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2 2 0 25 and P W 3 3 0 x 0 5 and 0 12 0 for all 0 55 and P X 3 36 pY 5 x 0 otherwise 4 36 pY 6 5 36 3 36 pY 11 2 36 0 otherwise 2 36 pW 4 1 36 pW 10 2 36 pW 20 3 36 pW 5 2 36 pW 12 2 36 pW 24 2 36 and pW 36 1 36, with pW 0 1 0 35 2 0 65 8 Appendix E: Answers to OddNumbered Exercises 731 2.3.17 (a) Hypergeometric 9 4 2 (b) Hypergeometric 9 5 2 2.3.19 P X 5 100 1000 5 5! exp 100 1000 2.4.1 (a) 0 (b) 0 (c) 0 (d) 2 3 (e) 2 3 (f) 1 (g) 2.4.3 (a) e 20 (b) 1 (c) e 12 (d) e 4 25 1 4 2.4.5 No 1 3 M3 2.4.7 c 2.4.9 2 2.4.11 Yes 1 f x dx 2 1 g x dx 2.4.13 P Y P X du 3 2 3 2 1 2 exp y 1 2 2 dy 2 2 1 2 exp u2 2 2.5.1 Properties (a) and (b) follow by inspection. Properties (c) and (d) follow since FX x 1 for x 1 6, and FX x 2.5.3 (a) No (b) Yes (c) Yes (d) No (e) Yes (f) Yes (g) No 0 for x 1. 2.5.5 Hence: (a) 0 933 (b) 0 00135 (c) 1 90 10 8 2.5.7 (a) 1 9 (b) 3 16 (c) 12 25 (d) 0 (e) 1 (f) 0 (g) 1 (h) 0 2.5.9 (b) No 2.5.11 (b) Yes 2.5.13 (b) The function F is nondecreasing, limx 1. (c) P X 1 4 4 5 0, P X 0 and limx 2 5 F x 5 12, P X 2e 16 25 3 (b) P 1 Z 5 36 (d) P Z 4 5 1 12 (e)
P Z 1 2 0 11 12 1 9 (f) P Z 2e 1 2 3 (c) 1 2 4 5 2.5.15 (a) P Z P Z 11 12 2.6.1 fY y equals 1 R 2 5 2e 1 2 3 L c for L y d c R and otherwise equals 0 c y1 3 2 for y 0 and otherwise equals 0 2.6.3 fY y 2.6.5 fY y equals 2.6.7 fY y 2.6.9 (a) fY y 2.6.11 fY y 2.6.13 fY y 2.7.1 e [y d c ]2 2c2 2 3 y 2 3e 1 6y1 2 for 0 y 8 (b) f Z z y 9 z7 2 for 0 y 1 2 sin y1 2 4 for 0 1 2 3 y 2 3 2 1 exp y y 2 3 2 2 z 2 and 0 otherwise FX Y x y 0 1 3 1 min[x y 0 min[x y min[x 2 4] y 2 4] 0 2 4] 1 1 2.7.3 (a) pX 2 otherwise (b) pY 3 0 otherwise (c) P Y pX 3 pX pX 3 pY 3 5 (d) P Y 2 2 pY X pX 17 1 5, with pX x 0 3 pY 19 0 (e) P XY 1 5, with pY y 0 0 pY 2 X 2.7.5 X x Y y X x and X x Y y Y y 732 Appendix E: Answers to OddNumbered Exercises c 1 cos 2x x for 0 x 1 and 0 otherwise (b) fY y 0 2 and 0 otherwise (b) fY y 5 48 1 2 otherwise, pX x 1 0 (b) 1 and f X x 0 otherwise (b) 0 otherwise (c) No 1 2 (c, 1 and 0 fY X y x = 0), thus, X and C y5 6 y 2 for fY y x 5 6 x 5 y5 x 2 (other C 500,000x 5 3 10, 0 (c) Yes x 1 and fY y 2 X 5 Y 2y 3 for 0 cos y 2.7.7 (a) f X x c 1 2.7.9 (a) f X x y
3 2.8.1 (a) pX pY 3 2.8.3 (a) f X x 48y2 fY y 2.8.5 (a) P Y y for 0 4 2 and 0 otherwise y 2x 3 8 for x 3x 2 0 2 and 0 otherwise (c) P Y 3y2 12 for y 2 1 4 pX 9 1 4 pX 13 2 3, pY 5 1 3 otherwise, pY y 40 49 for 0 18x 49 30 49 for 0 y 6y 4 X x y 4 9 2 0 (d) P Y 9 2 X fY X y x fY y x 5 y5 x 2 2x 2 y 1 (e) P X 4y5 2 3 (otherwiseb) P Y 5 2 3 fY y 2.8.7 (a) f X x x 2 4y5 fY X y x Y are not independent. (b) f X x 0 1, 1 and 0 wise, fY X y x = 0) X and Y are not independent. (c) f X x 50x x y independent. (d) f X x and 0 y are independent. 2.8.9 P X 3 1 4 2.8.11 If X IB1 C P Y 2.8.13 (a) C is constant, then P X B2 C 500,000x 5 3 and fY y C 2048y5 3 500,000x 5 3 10, fY B1 2 1 x 4 and 0 8y for 0 50x (otherwise, fY X y x = 0), thus, X and Y are not 4 3y5 500,000 (otherwise, fY X y x = 0), X and Y C 2048y5 3 for 0 fY IB1 C and P X B1 Y B2 pY b) y 1 4 1 4 7 Others 0 0 5 Others pX c) X and Y are independent. 2.8.15 fY.9.1 h1 u1 h2 u1 2 x 2 y cos 2 u2 sin 2 u2 3 x 2 y3 2.9.3 (b) h x y at least for z u1 2 log 1 u1 h2 u2 u1 2 log 1 u1 x 2 y2 x 2 0 and z y2 (c) h 1 z 0 (d) f Z W z 3x
2 4 3y2 for 0 2x 3 for x x 2, and 0 otherwise (b) y and 0 otherwise (c) Not independent y h1 u2 2 sin 2 u2 2 log 1 u1 2 cos 2 u2 2 log 1 u1 z z e 2 z 2 2 z2 2, 2 for Appendix E: Answers to OddNumbered Exercises 733 z 2 0 and 1 z 2 4, i.e., for z 4 and max z z 64 z 4, and 0 otherwise 0 and 1 y4 x 4 (c) h 1 z z1 4 4, i.e., for 1 12 pZ 5 1 18 pZ 4 1 4 z1 4 (d) f Z W z 0 and 1 z 1 18 pZ 7 e 256, and 0 otherwise 1 24 pZ 8 1 72 1 4 1 4 pZ 11 3 8 pZ 12 1 8 pZ z 0 otherwise 2.9.5 (b) h x y 1 4 for 2.9.7 pZ 2 pZ 9 2.9.9 (a) z z W P Z ( 8,16) 1 5 ( 7,19) 1 5 ( 3,11) 1 5 ( 2,14) 1 5 (0,6) 1 5 otherwise 0 8 7 1 5 for z 6 11 14 16 19, and otherwise pW 7 if U 1 2, Z 3 2 if 1 2 U 5 6, and Z 2 0, and otherwise pZ z 0 0 (c) pW 5 if U 5 6 Exponential 3 3 2 and c2 5 4 for t y U 1 4 8 7 (b) E X 1 (b) E Y 4Y 8 1 427 3 (f) E XY 12 Y 1 2, for x 0 (b) pZ z 1 5 for 2.10.1 Z 2.10.3 Y 2.10.5 c1 2.10.7 (a) For x FX x because F 1 X and F 1 t X 1 3, for 2 FY y F 1 2.10.9 Y Z U 3.1.1 (a) E X 3.1.3 (a) E X Y 2 3.1.5 E 8X 3.1.7 E XY 3.1.9 E X 3.1.11 (a) E Z 3.1.13 E Y 3.2.1 (a) C E X 3
.2.3 (a) E X E Y 4 3.2.5 E 3.2.7 E Y 3.2.9 Let 3.2.11 334 3.2.13 E Y 3.2.15 Yes 30 Z 6 k 7 (b) E W 7 4 1 4 E X 8645 2062 17 24 (b) E Y 216 7 (f) E X 2Y 3 6Y 5X 77 3 17 72 E X k 214 1 1 FX x 4, FX x. F 1 t X 1 2 1]. 4, FY y 0, for 1 x 2, FX x 1 3, for 2 4, 1 (b) The range of t must be restricted on 0 1] 1 3 1 2], t 1 for t 2, (c) For y y x 0 1 3], F 1 X 1, FY y 4, FY y 2 for t 0, for 1 1. 1 2, for y 19 (d) E Y 2 370 3 (e) E X 2 8 1 (c) E X 11 (c) E X 2 113 2 p p 12 49 4 7 (b) C 1 16 E X 169 24 (c) C 5 3093, 17 8 (c) E X 2 11 20 (d) E Y 2 99 20 (e) 27 4 then 1 39 25, 2 64 25, 3 152 35 3.3.1 (a) Cov X Y 3.3.3 Corr X Y 3.3.5 E XY 3.3.7 (a) Cov X Z 3.3.9 E X X 1 n n 1 3.3.11 E X 3.3.13 Cov Z W 3.3.15 Cov X Y 3.4.1 (a) rZ t Var Z 2, m Z t 2 734 Appendix E: Answers to OddNumbered Exercises 2 3 (b) Var X 0 18292 2, Var Y 32 9 (c) Corr b) Corr X Z E X 2 E X, when X 1 46 Binomial XY 329 12 Cov X Y 35 12 0 0 Corr Z W 35 24 t rZ t et 2 et m Z t 2 2 t 2 t 2 rZ t 4 t 2et 2 et 2 m Z t ese es 1, so mY s 2 2 2, Var Y 2 3 (b) E Z 2et 2 et, mY s 2, 2
et es mY 0 3 22 3 (c) E X Y 1 3 22 3 36 7 (c) E Y X 4. 2 2 25 4 1 3es e es 1, mY s 3.4.3 mY s 2e2s e es 1, so mY s e4sm X 3s 3.4.5 mY s e es 1 es 3.4.7 mY s 3 3.5.1 (ad) E Y X 3.5.3 (a) E Y X whenever X 3.5.7 E Z W 4 3.5.9 (a) E X Y 1 3.5.11 (a) E X 0 1 3 (e) E Y X 6 6 and E Y X 17 14 3 (b) E W Z 1 (b) E X Y 2 27 19 (b 25 4 (b) E Y X 5 2 (b) E Y X e2s 36 7 whenever X 10 3 4 2 (c) E Y X 1 3 2 3 (f) E Y X 52 95 (c) E X Y 1 4 (e) E[E X Y ] 0 1 (g) E Y X 3 2 y 0 (d) E Y X X 3 4 3 2 y3 4 3y3 3y3 (d) 4 19 4 y3 3y3 dy 27 19 (f) E[E Y X ] 6 19 x 2 1 4 dx 52 95 3.6.1 3 7 3.6.3 (a) 1 9 (b) 1 2 (c) 2 (d) The upper bound in part (b) is smaller and thus more useful than that in part (c). 3.6.5 1 4 3.6.7 (a) 10,000 (b) 12,100 3.6.9 (a) 1 (b) 1 4 3.6.11 (a) E Z 3.6.13 7 16 3.7.1 E X1 3.7.3 P X for t C 3.7.5 E X 3.7.7 E W 0 E Y 0, while P X 3 E X2 0 for t 8 5 (b) 32 75 C and P X 2 1 5 1 for 0 3 5 t t t t 0 Appendix E: Answers to OddNumbered Exercises 735 1 8 P Y3 3 64 P Y3 21 2 1 31 3
2 3 16 P Y3 181 3 3.7.9 E W 4.1.1 P Y3 3 16 P Y3 121 3 4.1.3 If Z is the sample mean, then P Z p 2. P Z 4.1.5 For 1 6, P max 4.1.7 If W XY then 1 j 1 j 41 3 3 64 P Y3 0 j 6 20 1 64 P Y3 3 3 32 P Y3 61 3 3 16 1 64 P Y3 91 3 21 3 3 32 P Y3 p2, P Z 0 5 2 p 1 p, and j 1 6 20. P W 1 36 1 18 1 12 1 9 0 1 9 16 25 36 2 3 5 8 10 15 18 20 24 30 4 6 12 if if if if otherwise. 1 2 otherwise, pY y 0 1 5n2 1 4.1.9 pY y 4.2.1 Note that Zn 1 2 for y P 7 U 7 Z unless 7 U 7 1 n2. Hence, for any 1 n2 0 as n. 0, P Zn Z 4.2.3 P W1 Wn n 2 P W1 Wn n 2 1 P 1 n W1 1 3 n 1 1 7 1 U ln ln 9n Fn Wn Hn P Hn Yn 1 P Hn P Hn Xn n Z unless 7 P Hn 1 P Hn n 2 Hn and P Xn 1 8 P e Hn 4.2.5 P X1 4.2.7 For all n 2 1 P 1 Hn P X1 2 ln, P Xn 1 6 Xn 0 and n n 2 4.2.9 By deﬁnition, Hn 1 Fn Hn 1 n 2 4.2.11 r 4.3.1 Note that Zn whenever 1 n2 4.3.3 W1 n W1 4.3.5 P Xn X P Yn 4.3.7 m 5 4.3.9 r 4.3.11 (a) Suppose there is no such m and from this get a contradiction to the strong law of large numbers. (b) No 4.4.1 limn 4.4.3 Here, P Zn for z 1. 4.4.5 P S 4.4.7 P S 1 n2. Also, if U Z Wn n
1 be i.i.d. and set X N 43 629 V 3 C3 Nn D1 0 ey h x y I[0 eYi h Xi Yi Yi 12 and Ni Dn. N1 y e y dy I[0 1] x dx (b) Generate Xi and Yi 4.5.13 (a) J Tn n. appropriately, set Ti 1 y 5e 5y dy I[0 1] x dx (d) As in part (b). (e) The (c) J 0 estimator having smaller variance is better. So use sample variances to choose between them. 4.6.1 (a) U 4.6.3 C1 4.6.5 Let Z1 W1 and Y n 4.6.7 C 7 C5 4.6.9 C1 2 5 C2 1 671 (c) a 4.6.11 (a) m 60 K 5.1.1 The mean survival times for the control group and the treatment group are 93.2 days and 356.2 days, respectively. 5.1.3 For those who are still alive, their survival times will be longer than the recorded values, so these data values are incomplete. 5.1.5 x 5.1.7 Use the difference x 5.2.1 In Example 5.2.1, the mode is 0 In Example 5.2.2, the mode is 1 5.2.3 The mixture has density 5 2 2 for x 5.2.5 x 1 60 (d) 1 C7 1 c 61 (b) y 4 2 2 0 1375 2 C6 3 C3 2 C4 61 b 4 00 4 2 exp exp Zn Z1 10 y. 5 2 x x 2 2 Appendix E: Answers to OddNumbered Exercises 737 5.2.7 The mode is 1 3. 5.2.9 The mode is x 0. 5.3.1 The statistical model for a single response consists of three probability functions Bernoulli 1 2 Bernoulli 1 3 Bernoulli 2 3 5.3.3 The sample X1 Xn is a sample from an N 8 3. Both the population mean and variance uniquely 2 distribution, where 10 2 2 identify the population distribution. 5.3.5 A single observation is from an Exponential [0 cient of variation.. We can parameterize this model by the mean or
variance but not by the coefﬁ distribution, where 5.3.7 (a) A and A B (b) The value X B are possible. 1 is observable only when A. (c) Both 5.3.9 P1 5.4.1 FX x 0 4 10 7 10 9 10 10 3 10 2 10 1 10 22.4.3 (a) Yes (b) Use Table D.1 by selecting a row and reading off the ﬁrst three single numbers (treat 0 in the table as a 10). (c) Using row 108 of Table D.1 (treating 0 as 10): First sample — we obtain random numbers 6 0 9 and so compute X X so compute X random numbers 2 0 2 (note we do not skip the second 2), and so compute X X 3 0 Second sample — we obtain random numbers 4 0 7 and 2 666 7 Third sample — we obtain X 2 0 X X X 10 10 3 3 3 6 9 6 9 6 10 9 5.4.5 (c) The shape of a histogram depends on the intervals being used. 5.4.7 It is a categorical variable. 0 625. Let Yi be the answer from student i. Then Y 5.4.9 (a) Students are more likely to lie when they have illegally downloaded music, so the results of the study will be awed. (b) Under anonymity, students are more likely to tell the truth, so there will be less error. (c) The probability a student tells the truth 1 is is p recorded as an estimate of the proportion of students who have ever downloaded music illegally. 5.5.1 (a) f X 0 (b) FX 0 1 000 (d) The mean x the I Q R 5.5.3 (a 4667 FX 2 1 667 and the variance s2 3. According to the 1.5 I Q R rule, there are no outliers. 0 2667 f X 3 0 7333 FX 3 1 952 (e) The median is 2 and f X 4 0 8667 FX 4 0 2667 FX 1 0 2667 f X 1 25 82 f X 2 35 82 f X 3 22 82 (b) No 0 1333 2 p p 1 6.1.3 L 6.1.5 c x1 10 4 x1 6.1.11 No 6.1.13 No 1 6.2
.1 6.2.3 6.2.5 0 x n 738 Appendix E: Answers to OddNumbered Exercises 2 125. We estimate FX 1 by FX 1 5.5.5 The sample median is 0, ﬁrst quartile is I Q R 5.5.7 5.5.9 0z0 25 where z0 25 satisﬁes 3 0 1 150, third quartile is 0 975, and the 17 20 0 85 z0 25 0 25 2 3 2 1 5.5.11 5.5.13 5.5.15 6.1.1 The appropriate statistical model is Binomial(n, probability of having this antibody in the blood. The likelihood function is L 10 3, where 3 1 7 0 2 [0 1] is the 3 x20 20 exp 20x and x is a sufﬁcient statistic. 9 3 6.1.7 L sufﬁcient statistic. 6.1.9 L 1 0 L 2 0 xn n i 1 xi e xi! nx e n xi! and x is a minimal 4 4817, the distribution f1 is 4 4817 times more likely than f2. a 2 3 b 2 is 1–1, and so b 4 x1 a xn x 2 is the MLE. n xi n i 1 ln xi n i 1 ln 1 32 768 cm3 is the MLE 6.2.7 6.2.9 6.2.11 3 6.2.13 A likelihood function cannot take negative values. 6.2.15 Equivalent loglikelihood functions differ by an additive constant. 6.3.1 Pvalue 6.3.3 Pvalue 6.3.5 Pvalue mum required sample size is 2. 6.3.7 Pvalue 1 05 6.3.9 Pvalue 6.3.11 Pvalue n i 1 xi 0 05 is well within the range of practical signiﬁcance. nx 2 (b) The plugin estimator is 1 0 592 and 0.95conﬁdence interval is 4 442 5 318. 0 000 and 0.95conﬁdence interval is 63 56 67 94. 0 00034 and 0.95conﬁdence interval is [47 6
17 56 383]. The mini 6.3.13 (a) so 2 s2 n 6.3.15 (a) Yes (b) No 6.3.17 The Pvalue 0 22 does not imply the null hypothesis is correct. It may be that we have just not taken a large enough sample size to detect a difference. x 1 x 0 as n 2 n 0 527 0 014 x 2 1 n (c) bias 0 1138 so not statistically signiﬁcant and the observed difference of n i 1 x 2 i 2 2 Appendix E: Answers to OddNumbered Exercises 739 6.4.1 m3 z 1 2s3 n 26 027 151 373 6.4.3 The method of moments estimator is m2 m2 c2 Var X cE X and Var Y 6.4.5 From the mgf, m X 0 m1 m3 6.4.7 The sample median is estimated by 3 2 1 while the method of moments estimator of 3. The plugin estimator is 3 is m3 1 m1 If Y cX then E Y 3 1 n 3 m2 m2 1 x 3 i 0 03 and the estimate of the ﬁrst quartile is 1 28 and for the third quartile is 0 98. Also F 2 F 1 36 0 90 j. The bootstrap sample range y n 6.4.9 The bootstrap procedure is sampling from a discrete distribution and by the CLT the distribution of the bootstrap mean is approximately normal when n and m are large. The delta theorem justiﬁes the approximate normality of functions of the bootstrap mean under conditions. 1 2. Here, 0 6.4.11 The maximum number of possible values is 1 y 1 has the largest possible is obtained when i x 1 and smallest possible value of 0. If there are many repeated xi values value x n in the bootstrap sample, then the value 0 will occur with high probability for y n y 1 and so the bootstrap distribution of the sample range will not be approximately normal. 6.5.1 n 2 4 2 6.5.3 n 2 x z0 95 6.5.5 2 x 6.5.7 n z 1 this does not contain 0 18123 0 46403 as a 0 95conﬁdence interval, and 10 4 1 5045 9 5413 10 3 n
n 2n n 2 1 1 x 6.5.9 [0 min 1 4 45 7.1.1 Based on m 1 1 6 80 91 360 the posterior probability distributions for each of the four possible samples are as follows: x 2 45 m 1 2 m 2 2 x 23 72 m 1 2 8 45 77 360 m 2 1 x 1 18 80 1 20 2 80 1 20 1 ] 1z 2 1 1 25 n 1 2 1 1 20 1 04 1 sample 1 2 3 sample 16 2 5 m 1 1 18 115 16 115 81 115 16 2 5 m 1 2 18 77 32 77 27 77 16 2 5 m 2 1 18 77 32 77 27 77 16 2 5 m 2 2 is positive is 0.5, and the posterior probability is 18 91 64 91 9 91 7.1.3 The prior probability that 0 9992 7.1.5 n 1e 7.1.7 2 x1 n 1e d I[x n xn 1 n I[0 4 0 6] x n N 5 5353 4 81 0 6n 1 2 x1 2 1 xn Gamma 11 41 737 0 4n 1 (b) No (c) The prior must be greater 7.1.9 (a) n than 0 on any parameter values that we believe are possible. 7.1.11 (a) 1 6, so 2 is not uniformly distributed on 0 1 2 3. (b) No 1 3 1 6 0 1 1 3 3 740 Appendix E: Answers to OddNumbered Exercises 7.2.1 nx n nx m n m 7.2.3 E 1 n 2 0 2 x1 1 x xn 0 n 2 x and the posterior mode is 1 2 7.2.5 As in Example 7.2.4, the posterior distribution of so E 1 x1 xn fk 1 and maximizes ln n f1 k i 1 1 1 1 1 is Beta f1 f1 k i 2 fi k i 1 f2 1 1 i f1 1 for the 1 1 i 2 k n i 1 posterior mode. 1 xn 0 2 0 7.2.7 Recall that the posterior distribution of 1 in Example 7.2.2 is Beta f1 1 x1 k. Find the second moment and use Var 2 Now 0 k i 1 xn n 1 1, so Var k i 1 E 1 x1 fi n 1 n 1 f1 n i n i 1
