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Solution Manual For Introduction To Mathematical Statistics, 7th Edition

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    S OLUTIONS M ANUAL I NTRODUCTION TO M ATHEMATICAL S TATISTICS S EVENTH E DITION Robert Hogg University of Iowa Joseph McKean Western Michigan University Allen Craig

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    Contents 1 Probability and Distributions 1 2 Multivariate Distributions 11 3 Some Special Distributions 19 4 Some Elementary Statistical Inferences 31 5 Consistency and Limiting Distributions 49 6 Maximum Likelihood Methods 53 7 Sufficiency 65 8 Optimal Tests of Hypotheses 77 9 Inferences about Normal Models 83 10 Nonparametric and Robust Statistics 93 11 Bayesian Statistics 103 iii

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    Chapter 1 Probability and Distributions 1.2.1 Part (c): C 1 ∩ C 2 = { ( x, y ) : 1 < x < 2 , 1 < y < 2 } . 1.2.3 C 1 ∩ C 2 = { mary,mray } . 1.2.6 C k = { x : 1 /k ≤ x ≤ 1 − (1 /k ) } . 1.2.7 C k = { ( x, y ) : 0 ≤ x ≤ 1 /k, 0 ≤ y ≤ 1 /k } . 1.2.8 lim k →∞ C k = { x : 0 < x < 3 } . Note: neither the number 0 nor the number 3 is in any of the sets C k , k = 1 , 2 , 3 , . . . 1.2.9 Part (b): lim k →∞ C k = φ , because no point is in all the sets C k , k = 1 , 2 , 3 , . . . 1.2.11 Because f ( x ) = 0 when 1 ≤ x < 10, Q ( C 3 ) = ∫ 10 0 f ( x ) dx = ∫ 1 0 6 x (1 − x ) dx = 1 . 1.2.13 Part (c): Draw the region C carefully, noting that x < 2 / 3 because 3 x/ 2 < 1. Thus Q ( C ) = ∫ 2 / 3 0 [ ∫ 3 x/ 2 x/ 2 dy ] dx = ∫ 2 / 3 0 x dx = 2 / 9 . 1.2.16 Note that 25 = Q ( C ) = Q ( C 1 ) + Q ( C 2 ) − Q ( C 1 ∩ C 2 ) = 19 + 16 − Q ( C 1 ∩ C 2 ) . Hence, Q ( C 1 ∩ C 2 ) = 10. 1.2.17 By studying a Venn diagram with 3 intersecting sets, it should be true that 11 ≥ 8 + 6 + 5 − 3 − 2 − 1 = 13 . It is not, and the accuracy of the report should be questioned. 1

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    2 Probability and Distributions 1.3.3 P ( C ) = 1 2 + 1 4 + 1 8 + · · · = 1 / 2 1 − (1 / 2) = 1 . 1.3.6 P ( C ) = ∫ ∞ −∞ e −| x | dx = ∫ 0 −∞ e x dx + ∫ ∞ 0 e − x dx = 2 6 = 1 . We must multiply by 1 / 2. 1.3.8 P ( C c 1 ∪ C c 2 ) = P [( C 1 ∩ C 2 ) c ] = P ( C ) = 1 , because C 1 ∩ C 2 = φ and φ c = C . 1.3.11 The probability that he does not win a prize is ( 990 5 ) / ( 1000 5 ) . 1.3.13 Part (a): We must have 3 even or one even, 2 odd to have an even sum. Hence, the answer is ( 10 3 )( 10 0 ) ( 20 3 ) + ( 10 1 )( 10 2 ) ( 20 3 ) . 1.3.14 There are 5 mutual exclusive ways this can happen: two “ones”, two “twos”, two “threes”, two “reds”, two “blues.” The sum of the corresponding proba- bilities is ( 2 2 )( 6 0 ) + ( 2 2 )( 6 0 ) + ( 2 2 )( 6 0 ) + ( 5 2 )( 3 0 ) + ( 3 2 )( 5 0 ) ( 8 2 ) . 1.3.15 (a) 1 − ( 48 5 )( 2 0 ) ( 50 5 ) (b) 1 − ( 48 n )( 2 0 ) ( 50 n ) ≥ 1 2 , Solve for n . 1.3.20 Choose an integer n 0 > max { a − 1 , (1 − a ) − 1 } . Then { a } = ∩ ∞ n = n 0 ( a − 1 n , a + 1 n ) . Hence by (1.3.10), P ( { a } ) = lim n →∞ P [( a − 1 n , a + 1 n )] = 2 n = 0 . 1.4.2 P [( C 1 ∩ C 2 ∩ C 3 ) ∩ C 4 ] = P [ C 4 | C 1 ∩ C 2 ∩ C 3 ] P ( C 1 ∩ C 2 ∩ C 3 ) , and so forth. That is, write the last factor as P [( C 1 ∩ C 2 ) ∩ C 3 ] = P [ C 3 | C 1 ∩ C 2 ] P ( C 1 ∩ C 2 ) .

