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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,8.1.1 8.1.3 8.2.2 8.3.2 8.3.3 and 9.1.2 5 © σ2 n. Realizing that σ2
X
X
£ ¤ σ2
X
9
£ ¦ Var M9 X 25, we obtain
25
9 £ ¥
£¦
¡¡
¢¢¡ § x £ ¨ (a) Using Theorem 8.1, σ2 n
M £ Problem 8.1.1
Recall that X1 X2 Xn are independent exponential random variables with mean value µX
so that for x 0, FX x
1 e x 5. ¥ (b)
e 75 X9 !¡¡¡
#¢¢"! P X1 0 247 63 ¦ ¥ ¡¡ ¢¢¡ 1 75 X9 . 7 e § § ¥ £ £ ¦ ¥ § 7 can be approximated using the Central Limit Theorem
9µX
9σX 1
£ § 63 Φ § & § Φ65 0 1151.
¡
$¦ !¡¡¡
%¢¢"! ¤ £ ¥ £ ¦ ¥ ¥ ¥ § £ ¦ ¥ ¦ ¥ £ Problem 8.1.3
X1 X2 Xn are independent uniform random variables with mean value µX 7 and σ2
X
£ ¦ 7 1 ¥
¦ § Consulting with Table 4.1 yields P M9 X 63 X9 ¦ P X1 £¦ 1 § 7 P M9 X ¥ 1 1
§ 7 1 ¨ 7 in terms of X1 Now the probability that M9 X
(CLT).
P M9 X FX 7 © P M9 X 1 £¦ (c) First we express P M9 X 7 ¨ P X1 © 1 7 ¡ P X1 3 ¡¡
¢¢¡ (a) Since X1 is a uniform random variable, it must have a uniform PDF over an interval a b . From
a b 2 and that Var X
b a 2 12. Hence, given
Appendix A, we can look up that µX
the mean and variance, we obtain the following equations for a and b.
¦ £ ¦ ! ¥ ! £
£ £ ¥
£ '¦ Var X
16 ¦
£¦ §
£ ¥ ¥ (b) From Theorem 8.1, we know that
£ 1 7 10 from which we can state the distribution of 1 6 4 x 10
0
otherwise Var M16 X £ fX x b2 ¥ 4 and b a § 3 ¦ 12 Solving these equations yields a
X. 2 a b 3
16 ¥ (c)
∞ 10 1 6 dx
¥ (
)£ 9 ¦ ¥ ¤ (
)£ 9 ¦ fX1 x dx 16
£ 9 P X1
(d) The variance of M16 X is much less than Var X1 . Hence, the PDF of M16 X should be much
9 to
more concentrated about E X than the PDF of X1 . Thus we should expect P M16 X
be much less than P X1 9 .
¦ ¦ ¥ ¦ ¥ 1 P X1
§ 9 ¥ £ ¦ P M16 X 1 Φ 2 66 X16 !000
#¢¢"! ¥ § 1 144 ¦ 9 ¥ £ ¦ P M16 X
¥ By a Central Limit Theorem approximation,
£ § & 0 0039 9.
§ P X1 § 9 16µX
16σX ¥ 144 Φ ¡ 1
2 ¦ ¦ ¥ As we predicted, P M16 X 1 £¦ 9 ¡ P M16 X
¥ Problem 8.2.2
We know from the Chebyshev inequality that
¤ 3
5 § 3
4 σ2
X
c2 c kσX , we obtain
3
% EX 1
k2 kσ
¤ § PX £ Choosing c EX PX 3
6 Problem 8.3.2
are iid random variables each with mean 75 and standard deviation 15.
X1 X2
¡¡
¢¢¡ (a) We would like to ﬁnd the value of n such that
76 0 99 £ ¦ Mn X ¡ ¥ P 74
When we know only the mean and variance of Xi , our only real tool is the Chebyshev inequality
which says that
¤ ¤ 1 1 Var X
n
§ 3
% EX §
7¦ P Mn X £ 3
6 ¥ § 1 1
§ £ ¦ ¥ ¤ This yields n 76 225
n
¤ Mn X 0 99
¡ P 74 22500. (b) If each Xi is a Gaussian, the sample mean, Mn X will also be Gaussian with mean and variance
¦ £
£ ¦
£ ¦ Var X n
2 75 £ ¥8 Var Mn X EX 225 n
¥ E Mn X
¥8 In this case, 0 99
£ 1 n 15
¡ §
7¦ £ §
7¦ £ & 9 ¥ & n 15 §¥ Φ σ
& n 15 µ 9 2Φ σ 74 Φ ¦ £ Φ µ
§ 76 Φ 9 ¥ 76 § ¦ ¥8 9 Thus, n Mn X § P 74 1521. Since even under the Gaussian assumption, the number of samples n is so large that even if the Xi
are not Gaussian, the sample mean may be approximated by a Gaussian. Hence, about 1500 samples
probably is about right. However, in the absence of any information about the PDF of Xi beyond
the mean and variance, we cannot make any guarantees stronger than that given by the Chebyshev
inequality.
9 Problem 8.3.3
Both questions can be answered using the following equation from Example 8.8:
PA 1 PA
nc2 3
% § 3
4 £ 9900
n £
¡ £ ¨ £ £ @
¦ ¡ 99 ¨ 0 01 or n
£ 3
4 ¡ § £ ¥ 107 n 99
¡ § ¤ 3
5 § 3
4 A The conﬁdence level 0 01 is met if 9 9 0 0099
n10 10 A 3
5 PA 1 PA
nc2 c £ ¡ PA 10 5 . This implies
¡ 3
4 P Rn 10 3 P A A § (b) In this case, we meet the requirement by choosing c 0 01. This requires n
¨ 3
% Thus to have conﬁdence level 0 01, we require that 9900 n
¤ 0 001 ¡ PA 0 001 yielding
¡ ¤ c (a) In this part, we meet the requirement by choosing c
P Rn 990 000. 0 0099
nc2
¡ PA ¤ P Rn 0 01, we ¤ The unusual part of this problem is that we are given the true value of P A . Since P A
can write ¡ ¥ c § PA ¦
@ P Rn 107
n 109 . ¡ ¡ Problem 9.1.2
(a) We wish to develop a hypothesis test of the form
c £ EK 0 05
¡ 3
C5 § PK
3
B to determine if the coin we’ve been ﬂipping is indeed a fair one. We would like to ﬁnd the
value of c, which will determine the upper and lower limits on how many heads we can get 3 away from the expected number out of 100 ﬂips and still accept our hypothesis. Under our fair
coin hypothesis, the expected number of heads, and the standard deviation of the process are
EK 50 £ 100 1 2 1 2 5
£ 0 £ 0 σK Now in order to ﬁnd c we make use of the central limit theorem and divide the above inequality
through by σK to arrive at
EK
σk c
σK 0 05
£ 3
5
§ K ¡ 3 P
Taking the complement, we get EK
σk c
σK K 0 95
£ § c
σK ¡ § P Using the Central Limit Theorem we can write
c
σK 1
§ 2Φ 0 95
£ c
σK Φ ¡ § £ c
σK
§ Φ This implies Φ c σK
0 975 or c 5 1 96. That is, c 9 8 ﬂips. So we see that if we
observe more then 50 10 60 or less then 50 10 40 heads, then with signiﬁcance level
α 0 05 we should reject the hypothesis that the coin is fair.
£ ¡ £ ¡ £ £ § £¦ ¡ ¥ ! ¡ (b) Now we wish to develop a test of the form
c 0 01 £ ¡ PK
Thus we need to ﬁnd the value of c that makes the above probability true. This value will tell
us that if we observe more than c heads, then with signiﬁcance level α 0 01, we should reject
the hypothesis that the coin is fair. To ﬁnd this value of c we look to evaluate the CDF
£ 0
01
100
£
k
k
k 100 D ¦ 12 100
0i ¡¡ ¢¢¡ 0
∑k
i
1 ¡ ¥ £¦ FK k 100 E ¥ Computation reveals that c 62 ﬂips. So if we observe 62 or greater heads, then with a signiﬁcance level of 0.01 we should reject the fair coin hypothesis. Another way to obtain this
result is to use a Central Limit Theorem approximation. First, we express our rejection region
in terms of a zero mean, unit variance random variable.
EK
σK
§ K c P § 1 § c EK
σK
£ § 0 01 5, the CLT approximation is
c ¡ 4 £ £ From Table 4.1, we have c 50 5 2 35 or c
the hypothesis if we observe 62 or more heads. 50
5 0 01 61 75. Once again, we see that we reject
¡ Φ
§ 1 § c £ PK ¡ £ £ 50 and σK PK 1 £ Since E K c ¡ PK £ ¦ § ¥ ...
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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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