x1 i n 2 1 f2 xn E 2 fk 1 x1 f1 n as n xn 2 1 k i 2 7.2.9 The posterior predictive density of xn 1 is obtained by averaging the N x 1 n this is also the posterior predictive distribution. 0 density with respect to the posterior density of so we must have that 1 2 2 0 7.2.11 The posterior predictive distribution of t So the posterior mode is t and the posterior variance is nx xn 1 is nx 0 the posterior expectation is nx Pareto n n 0 0 2 ]. 0 0 1 7.2.13 (a) The posterior distribution of x n 1 s2 2 n x 2 2 0 1 (c) To assess the hypothesis H0 : 1 cdf. 1 G 2 x 2 0 x1 xn 2 0 2 0 2 is inverse Gamma n 2 0 (b) E 2 x1 2 2 0 compute the probability n is the n where G 2 0 xn 0 x where.2.15 (a) The odds in favor of A 1 odds in favor of Ac (b) B F A 1 B F Ac 7.2.17 Statistician I’s posterior probability for H0 is 0 0099. Statistician II’s posterior probability for H0 is 0 0292. Hence, Statistician II has the bigger posterior belief in H0. 7.2.19 The range of a Bayes factor in favor of A ranges in [0 probability equal to 0, then the Bayes factor will be 0. If A has posterior 7.3.1 3 2052 4 4448 7.3.3 The posterior mode is 2 n 0 z 1 2 0 1 2 nt Hence, the asymptotic credible interval is 2 0 and 2 x1 z 1 xn 2 7.3.5 For a sample x1 [0 1]. A simple Monte Carlo algorithm for the posterior distribution is 1. Generate from N x 1 n 2. Accept value is not in [0 1], then the acceptance rate will be very small for large n if it is in [0 1] and return to step 1 otherwise. If the true the posterior distribution is N x 1 n restricted to xn 7.4.1 The posterior density is proportional to n 1 exp ln 1 xi. Appendix E: Answers to OddNumbered Exercises 7.4.3 (
a) The maximum value of the prior predictive is obtained when posterior of given 1 is 741 1 (b) The 1 1 1 3 1 2 1 3 3 59 1728 1 2 1 2 2 1 8 59 1728 32 59 27 59 a b 7.4.5 The prior predictive is given by m Based on the prior predictive, we would select the prior given by 1 2 1 7.4.7 Jeffreys’ prior is 1 2 The posterior distribution of x1 xn nx 1 n 1 x n 1 is Beta nx.4.9 The prior distribution is 7.4.11 Let the prior be 8.1.1 L 1 tional distributions of s are as follows. 3 2 L 2 N 66 Exponential 101 86. 0 092103. so by Section 6.1.1 T is a sufﬁcient statistic. The condi 2 with 2 with fa s T fb fa s T fb s T fa s T fb.1.3 x 2 8.1.5 UMVU for 5 8.1.7 x 8.1.9 n 1 8.1.11 Yes 8.2.1 When power of the test is 23 120 When 3 5 The power of the test is 1 10 8.2.3 By (8.2.6) the optimal 0 01 test is of the form 0 05 c0 3 2 and 0 1 c0 1 1 Xi 1 10 1 12 1 2 1 20 0 1 30 The 1 12 2 and 0 x 1 0 x x 1 1 2 10 2 10 2 3263 2 3263 1 0 x x 2 0404 2 0404 8.2.5 (a) 0 (b) Suppose 4 8.2.7 n 1. The power function is 1 1. 742 Appendix E: Answers to OddNumbered Exercises 2 2 2 1 2 8.2.9 The graph of the power function of the UMP size graph of the power function of any other size test 8.3.1 2 5 we accept H0 : 8.3.3 The Bayes rule is given by 1 to x as 8.3.5 The Bayes rule is given by n 0 numbers this converges in probability to 8.3.7 The Bayes rule rejects whenever 0 as n 3 5 so nx 2 0 2 0 n 0 1 0 test function lies above the function. 2 2 1 2 and 2 0 nx
2 0 which converges 0 and by the weak law of large B FH0 exp exp nx 2 2 0 the denominator converges to 0, so in the limit p0 p0 As is less than 1 we never reject H0 8.4.1 The model is given by the collection of probability functions 2 0 n nx : xn of 0’s and 1’s. The action space is nx 1 a n 2 0 1 n i 1 xi R1 the correct action function is A and the loss function is L x Var 1 2 exp [0 1] on the set of all sequences x1 [0 1] the correct action function is A a 2 The risk function for T is RT 8.4.3 The model is the set of densities 2 R1 on Rn. The action space is and the loss function is L a 2 0 n x Var 8.4.5 (a) Rd a function d given by d 9.1.1 The observed discrepancy statistic is given by D r P D R 9.1.3 (c) The plots suggest that the normal assumption seems reasonable. 9.1.5 The observed counts are given in the following table. a 2 The risk function for T is RT 1 2 Rd b 0 248 which doesn’t suggest evidence against the model. 2 2 2 22 761 22 761 and the Pvalue is 3 4 (b) No. Consider the risk function of the decision Interval 0 0 0 2] 0 2 0 4] 0 4 0 6] 0 6 0 8] 0 8 1] Count 4 7 3 4 2 2 4 ) 0 4779 Therefore, we have no evidence against the Uniform model The chisquared statistic is equal to 3.50 and the Pvalue is given by (X 2 P X 2 3 5 being correct. 9.1.7 (a) The probability of the event s having S as its support. The most appropriate Pvalue is 0. (b) 0 729 3 is 0 based on the probability measure P Appendix E: Answers to OddNumbered Exercises 743 9.1.9 No 9.1.11 (a) The conditional probability function of x1 xn is nx 1 n 1 x n nx nx 1 n 1 x 1 n nx y x 2 0 x1 x1 Y x2 P Y z where z x2 Y y y X x2
P Y x1 P X y For the converse, show P Y x will change with x whenever 2 0 n is independent of and summing this over x1 leads to P X (b) Hypergeometric n n 2 nx0 (c) 0 0476 9.2.1 (a) No (b) Pvalue is 1 10 so there is little evidence of a prior–data conict. (c) Pvalue is 1 300 so there is some evidence of a prior–data conict. 9.2.3 We can write x N 0 9.2.5 The Pvalue for checking prior–data conict is 0 Hence, there is deﬁnitely a prior–data conict. 10.1.1 For any x1 x2 (that occur with positive probability) and y we have P Y x2 Y y X x2 Thus P X y P X y P X 10.1.3 X and Y are related. 10.1.5 The conditional distributions P Y X is not degenerate. 10.1.7 If the conditional distribution of lifelength given various smoking habits changes, then we can conclude that these two variables are related, but we cannot conclude that this relationship is a cause–effect relationship due to the possible existence of con founding variables. 10.1.9 The researcher should draw a random sample from the population of voters and ask them to measure their attitude toward a particular political party on a scale from favorably disposed to unfavorably disposed. Then the researcher should randomly select half of this sample to be exposed to a negative ad, while the other half is exposed to a positive ad. They should all then be asked to measure their attitude toward the particular political party on the same scale. Next compare the conditional distribution of the response variable Y (the change in attitude from before seeing the ad to after), given the predictor X (type of ad exposed to), using the samples to make inference about these distributions. 10.1.11 (a) (b) A sample has not been taken from the population of interest. The individuals involved in the study have volunteered and, as a group, they might be very different from the full population. (c) We should group the indi viduals according to their initial weight W into homogenous groups (blocks) and then randomly apply the treatments to the individuals in each block. 10.
1.13 (a) The response variable could be the number of times an individual has watched the program. A suitable predictor variable is whether or not they received the brochure. (b) Yes, as we have controlled the assignment of the predictor variable. 10.1.15 W has a relationship with Y and X has a relationship with Y 10.1.17 (a) 0 100 1 100 744 Appendix E: Answers to OddNumbered Exercises (b) (c) (d) X 0 X 1 Rel. Freq. 0.5 0.5 Y 0 Y 1 Rel. Freq. 0.7 0.3 Rel. Freq. X 0 X 1 0 Y Y 1 sum 0.3 0.2 0.5 0.4 0.1 0.5 Sum 1.0 Sum 1.0 Sum 0.7 0.3 1..6 0.8 0.4 0.2 Sum 1.0 1.0 5 7143 (e) Yes 10.1.19 X and Y are related. We see that only the variance of the conditional distribu tion changes as we change X 10.1.21 The correlation is 0 but X and Y are related. 2 2, the P 5 7143 and, with X 2 10.2.1 The chisquared statistic is equal to X 2 0 value equals P X 2 0 05743. Therefore, we don’t have evidence against the null hypothesis of no difference in the distributions of thunderstorms between the two years, at least at the 0 05 level. 10.2.3 The chisquared statistic is equal to X 2 0 Pvalue equals P X 2 the null hypothesis of no relationship between the two digits. 10.2.5 (a) The chisquared statistic is equal to X 2 10 4674 and, with X 2 0 the Pvalue equals P X 2 0 03325. Therefore, we have some evidence against the null hypothesis of no relationship between hair color and gender. (c) The standardized residuals are given in the following table. They all look reasonable, so nothing stands out as an explanation of why the model of independence does not ﬁt. Overall, it looks like a large sample size has detected a small difference. 0 10409 and, with X 2 the 0 74698. Therefore, we have no evidence against 10 4674 4 8105 2 1 2 4 X m
statistic is F 0 05 from Table D.5. Hence, we conclude there is evidence against the null hypothesis of no linear relationship between the response and the predictor. (g) R2 0 3499 so, almost 35% of the observed variation in the response is explained by changes in the predictor. 10.3.9 In general, E Y X since it cannot be written in the form E Y X 1 variable and the i are unobserved parameter values. E Y X 2 in this case and E Y X 2 10.3.11 We can write E Y X so this is a simple linear regression model but the predictor is X 2 not X 10.3.13 R2 the response, so the model will not have much predictive power. 10.4.1 (b) Both plots look reasonable, indicating no serious concerns about the correct ness of the model assumptions. (c) The ANOVA table for testing H0 : is given below. 2 X is not a simple linear regression model 2V where V is an observed 0 05 indicates that the linear model explains only 5% of the variation in 2 X 2 3 1 2 1 Source Df 2 9 11 A Error Total SS 4.37 18.85 23.22 MS 2.18 2.09 1 0431 with Pvalue The F statistic for testing H0 is given by F P F 0 39135 Therefore, we don’t have evidence against the null hy pothesis of no difference among the conditional means of Y given X. (d) Since we 2 18 2 09 1 0431 746 Appendix E: Answers to OddNumbered Exercises did not ﬁnd any relationship between Y and X there is no need to calculate these conﬁdence intervals. 10.4.3 (b) Both plots indicate a possible problem with the model assumptions. (c) The ANOVA table for testing H0 : 2 is given below. 1 Source Df 1 Cheese 10 Error 11 Total SS 0.114 26.865 26.979 MS 0.114 2.686 04 0 04 and with the P 0 114 2 686 0 841. Therefore, we do not have any evidence against the null The F statistic for testing H0 is given by F value P F hypothesis of no difference among the conditional means of Y given Cheese. 10.4.5 (b) Both plots look reasonable, indicating no concerns about the correctness of the model assumptions.
(c) The ANOVA table for testing H0 : follows. 4 1 2 3 Source Treatment Error Total Df 3 20 23 SS 19.241 11.788 31.030 MS 6.414 0.589 10 89 10 89 and with P 6 414 0 589 The F statistic for testing H0 is given by F value P F 0 00019 Therefore, we have strong evidence against the null hypothesis of no difference among the conditional means of Y given the predictor. (d) The 0.95conﬁdence intervals for the difference (column level mean)(row level mean) between the means are given in the following table. 1 2 3 2 3 4 3913 1 4580 2 2746 2 5246 0 4254 0 6754 2 8080 3 0580 0 9587 1 2087 1 1746 0 6746 10.4.7 (b) Treating the marks as separate samples, the ANOVA table for testing any difference between the mean mark in Calculus and the mean mark in Statistics follows. Source Df 1 Course 18 Error 19 Total SS 36.45 685.30 721.75 MS 36.45 38.07 0 95745 36 45 38 07 The F statistic for testing H0 : with the Pvalue equal to P F 0 3408 Therefore, we do not have any evidence against the null hypothesis of no difference among the conditional means of Y given Course. 2 is given by F 1 0 95745 Both residual plots look reasonable, indicating no concerns about the correctness (c) Treating these data as repeated measures, the mean of the model assumptions. difference between the mark in Calculus and the mark in Statistics is given by d Appendix E: Answers to OddNumbered Exercises 747 1 2 7 with standard deviation s 2 00250 The Pvalue for testing H0 : 0 944155 2 Var Y1 Cov Y1 Y2 2 is 0 0021, so we have strong evidence against the null. Hence, we conclude that there is a difference between the mean mark in Calculus and the mean mark in Statistics. A normal probability plot of the data does not indicate any reason to doubt model assumptions. (d) rx y 10.4.9 When Y1 and Y2 are measured on the same individual, we have that Var Y1 Y2 0 If we had mea sured Y1 and Y2 on independently randomly selected individuals, then we would have that Var
Y1 10.4.11 The difference of the two responses Y1 and Y2 is normally distributed, i.e., Y1 10.4.13 (1) The conditional distribution of Y given X1 X2 only through E Y X1 X2 X1 X2 do not interact. depends on X1 X2 E Y X1 X2 is independent of E Y X1 X2 is normally distributed. (3) X1 and X2 and the error Z Y 2Var Y1 since Cov Y1 Y2 (2) The error Z 2Var Y1 Y2 Y2 N Y 2 x 1 1 x 2 dt ln p p 1 1 as e x F x p 1 and implies x 1 1 p 1 1 2 l p el 1 X1 ln p 1 p and substitute l xk Xk x1 1 1 p be the log odds so el 10.5.1 F x x 10.5.3 Let l Hence, el 1 10.5.5 P Y 11.1.1 (a) 0 (b) 0 (c) 1 3 (d) 2 3 (e) 0 (f) 4 9 (g) 0 (h) 1 9 (i) 0 (j) 0 (k) 0 00925 (l) 0 (m) 0 0987 11.1.3 (a) 5 108 (b) 5 216 (c) 5 72 (d) By the law of total probability, P X3 8 X3 6 X3 P X1 11.1.5 (a) Here, P c 0 89819. That is, if you start with $9 and repeatedly make $1 bets having probability 0 499 of winning each bet, then the probability you will reach $10 before going broke is equal to 0 89819 (b) 0 881065 (c) 0 664169 (d) 0 0183155 (e) 4 2x arctan 10 18 (f) 2 10 174 P X1 1x1 k xk 1. 8 8 8 0 11.1.7 We use Theorem 11.1.1. (a) 1 4 (b) 3 4 (c) 0 0625 (d) 1 4 (e) 0 (f) 1 (g) We know that the initial fortune is 5, so to get to 7 in two steps, the walk must have been at 6 after the ﬁrst step.