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    3 1.4.5 [( 4 3 )( 48 10 ) + ( 4 4 )( 48 9 )] / ( 52 13 ) [( 4 2 )( 48 11 ) + ( 4 3 )( 48 10 ) + ( 4 4 )( 48 9 )] / ( 52 13 ) . 1.4.10 P ( C 1 | C ) = (2 / 3)(3 / 10) (2 / 3)(3 / 10) + (1 / 3)(8 / 10) = 3 7 < 2 3 = P ( C 1 ) . 1.4.12 Part (c): P ( C 1 ∪ C c 2 ) = 1 − P [( C 1 ∪ C c 2 ) c ] = 1 − P ( C ∗ 1 ∩ C 2 ) = 1 − (0 . 4)(0 . 3) = 0 . 88 . 1.4.14 Part (d): 1 − (0 . 3) 2 (0 . 1)(0 . 6) . 1.4.16 1 − P ( T T ) = 1 − (1 / 2)(1 / 2) = 3 / 4, assuming independence and that H and T are equilikely. 1.4.19 Let C be the complement of the event; i.e., C equals at most 3 draws to get the first spade. (a) P ( C ) = 1 4 + 3 4 1 4 + ( 3 4 ) 2 1 4 . (b) P ( C ) = 1 4 + 13 51 39 52 + 13 50 38 51 39 52 . 1.4.22 The probability that A wins is ∑ ∞ n =0 ( 5 6 4 6 ) n 1 6 = 3 8 . 1.4.27 Let Y denote the bulb is yellow and let T 1 and T 2 denote bags of the first and second types, respectively. (a) P ( Y ) = P ( Y | T 1 ) P ( T 1 ) + P ( Y | T 2 ) P ( T 2 ) = 20 25 . 6 + 10 25 . 4 . (b) P ( T 1 | Y ) = P ( Y | T 1 ) P ( T 1 ) P ( Y ) . 1.4.30 Suppose without loss of generality that the prize is behind curtain 1. Con- dition on the event that the contestant switches. If the contestant chooses curtain 2 then she wins, (In this case Monte cannot choose curtain 1, so he must choose curtain 3 and, hence, the contestant switches to curtain 1). Like- wise, in the case the contestant chooses curtain 3. If the contestant chooses curtain 1, she loses. Therefore the conditional probability that she wins is 2 3 . 1.4.31 (1) The probability is 1 − ( 5 6 ) 4 . (2) The probability is 1 − [ ( 5 6 ) 2 + 10 36 ] 24 .

