{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

homework3_solutions

# homework3_solutions - 7.8 Let ten intervals be deﬁned...

This preview shows pages 1–3. Sign up to view the full content.

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

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 {71-80} 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 1-3 < 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 ){lnj-I-l. = 157 X1“; Illud 32363 git-2.34.1 = 1-16 X2“; 111.011 31.727 X3.j+1 = 1-12 X3“; 1110(1 31.657 Xj+1 = (X1.j+1 — 3(24-1-1 + {Y3J+1) 1119(132362 R3“ = { é—‘ggﬁ . if, J+1 2: o 33% = 0.999 ,ifXJ-_,.1 = 0 The first 5 rand-3111 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 = 1507-1 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 = 0-5387 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 Chi-Square test to check whether the data stream are uniformly.r distributed. Calculation for Chi-Square Iesl Mien-310.25) 0i Ei Oi-Ei (0i-Ei)"'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+Liners-inauro) + (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(5-1- 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)(2-r + 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 look-up 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 Tet-Jig > 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—Awe-Mi‘ 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. ...
View Full Document

{[ snackBarMessage ]}

### Page1 / 3

homework3_solutions - 7.8 Let ten intervals be deﬁned...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online