11.1.9 (a) 18 38 (b) 0 72299 (c) 0 46056 (d) 0 (e) In the long run, the gambler loses money. 1 1 0 0 72 P1 X2 0 73 (b) P0 X3 0 28 P0 X2 0 728 11.2.1 (a) 0 7 (b) 0 1 (c) 0 2 (d) 1 4 (e) 1 4 (f) 1 2 (g) 0 3 11.2.3 (a) P0 X2 1 11.2.5 (a) 1 2 (b) 0 (c) 1 2 (d) 1 2 (e) 1 10 (f) 2 5 (g) 37 100 (h) 11 20 (i) 0 (j) 0 (k) 0 (l) 0 (m) No 1 for all j. Hence, we must 11.2.7 This chain is doubly stochastic, i.e., has have the uniform distribution ( 1 1 4) as a stationary distribution. 11.2.9 (a) By either increasing or decreasing one step at a time, we see that for all i and j, we have p n d. (b) Each state has period 2. (c) If i i j 0 for some n 0 27 P1 X2 i pi j 4 0 3 2 748 Appendix E: Answers to OddNumbered Exercises X2 3 9 i 13 4 1 or i 1 let 2 1 3. 2 9 2 If! d i pi j 1!. 0. 1 2d i j 13 4 i, let Yn 1 p ji j p j i 4 9 (d) limn Yn 1 and let P1 Xn 0 54 1, then i min, let Yn 1 min 1 e j 8 7 9 e i 4 min so P1 X500 j with probability i j, otherwise let Xn 1 1 with probability 2 9. Let j j 6 1,. i with probability and j are two or more apart, then pi j 1 2d 1!, while 11.2.11 (a) This chain is irreducible. (b) The chain is aperiodic. (c) 3 9 3 11.2.13 P1 X1 11.2.15 (a) The chain is irreducible. (b) The chain is
not aperiodic. 11.3.1 First, choose any initial value X0. Then, given Xn with probability 1 2 each. Let j Then let Xn 1 1 i j. 11.3.3 First, choose any initial value X0. Then, given Xn with probability 7 9 or Yn 1 j then let probability i j, otherwise let Xn 1 11.3.5 Let Zn be i.i.d. x, let Yn 1 Xn Xn y6 x 4 y8 Xn 1 11.4.1 C 11.4.3 p 11.4.5 P Xn Xn. (b) T is non 11.4.7 (a) Here, E Xn 1 Xn negative, integervalued, and does not look into the future, so it is a stopping time. (c) E X T 27 40 27. (d) P X T 11.5.1 (a) 1 2 (b) 0 (c) 1 4 (d) We have P Y M Hence, P Y 1 1 5 16 11.5.3 (a) P B2 2 4 1 6 N 0 1. First, choose any initial value X0. Then, given 10 Zn 1. Let y. Then let Xn 1 x y. i 1 Yn 1 and, if i 1 j with 1 2 4 2 (d) P B26 26 3 (f) P B26 3 x 6 x with probability 1 2 9 i with probability 1 y with probability x y, otherwise let 3 (c) P B9 B5 15 (e) P B26 3 7 9 or, if j Then let Xn b) P B3 B11 P Y M 1 x y min 1 exp Yn 1 and let 12 5 1 3 3 4 Xn 3 1 4 3Xn M M. 26 3 5 8 9 8 X0 x 8 y4 11.5.5 E B13 B8 11.5.7 (a) 3 4 (b) 1 4 (c) The answer in part (a) is larger because 5 is closer to B0 than 15 is, whereas 15 is farther than 5 is. (d) 1 4 (e) We have 3 4 it must since the events in parts (a) and (d) are complementary events. 11.5.9 E X3 X5 11.5.11 (a)
P X10 P X10 11.6.1 (a) e 141413 13! (b) e 35353 3! (c) e 424220 20! (d) e 350350340 340! (e) 0 (f) e 141413 13! 6 11.6.3 P N2 e 21217 7! (g) 0 e 2 3 2 3 6 6! P N3 20 10 10 (d) P X10 20 1 10 (b) P X10 0 1, which e 3 3 3 3 5 5! 61 75 250 20 4 10 (c) 20 100 10 250 250 250 1 4 5 Appendix E: Answers to OddNumbered Exercises 749 11.6.5 P N2 6 2 N2 9 2 2 6 2 9 2 Index 0–1 loss function, 467 a priori, 374 abs command, 685 absolutely continuous jointly, 85 absolutely continuous random variable, 52 acceptance probability, 644 action space, 464 additive, 5 additive model, 593 adjacent values, 287 admissible, 470 admissiblity, 470 alternative hypothesis, 448 analysis of covariance, 595 analysis of variance (ANOVA), 545 analysis of variance (ANOVA) table, 545 ancillary, 481 anova command, 690 ANOVA (analysis of variance), 545 ANOVA test, 548 aov command, 690 aperiodicity, 635 balance, 520 ball, 626 bar chart, 288 barplot command, 688 basis, 559 batching, 414 Bayes factor, 397 Bayes risk, 471 Bayes rule, 460, 471 Bayes’ Theorem, 22 Bayesian decision theory, 471 Bayesian model, 374 Bayesian Pvalue, 395 Bayesian updating, 383 bellshaped curve, 56 Bernoulli distribution, 42, 131 bestﬁtting line, 542 beta command, 686 beta distribution, 61 beta function, 61 bias, 271, 322 binom command, 686 binomial coefﬁcient, 17 binomial distribution, 43, 116, 131, 162, 163, 167 binomial theorem, 131 birthday problem, 19 bivariate normal distribution, 89 blinding, 521 blocking variable, 523, 594 blocks, 523 bootstrap mean, 353 bootstrap percentile conﬁdence interval, 355 bootstrap samples, 353 bootstrap standard error, 353 boot
strap t conﬁdence interval, 355 bootstrapping, 351, 353 Borel subset, 38 boxplot, 287 boxplot command, 688 Brown, R., 657 Brownian motion, 657, 659 properties, 659 Buffon’s needle, 234 burnin, 643 calculus, 675 fundamental theorem of, 676 categorical variable, 270 751 752 Cauchy distribution, 61, 240 Cauchy–Schwartz inequality, 186, 197 cause–effect relationship, 516 cbind command, 698 cdf (cumulative distribution function), 62 inverse, 120 ceiling command, 685 ceiling (least integer function), 295 census, 271 central limit theorem, 215, 247 chain rule, 676 characteristic function, 169 Chebychev’s inequality, 185 Chebychev, P. L., 2 chisquared distribution, 236 chisquared goodness of ﬁt test, 490, 491 chisquared statistic, 491 chisquared n, 236 2 n, 236 chisq command, 686 chisq.test command, 688 classiﬁcation problems, 267 coefﬁcient of determination (R2), 546 coefﬁcient of variation, 267, 360 combinatorics, 15 complement, 7, 10 complement of B in A, 7 complementing function, 315 completely crossed, 522 completeness, 438 composite hypothesis, 466 composition, 676 conditional density, 96 conditional distribution, 94, 96 conditional expectation, 173 conditional probability, 20 conditional probability function, 95 conditionally independent, 184 conﬁdence interval, 326 conﬁdence level, 326 conﬁdence property, 326 conﬁdence region, 290 confounds, 517 conjugate prior, 422 consistency, 325 consistent, 200, 325 constant random variable, 42 continuity properties, 28 continuous random variable, 51–53 continuoustime stochastic process, 658, 666 control, 254 control treatment, 521 convergence almost surely, 208 in distribution, 213 in probability, 204, 206, 210, 211, 246 with probability 1, 208, 210, 211, 246 convolution, 113 correct action function, 464 correction for continuity, 219, 358 correlation, 89, 156 cos command, 685 countably additive, 5 counts, 102 covariance, 152, 153 covariance inequality, 440 C
matrix, 560 if command, 691 importance sampler, 233 importance sampling, 233 improper prior, 425 inclusion–exclusion, principle of, 12, 14 increasing sequence, 28 independence, 24, 98, 101, 137 pairwise, 24 independent and identically distributed (i.i.d.), 101 indicator function, 35, 210 indicator function, expectation, 131 individual error rate, 581 inequality Cauchy–Schwartz, 186, 197 Chebychev’s, 185 Jensen’s, 187 Markov’s, 185 inference, 258 inﬁnite series, 677 information inequality, 441 initial distribution, 623 integral, 676 intensity, 666 interaction, 522, 587 intercept term, 562 interpolation, 543 interquartile range (IQR), 286 intersection, 7 inverse cdf, 120 inverse Gamma, 380 inverse normalizing constant, 376 inversion method for generating random variables, 121 IQR (interquartile range), 286 irreducibility, 634 Jacobian, 110 Jeffreys’ prior, 426 Jensen’s inequality, 187 joint cumulative distribution function, 80 joint density function, 85 joint distribution, 80 jointly absolutely continuous, 85 kth cumulant, 173 Kolmogorov, A. N., 2 Kolmogorov–Smirnov test, 495 kurtosis statistic, 483 Laplace distribution, 61 large numbers law of, 206, 211 largestorder statistic, 104 latent variables, 414, 415 law of large numbers strong, 211 weak, 206 law of total probability, 11, 21 least integer function (ceiling), 295 least relative suprise estimate, 406 leastsquares estimate, 538, 560 leastsquares line, 542 leastsquares method, 538 leastsquares principle, 538 Lehmann–Scheffé theorem, 438 length command, 685 levels, 520 lgamma command, 685 755 likelihood, 298 likelihood function, 298 likelihood principle, 299 likelihood ratios, 298 likelihood region, 300 Likert scale, 279 linear independence property, 559 linear regression model, 558 linear subspace, 559 linearity of expected value, 135, 144, 192 link function, 603 Lipschitz function, 665 lm command, 689 location, 136 location mixture, 69 log command, 685 log odds, 603 loggamma function, 383 loglikelihood function,
310 lognormal distribution, 79 logistic distribution, 61, 606 logistic link, 603 logistic regression model, 603 logit, 603 loss function, 465 lower limit, 287 ls command, 686 lurking variables, 518 macro, 700 MAD (mean absolute deviation), 469 margin of error, 329 marginal distribution, 82 Markov chain, 122, 623 Markov chain Monte Carlo, 643 Markov’s inequality, 185 Markov, A. A., 2 martingale, 650 matrix, 559, 678 matrix inverse, 560 matrix product, 560 max command, 685 maximum likehood estimator, 308 maximum likelihood estimate (MLE), 308 maximum of random variables, 104 mean command, 688 756 mean absolute deviation (MAD), 469 mean value, 129, 130 meansquared error (MSE), 321, 434, 469 measurement, 270 measuring surprise (Pvalue), 332 median, 284 median command, 688 memoryless property, 61 Méré, C. de, 2 method of composition, 125 method of least squares, 538 method of moments, 349 method of moments principle, 350 method of transformations, 496 Metropolis–Hastings algorithm, 644 min command, 685 minimal sufﬁcient statistic, 304 minimax decision function, 471 Minitab, 699 mixture distribution, 68 location, 69 scale, 70 MLE (maximum likelihood estimate), 308 mode of a density, 260 model checking, 266, 479 model formula, 688 model selection, 464 moment, 164 momentgenerating function, 165 monotonicity of expected value, 137, 146, 192 monotonicity of probabilities, 11 Monte Carlo approximation, 225 Monty Hall problem, 27, 28 MSE (meansquared error), 321, 434, 469 multicollinearity, 515 multinomial coefﬁcient, 18 multinomial distributions, 102 multinomial models, 302, 305 multiple comparisons, 510, 581 multiple correlation coefﬁcient, 565 multiplication formula, 21 multiplication principle, 15 multivariate measurement, 271 multivariate normal, 500 2, 57 N 0 1, 57 N NA (not available in R), 686 nbinom command, 686 ncol command, 698 negativebinomial distribution, 44, 116 Neyman–Pearson theorem, 450 noninformative prior, 425 nonrandomized
decision function, 467 nonresponse error, 277 norm command, 686 normal distribution, 57, 89, 116, 142, 145, 234 normal probability calculations, 66 normal probability plot, 488 normal quantile plot, 488 normal score, 488 nrow command, 698 nuisance parameter, 338 null hypothesis, 332 observational study, 269 observed Fisher information, 364 observed relative surprise, 406 odds in favor, 397 onesided conﬁdence intervals, 347 onesided hypotheses, 347 onesided tests, 337 onetoone function, 110 oneway ANOVA, 577 optimal decision function, 470 optimal estimator, 434 optional stopping theorem, 653 order statistics, 103, 284 ordered partitions, 17 orthogonal rows, 236 outcome, 4 outliers, 288 overﬁtting, 481 pX, 42 Pvalue, 332 paired comparisons, 585 pairwise independence, 24 parameter, 262 parameter space, 262 757 Pareto distribution, 61 partial derivative, 678 partition, 11 Pascal’s triangle, 2, 632 Pascal, B., 2 pen, 626 percentile, 283 period of Markov chain, 635 permutations, 16, 66 Pi, 627 placebo effect, 521 plot command, 688 plugin Fisher information, 366 plugin MLE, 315 point distribution, 42 point hypothesis, 466 point mass, 42 pois command, 686 Poisson distribution, 45, 132, 162, 164 Poisson process, 50, 666 polling, 276 pooling, 593 population, 270 population cumulative distribution, 270 population distribution, 270 population interquartile range, 286 population mean, 285 population relative frequency function, 274 population variance, 285 posterior density, 376 posterior distribution, 376 posterior mode, 387 posterior odds, 397 posterior predictive, 400 posterior probability function, 376 power, 341 power function, 341, 449, 469 power transformations, 496 practical signiﬁcance, 335 prediction, 258, 400 prediction intervals, 576 prediction region, 402 predictor variable, 514 principle of conditional probability, 259 principle of inclusion–exclusion, 12, 14 prior elicitation, 422 prior odds, 397 prior predictive distribution, 375 prior probability distribution, 374 prior risk, 471 prior–data conict, 503 probability, 1 conditional, 20 law of total, 11, 21 probability function, 42 conditional, 95 probability measure, 5 probability model, 5 probability
714 tail probability, 259 tan command, 685 Taylor series, 677 test function, 449, 469 test of hypothesis, 332 test of signiﬁcance, 332 theorem of total expectation, 177 total expectation, theorem of, 177 total probability, law of, 11, 21 total sum of squares, 544 training set, 495 transition probabilities, 623 higherorder, 628 transpose, 560 treatment, 520 twosample tconﬁdence interval, 580 twosample tstatistic, 580 twosample ttest, 580 twosided tests, 337 twostage systems, 22 twoway ANOVA, 586 type I error, 448 type II error, 448 types of inferences, 289 UMA (uniformly most accurate), 460 UMP (uniformly most powerful), 449 UMVU (uniformly minimum variance un biased), 437 unbiased, 437 unbiased estimator, 322, 436 760 unbiasedness, hypothesis testing, 453 unbounded random variable, 36 underﬁtting, 481 unif command, 686 uniform distribution, 7, 53, 141, 142 uniformly minimum variance unbiased (UMVU), 437 uniformly most accurate (UMA), 460 uniformly most powerful (UMP), 449 union, 8 upper limit, 287 utility function, 134, 141 utility theory, 134, 141 validation set, 495 var command, 688 variance, 149 variance stabilizing transformations, 362 Venn diagrams, 7 volatility parameter, 662 von Savant, M., 28 weak law of large numbers (WLLN), 206 Weibull distribution, 61 whiskers, 287 Wiener process, 657, 659 Wiener, N., 2, 657 WLLN (weak law of large numbers), 206 zconﬁdence intervals, 328 zstatistic, 328 ztest, 333measure P (A/B) satisﬁes all three axioms of a probability measure. That is, 0 for all event A (CP1) P (A/B) (CP2) P (B/B) = 1 (CP3) If A1, A2,..., Ak,... are mutually exclusive events, then 1 P ( Ak/B) = [k=1 1 Xk=1 P (Ak/B). Thus, it is a probability measure with respect
to the new sample space B. Example 2.1. A drawer contains 4 black, 6 brown, and 8 olive socks. Two socks are selected at random from the drawer. (a) What is the probability that both socks are of the same color? (b) What is the probability that both socks are olive if it is known that they are of the same color? Answer: The sample space of this experiment consists of S = {(x, y) \| x, y Bl, Ol, Br}. 2 The cardinality of S is N (S) = 18 2 ✓ ◆ = 153. 6 Probability and Mathematical Statistics 29 Let A be the event that two socks selected at random are of the same color. Then the cardinality of A is given by N (A + 15 + 28 + 8 2 ✓ ◆ Therefore, the probability of A is given by = 49. P (A) = 49 18 2 = 49 153. Let B be the event that two socks selected at random are olive. Then the cardinality of B is given by N (B) = 8 2 ◆ ✓ P (B) = = 28 153. 8 2 18 2 and hence Notice that B A. Hence, ⇢ P (B/A) = = = = P (A B) \ P (A) P (B) P (A) 28 153 ✓ 28 49 ◆ ✓ 4. 7 = 153 49 ◆ Let A and B be two mutually disjoint events in a sample space S. We want to ﬁnd a formula for computing the probability that the event A occurs before the event B in a sequence trials. Let P (A) and P (B) be the probabilities that A and B occur, respectively. Then the probability that neither A P (B). Let us denote this probability by r, that nor B occurs is 1 is r = 1 P (A) P (B). P (A) In the ﬁrst trial, either A occurs, or B occurs, or neither A nor B occurs. In the ﬁrst trial if A occurs, then the probability of A occurs before B is 1. Conditional Probability and Bayes’ Theorem 30 If B occurs in the ﬁrst trial, then the probability of A occurs before B is 0. If neither A nor B occurs in the
ﬁrst trial, we look at the outcomes of the second trial. In the second trial if A occurs, then the probability of A occurs before B is 1. If B occurs in the second trial, then the probability of A occurs before B is 0. If neither A nor B occurs in the second trial, we look at the outcomes of the third trial, and so on. This argument can be summarized in the following diagram. A before B 1 P(A) 0 P(B) P(A) r A before B 1 0 P(B) P(A) r A before B 1 0 P(B) P(A) r = 1-P(A)-P(B) r A before B 1 0 P(B) r Hence the probability that the event A comes before the event B is given by P (A before B) = P (A) + r P (A) + r2 P (A) + r3 P (A) + · · · + rn P (A) + · · · = P (A) [1 + r + r2 + · · · + rn + · · · ] = P (A) 1 r 1 = P (A) [1 1 P (A) P (A) + P (B) . = 1 P (A) P (B)] The event A before B can also be interpreted as a conditional event. In this interpretation the event A before B means the occurrence of the event A given that A B has already occurred. Thus we again have [ P (A/A B) = [ = P (A (A B)) [ B) \ P (A [ P (A) P (A) + P (B). Example 2.2. A pair of four-sided dice is rolled and the sum is determined. What is the probability that a sum of 3 is rolled before a sum of 5 is rolled in a sequence of rolls of the dice? Probability and Mathematical Statistics 31 Answer: The sample space of this random experiment is S = {(1, 1) (2, 1) (3, 1) (4, 1) (1, 2) (2, 2) (3, 2) (4, 2) (1, 3) (2, 3) (3, 3) (4, 3) (1, 4) (2, 4) (3
, 4) (4, 4)}. Let A denote the event of getting a sum of 3 and B denote the event of getting a sum of 5. The probability that a sum of 3 is rolled before a sum of 5 is rolled can be thought of as the conditional probability of a sum of 3, given that a sum of 3 or 5 has occurred. That is, P (A/A B). Hence [ P (A/A B) = [ = = = = P (A (A B)) [ B) \ P (A [ P (A) P (A) + P (B) N (A) N (A) + N (B) 2 2 + 4 1 3. Example 2.3. If we randomly pick two television sets in succession from a shipment of 240 television sets of which 15 are defective, what is the probability that they will be both defective? Answer: Let A denote the event that the ﬁrst television picked was defective. Let B denote the event that the second television picked was defective. Then A B will denote the event that both televisions picked were defective. Using the conditional probability, we can calculate \ P (A \ B) = P (A) P (B/A) 14 239 ◆ = = ✓ 15 240 7 1912 ◆ ✓. In Example 2.3, we assume that we are sampling without replacement. Deﬁnition 2.2. If an object is selected and then replaced before the next object is selected, this is known as sampling with replacement. Otherwise, it is called sampling without replacement. Conditional Probability and Bayes’ Theorem 32 Rolling a die is equivalent to sampling with replacement, whereas dealing a deck of cards to players is sampling without replacement. Example 2.4. A box of fuses contains 20 fuses, of which 5 are defective. If 3 of the fuses are selected at random and removed from the box in succession without replacement, what is the probability that all three fuses are defective? Answer: Let A be the event that the ﬁrst fuse selected is defective. Let B be the event that the second fuse selected is defective. Let C be the event that the third fuse selected is defective. The probability that all three fuses selected are defective is P (A C). Hence B \ \ P (A B \ \ B) \ C) = P (A) P (B/A) P (C/A 4 19
3 18 ◆ ◆ ✓ = = 5 20 ✓ 1 114 ◆ ✓. Deﬁnition 2.3. Two events A and B of a sample space S are called independent if and only if P (A \ B) = P (A) P (B). Example 2.5. The following diagram shows two events A and B in the sample space S. Are the events A and B independent? S B A Answer: There are 10 black dots in S and event A contains 4 of these dots. So the probability of A, is P (A) = 4 10. Similarly, event B contains 5 black dots. Hence P (B) = 5 10. The conditional probability of A given B is P (A/B) = P (A B) \ P (B) = 2 5. Probability and Mathematical Statistics 33 This shows that P (A/B) = P (A). Hence A and B are independent. Theorem 2.1. Let A, B then ✓ S. If A and B are independent and P (B) > 0, P (A/B) = P (A). Proof: P (A/B) = = P (A B) \ P (B) P (A) P (B) P (B) = P (A). Theorem 2.2. independent. Similarly A and Bc are independent. If A and B are independent events. Then Ac and B are Proof: We know that A and B are independent, that is P (A \ B) = P (A) P (B) and we want to show that Ac and B are independent, that is P (Ac \ B) = P (Ac) P (B). Since P (Ac \ B) = P (Ac/B) P (B) = [1 P (A/B)] P (B) = P (B) P (A/B)P (B) = P (B) P (A B) \ P (A) P (B) = P (B) = P (B) [1 = P (B)P (Ac), P (A)] the events Ac and B are independent. Similarly, it can be shown that A and Bc are independent and the proof is now complete. Remark 2.1. The concept of independence is fundamental. In fact, it is this concept that just
iﬁes the mathematical development of probability as a separate discipline from measure theory. Mark Kac said, “independence of events is not a purely mathematical concept.” It can, however, be made plausible Conditional Probability and Bayes’ Theorem 34 that it should be interpreted by the rule of multiplication of probabilities and this leads to the mathematical deﬁnition of independence. Example 2.6. Flip a coin and then independently cast a die. What is the probability of observing heads on the coin and a 2 or 3 on the die? Answer: Let A denote the event of observing a head on the coin and let B be the event of observing a 2 or 3 on the die. Then P (A \ B) = P (A) P (B Example 2.7. An urn contains 3 red, 2 white and 4 yellow balls. An ordered sample of size 3 is drawn from the urn. If the balls are drawn with replacement so that one outcome does not change the probabilities of others, then what is the probability of drawing a sample that has balls of each color? Also, ﬁnd the probability of drawing a sample that has two yellow balls and a red ball or a red ball and two white balls? Answer: and P (RW 243 4 9 ◆ ◆ ✓ ◆ ✓ P (Y Y R or RW = 20 243. 2 9 ◆ ◆ ✓ ◆ ✓ ◆ ✓ ◆ ✓ If the balls are drawn without replacement, then P (RW Y ) = ✓ P (Y Y R or RW 21. ◆ ✓ 84. 1 7 ◆ ◆ ✓ ◆ ✓ There is a tendency to equate the concepts “mutually exclusive” and “independence”. This is a fallacy. Two events A and B are mutually exclusive if A and they are called possible if P (A) = P (B). B = = 0 \ ; Theorem 2.2. Two possible mutually exclusive events are always dependent (that is not independent). 6 6 Probability and Mathematical Statistics 35 Proof: Suppose not. Then P (A B) = P (A) P (B) \ ) = P (A) P (B) P ( ; 0 = P (A) P (B). Hence, we get either P (A) = 0 or P (B) = 0. This is a contradiction to the fact that A and B are possible events. This completes
the proof. Theorem 2.3. Two possible independent events are not mutually exclusive. Proof: Let A and B be two independent events and suppose A and B are mutually exclusive. Then P (A) P (B) = P (A ) = P ( ; = 0. B) \ Therefore, we get either P (A) = 0 or P (B) = 0. This is a contradiction to the fact that A and B are possible events. The possible events A and B exclusive implies A and B are not indepen- dent; and A and B independent implies A and B are not exclusive. 2.2. Bayes’ Theorem There are many situations where the ultimate outcome of an experiment depends on what happens in various intermediate stages. This issue is resolved by the Bayes’ Theorem. Deﬁnition 2.4. Let S be a set and let P = {Ai}m of S. The collection P is called a partition of S if i=1 be a collection of subsets m (a) S = Ai (b) Ai \ i=1 [ Aj = ; for i = j. A2 A5 Sample Space A1 A3 A4 6 Conditional Probability and Bayes’ Theorem 36 Theorem 2.4. If the events {Bi}m space S and P (Bi) = 0 for i = 1, 2,..., m, then for any event A in S i=1 constitute a partition of the sample m P (A) = P (Bi) P (A/Bi). i=1 X Proof: Let S be a sample space and A be an event in S. Let {Bi}m any partition of S. Then i=1 be Thus A = m i=1 [ (A \ Bi). P (A) = m i=1 X m P (A Bi) \ = P (Bi) P (A/Bi). i=1 X Theorem 2.5. If the events {Bi}m space S and P (Bi) that P (A) = 0 i=1 constitute a partition of the sample = 0 for i = 1, 2,..., m, then for any event A in S such P (Bk/A) = P (Bk) P (A/Bk) m i=1 P (Bi) P (A/Bi) k = 1, 2,...