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    4 Probability and Distributions 1.5.2 Part (a): c [(2 / 3) + (2 / 3) 2 + (2 / 3) 3 + · · · ] = c (2 / 3) 1 − (2 / 3) = 2 c = 1 , so c = 1 / 2. 1.5.5 Part (a): p ( x ) = { ( 13 x )( 39 5 − x ) ( 52 5 ) x = 0 , 1 , . . . , 5 0 elsewhere . 1.5.9 Part (b): 50 ∑ x =1 x/ 5050 = 50(51) 2(5050) = 51 202 . 1.5.10 For Part (c): Let C n = { X ≤ n } . Then C n ⊂ C n +1 and ∪ n C n = R . Hence, lim n →∞ F ( n ) = 1. Let ǫ > 0 be given. Choose n 0 such that n ≥ n 0 implies 1 − F ( n ) < ǫ . Then if x ≥ n 0 , 1 − F ( x ) ≤ 1 − F ( n 0 ) < ǫ . 1.6.2 Part (a): p ( x ) = ( 9 x − 1 ) ( 10 x − 1 ) 1 11 − x = 1 10 , x = 1 , 2 , . . . 10 . 1.6.3 (a) p ( x ) = ( 5 6 ) x − 1 ( 1 6 ) , x = 1 , 2 , 3 , . . . (b) ∞ ∑ x =1 ( 5 6 ) x − 1 ( 1 6 ) = 1 / 6 1 − (25 / 36) = 6 11 . 1.6.8 D y = { 1 , 2 3 , 3 3 , . . . } . The pmf of Y is p ( y ) = ( 1 2 ) y 1 / 3 , y ∈ D y . 1.7.1 If √ x < 10 then F ( x ) = P [ X ( c ) = c 2 ≤ x ] = P ( c ≤ √ x ) = ∫ √ x 0 1 10 dz = √ x 10 . Thus f ( x ) = F ′ ( x ) = { 1 20 √ x 0 < x < 100 0 elsewhere . 1.7.2 C 2 ⊂ C c 1 ⇒ P ( C 2 ) ≤ P ( C c 1 ) = 1 − (3 / 8) = 5 / 8 .

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    5 1.7.4 Among other characteristics, ∫ ∞ −∞ 1 π (1 + x 2 ) dx = 1 π arctan x ∣ ∣ ∣ ∣ ∣ ∣ ∞ −∞ = 1 π [ π 2 − ( − π 2 )] = 1 . 1.7.6 Part (b): P ( X 2 < 9) = P ( − 3 < X < 3) = ∫ 3 − 2 x + 2 19 dx = 1 18 [ x 2 2 + 2 x ] 3 − 2 = 1 18 [ 21 2 − ( − 2) ] = 25 36 . 1.7.8 Part (c): f ′ ( x ) = 1 2 2 xe − x = 0; hence, x = 2 is the mode because it maximizes f ( x ). 1.7.9 Part (b): ∫ m 0 3 x 2 dx = 1 2 ; hence, m 3 = 2 − 1 and m = (1 / 2) 1 / 3 . 1.7.10 ∫ ξ 0 . 2 0 4 x 3 dx = 0 . 2 : hence, ξ 4 0 . 2 = 0 . 2 and ξ 0 . 2 = 0 . 2 1 / 4 . 1.7.13 x = 1 is the mode because for 0 < x < ∞ because f ( x ) = F ′ ( x ) = e − x − e − x + xe − x = xe − x f ′ ( x ) = − xe − x + e − x = 0 , and f ′ (1) = 0. 1.7.16 Since ∆ > 0 X > z ⇒ Y = X + ∆ > z. Hence, P ( X > z ) ≤ P ( Y > z ). 1.7.19 Since f ( x ) is symmetric about 0, ξ . 25 < 0. So we need to solve, ∫ ξ . 25 − 2 ( − x 4 ) dx = . 25 . The solution is ξ . 25 = − √ 2.

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    6 Probability and Distributions 1.7.20 For 0 < y < 27, x = y 1 / 3 , dx dy = 1 3 y − 2 / 3 g ( y ) = = 1 3 y 2 / 3 y 2 / 3 9 = 1 27 . 1.7.22 f ( x ) = 1 π , − π 2 < x < π 2 . x = arctan y, dx dy = 1 1 + y 2 , −∞ < y < ