u . In this case, u F x and u
F x . Since F x is nondecreasing,
F x implies that x x . Hence, we can write
Fx
© © ¦ © © ¦ " © £ ¦ 6 PU Fx
© x ¦¥
I £
U Fx
© © 1 " PF ¦
" ¦
H FX x
© © © ¦ " © ...
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 Spring '08
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