, m. Proof: Using the deﬁnition of conditional probability, we get P P (Bk/A) = P (A Bk) \ P (A). Using Theorem 1, we get P (Bk/A) = P (A Bk) \ m i=1 P (Bi) P (A/Bi). This completes the proof. P This Theorem is called Bayes Theorem. The probability P (Bk) is called prior probability. The probability P (Bk/A) is called posterior probability. Example 2.8. Two boxes containing marbles are placed on a table. The boxes are labeled B1 and B2. Box B1 contains 7 green marbles and 4 white 6 6 6 Probability and Mathematical Statistics 37 marbles. Box B2 contains 3 green marbles and 10 yellow marbles. The boxes are arranged so that the probability of selecting box B1 is 1 3 and the probability of selecting box B2 is 2 3. Kathy is blindfolded and asked to select a marble. She will win a color TV if she selects a green marble. (a) What is the probability that Kathy will win the TV (that is, she will select a green marble)? (b) If Kathy wins the color TV, what is the probability that the green marble was selected from the ﬁrst box? Answer: Let A be the event of drawing a green marble. The prior probabilities are P (B1) = 1 3 and P (B2) = 2 3. (a) The probability that Kathy will win the TV is P (A) = P (A B1) + P (A B2) \ \ = P (A/B1) P (B1) + P (A/B2) P (B2) 1 3 + ◆ ✓ 3 13 ◆ ✓ 2 3 ◆ = = = 7 11 ✓ 7 33 91 429 ◆ ✓ 2 13 + + 66 429 = 157 429. (b) Given that Kathy won the TV, the probability that the green marble was selected from B1 is 1/3 2/3 7/11 4/11 3/13 Selecting box B1 Selecting box B2 Green marble Not a green marble Green marble 10/13 Not a green marble Conditional Probability and Bayes’ Theorem 38 P (B1/A) = P (A/B1) P (
B1) P (A/B1) P (B1) + P (A/B2) P (B2) = = 7 11 1 3 1 3 3 13 + 2 3 7 11 91 157. Note that P (A/B1) is the probability of selecting a green marble from B1 whereas P (B1/A) is the probability that the green marble was selected from box B1. Example 2.9. Suppose box A contains 4 red and 5 blue chips and box B contains 6 red and 3 blue chips. A chip is chosen at random from the box A and placed in box B. Finally, a chip is chosen at random from among those now in box B. What is the probability a blue chip was transferred from box A to box B given that the chip chosen from box B is red? Answer: Let E represent the event of moving a blue chip from box A to box B. We want to ﬁnd the probability of a blue chip which was moved from box A to box B given that the chip chosen from B was red. The probability of choosing a red chip from box A is P (R) = 4 9 and the probability of choosing a blue chip from box A is P (B) = 5 9. If a red chip was moved from box A to box B, then box B has 7 red chips and 3 blue chips. Thus the probability of choosing a red chip from box B is 7 10. Similarly, if a blue chip was moved from box A to box B, then the probability of choosing a red chip from box B is 6 10. Box A red 4/9 blue 5/9 7/10 3/10 6/10 Box B 7 red 3 blue Box B 6 red 4 blue Red chip Not a red chip Red chip 4/10 Not a red chip Probability and Mathematical Statistics 39 Hence, the probability that a blue chip was transferred from box A to box B given that the chip chosen from box B is red is given by P (E/R) = P (R/E) P (E) P (R) 6 10 4 9 5 9 6 10 + 5 9 = = 7 10 15 29. Example 2.10. Sixty percent of new drivers
have had driver education. During their ﬁrst year, new drivers without driver education have probability 0.08 of having an accident, but new drivers with driver education have only a 0.05 probability of an accident. What is the probability a new driver has had driver education, given that the driver has had no accident the ﬁrst year? Answer: Let A represent the new driver who has had driver education and B represent the new driver who has had an accident in his ﬁrst year. Let Ac and Bc be the complement of A and B, respectively. We want to ﬁnd the probability that a new driver has had driver education, given that the driver has had no accidents in the ﬁrst year, that is P (A/Bc). P (A/Bc) = P (A Bc) \ P (Bc) = P (Bc/A) P (A) P (Bc/A) P (A) + P (Bc/Ac) P (Ac) = [1 [1 P (B/A)] P (A) + [1 P (B/A)] P (A) P (B/Ac)] [1 P (A)] = 40 100 60 100 92 100 = 0.6077. 95 100 60 100 + 95 100 Example 2.11. One-half percent of the population has AIDS. There is a test to detect AIDS. A positive test result is supposed to mean that you Conditional Probability and Bayes’ Theorem 40 have AIDS but the test is not perfect. For people with AIDS, the test misses the diagnosis 2% of the times. And for the people without AIDS, the test incorrectly tells 3% of them that they have AIDS. (a) What is the probability that a person picked at random will test positive? (b) What is the probability that you have AIDS given that your test comes back positive? Answer: Let A denote the event of one who has AIDS and let B denote the event that the test comes out positive. (a) The probability that a person picked at random will test positive is given by P (test positive) = (0.005) (0.98) + (0.995) (0.03) = 0.
0049 + 0.0298 = 0.035. (b) The probability that you have AIDS given that your test comes back positive is given by P (A/B) = = = favorable positive branches total positive branches (0.005) (0.98) (0.005) (0.98) + (0.995) (0.03) 0.0049 0.035 = 0.14. 0.005 AIDS 0.995 No AIDS 0.98 0.02 0.03 Test positive Test negative Test positive 0.97 Test negative Remark 2.2. This example illustrates why Bayes’ theorem is so important. What we would really like to know in this situation is a ﬁrst-stage result: Do you have AIDS? But we cannot get this information without an autopsy. The ﬁrst stage is hidden. But the second stage is not hidden. The best we can do is make a prediction about the ﬁrst stage. This illustrates why backward conditional probabilities are so useful. Probability and Mathematical Statistics 41 2.3. Review Exercises 1. Let P (A) = 0.4 and P (A B independent? [ B) = 0.6. For what value of P (B) are A and 2. A die is loaded in such a way that the probability of the face with j dots turning up is proportional to j for j = 1, 2, 3, 4, 5, 6. In 6 independent throws of this die, what is the probability that each face turns up exactly once? 3. A system engineer is interested in assessing the reliability of a rocket composed of three stages. At take oﬀ, the engine of the ﬁrst stage of the rocket must lift the rocket oﬀ the ground. If that engine accomplishes its task, the engine of the second stage must now lift the rocket into orbit. Once the engines in both stages 1 and 2 have performed successfully, the engine of the third stage is used to complete the rocket’s mission. The reliability of the rocket is measured by the probability of the completion of the mission. If the probabilities of successful performance of the engines of stages 1, 2 and 3 are 0.99, 0.97 and 0.98, respectively, ﬁnd the reliability of the rocket. 4. Identical twins come from the same egg and hence are of the same sex. Fraternal
twins have a 50-50 chance of being the same sex. Among twins the probability of a fraternal set is 1 3. If the next set of twins are of the same sex, what is the probability they are identical? 3 and an identical set is 2 5. In rolling a pair of fair dice, what is the probability that a sum of 7 is rolled before a sum of 8 is rolled? 6. A card is drawn at random from an ordinary deck of 52 cards and replaced. This is done a total of 5 independent times. What is the conditional probability of drawing the ace of spades exactly 4 times, given that this ace is drawn at least 4 times? 7. Let A and B be independent events with P (A) = P (B) and P (A 0.5. What is the probability of the event A? [ B) = 8. An urn contains 6 red balls and 3 blue balls. One ball is selected at random and is replaced by a ball of the other color. A second ball is then chosen. What is the conditional probability that the ﬁrst ball selected is red, given that the second ball was red? Conditional Probability and Bayes’ Theorem 42 9. A family has ﬁve children. Assuming that the probability of a girl on each birth was 0.5 and that the ﬁve births were independent, what is the probability the family has at least one girl, given that they have at least one boy? 10. An urn contains 4 balls numbered 0 through 3. One ball is selected at random and removed from the urn and not replaced. All balls with nonzero numbers less than that of the selected ball are also removed from the urn. Then a second ball is selected at random from those remaining in the urn. What is the probability that the second ball selected is numbered 3? 11. English and American spelling are rigour and rigor, respectively. A man staying at Al Rashid hotel writes this word, and a letter taken at random from his spelling is found to be a vowel. If 40 percent of the English-speaking men at the hotel are English and 60 percent are American, what is the probability that the writer is an Englishman? 12. A diagnostic test for a certain disease is said to be 90% accurate in that, if a person has the disease, the test will detect with probability 0.9. Also, if a person does not have the disease, the
test will report that he or she doesn’t have it with probability 0.9. Only 1% of the population has the disease in question. If the diagnostic test reports that a person chosen at random from the population has the disease, what is the conditional probability that the person, in fact, has the disease? 13. A small grocery store had 10 cartons of milk, 2 of which were sour. If you are going to buy the 6th carton of milk sold that day at random, ﬁnd the probability of selecting a carton of sour milk. 14. Suppose Q and S are independent events such that the probability that at least one of them occurs is 1 3 and the probability that Q occurs but S does not occur is 1 9. What is the probability of S? 15. A box contains 2 green and 3 white balls. A ball is selected at random from the box. If the ball is green, a card is drawn from a deck of 52 cards. If the ball is white, a card is drawn from the deck consisting of just the 16 pictures. (a) What is the probability of drawing a king? (b) What is the probability of a white ball was selected given that a king was drawn? Probability and Mathematical Statistics 43 16. Five urns are numbered 3,4,5,6 and 7, respectively. Inside each urn is n2 dollars where n is the number on the urn. The following experiment is performed: An urn is selected at random. If its number is a prime number the experimenter receives the amount in the urn and the experiment is over. If its number is not a prime number, a second urn is selected from the remaining four and the experimenter receives the total amount in the two urns selected. What is the probability that the experimenter ends up with exactly twentyﬁve dollars? 17. A cookie jar has 3 red marbles and 1 white marble. A shoebox has 1 red marble and 1 white marble. Three marbles are chosen at random without replacement from the cookie jar and placed in the shoebox. Then 2 marbles are chosen at random and without replacement from the shoebox. What is the probability that both marbles chosen from the shoebox are red? 18. A urn contains n black balls and n white balls. Three balls are chosen from the urn at random and without replacement. What is the value of n if
the probability is 1 12 that all three balls are white? 19. An urn contains 10 balls numbered 1 through 10. Five balls are drawn at random and without replacement. Let A be the event that “Exactly two odd-numbered balls are drawn and they occur on odd-numbered draws from the urn.” What is the probability of event A? I have ﬁve envelopes numbered 3, 4, 5, 6, 7 all hidden in a box. 20. I pick an envelope – if it is prime then I get the square of that number in dollars. Otherwise (without replacement) I pick another envelope and then get the sum of squares of the two envelopes I picked (in dollars). What is the probability that I will get $25? Conditional Probability and Bayes’ Theorem 44 Probability and Mathematical Statistics 45 Chapter 3 RANDOM VARIABLES AND DISTRIBUTION FUNCTIONS 3.1. Introduction In many random experiments, the elements of sample space are not necessarily numbers. For example, in a coin tossing experiment the sample space consists of S = {Head, Tail}. Statistical methods involve primarily numerical data. Hence, one has to ‘mathematize’ the outcomes of the sample space. This mathematization, or quantiﬁcation, is achieved through the notion of random variables. Deﬁnition 3.1. Consider a random experiment whose sample space is S. A random variable X is a function from the sample space S into the set of real numbers IR such that for each interval I in IR, the set {s I} is an event in S. S \| X(s) 2 2 In a particular experiment a random variable X would be some function that assigns a real number X(s) to each possible outcome s in the sample space. Given a random experiment, there can be many random variables. This is due to the fact that given two (ﬁnite) sets A and B, the number of distinct functions one can come up with is \|B\|\|A\|. Here \|A\| means the cardinality of the set A. Random variable is not a variable. Also, it is not random. Thus someone named it inappropriately. The following analogy speaks the role of the random variable. Random variable is like the Holy Roman Empire – it was Random Variables and Distribution Functions 46 not holy, it was not Roman, and it was not an empire. A random variable is neither
random nor variable, it is simply a function. The values it takes on are both random and variable. Deﬁnition 3.2. The set {x random variable X. 2 IR \| x = X(s), s 2 S} is called the space of the The space of the random variable X will be denoted by RX. The space of the random variable X is actually the range of the function X : S IR.! Example 3.1. Consider the coin tossing experiment. Construct a random variable X for this experiment. What is the space of this random variable X? Answer: The sample space of this experiment is given by S = {Head, Tail}. Let us deﬁne a function from S into the set of reals as follows X(Head) = 0 X(T ail) = 1. Then X is a valid map and thus by our deﬁnition of random variable, it is a random variable for the coin tossing experiment. The space of this random variable is RX = {0, 1}. Tail Head X Sample Space 0 1 Real line X(head) = 0 and X(tail) = 1 Example 3.2. Consider an experiment in which a coin is tossed ten times. What is the sample space of this experiment? How many elements are in this sample space? Deﬁne a random variable for this sample space and then ﬁnd the space of the random variable. Probability and Mathematical Statistics 47 Answer: The sample space of this experiment is given by S = {s \| s is a sequence of 10 heads or tails}. The cardinality of S is \|S\| = 210. Let X : S deﬁned as follows:! IR be a function from the sample space S into the set of reals IR X(s) = number of heads in sequence s. Then X is a random variable. This random variable, for example, maps the sequence HHT T T HT T HH to the real number 5, that is X(HHT T T HT T HH) = 5. The space of this random variable is RX = {0, 1, 2,..., 10}. Now, we introduce some notations. By (X = x) we mean the event {s S \| X(s) = x}. Similarly, (a < X < b) means the event {s of the sample space S. These are illustrated in the following diagrams} S A X
Sample Space S B X Real line x Sample Space a b Real line (X=x) means the event A (a<X<b) means the event B There are three types of random variables: discrete, continuous, and mixed. However, in most applications we encounter either discrete or continuous random variable. In this book we only treat these two types of random variables. First, we consider the discrete case and then we examine the continuous case. Deﬁnition 3.3. If the space of random variable X is countable, then X is called a discrete random variable. Random Variables and Distribution Functions 48 3.2. Distribution Functions of Discrete Random Variables Every random variable is characterized through its probability density function. Deﬁnition 3.4. Let RX be the space of the random variable X. The function f : RX! IR deﬁned by f (x) = P (X = x) is called the probability density function (pdf) of X. Example 3.3. In an introductory statistics class of 50 students, there are 11 freshman, 19 sophomores, 14 juniors and 6 seniors. One student is selected at random. What is the sample space of this experiment? Construct a random variable X for this sample space and then ﬁnd its space. Further, ﬁnd the probability density function of this random variable X. Answer: The sample space of this random experiment is Deﬁne a function X : S S = {F r, So, Jr, Sr}. IR as follows:! X(F r) = 1, X(So) = 2 X(Jr) = 3, X(Sr) = 4. Then clearly X is a random variable in S. The space of X is given by RX = {1, 2, 3, 4}. The probability density function of X is given by f (1) = P (X = 1) = f (2) = P (X = 2) = f (3) = P (X = 3) = f (4) = P (X = 4) = 11 50 19 50 14 50 6 50. Example 3.4. A box contains 5 colored balls, 2 black and 3 white. Balls are drawn successively without replacement. If the random variable X is the Probability and Mathematical Statistics 49 number of draws until the last black ball is obtained, ﬁnd the probability density function for the random
variable X. Answer: Let ‘B’ denote the black ball, and ‘W’ denote the white ball. Then the sample space S of this experiment is given by (see the ﬁgure below) 2B 3W = { BB, BW B, W BB, BW W B, W BW B, W W BB, BW W W B, W W BW B, W W W BB, W BW W B}. Hence the sample space has 10 points, that is \|S\| = 10. It is easy to see that the space of the random variable X is {2, 3, 4, 5}. X BB BWB WBB BWWB WBWB WWBB BWWWB WWBWB WWWBB 2 3 4 5 WBWWB Sample Space S Real line Therefore, the probability density function of X is given by f (2) = P (X = 2) = f (4) = P (X = 4) = 1 10 3 10,, f (3) = P (X = 3) = f (5) = P (X = 5) = 2 10 4 10. Random Variables and Distribution Functions 50 Thus f (x) = 1, x 10 x = 2, 3, 4, 5. Example 3.5. A pair of dice consisting of a six-sided die and a four-sided die is rolled and the sum is determined. Let the random variable X denote this sum. Find the sample space, the space of the random variable, and probability density function of X. Answer: The sample space of this random experiment is given by S = {(1, 1) (2, 1) (3, 1) (4, 1) (1, 2) (2, 2) (3, 2) (4, 2) (1, 3) (2, 3) (3, 3) (4, 3) (1, 4) (2, 4) (3, 4) (4, 4) (1, 5) (2, 5) (3, 5) (4, 5) (1, 6) (2, 6) (3, 6) (4, 6)} The space of the random variable X is given by RX = {2, 3, 4, 5, 6, 7, 8, 9, 10}. Therefore, the probability density function of X is given by f (2) = P (X = 2
) = f (4) = P (X = 4) = f (6) = P (X = 6) = f (8) = P (X = 8) = 1 24 3 24 4 24 3 24,,,, f (3) = P (X = 3) = f (5) = P (X = 5) = f (7) = P (X = 7) = f (9) = P (X = 9) = 2 24 4 24 4 24 2 24 f (10) = P (X = 10) = 1 24. Example 3.6. A fair coin is tossed 3 times. Let the random variable X denote the number of heads in 3 tosses of the coin. Find the sample space, the space of the random variable, and the probability density function of X. Answer: The sample space S of this experiment consists of all binary sequences of length 3, that is S = {T T T, T T H, T HT, HT T, T HH, HT H, HHT, HHH}. Probability and Mathematical Statistics 51 TTT TTH THT HTT THH HTH HHT HHH X 0 1 2 3 Sample Space S Real line The space of this random variable is given by RX = {0, 1, 2, 3}. Therefore, the probability density function of X is given by f (0) = P (X = 0) = f (1) = P (X = 1) = f (2) = P (X = 2) = f (3) = P (X = 3. This can be written as follows: f (x, 1, 2, 3. The probability density function f (x) of a random variable X completely characterizes it. Some basic properties of a discrete probability density function are summarized below. Theorem 3.1. If X is a discrete random variable with space RX and probability density function f (x), then (a) f (x) (b) 0 for all x in RX, and f (x) = 1. RX Xx 2 Example 3.7. If the probability of a random variable X with space RX = {1, 2, 3,..., 12} is given by f (x) = k (2x 1), Random Variables and Distribution Functions 52 then, what is the value of the constant k? Answer: 1 = f (x) = RX Xx
2 RX Xx 2 12 k (2x 1) = k (2x 1) x= 144. 12 x x=1 X (12)(13) 2 12 # 12 Hence k = 1 144. Deﬁnition 3.5. The cumulative distribution function F (x) of a random variable X is deﬁned as for all real numbers x. F (x) = P (X x)  Theorem 3.2. If X is a random variable with the space RX, then F (x) = f (t) x Xt  for x RX. 2 Example 3.8. If the probability density function of the random variable X is given by 1 144 (2x 1) for x = 1, 2, 3,..., 12 then ﬁnd the cumulative distribution function of X. Answer: The space of the random variable X is given by RX = {1, 2, 3,..., 12}. Probability and Mathematical Statistics 53 Then F (1) = f (t) = f (1) = 1 Xt  1 144 F (2) = f (t) = f (1) + f (2) = 2 Xt  1 144 + 3 144 = 4 144 F (3) = f (t) = f (1) + f (2) + f (3) = Xt 3 .................... 1 144 + 3 144 + 5 144 = 9 144 F (12) = f (t) = f (1) + f (2) + · · · + f (12) = 1. 12 Xt  F (x) represents the accumulation of f (t) up to t x.  Theorem 3.3. Let X be a random variable with cumulative distribution function F (x). Then the cumulative distribution function satisﬁes the followings: ) = 0, ) = 1, and (a) F ( (b) F ( (c) F (x) is an increasing function, that is if x < y, then F (x) 1 1 all reals x, y. F (y) for  The proof of this theorem is trivial and we leave it to the students. Theorem 3.4. If the space RX of the random variable
X is given by RX = {x1 < x2 < x3 < · · · < xn}, then f (x1) = F (x1) f (x2) = F (x2) f (x3) = F (x3).................... F (x1) F (x2) f (xn) = F (xn) F (xn 1). Random Variables and Distribution Functions 54 F(x4) 1 F(x3) F(x2) F(x1) 0 f(x4) f(x3) f(x2) f(x1) x1 x2 x3 x4 x Theorem 3.2 tells us how to ﬁnd cumulative distribution function from the probability density function, whereas Theorem 3.4 tells us how to ﬁnd the probability density function given the cumulative distribution function. Example 3.9. Find the probability density function of the random variable X whose cumulative distribution function is F (x) = 0.00 if x < 1 0.25 if 1  x < 1 0.50 if 1 0.75 if 3 1.00 if >>>>>>>>>>>< >>>>>>>>>>>: Also, ﬁnd (a) P (X  3), (b) P (X = 3), and (c) P (X < 3). Answer: The space of this random variable is given by RX = { 1, 1, 3, 5}. By the previous theorem, the probability density function of X is given by f ( 1) = 0.25 f (1) = 0.50 f (3) = 0.75 f (5) = 1.00 0.25 = 0.25 0.50 = 0.25 0.75 = 0.25. The probability P (X Hence  3) can be computed by using the deﬁnition of F. P (X  3) = F (3) = 0.75. Probability and Mathematical Statistics 55 The probability P (X = 3) can be computed from P (X = 3) = F (3) F (1) = 0.75 0.50 = 0.25. Finally, we get P (X < 3) from P (X
< 3) = P (X 1) = F (1) = 0.5.  We close this section with an example showing that there is no one-toone correspondence between a random variable and its distribution function. Consider a coin tossing experiment with the sample space consisting of a head and a tail, that is S = { head, tail }. Deﬁne two random variables X1 and X2 from S as follows: X1( head ) = 0 and X1( tail ) = 1 and X2( head ) = 1 and X2( tail ) = 0. It is easy to see that both these random variables have the same distribution function, namely FXi(x) = ( 0 1 2 1 if x < 0 if 0 if 1   x < 1 x for i = 1, 2. Hence there is no one-to-one correspondence between a random variable and its distribution function. 3.3. Distribution Functions of Continuous Random Variables A random variable X is said to be continuous if its space is either an interval or a union of intervals. The folllowing deﬁnition formally deﬁnes a continuous random variable. Deﬁnition 3.6. A random variable X is said to be a continuous random ) such that for variable if there exists a continuous function f : IR every set of real numbers A [0,! 1 P (X 2 A) = f (x) dx. ZA (1) Deﬁnition 3.7. The function f in (1) is called the probability density function of the continuous random variable X. Random Variables and Distribution Functions 56 It can be easily shown that for every probability density function f, f (x)dx = 1. 1 Z 1 Example 3.10. Is the real valued function f : IR IR deﬁned by! f (x) = ⇢ 2 2 x 0 if 1 < x < 2 otherwise, a probability density function for some random variable X? Answer: We have to show that f is nonnegative and the area under f (x) is unity. Since the domain of f is the interval (0, 1), it is clear that f is nonnegative. Next, we calculate 1 f (x) dx = Z 1 2 1 Z 2 x 2 dx.   Thus f is a probability density function.
Example 3.11. Is the real valued function f : IR IR deﬁned by! f (x) = 1 + \|x\| 0 ⇢ 1 < x < 1 if otherwise, a probability density function for some random variable X? Probability and Mathematical Statistics 57 Answer: It is easy to see that f is nonnegative, that is f (x) 0, since f (x) = 1 + \|x\|. Next we show that the area under f is not unity. For this we compute 1 1 f (x) dx = (1 + \|x\|) dx 1 1 x) dx + (1 + x) dx 0 Z + x +  1 2 x2 x2. Thus f is not a probability density function for some random variable X. Example 3.12. For what value of the constant c, the real valued function f : IR IR given by! f (x) = c 1 + (x ✓)2, < x <, 1 1 where ✓ is a real parameter, is a probability density function for random variable X? Answer: Since f is nonnegative, we see that c 0. To ﬁnd the value of c, Random Variables and Distribution Functions 58 we use the fact that for pdf the area is unity, that is 1 = 1 f (x) dx ✓)2 dx Z 1 1 Z 1 1 = = c 1 + (x c 1 + z2 dz 1 z 1( 1 1 tan = c tan ⇡. tan 1( ) 1 ⇤ Hence c = 1 ⇡ and the density function becomes f (x) = 1 ⇡ [1 + (x, ✓)2] < x <. 1 1 This density function is called the Cauchy distribution function with parameter ✓. If a random variable X has this pdf then it is called a Cauchy random variable and is denoted by X CAU (✓). ⇠ This distribution is symmetrical about ✓. Further, it achieves it maximum at x = ✓. The following ﬁgure illustrates the symmetry of the distribution for ✓ = 2. Example 3.13. For what value of the constant c, the real valued function f : IR IR given by! f (x) = c if a x b   (
0 otherwise, Probability and Mathematical Statistics 59 where a, b are real constants, is a probability density function for random variable X? Answer: Since f is a pdf, k is nonnegative. Further, since the area under f is unity, we get 1 = 1 f (x) dx Z 1 b = c dx a Z = c [x]b a = c [b a]. Hence c = 1 a, and the pdf becomes b 1 f (x) = b a ( 0 if a x b   otherwise. This probability density function is called the uniform distribution on If a random variable X has this pdf then it is called a the interval [a, b]. uniform random variable and is denoted by X U N IF (a, b). The following is a graph of the probability density function of a random variable on the interval [2, 5]. ⇠ Deﬁnition 3.8. Let f (x) be the probability density function of a continuous random variable X. The cumulative distribution function F (x) of X is deﬁned as x F (x) = P (X x) =  f (t) dt. Z 1 The cumulative distribution function F (x) represents the area under the, x) (see ﬁgure below). probability density function f (x) on the interval ( 1 Random Variables and Distribution Functions 60 Like the discrete case, the cdf is an increasing function of x, and it takes value 0 at negative inﬁnity and 1 at positive inﬁnity. Theorem 3.5. If F (x) is the cumulative distribution function of a continuous random variable X, the probability density function f (x) of X is the derivative of F (x), that is Proof: By Fundamental Theorem of Calculus, we get d dx F (x) = f (x). d dx (F (x)) = d dx x f (t) dt ◆ = f (x) = f (x). 1 ✓Z dx dx This theorem tells us that if the random variable is continuous, then we can ﬁnd the pdf given cdf by taking the derivative of the cdf. Recall that for discrete random variable, the pdf at a point in space of the random variable can be obtained from the cdf by taking
the diﬀerence between the cdf at the point and the cdf immediately below the point. Example 3.14. What is the cumulative distribution function of the Cauchy random variable with parameter ✓? Answer: The cdf of X is given by x F (x) = f (t) dt Z 1 x Z 1 x Z 1 1 ⇡ = = = 1 ⇡ [1 + (t ✓ 1 ⇡ [1 + z2] dt ✓)2] dz tan 1 (x ✓) + 1 2. Probability and Mathematical Statistics 61 Example 3.15. What is the probability density function of the random variable whose cdf is F (x) = 1 1 + e x, < x <? 1 1 Answer: The pdf of the random variable is given by f (x) = = = d dx d dx d dx F (x) 1 x 1 + e ✓ x 1 + e) 1) (1 + e 2 d dx = x e (1 + e x)2. x 1 + e Next, we brieﬂy discuss the problem of ﬁnding probability when the cdf is given. We summarize our results in the following theorem. Theorem 3.6. Let X be a continuous random variable whose cdf is F (x). Then followings are true: (a) P (X < x) = F (x), (b) P (X > x) = 1 (c) P (X = x) = 0, and (d) P (a < X < b) = F (b) F (x), F (a). 3.4. Percentiles for Continuous Random Variables In this section, we discuss various percentiles of a continuous random variable. If the random variable is discrete, then to discuss percentile, we have to know the order statistics of samples. We shall treat the percentile of discrete random variable in Chapter 13. Deﬁnition 3.9. Let p be a real number between 0 and 1. A 100pth percentile of the distribution of a random variable X is any real number q satisfying P (X q) p   and P (X > q) p. 1  A 100pth percentile is a
under f (x) should be unity. Hence a 1 = f (x) dx 1 4a x dx 0 Z a = = = 0 Z 1 8a a 8. a2 Thus a = 8. Hence the probability density function of X is f (x) = 1 32 x. Now we want to ﬁnd the 25th percentile. 25 100 = = = q 0 Z q 0 Z 1 64 f (x) dx x dx 1 32 q2. Hence q = p16, that is the 25th percentile of the above distribution is 4. Deﬁnition 3.10. The 25th and 75th percentiles of any distribution are called the ﬁrst and the third quartiles, respectively. Probability and Mathematical Statistics 65 Deﬁnition 3.11. The 50th percentile of any distribution is called the median of the distribution. The median divides the distribution of the probability mass into two equal parts (see the following ﬁgure). If a probability density function f (x) is symmetric about the y-axis, then the median is always 0. Example 3.19. A random variable is called standard normal if its probability density function is of the form f (x) = 1 p2⇡ e 1 2 x2, < x <. 1 1 What is the median of X? Answer: Notice that f (x) = f ( is symmetric about the y-axis. Thus the median of X is 0. x), hence the probability density function Deﬁnition 3.12. A mode of the distribution of a continuous random variable X is the value of x where the probability density function f (x) attains a relative maximum (see diagram). y 0 Relative Maximum f(x) mode mode x Random Variables and Distribution Functions 66 A mode of a random variable X is one of its most probable values. A random variable can have more than one mode. Example 3.20. Let X be a uniform random variable on the interval [0, 1], that is X U N IF (0, 1). How many modes does X have? ⇠ Answer: Since X ⇠ U N IF (0, 1), the probability density function of X is f (x) = 1 if 0 x 1   ( 0 otherwise. Hence the derivative of f (x) is f 0(x) = 0 (
0, 1). x 2 Therefore X has inﬁnitely many modes. Example 3.21. Let X be a Cauchy random variable with parameter ✓ = 0, that is X CAU (0). What is the mode of X? ⇠ Answer: Since X ⇠ CAU (0), the probability density function of f (x) is f (x) = 1 ⇡ (1 + x2) 1 < x <. 1 Hence f 0(x) = 2x ⇡ (1 + x2)2. Setting this derivative to 0, we get x = 0. Thus the mode of X is 0. Example 3.22. Let X be a continuous random variable with density function x2 e bx f (x) = 8 < 0 where b > 0. What is the mode of X? : for x 0 otherwise, x2be bx 0 = df dx = 2xe bx bx)x = 0. = (2 x = 0 or x = 2 b. Answer: Hence Probability and Mathematical Statistics 67 Thus the mode of X is 2 b. The graph of the f (x) for b = 4 is shown below. Example 3.23. A continuous random variable has density function f (x) = 3x2 ✓3 8 < 0 for 0 x ✓   otherwise, for some ✓ > 0. What is the ratio of the mode to the median for this distribution? : Answer: For ﬁxed ✓ > 0, the density function f (x) is an increasing function. Thus, f (x) has maximum at the right end point of the interval [0, ✓]. Hence the mode of this distribution is ✓. Next we compute the median of this distribution x3 ✓3 q3 ✓3   f (x) dx 3x2 ✓3 dx q 0 . Hence q = 2 1 3 ✓. Thus the ratio of the mode of this distribution to the median is mode median = ✓ 1 3 ✓ 2 = 3p2. Random Variables and Distribution Functions 68 Example 3.24. A continuous random variable has density function f (x) = 3x2 ✓3 8 < 0 for 0 x ✓   otherwise, for some ✓ > 0. What is the probability of X less than the ratio of
the mode to the median of this distribution? : Answer: In the previous example, we have shown that the ratio of the mode to the median of this distribution is given by a := mode median = 3p2. Hence the probability of X less than the ratio of the mode to the median of this distribution is a P (X < a) = f (x) dx 0 Z a 3x2 ✓3 dx a 0 0 Z x3 ✓3  a3 ✓3 3p2 ✓3 = 3 2 ✓3. = = = = 3.5. Review Exercises 1. Let the random variable X have the density function f (x) = 8 < 0 k x for 0 x   2 k q elsewhere. If the mode of this distribution is at x = p2 : 4, then what is the median of X? 2. The random variable X has density function f (x) = c xk+1 (1 x)k for 0 < x < 1 8 < 0 otherwise, : where c > 0 and 1 < k < 2. What is the mode of X? Probability and Mathematical Statistics 69 3. The random variable X has density function (k + 1) x2 for 0 < x < 1 f (x) = 8 < 0 otherwise, where k is a constant. What is the median of X? : 4. What are the median, and mode, respectively, for the density function f (x) = 1 ⇡ (1 + x2), < x <? 1 1 5. What is the 10th percentile of the random variable X whose probability density function is 1 ✓ e x ✓ if x 0, ✓ > 0 f (x) = ( 0 elsewhere? 6. What is the median of the random variable X whose probability density function is 1 2 e x 2 if x 0 f (x) = 7. A continuous random variable X has the density ( 0 elsewhere? f (x) = 3 x2 8 8 < 0 for 0 x 2   otherwise. What is the probability that X is greater than its 75th percentile? : 8. What is the probability density function of the random variable X if its cumulative distribution function is given by F (x) = 8 >< 0.0 if x < 2 0.5 if 2 0.7 if 3 1.0 if
x x < 3 x < ⇡ ⇡?   9. Let the distribution of X for x > 0 be >: F (x) = 1 3 Xk=0 xk e x k!. Random Variables and Distribution Functions 70 What is the density function of X for x > 0? 10. Let X be a random variable with cumulative distribution function F (x) = What is the P 0 eX   4? x e for x > 0 for x 0.  1 0 8 < : 11. Let X be a continuous random variable with density function f (x) = a x2 e 10 x for x 0 8 < 0 otherwise, where a > 0. What is the probability of X greater than or equal to the mode of X? 12. Let the random variable X have the density function : k x for 0 x   2 k q elsewhere. f (x) = 8 < 0 : If the mode of this distribution is at x = p2 X less than the median of X? 4, then what is the probability of 13. The random variable X has density function (k + 1) x2 for 0 < x < 1 f (x) = 8 < 0 otherwise, where k is a constant. What is the probability of X between the ﬁrst and third quartiles? : 14. Let X be a random variable having continuous cumulative distribution function F (x). What is the cumulative distribution function Y = max(0, X)? 15. Let X be a random variable with probability density function f (x) = 2 3x for x = 1, 2, 3,.... What is the probability that X is even? Probability and Mathematical Statistics 71 16. An urn contains 5 balls numbered 1 through 5. Two balls are selected at random without replacement from the urn. If the random variable X denotes the sum of the numbers on the 2 balls, then what are the space and the probability density function of X? 17. A pair of six-sided dice is rolled and the sum is determined. If the random variable X denotes the sum of the numbers rolled, then what are the space and the probability density function of X? 18. Five digit codes are selected at random from the set {0, 1, 2,..., 9} with replacement. If
the random variable X denotes the number of zeros in randomly chosen codes, then what are the space and the probability density function of X? 19. A urn contains 10 coins of which 4 are counterfeit. Coins are removed from the urn, one at a time, until all counterfeit coins are found. If the random variable X denotes the number of coins removed to ﬁnd the ﬁrst counterfeit one, then what are the space and the probability density function of X? 20. Let X be a random variable with probability density function f (x) = 2c 3x for x = 1, 2, 3, 4,..., 1 for some constant c. What is the value of c? What is the probability that X is even? 21. If the random variable X possesses the density function f (x) = cx 0 ⇢ x if 0   otherwise, 2 then what is the value of c for which f (x) is a probability density function? What is the cumulative distribution function of X. Graph the functions f (x) and F (x). Use F (x) to compute P (1  22. The length of time required by students to complete a 1-hour exam is a random variable with a pdf given by 2). X  f (x) = cx2 + x 0 ⇢ x if 0   otherwise, 1 then what the probability a student ﬁnishes in less than a half hour? Random Variables and Distribution Functions 72 23. What is the probability of, when blindfolded, hitting a circle inscribed on a square wall? 24. Let f (x) be a continuous probability density function. Show that, for is also a probability every < µ < 1 density function. and > 0, the function 1 f 1 µ x 25. Let X be a random variable with probability density function f (x) and cumulative distribution function F (x). True or False? (a) f (x) can’t be larger than 1. (b) F (x) can’t be larger than 1. (c) f (x) can’t decrease. (d) F (x) can’t decrease. (e) f (x) can’t be negative. (f) F (x) can’t be negative
. (g) Area under f must be 1. (h) Area under F must be 1. (i) f can’t jump. (j) F can’t jump. Probability and Mathematical Statistics 73 Moments of Random Variables and Chebychev Inequality 74 Chapter 4 MOMENTS OF RANDOM VARIABLES AND CHEBYCHEV INEQUALITY 4.1. Moments of Random Variables In this chapter, we introduce the concepts of various moments of a random variable. Further, we examine the expected value and the variance of random variables in detail. We shall conclude this chapter with a discussion of Chebychev’s inequality. Deﬁnition 4.1. The nth moment about the origin of a random variable X, as denoted by E(X n), is deﬁned to be xn f (x) if X is discrete E (X n) = 8 >< RX Xx 2 1 1 R >: xn f (x) dx if X is continuous for n = 0, 1, 2, 3,...., provided the right side converges absolutely. If If n = 1, then E(X) is called the ﬁrst moment about the origin. n = 2, then E(X 2) is called the second moment of X about the origin. In general, these moments may or may not exist for a given random variable. If for a random variable, a particular moment does not exist, then we say that the random variable does not have that moment. For these moments to exist one requires absolute convergence of the sum or the integral. Next, we shall deﬁne two important characteristics of a random variable, namely the expected value and variance. Occasionally E (X n) will be written as E [X n]. Probability and Mathematical Statistics 75 4.2. Expected Value of Random Variables A random variable X is characterized by its probability density function, which deﬁnes the relative likelihood of assuming one value over the others. In Chapter 3, we have seen that given a probability density function f of a random variable X, one can construct the distribution function F of it through summation or integration. Conversely, the density function f (x) can be obtained as the marginal value or derivative of F (x). The density function can be used to infer a number of characteristics of the underlying random variable. The two
most important attributes are measures of location and dispersion. In this section, we treat the measure of location and treat the other measure in the next section. Deﬁnition 4.2. Let X be a random variable with space RX and probability density function f (x). The mean µX of the random variable X is deﬁned as x f (x) if X is discrete RX Xx 2 1 1 if the right hand side exists. R µX = 8 >< >: x f (x) dx if X is continuous The mean of a random variable is a composite of its values weighted by the corresponding probabilities. The mean is a measure of central tendency: the value that the random variable takes “on average.” The mean is also called the expected value of the random variable X and is denoted by E(X). The symbol E is called the expectation operator. The expected value of a random variable may or may not exist. Example 4.1. If X is a uniform random variable on the interval (2, 7), then what is the mean of X? E(X) = (2+7)/2 Moments of Random Variables and Chebychev Inequality 76 Answer: The density function of X is f (x) = 1 5 ( 0 if 2 < x < 7 otherwise. Thus the mean or the expected value of X is µX = E(X) 1 = x f (x) dx Z 1 = = = = 7 x 1 5 dx 2 Z 1 10 x2 7 2  (49 4) 1 10 45 10 = 9 2 = 2 + 7 2. In general, if X ⇠ U N IF (a, b), then E(X) = 1 2 (a + b). Example 4.2. If X is a Cauchy random variable with parameter ✓, that is X CAU (✓), then what is the expected value of X? ⇠ Answer: We want to ﬁnd the expected value of X if it exists. The expected value of X will exist if the integral IR xf (x)dx converges absolutely, that is R \|x f (x)\| dx <. 1 Z IR If this integral diverges, then the expected value of X does not exist. Hence, let us ﬁnd out if IR \|x f (x)\| dx converges or not. R Probability and Mathemat
� Example 4.4. A couple decides to have 3 children. If none of the 3 is a girl, they will try again; and if they still don’t get a girl, they will try once more. If the random variable X denotes the number of children the couple will have following this scheme, then what is the expected value of X? Answer: Since the couple can have 3 or 4 or 5 children, the space of the random variable X is RX = {3, 4, 5}. Probability and Mathematical Statistics 79 The probability density function of X is given by f (3) = P (X = 3) = P (at least one girl) P (no girls) P (3 boys in 3 tries) (P (1 boy in each try))4) = P (X = 4) =. = P (3 boys and 1 girl in last try) = (P (1 boy in each try))3 P (1 girl in last try) = = f (5) = P (X = 5 16 ◆. = P (4 boys and 1 girl in last try) + P (5 boys in 5 tries) = P (1 boy in each try)4 P (1 girl in last try) + P (1 boy in each try) 16 ◆. Hence, the expected value of the random variable is E(X) = x f (x) RX Xx 2 5 = x f (x) x=3 X = 3 14 16 = 3 f (3) + 4 f (4) + 5 f (5) 1 16 42 + 4 + 5 16 1 16 + 5 + 4 = = 51 16 = 3 3 16. Moments of Random Variables and Chebychev Inequality 80 Remark 4.2. We interpret this physically as meaning that if many couples have children according to this scheme, it is likely that the average family size would be near 3 3 16 children. Example 4.5. A lot of 8 TV sets includes 3 that are defective. If 4 of the sets are chosen at random for shipment to a hotel, how many defective sets can they expect? Answer: Let X be the random variable representing the number of defective TV sets in a shipment of 4. Then the space of the random variable X is RX = {0, 1, 2, 3}. Then the probability density function of X is given by f (x) = P (X = x) = P (x defective TV sets
in a shipment of four Hence, we have x = 0, 1, 2, 3. f (0) = f (1) = f (2) = f (3 70 30 70 30 70 5 70. Therefore, the expected value of X is given by E(X) = x f (x) RX Xx 2 3 = x f (x) 0 X = = f (1) + 2 f (2) + 3 f (3) 30 70 5 70 + 3 + 2 30 70 30 + 60 + 15 70 = = 105 70 = 1.5. Probability and Mathematical Statistics 81 Remark 4.3. Since they cannot possibly get 1.5 defective TV sets, it should be noted that the term “expect” is not used in its colloquial sense. Indeed, it should be interpreted as an average pertaining to repeated shipments made under given conditions. Now we prove a result concerning the expected value operator E. Theorem 4.1. Let X be a random variable with pdf f (x). If a and b are any two real numbers, then E(aX + b) = a E(X) + b. Proof: We will prove only for the continuous case. E(aX + b) = = 1 Z 1 1 (a x + b) f (x) dx a x f (x) dx + 1 b f (x) dx Z = a 1 1 Z x f (x) dx + b 1 Z 1 = aE(X) + b. To prove the discrete case, replace the integral by summation. This completes the proof. 4.3. Variance of Random Variables The spread of the distribution of a random variable X is its variance. Deﬁnition 4.4. Let X be a random variable with mean µX. The variance of X, denoted by V ar(X), is deﬁned as V ar(X) = E [ X µX ]2. It is also denoted by 2 X. The positive square root of the variance is called the standard deviation of the random variable X. Like variance, the standard deviation also measures the spread. The following theorem tells us how to compute the variance in an alternative way. Theorem 4.2. If X is a random variable with
mean µX and variance 2 X, then X = E(X 2) 2 ( µX )2. Moments of Random Variables and Chebychev Inequality 82 Proof: (X 2) = E(X 2) = E(X 2) µX ]2 2 µX X + µ2 X 2 µX E(X) + ( µX )2 2 µX µX + ( µX )2 ( µX )2. Theorem 4.3. If X is a random variable with mean µX and variance 2 X, then V ar(aX + b) = a2 V ar(X), where a and b are arbitrary real constants. Proof: V ar(a X + b) = E [ (a X + b) µaX+b ]2 ⌘ E(a X + b) ] µX+ µX ]2 µX ]2 ⌘ ⌘ = E ⇣ a2 [ X ⇣ = a2 E [ X ⇣ = a2 V ar(X). ⌘ b ]2 ⌘ Example 4.6. Let X have the density function f (x) = 2 x k2 for 0 x k   ( 0 otherwise. For what value of k is the variance of X equal to 2? Answer: The expected value of X is k E(X) = x f (x) dx x 2 x k2 dx 0 Z k 0 Z 2 3 k. = = Probability and Mathematical Statistics 83 E(X 2) = k 0 Z k x2 f (x) dx x2 2 x k2 dx = = 0 Z 2 4 k2. Hence, the variance is given by V ar(X) = E(X 2) = = 2 4 1 18 k2 k2. ( µX )2 k2 4 9 Since this variance is given to be 2, we get 1 18 k2 = 2 and this implies that k = ±6. But k is given to be greater than 0, hence k must be equal to 6. Example 4.7. If the probability density function of the random variable is f (x) = 1 0 8 < \|x\| for \|x\| < 1
otherwise, then what is the variance of X? : Answer: Since V ar(X) = E(X 2) moments of X. The ﬁrst moment of X is given by µ2 X, we need to ﬁnd the ﬁrst and second µX = E(X) 1 x f (x) dx Z 1 1 x (1 \|x\|) dx x (1 + x) dx + x (x + x2) dx +. x) dx 0 Z 1 0 Z x2) dx (x Moments of Random Variables and Chebychev Inequality 84 The second moment E(X 2) of X is given by E(X 2) = 1 x2 f (x) dx Z 1 1 x2 (1 \|x\|) dx . x2 (1 + x) dx + 1 x2 (1 x) dx (x2 + x3) dx + (x2 x3) dx 0 Z 1 0 Z Thus, the variance of X is given by V ar(X) = E(X 2) µ2. Example 4.8. Suppose the random variable X has mean µ and variance 2 > 0. What are the values of the numbers a and b such that a + bX has mean 0 and variance 1? Answer: The mean of the random variable is 0. Hence 0 = E(a + bX) = a + b E(X) = a + b µ. Thus a = b µ. Similarly, the variance of a + bX is 1. That is 1 = V ar(a + bX) = b2 V ar(X) = b2 2. Probability and Mathematical Statistics 85 Hence or b = 1 b = 1 and a = µ and a = µ . Example 4.9. Suppose X has the density function f (x) = 3 x2 0 ⇢ for 0 < x < 1 otherwise. What is the expected area of a random isosceles right triangle with hypotenuse X? Answer: Let ABC denote this random isosceles right triangle. Let AC = x. Then x2 4 The expected area of this random triangle is given by Area of ABC = x p2 1 2 = AB = BC = x p2 x
is P (\|X µ\|  2 ) 1 1 4 = 0.75 Hence, Chebychev inequality tells us that if we do not know the distribution of X, then P (\|X 2 ) is at least 0.75. µ\|  Lower the standard deviation, and the smaller is the spread of the distribution. If the standard deviation is zero, then the distribution has no spread. This means that the distribution is concentrated at a single point. In the literature, such distributions are called degenerate distributions. The above ﬁgure shows how the spread decreases with the decrease of the standard deviation. 4.5. Moment Generating Functions We have seen in Section 3 that there are some distributions, such as geometric, whose moments are diﬃcult to compute from the deﬁnition. A Moments of Random Variables and Chebychev Inequality 92 moment generating function is a real valued function from which one can generate all the moments of a given random variable. In many cases, it is easier to compute various moments of X using the moment generating function. Deﬁnition 4.5. Let X be a random variable whose probability density function is f (x). A real valued function M : IR IR deﬁned by! M (t) = E et X is called the moment generating function of X if this expected value exists h < t < h for some h > 0. for all t in the interval In general, not every random variable has a moment generating function. But if the moment generating function of a random variable exists, then it is unique. At the end of this section, we will give an example of a random variable which does not have a moment generating function. Using the deﬁnition of expected value of a random variable, we obtain the explicit representation for M (t) as et x f (x) if X is discrete et x f (x) dx RX Xx 2 1 1 R if X is continuous. M (t) = 8 >< >: Example 4.12. Let X be a random variable whose moment generating function is M (t) and n be any natural number. What is the nth derivative of M (t) at t = 0? Answer: Similarly, d dt M (t) = d dt E et X et X d
by E X 3 = 24 (1 0)5 = 24. Theorem 4.5. Let M (t) be the moment generating function of the random variable X. If M (t) = a0 + a1 t + a2 t2 + · · · + an tn + · · · (4.3) is the Taylor series expansion of M (t), then E (X n) = (n!) an for all natural number n. Proof: Let M (t) be the moment generating function of the random variable X. The Taylor series expansion of M (t) about 0 is given by M (t) = M (0) + M 0(0) 1! t + M 00(0) 2! t2 + M 000(0) 3! t3 + · · · + M (n)(0) n! tn + · · · Since E(X n) = M (n)(0) for n 1 and M (0) = 1, we have E(X 2) 2! M (t) = 1 + E(X) 1! t + t2 + E(X 3) 3! t3 + · · · + E(X n) n! tn + · · · (4.4) From (4.3) and (4.4), equating the coeﬃcients of the like powers of t, we obtain an = E (X n) n! which is This proves the theorem. E (X n) = (n!) an. Probability and Mathematical Statistics 97 Example 4.17. What is the 479th moment of X about the origin, if the 1 1+t? moment generating function of X is Answer The Taylor series expansion of M (t) = 1 long division (a technique we have learned in high school). 1+t can be obtained by using M (t) + ( t)2 + ( t)3 + · · · + ( t)n + · · · t + t2 t3 + t4 + · · · + ( 1)ntn + · · · Therefore an = ( 1)n and from this we obtain a479 = 1. By Theorem 4.5, E X 479 = (479!) a479 = 479!
