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Unformatted text preview: 7.8 Let ten intervals be deﬁned each from {1017—9) to [1017} where 1'. = 1,2 ..... 10. By counting the numbers
that fall within each interval and mmparing this to the expected value for eaeh interval. E, = 10, the
following table is generated: Interval 01 (01 — Ei)2/E§ (01 10} 9 0.1
(11 20} 9 0.1
(21 30} 9 0.1
(31 40} 5 1.6
(41 so} 17 4.9
(51 60} 5 2.5
(61 70} 10 0.0
{7180} 12 0.4
(81 90} 7 0.9
('91 00} 16 3.6 ﬁ—
100 142— kg
From Table A.6. £053 = 16.9. Since 13 < 1353, then the null hypothesis of no (hiferenee between the
sample distribution and the uniform distribution is not. rejected. 7.16 X“, = 109. X21, = 300. X10 = 500
The generator is ){lnjIl. = 157 X1“; Illud 32363
git2.34.1 = 116 X2“; 111.011 31.727
X3.j+1 = 112 X3“; 1110(1 31.657
Xj+1 = (X1.j+1 — 3(2411 + {Y3J+1) 1119(132362
R3“ = { é—‘ggﬁ . if, J+1 2: o
33% = 0.999 ,ifXJ_,.1 = 0 The first 5 rand3111 nlunbers: are XL]. = [157 x 100] [laud 32363 15700 X11 = [146 x 300] 1nod 31727 12073 3711 = [142 x 500] 113°C]. 31557 = 7686 X1 = [15700 — 12073 + 7686] luoel 32362 = 1.1313
R1 = 11313/32363 = 0.3496 X13 = 5312
X232 = 17673
X32 = 15071
X2 = 2713
R2 = 0.0838
X13 = 24909
X23 = 10371
3733 = 19—189
X3 = 1665
R3 = 0.0515 X1“; = 27153
X2_4 = 22997 X3,4 = 13279 X4 = 17435
R4 = 05387 X195 = 23468
X1135 = 2622?
X3 5 = 17855 X3: 15096
R5 = 0.4665 7.20 Use Kolmogorov—Smimos test to compare the data stream with the uniform distribution Calculation for Kolmogoror—Smirnov Test 0007 0.055 0.097 0.127 0.182 0.227 0.262 0.351 0.442 0.474
0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5
0.043 0.045 0.053 0.073 0.068 0.073 0.088 0.049 0.008 0.026 12 1 3 14 1 5 1 6 17 18 19 20 0.507 0.515 0.549 0.788 0.797 0.798 0.825 0.852 0.928 0.929
0.55 0.6 0.65 0.7 0.75 0.8 0.85 0.9 0.95 0.043 0.085 0.101 0.088 0.047 0.002 0.025 0.048 0.022 0.071 Since 0.088 c: 0.29:1, the critical value when N = 20 and or = 0.05. the hypothesis that. the distribution
of the generated numbers is the uniform distribution is not rejected. Use ChiSquare test to check whether the data stream are uniformly.r distributed. Calculation for ChiSquare Iesl Mien310.25) 0i Ei OiEi (0iEi)"'2 (Oiin}"2.'Ei
1 6 5 1 1 0.2
2 4 5 l 1 0.2
3 3 5 2 l 0.8
4 l 2 3 4 LE
20 20 2 Since 2 <.' 7.81. the value of ran”, the null hypothesis of a uniform distribution is not rejected.
Please note it = 4 is used here because it is recommended that n and N be chosen so that each E; 2 5
Test autocorl‘elation Let i = 1 and m. = 3. Then :1! = 5 (largest integer such that 1+(M+1)'1 6. 20) 513 r+Linersinauro) + (0.055)(0.262) + (0.262)(0.412) +
._) (0.442)(0.227) + (0.227)(0.325) + (usesxnaeen — 0.25
41.047 _ _V13(5)+7 = 0.117351 a».
ma 12(51 1) _ﬁra_ 41047 _ . k y
20 _ or?” _ (1117351 — ‘0 '1 : (—41.0259 ~o.o25) Do not reject the null hypothesis of independence. 8.4 Triangular distribution with a. = 1. c = 10 and E(X) = 4. Since (a + b + c) f3 = ELY). the mode is at
b = 1. Thus. the height of the triangular pdf is h = 230. (See solution to previous problem. Note that
the triangle here is a right triangle.) Step 1: Find cdf F(r) = total area from 1 to 3:. =1— (total area from .r' to 10). By similar triangles. mm = (10 _ r);'{10 _ 1), so HI]:1—(10—.r)f(r),r’2=1—(10—I)2/81. 1S :1: S 10.
Step 2: SetF(X)=Ron1<_IX S 10. Step 3: X = 10 — «31(1 — R). o <_: R 51 8.9 Use Inequality (8.14] to conclude that, for R given, X will assume the value .11 in R3 = {1.2.3. =1}
provided
(.7: — 1).r{2:r. — 1) .r(.r +1)(2r + 1)
180 ‘ R 3 130 By direct computation. F“) = 6,3180 = .03, 17(2) = 30,5130 = .167. F(3} = 42f130 = .233, Flil) = 1.
Thus. X can be generated by the table lookup procedure using the following table: F(.r _ 1) = = He) 1' 1 2 3 4
17(1) .033 .137 .233 1 31 = 0.33 — X = 4
R3 = 0.24 — X = 4
33 = 0.57 — X = 4 8.15 The mean is (1/3)) — 1 = 2.5, so p = 2;"? . By Equation (9.21),
X = [—2.971n(1 — R) — 1]
8.20 Method 1: Generate Xl ~ U{D,Sj and X2 ~ Ugo, 8). Set Y = min[X1,X2]l. Method 2: The cdf on is Flyl=P(YSttJ 1—P(Y}el
l—P{X1 TetJig > y) = 1(1—yf8)2,0£y38 by independence of X1 and X2. Fm = 1 — (1 — me)? = R implies
Y=8—8\/1—R, 0g R31. 8.21 Assume X, is exponentially distributed with mean 1,05, where 1M1 = 2 hours and 1,“; = 6 hours.
Method 1 is similar to that in Exercise 20. Method 2: The cdf of Y is
HI!) = P(Y S y) 1—P(Y } y)
1 PUG > y‘Xe : y} 1_ c—AweMi‘
1 _ c—(X1+.\a)y Therefore Y is exponential with parameter A1 + A2 = 132 +1j6 = 2,53.
Generate l" = —1.5h1R. Clearly, method 2 is twice as efﬁcient as method 1. ...
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 Winter '10
 LAMBADARIS

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