Example 4.18. If the moment generating of a random variable X is M (t) = 1) e(t j j!, 1 j=0 X then what is the probability of the event X = 2? Answer: By examining the given moment generating function of X, it is easy to note that X is a discrete random variable with space RX = {0, 1, 2, · · ·, }. Hence by deﬁnition, the moment generating function of X is 1 (4.5) But we are given that M (t) = M (t) = = et j f (j). 1) e(t j j! 1 e j! et j. 1 j=0 X 1 j=0 X 1 j=0 X From (4.5) and the above, equating the coeﬃcients of etj, we get f (j) = 1 e j! for j = 0, 1, 2,...,. 1 Moments of Random Variables and Chebychev Inequality 98 Thus, the probability of the event X = 2 is given by P (X = 2) = f (2) = 1 e 2! = 1 2 e. Example 4.19. Let X be a random variable with E (X n) = 0.8 for n = 1, 2, 3,...,. 1 What are the moment generating function and probability density function of X? Answer: M (t) = M (0) + = M (0) + = 1 + 0.8 1 n=1 X 1 n=1 X 1 M (n)(0) E (X n) tn n! ◆ ✓ tn n! ✓ ◆ tn n! ◆ 1 n=1 ✓ X tn n! ◆ = 0.2 + 0.8 + 0.8 = 0.2 + 0.8 1 n=0 ✓ X = 0.2 e0 t + 0.8 e1 t. n=1 ✓ X tn n! ◆ Therefore, we get f (0) = P (X = 0) = 0.2 and f (1) = P (X = 1) = 0.8. Hence the moment generating function of X is M (t) = 0.2 + 0.8 et, and the probability density function of X is f (x)
= 0.2\| \|x for x = 0, 1 ( 0 otherwise. Example 4.20. If the moment generating function of a random variable X is given by M (t) = 5 15 et + 4 15 e2 t + 3 15 e3 t + 2 15 e4 t + 1 15 e5 t, Probability and Mathematical Statistics 99 then what is the probability density function of X? What is the space of the random variable X? Answer: The moment generating function of X is given to be M (t) = 5 15 et + 4 15 e2 t + 3 15 e3 t + 2 15 e4 t + 1 15 e5 t. This suggests that X is a discrete random variable. Since X is a discrete random variable, by deﬁnition of the moment generating function, we see that M (t) = et x f (x) RX Xx 2 = et x1 f (x1) + et x2 f (x2) + et x3 f (x3) + et x4 f (x4) + et x5 f (x5). Hence we have f (x1) = f (1) = f (x2) = f (2) = f (x3) = f (3) = f (x4) = f (4) = f (x5) = f (5) = 5 15 4 15 3 15 2 15 1 15. Therefore the probability density function of X is given by f (x) = 6 x 15 for x = 1, 2, 3, 4, 5 and the space of the random variable X is RX = {1, 2, 3, 4, 5}. Example 4.21. variable X is If the probability density function of a discrete random f (x) = 6 ⇡2 x2, for x = 1, 2, 3,...,, 1 then what is the moment generating function of X? Moments of Random Variables and Chebychev Inequality 100 Answer: If the moment generating function of X exists, then M (t) = = = = etx f (x) etx 2 p6 ⇡ x! 1 x=1 X 1 x=1 X 1 etx 6 ⇡2 x2 x=1 ✓ X 6 1 ⇡2 x=1 X ◆ etx x2. Now we show that the above inﬁnite series
diverges if t belongs to the interval ( h, h) for any h > 0. To prove that this series is divergent, we do the ratio test, that is lim n!1 ✓ an+1 an ◆ et (n+1) (n + 1)2 et n et (n + 1)2 = lim n!1 ✓ = lim n!1 ✓ ◆ n2 et n n2 et n ◆ 2 = lim n!1 et = et. ✓ n n + 1! ◆ For any h > 0, since et is not always less than 1 for all t in the interval h, h), we conclude that the above inﬁnite series diverges and hence for ( this random variable X the moment generating function does not exist. Notice that for the above random variable, E [X n] does not exist for any natural number n. Hence the discrete random variable X in Example 4.21 has no moments. Similarly, the continuous random variable X whose Probability and Mathematical Statistics 101 probability density function is f (x) = 1 x2 8 < 0 for 1 x <  1 otherwise, has no moment generating function and no moments. : In the following theorem we summarize some important properties of the moment generating function of a random variable. Theorem 4.6. Let X be a random variable with the moment generating function MX (t). If a and b are any two real constants, then MX+a(t) = ea t MX (t) Mb X (t) = MX (b t) M X+a b (t) = e a b t MX t b. ◆ ✓ (4.6) (4.7) (4.8) Proof: First, we prove (4.6). MX+a(t) = E et (X+a) ⌘ = E ⇣ et X+t a et X et a = E et X = et a E = et a MX (t). Similarly, we prove (4.7). Mb X (t) = E et (b X) = E ⇣ e(t b) X ⇣ = MX (t b). ⌘ ⌘ By using (4.6) and (4.7), we easily get (4.8). M X+a b (t) = M X
(t MX (t) t b. ◆ ✓ Moments of Random Variables and Chebychev Inequality 102 This completes the proof of this theorem. Deﬁnition 4.6. The nth factorial moment of a random variable X is E(X(X 2) · · · (X n + 1)). 1)(X Deﬁnition 4.7. The factorial moment generating function (FMGF) of X is denoted by G(t) and deﬁned as G(t) = E tX. It is not diﬃcult to establish a relationship between the moment generating function (MGF) and the factorial moment generating function (FMGF). The relationship between them is the following: G(t) = E tX = E eln tX = E eX ln t = M (ln t). ⇣ ⌘ Thus, if we know the MGF of a random variable, we can determine its FMGF and conversely. Deﬁnition 4.8. Let X be a random variable. The characteristic function (t) of X is deﬁned as (t) = E ei t X = E ( cos(tX) + i sin(tX) ) = E ( cos(tX) ) + i E ( sin(tX) ). The probability density function can be recovered from the characteristic function by using the following formula f (x) = 1 2 ⇡ 1 e i t x (t) dt. Z 1 Unlike the moment generating function, the characteristic function of a random variable always exists. For example, the Cauchy random variable X with probability density f (x) = ⇡(1+x2) has no moment generating function. However, the characteristic function is 1 (t) = E ei t X eitx ⇡(1 + x2) dx 1 = Z = e 1 \|t\|. Probability and Mathematical Statistics 103 To evaluate the above integral one needs the theory of residues from the complex analysis. The characteristic function (t) satisﬁes the same set of properties as the moment generating functions as given in Theorem 4.
6. The following integrals 1 xm e x dx = m! if m is a positive integer 0 Z and 1 px e x dx = p⇡ 2 0 Z are needed for some problems in the Review Exercises of this chapter. These formulas will be discussed in Chapter 6 while we describe the properties and usefulness of the gamma distribution. We end this chapter with the following comment about the Taylor’s series. Taylor’s series was discovered to mimic the decimal expansion of real numbers. For example 125 = 1 (10)2 + 2 (10)1 + 5 (10)0 is an expansion of the number 125 with respect to base 10. Similarly, 125 = 1 (9)2 + 4 (9)1 + 8 (9)0 is an expansion of the number 125 in base 9 and it is 148. Since given a function f : IR IR, f (x) is a real number and it can be expanded with respect to the base x. The expansion of f (x) with respect to base x will have a form IR and x! 2 f (x) = a0x0 + a1x1 + a2x2 + a3x3 + · · · which is f (x) = 1 akxk. Xk=0 If we know the coeﬃcients ak for k = 0, 1, 2, 3,..., then we will have the expansion of f (x) in base x. Taylor found the remarkable fact that the the coeﬃcients ak can be computed if f (x) is suﬃciently diﬀerentiable. He proved that for k = 1, 2, 3,... ak = f (k)(0) k! with f (0) = f (0). Moments of Random Variables and Chebychev Inequality 104 4.6. Review Exercises 1. In a state lottery a ﬁve-digit integer is selected at random. If a player bets 1 dollar on a particular number, the payoﬀ (if that number is selected) is $500 minus the $1 paid for the ticket. Let X equal the payoﬀ to the better. Find the expected value of X. 2. A discrete random variable X has probability density function of the form f (x) = c (8 ( 0 x) for x = 0, 1, 2
, 3, 4, 5 otherwise. (a) Find the constant c. (b) Find P (X > 2). (c) Find the expected value E(X) for the random variable X. 3. A random variable X has a cumulative distribution function F (x if 0 < x if a) Graph F (x). (b) Graph f (x). (c) Find P (X : 1.25). (f) Find P (X = 1.25). (e) Find P (X  0.5). (d) Find P (X 0.5).  4. Let X be a random variable with probability density function f (x) = 1 8 x ( 0 for x = 1, 2, 5 otherwise. (a) Find the expected value of X. (b) Find the variance of X. (c) Find the expected value of 2X + 3. (d) Find the variance of 2X + 3. (e) Find the expected value of 3X 5X 2 + 1. 5. The measured radius of a circle, R, has probability density function f (r) = 6 r (1 r) if 0 < r < 1 ( 0 otherwise. (a) Find the expected value of the radius. (b) Find the expected circumference. (c) Find the expected area. 6. Let X be a continuous random variable with density function ✓ x + 3 2 ✓ 3 2 x2 f (x) = 0 8 < : for 0 < x < 1 p✓ otherwise, Probability and Mathematical Statistics 105 where ✓ > 0. What is the expected value of X? 7. Suppose X is a random variable with mean µ and variance 2 > 0. For a X what value of a, where a > 0 is E minimized? 2 8. A rectangle is to be constructed having dimension X by 2X, where X is a random variable with probability density function ⇣⇥ ⌘ ⇤ 1 a f (x) = 1 2 ( 0 for 0 < x < 2 otherwise. What is the expected area of the rectangle? 9. A box is to be constructed so that the height is 10 inches and its base is X inches by X inches. If X has a uniform distribution over the interval [2, 8], then what is the expected volume of the box in cubic inches? 10. If X
is a random variable with density function 1.4 e 2x + 0.9 e 3x f (x) = 8 < 0 then what is the expected value of X? : for x > 0 elsewhere, 11. A fair coin is tossed. If a head occurs, 1 die is rolled; if a tail occurs, 2 dice are rolled. Let X be the total on the die or dice. What is the expected value of X? If velocities of the molecules of a gas have the probability density 12. (Maxwell’s law) f (v) = a v2e h2 v2 8 < 0 for v 0 otherwise, then what are the expectation and the variance of the velocity of the molecules and also the magnitude of a for some given h? : 13. A couple decides to have children until they get a girl, but they agree to stop with a maximum of 3 children even if they haven’t gotten a girl. If X and Y denote the number of children and number of girls, respectively, then what are E(X) and E(Y )? 14. In roulette, a wheel stops with equal probability at any of the 38 numbers 0, 00, 1, 2,..., 36. If you bet $1 on a number, then you win $36 (net gain is Moments of Random Variables and Chebychev Inequality 106 $35) if the number comes up; otherwise, you lose your dollar. What are your expected winnings? 15. If the moment generating function for the random variable X is MX (t) = 1 1+t, what is the third moment of X about the point x = 2? 16. If the mean and the variance of a certain distribution are 2 and 8, what are the ﬁrst three terms in the series expansion of the moment generating function? 17. Let X be a random variable with density function ax a e for x > 0 f (x) = 8 < 0 otherwise, where a > 0. If M (t) denotes the moment generating function of X, what is M ( 3a)? : 18. Suppose the random variable X has moment generating M (t) = 1 t)k, (1 What is the nth moment of X? for t < 1 . 19. Two balls are dropped in such a way that each ball is equally likely to fall into any one of four
holes. Both balls may fall into the same hole. Let X denote the number of unoccupied holes at the end of the experiment. What is the moment generating function of X? 20. If the moment generating function of X is M (t) = 1 t)2 for t < 1, then what is the fourth moment of X? (1 21. Let the random variable X have the moment generating function M (t) = e3t 1 t2, 1 < t < 1. What are the mean and the variance of X, respectively? 22. Let the random variable X have the moment generating function M (t) = e3t+t2. What is the second moment of X about x = 0? Probability and Mathematical Statistics 107 23. Suppose the random variable X has the cumulative density function c)2 is F (x). Show that the expected value of the random variable (X minimum if c equals the expected value of X. 24. Suppose the continuous random variable X has the cumulative density function F (x). Show that the expected value of the random variable \|X c\| is minimum if c equals the median of X (that is, F (c) = 0.5). 25. Let the random variable X have the probability density function f (x) = 1 2 \|x\| e 1 < x <. 1 What are the expected value and the variance of X? 26. If MX (t) = k (2 + 3et)4, what is the value of k? 27. Given the moment generating function of X as M (t) = 1 + t + 4t2 + 10t3 + 14t4 + · · · what is the third moment of X about its mean? 28. A set of measurements X has a mean of 7 and standard deviation of 0.2. For simplicity, a linear transformation Y = aX + b is to be applied to make the mean and variance both equal to 1. What are the values of the constants a and b? 29. A fair coin is to be tossed 3 times. The player receives 10 dollars if all three turn up heads and pays 3 dollars if there is one or no heads. No gain or loss is incurred otherwise. If Y is the gain of the player, what the expected value of Y? 30. If X has the probability density function f (x) = x e 0 ⇢ for x >
0 otherwise, then what is the expected value of the random variable Y = e 3 4 X + 6? 31. If the probability density function of the random variable X if f (x) = (1 8 < 0 p)x 1 p if x = 1, 2, 3,..., 1 otherwise, then what is the expected value of the random variable X : 1? Some Special Discrete Distributions 108 Chapter 5 SOME SPECIAL DISCRETE DISTRIBUTIONS Given a random experiment, we can ﬁnd the set of all possible outcomes which is known as the sample space. Objects in a sample space may not be numbers. Thus, we use the notion of random variable to quantify the qualitative elements of the sample space. A random variable is characterized by either its probability density function or its cumulative distribution function. The other characteristics of a random variable are its mean, variance and moment generating function. In this chapter, we explore some frequently encountered discrete distributions and study their important characteristics. 5.1. Bernoulli Distribution A Bernoulli trial is a random experiment in which there are precisely two possible outcomes, which we conveniently call ‘failure’ (F) and ‘success’ (S). We can deﬁne a random variable from the sample space {S, F } into the set of real numbers as follows: X(F ) = 0 X(S) = 1. Probability and Mathematical Statistics 109 Sample Space S F X X(F) = 0 1= X(S) The probability density function of this random variable is f (0) = P (X = 0) = 1 p f (1) = P (X = 1) = p, where p denotes the probability of success. Hence f (x) = px (1 p)1 x, x = 0, 1. Deﬁnition 5.1. The random variable X is called the Bernoulli random variable if its probability density function is of the form f (x) = px (1 p)1 x, x = 0, 1 where p is the probability of success. We denote the Bernoulli random variable by writing X BER(p). ⇠ Example 5.1. What is the probability of getting a score of not less than 5 in a throw of a six-sided die? Answer: Although there are six possible scores {1,
2, 3, 4, 5, 6}, we are grouping them into two sets, namely {1, 2, 3, 4} and {5, 6}. Any score in {1, 2, 3, 4} is a failure and any score in {5, 6} is a success. Thus, this is a Bernoulli trial with P (X = 0) = P (failure) = 4 6 and P (X = 1) = P (success) = 2 6. Hence, the probability of getting a score of not less than 5 in a throw of a six-sided die is 2 6. Some Special Discrete Distributions 110 Theorem 5.1. If X is a Bernoulli random variable with parameter p, then the mean, variance and moment generating functions are respectively given by µX = p 2 X = p (1 MX (t) = (1 p) p) + p et. Proof: The mean of the Bernoulli random variable is 1 µX = x f (x) x=0 X 1 = x px (1 x=0 X = p. Similarly, the variance of X is given by p)1 x 2 X = 1 (x x=0 X 1 = (x µX )2 f (x) p)2 px (1 p)1 x x=0 X = p2 (1 = p (1 = p (1 p). p) + p (1 p) [p + (1 p)2 p)] Next, we ﬁnd the moment generating function of the Bernoulli random variable M (t) = E etX 1 etx px (1 = x=0 X = (1 p) + et p. p)1 x This completes the proof. The moment generating function of X and all the moments of X are shown below for p = 0.5. Note that for the Bernoulli distribution all its moments about zero are same and equal to p. Probability and Mathematical Statistics 111 5.2. Binomial Distribution Consider a ﬁxed number n of mutually independent Bernoulli trails. Suppose these trials have same probability of success, say p. A random variable X is called a binomial random variable if it represents the
total number of successes in n independent Bernoulli trials. Now we determine the probability density function of a binomial random variable. Recall that the probability density function of X is deﬁned as f (x) = P (X = x). Thus, to ﬁnd the probability density function of X we have to ﬁnd the probability of x successes in n independent trails. If we have x successes in n trails, then the probability of each n-tuple with x successes and n x failures is px (1 p)n x. tuples with x successes and n n x x failures in n trials. However, there are Hence P (X = x) = n x ✓ ◆ px (1 p)n x. Therefore, the probability density function of X is f (x) = n x ✓ ◆ px (1 p)n x, x = 0, 1,..., n. Deﬁnition 5.2. The random variable X is called the binomial random variable with parameters p and n if its probability density function is of the form f (x) = n x ✓ ◆ px (1 p)n x, x = 0, 1,..., n Some Special Discrete Distributions 112 where 0 < p < 1 is the probability of success. We will denote a binomial random variable with parameters p and n as BIN (n, p). ⇠ X Example 5.2. Is the real valued function f (x) given by f (x) = n x ✓ ◆ px (1 p)n x, x = 0, 1,..., n where n and p are parameters, a probability density function? Answer: To answer this question, we have to check that f (x) is nonnegative and 0. We show that sum is one. n x=0 f (x) is 1. It is easy to see that f (x) P n n f (x) = x=0 X n x ◆ x=0 ✓ X = (p + 1 = 1. p)n x px (1 p)n Hence f (x) is really a probability density function. Example 5.3. On a ﬁve-question multiple-
choice test there are ﬁve possible answers, of which one is correct. If a student guesses randomly and independently, what is the probability that she is correct only on questions 1 and 4? Answer: Here the probability of success is p = 1 Therefore, the probability that she is correct on questions 1 and 4 is 5, and thus 1 p = 4 5. P (correct on questions 1 and 4) =p2(1 p)3 2 3 4 5 1 5 ✓ ✓ 64 55 = 0.02048. ◆ ◆ = = Probability and Mathematical Statistics 113 Example 5.4. On a ﬁve-question multiple-choice test there are ﬁve possible answers, of which one is correct. If a student guesses randomly and independently, what is the probability that she is correct only on two questions? 5. There diﬀerent ways she can be correct on two questions. Therefore, the Answer: Here the probability of success is p = 1 are probability that she is correct on two questions is 5, and thus correct on two questions) = 5 2 ◆ ✓ p2(1 p)3 = 10 1 5 ✓ 2 ◆ ✓ 4 5 3 ◆ = 640 55 = 0.2048. Example 5.5. What is the probability of rolling two sixes and three nonsixes in 5 independent casts of a fair die? Answer: Let the random variable X denote the number of sixes in 5 independent casts of a fair die. Then X is a binomial random variable with probability of success p and n = 5. The probability of getting a six is p = 1 6. Hence P (X = 2) = f (2 ◆ = 10 1 36 ✓ ◆ ✓ 125 216 ◆ = 1250 7776 = 0.160751. Example 5.6. What is the probability of rolling at most two sixes in 5 independent casts of a fair die? Answer: Let the random variable X denote number of sixes in 5 independent casts of a fair die. Then X is a binomial random variable with probability of success p and n = 5. The probability of getting a six is p = 1 6. Hence, the Some Special Discrete Distributions 114 probability of rolling at most two sixes is P (X  2) = F (2) = f (0) + f (1) + f (2 Xk=0 ✓ 1
�. Since, the random variable represents the number of successes in 2000 independent trials, the random variable X is a binomial with parameters p = ⇡ 4 and n = 2000, that is X BIN (2000, ⇡ 4 ). ⇠ Some Special Discrete Distributions 116 Hence by Theorem 5.2, µX = 2000 ⇡ 4 = 1570.8, and 2 X = 2000 1 The standard deviation of X is ⇣ ⇡ 4 ⇡ 4 ⌘ = 337.1. X = p337.1 = 18.36. Example 5.8. Let the probability that the birth weight (in grams) of babies in America is less than 2547 grams be 0.1. If X equals the number of babies that weigh less than 2547 grams at birth among 20 of these babies selected at random, then what is P (X 3)?  Answer: If a baby weighs less than 2547, then it is a success; otherwise it is a failure. Thus X is a binomial random variable with probability of success p and n = 20. We are given that p = 0.1. Hence 3 3) = P (X  20 k 20 k ◆ ✓ k 1 10 9 10 ◆ ✓ ◆ (from table). Xk=0 ✓ = 0.867 Example 5.9. Let X1, X2, X3 be three independent Bernoulli random variables with the same probability of success p. What is the probability density function of the random variable X = X1 + X2 + X3? Answer: The sample space of the three independent Bernoulli trials is S = {F F F, F F S, F SF, SF F, F SS, SF S, SSF, SSS}. Probability and Mathematical Statistics 117 The random variable X = X1 + X2 + X3 represents the number of successes in each element of S. The following diagram illustrates this. FFF FFS FSF SFF FSS SFS SSF SSSS Sum of three Bernoulli Trials X R x 0 1 2 3 p)3 Let p be the probability of success. Then f (0) = P (X = 0) = P (F F F ) = (1 f (1) = P (X = 1) = P (F F S) + P (F SF ) + P (SF
F ) = 3 p (1 f (2) = P (X = 2) = P (F SS) + P (SF S) + P (SSF ) = 3 p2 (1 f (3) = P (X = 3) = P (SSS) = p3. p)2 p) Hence Thus f (x) = 3 x ✓ ◆ px (1 p)3 x, x = 0, 1, 2, 3. X ⇠ BIN (3, p). In general, if Xi ⇠ BER(p), then n i=1 Xi ⇠ BIN (n, p) and hence and n P E = n p Xi! i=1 X V ar n i=1 X Xi! = n p (1 p). The binomial distribution can arise whenever we select a random sample of n units with replacement. Each unit in the population is classiﬁed into one of two categories according to whether it does or does not possess a certain property. For example, the unit may be a person and the property may be Some Special Discrete Distributions 118 whether he intends to vote “yes”. If the unit is a machine part, the property may be whether the part is defective and so on. If the proportion of units in the population possessing the property of interest is p, and if Z denotes the number of units in the sample of size n that possess the given property, then BIN (n, p). Z ⇠ 5.3. Geometric Distribution If X represents the total number of successes in n independent Bernoulli trials, then the random variable X ⇠ BIN (n, p), where p is the probability of success of a single Bernoulli trial and the probability density function of X is given by f (x) = n x ✓ ◆ px (1 p)n x, x = 0, 1,..., n. Let X denote the trial number on which the ﬁrst success occurs. Geometric Random Variable X FFFFFFS FFFFFS FFFFS Sample Space 1 2 3 4 5 6 7 Space of the random variable Hence the probability that the ﬁrst success occurs on xth trial is given by f (x) = P (X = x) = (1 p)x 1 p. Hence, the probability
density function of X is f (x) = (1 p)x 1 p x = 1, 2, 3,...,, 1 where p denotes the probability of success in a single Bernoulli trial. Probability and Mathematical Statistics 119 Deﬁnition 5.3. A random variable X has a geometric distribution if its probability density function is given by f (x) = (1 p)x 1 p x = 1, 2, 3,...,, 1 where p denotes the probability of success in a single Bernoulli trial. If X has a geometric distribution we denote it as X GEO(p). ⇠ Example 5.10. Is the real valued function f (x) deﬁned by f (x) = (1 p)x 1 p x = 1, 2, 3,..., 1 where 0 < p < 1 is a parameter, a probability density function? Answer: It is easy to check that f (x) is one. 0. Thus we only show that the sum f (x) = 1 (1 p)x 1 p 1 x=1 X x=1 X 1 = p = p y=0 X 1 1 (1 p) = 1. (1 p)y, where y = x 1 Hence f (x) is a probability density function. Example 5.11. The probability that a machine produces a defective item is 0.02. Each item is checked as it is produced. Assuming that these are independent trials, what is the probability that at least 100 items must be checked to ﬁnd one that is defective? Some Special Discrete Distributions 120 Answer: Let X denote the trial number on which the ﬁrst defective item is observed. We want to ﬁnd P (X 100) = = 1 f (x) x=100 X 1 (1 x=100 X p)x 1 p (1 p)y p = (1 p)99 1 y=0 X p)99 = (1 = (0.98)99 = 0.1353. Hence the probability that at least 100 items must be checked to ﬁnd one that is defective is 0.1353. Example 5.12. A gambler plays roulette at Monte
2 (1 p) et = = Hence p et [1 (1 [1 p) et + (1 p)et]2 (1 p) et] p et (1 [1 p)et]2. µX = E(X) = M 0(0) = 1 p. Similarly, the second derivative of M (t) can be obtained from the ﬁrst derivative as M 00(t) = (1 [1 p) et]2 p et + p et 2 [1 [1 (1 (1 p)et]4 p) et] (1 p) et. Hence M 00(0) = p3 + 2 p2(1 p4 p) 2 = p. p2 Therefore, the variance of X is ( M 0(0) )2 1 p2 2 X = M 00(0) p 2 p2 p = 1 =. p2 This completes the proof of the theorem. Theorem 5.4. The random variable X is geometric if and only if it satisﬁes the memoryless property, that is P (X > m + n / X > n) = P (X > m) for all natural numbers n and m. Proof: It is easy to check that the geometric distribution satisﬁes the lack of memory property P (X > m + n / X > n) = P (X > m) Probability and Mathematical Statistics 123 which is P (X > m + n and X > n) = P (X > m) P (X > n). (5.1) If X is geometric, that is X (1 ⇠ p)x 1 p, then P (X > n + m) = 1 (1 p)x 1 p x=n+m+1 X = (1 = (1 p)n+m p)n (1 p)m = P (X > n) P (X > m). Hence the geometric distribution has the lack of memory property. Let X be a random variable which satisﬁes the lack of memory property, that is P (X >
m + n and X > n) = P (X > m) P (X > n). We want to show that X is geometric. Deﬁne g : N IR by! g(n) := P (X > n) Using (5.2) in (5.1), we get g(m + n) = g(m) g(n) m, n N, 2 8 (5.2) (5.3) since P (X > m + n and X > n) = P (X > m + n). Letting m = 1 in (5.3), we see that g(n + 1) = g(n) g(1) = g(n = g(n = · · · 1) g(1)2 2) g(1)3 · · · 1)) g(1)n = g(n (n = g(1)n+1 = an+1, where a is an arbitrary constant. Hence g(n) = an. From (5.2), we get F (n) = P (X > n) = an 1 Some Special Discrete Distributions 124 and thus Since F (n) is a distribution function F (n) = 1 an. 1 = lim n!1 F (n) = lim!1 n (1 an). From the above, we conclude that 0 < a < 1. We rename the constant a as (1 p). Thus, F (n) = 1 (1 p)n. The probability density function of X is hence f (1) = F (1) = p f (2) = F (2) f (3) = F (3) · · · F (1) = 1 F (2) = 1 · · · (1 (1 p)2 p)3 1 + (1 1 + (1 f (x) = F (x) F (x 1) = (1 p)x 1 p. p) = (1 p)2 = (1 p) p p)2 p Thus X is geometric with parameter p. This completes the proof. The diﬀerence between the
binomial and the geometric distributions is the following. In binomial distribution, the number of trials was predetermined, whereas in geometric it is the random variable. 5.4. Negative Binomial Distribution Let X denote the trial number on which the rth success occurs. Here r is a positive integer greater than or equal to one. This is equivalent to saying that the random variable X denotes the number of trials needed to observe the rth successes. Suppose we want to ﬁnd the probability that the ﬁfth head is observed on the 10th independent ﬂip of an unbiased coin. This is a case of ﬁnding P (X = 10). Let us ﬁnd the general case P (X = x). P (X = x) = P (ﬁrst x 1 trials contain x r failures and r 1 successes) · P (rth success in x th trial pr (1 1 (1 p)x p)x r p pr r, r + 1,...,. 1 Probability and Mathematical Statistics 125 Hence the probability density function of the random variable X is given by f (x) = x r ✓ 1 1 ◆ pr (1 p)x r, x = r, r + 1,...,. 1 Notice that this probability density function f (x) can also be expressed as f (x pr (1 ◆ p)x, x = 0, 1,...,. 1 X SSSS FSSSS SFSSS SSFSS SSSFS FFSSSS FSFSSS FSSFSS FSSSFS X is NBIN(4,P Deﬁnition 5.4. A random variable X has the negative binomial (or Pascal) distribution if its probability density function is of the form f (x) = x r ✓ 1 1 ◆ pr (1 p)x r, x = r, r + 1,...,, 1 where p is the probability of success in a single Bernoulli trial. We denote the random variable X whose distribution is negative binomial distribution by writing X N BIN (r, p). ⇠ We need the following technical result to show that the above function is really a probability density function. The technical result we are going to establish is called the negative binomial theorem. Some Special Disc
Mathematical Statistics 131 Example 5.17. A random sample of 5 students is drawn without replacement from among 300 seniors, and each of these 5 seniors is asked if she/he has tried a certain drug. Suppose 50% of the seniors actually have tried the drug. What is the probability that two of the students interviewed have tried the drug? Answer: Let X denote the number of students interviewed who have tried the drug. Hence the probability that two of the students interviewed have tried the drug is P (X = 2) = 150 2 150 3 300 5 = 0.3146. Example 5.18. A radio supply house has 200 transistor radios, of which 3 are improperly soldered and 197 are properly soldered. The supply house randomly draws 4 radios without replacement and sends them to a customer. What is the probability that the supply house sends 2 improperly soldered radios to its customer? Answer: The probability that the supply house sends 2 improperly soldered Some Special Discrete Distributions 132 radios to its customer is P (X = 2) = 3 2 197 2 200 4 = 0.000895. Theorem 5.7. If X HY P (n1, n2, r), then ⇠ E(X) = r V ar(X) = r n1 n1 + n2 n1 n2 n1 + n2 ◆ ✓ n1 + n2 ◆ ✓ ✓ n1 + n2 n1 + n2 r 1. ◆ ⇠ Proof: Let X HY P (n1, n2, r). We compute the mean and variance of X by computing the ﬁrst and the second factorial moments of the random variable X. First, we compute the ﬁrst factorial moment (which is same as the expected value) of X. The expected value of X is given by r E(X) = x f (x) n2 x r n1+n2 r x)! = x=0 X r x=0 X x r = n1 x=1 X r = n1 n2 x r n1 x n1+n2 r (x (n1 1)! 1)! (n1 1 n
f (x) = x e x! Hence Therefore f (x) = 3x e 3 x!, x = 0, 1, 2,... P (1 X   3) = f (1) + f (2) + f (3) 27 6 = 3 e 3 + 3 + 9 2 e 3 e = 12 e 3. Example 5.21. The number of traﬃc accidents per week in a small city has a Poisson distribution with mean equal to 3. What is the probability of exactly 2 accidents occur in 2 weeks? Answer: The mean traﬃc accident is 3. Thus, the mean accidents in two weeks are = (3) (2) = 6. Some Special Discrete Distributions 136 Since we get f (x) = x e x! f (2) = 62 e 6 2! = 18 e 6. Example 5.22. Let X have a Poisson distribution with parameter = 1. What is the probability that X 2 given that X 4?  Answer: P (X 2 / X 4) =  P (2 X   4) = = = P (2 X  P (X 4).  4) 4  x e x! 1 x! 4 x=2 X 1 e x=2 X. 17 24 e Similarly P (X 4) =  1 e 4 1 x! x=0 X. 65 24 e = Therefore, we have P (X 2 / X 4) =  17 65. Example 5.23. If the moment generating function of a random variable X is M (t) = e4.6 (et 1), then what are the mean and variance of X? What is the probability that X is between 3 and 6, that is P (3 < X < 6)? Probability and Mathematical Statistics 137 Answer: Since the moment generating function of X is given by M (t) = e4.6 (et 1) we conclude that X ⇠ P OI() with = 4.6. Thus, by Theorem 5.8, we get E(X) = 4.6 = V ar(
X). P (3 < X < 6) = f (4) + f (5) = F (5) = 0.686 F (3) 0.326 5.7. Riemann Zeta Distribution = 0.36. The zeta distribution was used by the Italian economist Vilfredo Pareto (1848-1923) to study the distribution of family incomes of a given country. Deﬁnition 5.7. A random variable X is said to have Riemann zeta distribution if its probability density function is of the form f (x) = 1 ⇣(↵ + 1) x (↵+1), x = 1, 2, 3,..., 1 where ↵ > 0 is a parameter and ⇣(s ◆ + · · · is the well known the Riemann zeta function. A random variable having a RIZ(↵). Riemann zeta distribution with parameter ↵ will be denoted by X ⇠ The following ﬁgures illustrate the Riemann zeta distribution for the case ↵ = 2 and ↵ = 1. Some Special Discrete Distributions 138 The following theorem is easy to prove and we leave its proof to the reader. Theorem 5.9. If X RIZ(↵), then ⇠ E(X) = V ar(X) = ⇣(↵) ⇣(↵ + 1) ⇣(↵ 1)⇣(↵ + 1) (⇣(↵ + 1))2 (⇣(↵))2. Remark 5.1. If 0 < ↵ parameter ↵ 1, then ⇣(↵) = 1, then the variance of X is inﬁnite. 1 . Hence if X  RIZ(↵) and the ⇠ 5.8. Review Exercises 1. What is the probability of getting exactly 3 heads in 5 ﬂips of a fair coin? 2. On six successive ﬂips of a fair coin, what is the probability of observing 3 heads and 3 tails? 3. What is the probability that in 3 rolls of a pair of six-sided dice, exactly one total of 7 is rolled? 4. What is the probability of getting exactly four “sixes” when a
die is rolled 7 times? 5. In a family of 4 children, what is the probability that there will be exactly two boys? 6. If a fair coin is tossed 4 times, what is the probability of getting at least two heads? 7. In Louisville the probability that a thunderstorm will occur on any day during the spring is 0.05. Assuming independence, what is the probability that the ﬁrst thunderstorm occurs on April 5? (Assume spring begins on March 1.) 8. A ball is drawn from an urn containing 3 white and 3 black balls. After the ball is drawn, it is then replaced and another ball is drawn. This goes on indeﬁnitely. What is the probability that, of the ﬁrst 4 balls drawn, exactly 2 are white? 9. What is the probability that a person ﬂipping a fair coin requires four tosses to get a head? 10. Assume that hitting oil at one drilling location is independent of another, and that, in a particular region, the probability of success at any individual Probability and Mathematical Statistics 139 location is 0.3. Suppose the drilling company believes that a venture will be proﬁtable if the number of wells drilled until the second success occurs is less than or equal to 7. What is the probability that the venture will be proﬁtable? 11. Suppose an experiment consists of tossing a fair coin until three heads occur. What is the probability that the experiment ends after exactly six ﬂips of the coin with a head on the ﬁfth toss as well as on the sixth? 12. Customers at Fred’s Cafe wins a $100 prize if their cash register receipts show a star on each of the ﬁve consecutive days Monday, Tuesday,..., Friday in any one week. The cash register is programmed to print stars on a randomly selected 10% of the receipts. If Mark eats at Fred’s Cafe once each day for four consecutive weeks, and if the appearance of the stars is an independent process, what is the probability that Mark will win at least $100? 13. If a fair coin is tossed repeatedly, what is the probability that the third head occurs on the nth toss? 14. Suppose 30 percent of all electrical fuses manufactured by a certain company fail to meet municipal building standards. What is the probability that in a random sample of 10 fuses, exactly 3 will fail to meet municipal
building standards? 15. A bin of 10 light bulbs contains 4 that are defective. If 3 bulbs are chosen without replacement from the bin, what is the probability that exactly k of the bulbs in the sample are defective? 16. Let X denote the number of independent rolls of a fair die required to 6)? obtain the ﬁrst “3”. What is P (X 17. The number of automobiles crossing a certain intersection during any time interval of length t minutes between 3:00 P.M. and 4:00 P.M. has a Poisson distribution with mean t. Let W be time elapsed after 3:00 P.M. before the ﬁrst automobile crosses the intersection. What is the probability that W is less than 2 minutes? 18. In rolling one die repeatedly, what is the probability of getting the third six on the xth roll? 19. A coin is tossed 6 times. What is the probability that the number of heads in the ﬁrst 3 throws is the same as the number in the last 3 throws? Some Special Discrete Distributions 140 20. One hundred pennies are being distributed independently and at random into 30 boxes, labeled 1, 2,..., 30. What is the probability that there are exactly 3 pennies in box number 1? 21. The density function of a certain random variable is (0.2)4x (0.8)22 4x f (x) = 22 4x ( 0 if x = 0, 1 4, 2 4, · · ·, 22 4 otherwise. What is the expected value of X 2? 22. If MX (t) = k (2 + 3et)100, what is the value of k? What is the variance of the random variable X? 3, what is the value of k? What is the variance of et 23. If MX (t) = k 5et the random variable X? 7 ⇣ ⌘ 24. If for a Poisson distribution 2f (0) + f (2) = 2f (1), what is the mean of the distribution? 25. The number of hits, X, per baseball game, has a Poisson distribution. If the probability of a no-hit game is 1 3, what is the probability of having 2 or more hits in speciﬁed game? 26. Suppose X has a Poisson distribution with a standard
deviation of 4. 1? What is the conditional probability that X is exactly 1 given that X 27. A die is loaded in such a way that the probability of the face with j dots turning up is proportional to j2 for j = 1, 2, 3, 4, 5, 6. What is the probability of rolling at most three sixes in 5 independent casts of this die? 28. A die is loaded in such a way that the probability of the face with j dots turning up is proportional to j2 for j = 1, 2, 3, 4, 5, 6. What is the probability of getting the third six on the 7th roll of this loaded die? Probability and Mathematical Statistics 141 Some Special Continuous Distributions 142 Chapter 6 SOME SPECIAL CONTINUOUS DISTRIBUTIONS In this chapter, we study some well known continuous probability density functions. We want to study them because they arise in many applications. We begin with the simplest probability density function. 6.1. Uniform Distribution  Let the random variable X denote the outcome when a point is selected at random from an interval [a, b]. We want to ﬁnd the probability of the event X x, that is we would like to determine the probability that the point selected from [a, b] would be less than or equal to x. To compute this probability, we need a probability measure µ that satisﬁes the three axioms of Kolmogorov (namely nonnegativity, normalization and countable additivity). For continuous variables, the events are interval or union of intervals. The length of the interval when normalized satisﬁes all the three axioms and thus it can be used as a probability measure for one-dimensional random variables. Hence P (X x) =  length of [a, x] length of [a, b]. Thus, the cumulative distribution function F is x) = F (x) = P (X where a and b are any two real constants with a < b. To determine the probability density function from cumulative density function, we calculate the derivative of F (x). Hence    bx) = d dx F (x. Probability and Mathematical Statistics 143 Deﬁnition 6.1. A random variable X is said to be uniform on the interval [a, b] if its probability density function is of
the form f (x, 1 where a and b are constants. We denote a random variable X with the uniform distribution on the interval [a, b] as X U N IF (a, b). ⇠ The uniform distribution provides a probability model for selecting points at random from an interval [a, b]. An important application of uniform distribution lies in random number generation. The following theorem gives the mean, variance and moment generating function of a uniform random variable. Theorem 6.1. If X is uniform on the interval [a, b] then the mean, variance and moment generating function of X are given by E(X) = b + a 2 V ar(X) = M (t) = (b 8 < : a)2 12 1 etb t (b eta a), b if t = 0 if t = 0 E(X) = x f (x) dx Proofb + a). b 1 2 1 a x2 2 dx b a = = = 6 Some Special Continuous Distributions 144 E(X 2 x2 f (x) dx x2 b dx b 1 a x3 3 3 (b  b3 a a3 a) (b2 + ba + a2). 3 (b 1 3 a) (b2 + ba + a2) Hence, the variance of X is given by V ar(X) = E(X 2) ( E(X) )2 (b2 + ba + a2) = = = = 1 3 1 12 1 12 1 12 4b2 + 4ba + 4a2 2ba + a2 ⇥ b2 ⇥ (b a)2. ⇤ (b + a)2 4 3a2 3b2 6ba ⇤ Next, we compute the moment generating function of X. First, we handle the case t = 0. Assume t = 0. Hence M (t) = E etX b etx 1 b a etx t dx etb t (b a  eta a). If t = 0, we have know that M (0) = 1, hence we get 1 if t = 0 etb t (b eta a),
if t = 0 M (t) = 8 < : 6 6 6 Probability and Mathematical Statistics 145 and this completes the proof. Example 6.1. Suppose Y probability density function of X? ⇠ U N IF (0, 1) and Y = 1 4 X 2. What is the Answer: We shall ﬁnd the probability density function of X through the cumulative distribution function of Y. The cumulative distribution function of X is given by x) x2 = P F (x) = P ( x2 ◆ ◆  x2 4 f (y) dy dy ✓ x2 4 0 Z x2 4 0 Z x2 4. = = = Thus f (x) = d dx F (x) = x 2. Hence the probability density function of X is given by f (x) = x 2 ( 0 for 0 x 2   otherwise. Some Special Continuous Distributions 146 Example 6.2. If X has a uniform distribution on the interval from 0 to 10, then what is P X + 10 7? X U N IF (0, 10), the probability density function of X is 10. Hence Answer: Since X f (x) = 1 10 for 0 ⇠ x   P X + ✓ 10 X 7 ◆ = P X 2 + 10 7 X 7 X + 10 0 5) 5) (X 2) 0) 5) X   = P (X 2 or X = P X 2 = P (( 10 f (x) dx 1 10 dx = 7 10. Example 6.3. If X is uniform on the interval from 0 to 3, what is the probability that the quadratic equation 4t2 + 4tX + X + 2 = 0 has real solutions? Answer: Since X ⇠ U N IF (0, 3), the probability density function of X is f (x) = 1 3 ( 0 0 x 3   otherwise. Probability and Mathematical Statistics 147 The quadratic equation 4t2 + 4tX + X + 2 = 0 has real solution if the discriminant of this equation is positive. That is which is From this, we get 16X 2 16(X + 2) 0, X 2 X 2
0. (X 2) (X + 1) 0. The probability that the quadratic equation 4t2 + 4tX + X + 2 = 0 has real roots is equivalent to P ( (X 2) (X + 1) 0 ) = P (X = P (X 1 = Z 1 = 0 + 1 or X 1) + P (X 2) 2)   3 f (x) dx + f (x) dx 2 Z 3 1 3 dx 2 Z = 0.3333. = 1 3 Theorem 6.2. If X is a continuous random variable with a strictly increasing cumulative distribution function F (x), then the random variable Y, deﬁned by Y = F (X) has the uniform distribution on the interval [0, 1]. 1(x) of F (x) exists. We Proof: Since F is strictly increasing, the inverse F want to show that the probability density function g(y) of Y is g(y) = 1. First, we ﬁnd the cumulative distribution G(y) function of Y. G(y) = P (Y y)  = P (F (X(y) = y. y) 1(y) Some Special Continuous Distributions 148 Hence the probability density function of Y is given by g(y) = d dy G(y) = d dy y = 1. The following problem can be solved using this theorem but we solve it without this theorem. Example 6.4. If the probability density function of X is f (x) = x e (1 + e x)2, < x <, 1 1 then what is the probability density function of Y = 1 1+e X? Answer: The cumulative distribution function of Y is given by G(y) = P (Y y)  1 1 + e ln y 1 ◆ y ◆ x)2 dx 1 ln y x e (1 + e ln 1 y y 1 ✓ = P X e ln 1  1 x 1 + e. Hence, the probability density function of Y is given by f (y) = 1 if 0 <
y < 1 ( 0 otherwise. Probability and Mathematical Statistics 149 Example 6.5. A box to be constructed so that its height is 10 inches and its base is X inches by X inches. If X has a uniform distribution over the interval (2, 8), then what is the expected volume of the box in cubic inches? Answer: Since X U N IF (2, 8), ⇠ f (x) = = 2 1 6 1 8 on (2, 8). The volume V of the box is V = 10 X 2. Hence E(V ) = E dx = 10 E = 10 = = 10 6 10 18 10X 2 X 2 8 x2 1 6 8 2 Z x3 3 2  ⇥ = (5) (8) (7) = 280 cubic inches. 83 23 ⇤ Example 6.6. Two numbers are chosen independently and at random from the interval (0, 1). What is the probability that the two numbers diﬀers by more than 1 2? Answer: See ﬁgure below: Choose x from the x-axis between 0 and 1, and choose y from the y-axis between 0 and 1. The probability that the two numbers diﬀer by more than Some Special Continuous Distributions 150 1 2 is equal to the area of the shaded region. Thus P \|.2. Gamma Distribution The gamma distribution involves the notion of gamma function. First, we develop the notion of gamma function and study some of its well known properties. The gamma function, Γ(z), is a generalization of the notion of factorial. The gamma function is deﬁned as Γ(z) := 1 xz 0 Z 1 e x dx, where z is positive real number (that is, z > 0). The condition z > 0 is assumed for the convergence of the integral. Although the integral does not converge for z < 0, it can be shown by using an alternative deﬁnition of gamma function that it is deﬁned for all z IR \ {0, 3,...}. 1, 2, The integral on the right side of the above expression is called Euler’s second integral, after the Swiss mathematician Leonhard Euler (1707-1783). The graph of the gamma function is shown below. Observe that the zero and negative
(✓) r cos(✓) ✓ ◆ = r. Some Special Continuous Distributions 152 Hence, we get r2 e J(r, ✓) dr d✓ ◆◆ r2 e r dr d✓ 1 r2 e 2r dr d✓ 1 r2 e dr2 d✓ 0 Z ⇡ 2 0 Z Γ(1) d = ⇡. Therefore, we get Γ 1 2 ✓ ◆ = p⇡. 1 2 = 2 p⇡. Lemma 6.4. Γ Proof: By Lemma 6.2, we get Γ (z) = (z 1) Γ (z 1) for all z IR \ {1, 0, 1, 2, 2 3,...}. Letting z = 1 2, we get which is ⇡. Example 6.7. Evaluate Γ 5 2. Answer⇡. Example 6.8. Evaluate Γ 7 2. Probability and Mathematical Statistics 153 Answer: Consider. ◆ ✓ Hence ◆ ✓ = 16 105 ◆ ✓ p⇡. ◆ ✓ Example 6.9. Evaluate Γ (7.8). ◆ ✓ ✓ Answer: Γ (7.8) = (6.8) (5.8) (4.8) (3.8) (2.8) (1.8) Γ (1.8) = (3625.7) Γ (1.8) = (3625.7) (0.9314) = 3376.9. Here we have used the gamma table to ﬁnd Γ (1.8) to be 0.9314. Example 6.10. If n is a natural number, then Γ(n + 1) = n!. Answer: Γ(n + 1) = n Γ(n) 1) 2) Γ(n 1) Γ(n 1) (n · · · 2) 1) (n 2) · · · (1) Γ(1) = n (n = n (n
Hence from above, we ﬁnd the expected value of X to be E(X) = M 0(0) = ↵ ✓. Similarly, M 00(t) = d dt Thus, the variance of X is V ar(X) = M 00(0) ↵ ✓ (1 (↵+1) ✓t) ⇣ = ↵ ✓ (↵ + 1) ✓ (1 = ↵ (↵ + 1) ✓2 (1 ⌘ (↵+2) ✓t) ✓t) (↵+2). (M 0(0))2 ↵2 ✓2 = ↵ (↵ + 1) ✓2 = ↵ ✓2 and proof of the theorem is now complete In ﬁgure below the graphs of moment generating function for various values of the parameters are illustrated. Example 6.11. Let X have the density function f (x) = 1 Γ(↵) ✓↵ x↵ 1 e 0 8 < : x ✓ if 0 < x < 1 otherwise, Some Special Continuous Distributions 156 where ↵ > 0 and ✓ > 0. If ↵ = 4, what is the mean of 1 X 3? Answer! ✓4 1 3! ✓3 1 3! ✓3 1 x3 f (x) dx 1 1 x3 Γ(4) ✓4 x3 e x ✓ dx 1 e x ✓ dx 0 Z 0 Z 1 1 ✓ e x ✓ dx since the integrand is GAM(✓, 1). Deﬁnition 6.3. A continuous random variable is said to be an exponential random variable with parameter ✓ if its probability density function is of the form 1 ✓ e x ✓ f (x) = 8 < 0 if x > 0 otherwise, where ✓ > 0. If a random variable X has an exponential density function with parameter ✓, then we denote it by writing X EXP (✓). : ⇠ An exponential distribution is a special case of the gamma distribution. If the parameter ↵ = 1, then the gamma distribution reduces to the exponential distribution. Hence most of the information about an exponential distribution can be obtained from the gamma distribution. Example 6.12. What is the cumulative density function of a random variable which has an exponential distribution with variance 25? Probability and Mathematical Statistics 157 Answer
: Since an exponential distribution is a special case of the gamma distribution with ↵ = 1, from Theorem 6.3, we get V ar(X) = ✓2. But this is given to be 25. Thus, ✓2 = 25 or ✓ = 5. Hence, the probability density function of X is x F (x) = f (t) dt t 5 dt 1 5 t 5 x 0 i 5 e h e x 5. Example 6.13. If the random variable X has a gamma distribution with parameters ↵ = 1 and ✓ = 1, then what is the probability that X is between its mean and median? Answer: Since X ⇠ GAM(1, 1), the probability density function of X is f (x) = x e if x > 0 ( 0 otherwise. Hence, the median q of X can be calculated from x dx x e q 0 e q. ⇤ Hence 1 2 = 1 q e Some Special Continuous Distributions 158 and from this, we get q = ln 2. The mean of X can be found from the Theorem 6.3. E(X) = ↵ ✓ = 1. Hence the mean of X is 1 and the median of X is ln 2. Thus P (ln 2 X   1) = 1 e x dx ln 2 Z = ⇥ = e x e ln 2 1 ln Example 6.14. If the random variable X has a gamma distribution with parameters ↵ = 1 and ✓ = 2, then what is the probability density function of the random variable Y = eX? Answer: First, we calculate the cumulative distribution function G(y) of Y. G(y) = P ( Y y ) y   ln y ) e x 2 dx  1 2 2 e x 2 ln y 0 = P eX = P ( X ln ln y 1 py. = 1 Hence, the probability density function of Y is given by g(y) = d dy G(y) = d dy 1 ✓ 1 py ◆ = 1 2ypy. Probability and Mathematical Statistics 159 Thus, if X ⇠ GAM(1, 2), then probability density function of eX is f (x)
= 1 2xpx if 1 x <  1 ( 0 otherwise. Deﬁnition 6.4. A continuous random variable X is said to have a chi-square distribution with r degrees of freedom if its probability density function is of the form 1 Γ x 2 f (x) = 8 < 0 if 0 < x < 1 otherwise, where r > 0. If X has a chi-square distribution, then we denote it by writing X 2(r). : ⇠ The gamma distribution reduces to the chi-square distribution if ↵ = r 2 and ✓ = 2. Thus, the chi-square distribution is a special case of the gamma distribution. Further, if r, then the chi-square distribution tends to the normal distribution.! 1 Some Special Continuous Distributions 160 The chi-square distribution was originated in the works of British Statistician Karl Pearson (1857-1936) but it was originally discovered by German physicist F. R. Helmert (1843-1917). Example 6.15. If X function of the random variable 2X? ⇠ GAM(1, 1), then what is the probability density Answer: We will use the moment generating method to ﬁnd the distribution of 2X. The moment generating function of a gamma random variable is given by (see Theorem 6.3) M (t) = (1 ✓ t) ↵, if t < 1 ✓. GAM(1, 1), the moment generating function of X is given by Since X ⇠ Hence, the moment generating function of 2X is MX (t) = 1, t t < 1. 1 M2X (t) = MX (2t) 1 = 1 = 2t 1 2t) 2 = MGF of 2(2). (1 2 Hence, if X is an exponential with parameter 1, then 2X is chi-square with 2 degrees of freedom. Example 6.16. If X 1.145 and 12.83? ⇠ 2(5), then what is the probability that X is between Answer: The probability of X between 1.145 and 12.83 can be calculated from the following: P (1.145 X 12.83)   = P (X  12.83 12.83) P (X  1.145 1.145
) = = 0 Z 0 Z 12.83 f (x) dx .050 = 0.975 = 0.925. f (x) dx 0 Z 5 2 x 1 e 1.145 x 2 dx x 2 dx x (from 2 table) Probability and Mathematical Statistics 161 These integrals are hard to evaluate and so their values are taken from the chi-square table. Example 6.17. If X b such that P (a < X < b) = 0.95? ⇠ 2(7), then what are values of the constants a and Answer: Since 0.95 = P (a < X < b) = P (X < b) P (X < a), we get P (X < b) = 0.95 + P (X < a). We choose a = 1.690, so that P (X < 1.690) = 0.025. From this, we get P (X < b) = 0.95 + 0.025 = 0.975 Thus, from the chi-square table, we get b = 16.01. Deﬁnition 6.5. A continuous random variable X is said to have a n-Erlang distribution if its probability density function is of the form f (x) = e 1 x (x)n (n 1)!, 8 < 0 if 0 < x < 1 otherwise, where > 0 is a parameter. : The gamma distribution reduces to n-Erlang distribution if ↵ = n, where n is a positive integer, and ✓ = 1 . The gamma distribution can be generalized to include the Weibull distribution. We call this generalized distribution the uniﬁed distribution. The form of this distribution is the following: f (x) = 8 >< ↵ ✓↵ Γ(↵ +1) 0 x↵ 1 e x (↵ ✓ ↵ 1) , if 0 < x < 1 otherwise, where ✓ > 0, ↵ > 0, and >: {0, 1} are parameters. If = 0, the uniﬁed distribution reduces 2 f (x) = ↵ ✓ x↵ 1e 0 8 < : x↵ ✓, if 0 < x
= = = = x) 1 dx Γ(↵ + ) Γ(↵) Γ() Γ(↵ + ) Γ(↵) Γ() Similarly, we can show that ↵ (↵ + 1) (↵ + + 1) (↵ + ). E X 2 = Therefore V ar(X) = E X 2 E(X) = ↵ (↵ + )2 (↵ + + 1) and the proof of the theorem is now complete. Example 6.18. The percentage of impurities per batch in a certain chemical product is a random variable X that follows the beta distribution given by f (x) = 60 x3 (1 x)2 for 0 < x < 1 ( 0 otherwise. What is the probability that a randomly selected batch will have more than 25% impurities? Some Special Continuous Distributions 166 Proof: The probability that a randomly selected batch will have more than 25% impurities is given by P (X 0.25) = 1 60 x3 (1 x)2 dx 0.25 Z 1 = 60 = 60 = 60 0.25 Z x4 4 657 40960  x3 2x5 5 2x4 + x5 dx + x6 6 1 0.25 = 0.9624. Example 6.19. The proportion of time per day that all checkout counters in a supermarket are busy follows a distribution f (x) = k x2 (1 x)9 for 0 < x < 1 ( 0 otherwise. What is the value of the constant k so that f (x) is a valid probability density function? Proof: Using the deﬁnition of the beta function, we get that 1 x2 (1 0 Z x)9 dx = B(3, 10). Hence by Theorem 6.4, we obtain B(3, 10) = Γ(3) Γ(10) Γ(13) = 1 660. Hence k should be equal to 660. The beta distribution can be generalized to any bounded interval [a, b]. This generalized distribution is called the generalized beta distribution. If a random variable X has this generalized beta distribution
�� = P (Z  = 0.8944 = 0.4944. Y Z  2.50) 1.25)  1.25) 0.5000 P (Z 0)  Probability and Mathematical Statistics 177 Example 6.29. If the amount of time needed to solve a problem by a group of students follows the lognormal distribution with parameters µ and 2, then what is the value of µ so that the probability of solving a problem in 10 minutes or less by any randomly picked student is 95% when 2 = 4? Answer: Let the random variable X denote the amount of time needed \(µ, 4). We want to ﬁnd µ so that to a solve a problem. Then X 10) = 0.95. Hence P (X ⇠  V 0.95 = P (X 10)  = P (ln(X) ln(10))  = P (ln(X) µ ln(10) = P = P  µ ln(X) 2 ln(10) 2  Z ln(10) µ ◆ where Z ⇠ Hence N (0, 1). Using the table for standard normal distribution, we get ln(10) 2 µ = 1.65. µ = ln(10) 2(1.65) = 2.3025 3.300 = 0.9975. 6.6. Inverse Gaussian Distribution If a suﬃciently small macroscopic particle is suspended in a ﬂuid that is in thermal equilibrium, the particle will move about erratically in response to natural collisional bombardments by the individual molecules of the ﬂuid. This erratic motion is called “Brownian motion” after the botanist Robert Brown (1773-1858) who ﬁrst observed this erratic motion in 1828. Independently, Einstein (1905) and Smoluchowski (1906) gave the mathematical description of Brownian motion. The distribution of the ﬁrst passage time in Brownian motion is the inverse Gaussian distribution. This distribution was systematically studied by Tweedie in 1945. The interpurchase times of toothpaste of a family, the duration of
variable Y has a chi-square distribution with 54 degrees of freedom, then what is the approximate 84th percentile of Y? 18. Let X be a continuous random variable with density function f (x) = 2 x2 ( 0 for 1 < x < 2 elsewhere. If Y = pX, what is the density function for Y where nonzero? 4 X 0, that is P 19. If X is normal with mean 0 and variance 4, then what is the probability of the event X 20. If the waiting time at Rally’s drive-in-window is normally distributed with mean 13 minutes and standard deviation 2 minutes, then what percentage of customers wait longer than 10 minutes but less than 15 minutes? 4 X X 0? 21. If X is uniform on the interval from the quadratic equation 100t2 + 20tX + 2X + 3 = 0 has complex solutions? 5 to 5, what is the probability that 22. If the random variable X function of the random variable Y = XpX? ⇠ Exp(✓), then what is the probability density 23. If the random variable X function of the random variable Y = ⇠ N (0, 1), then what is the probability density \|X\|? p Some Special Continuous Distributions 184 If the random variable X 24. density function of the random variable ln(X)? ⇠ V \(µ, 2), then what is the probability 25. If the random variable X 26. If the random variable X ⇠ ⇠ \(µ, 2), then what is the mode of X? \(µ, 2), then what is the median of X? V \(µ, 2), then what is the probability that V 27. If the random variable X the quadratic equation 4t2 + 4tX + X + 2 = 0 has real solutions? V ⇠ 28. Consider the Karl Pearson’s diﬀerential equation p(x) dy where p(x) = a + bx + cx2 and q(x) = x b > 0, d > then y(x) is beta. b, then y(x) is gamma; and if a = 0, b = dx + q(x) y = 0 d. Show that if a = c = 0, 1 1, b < 1,

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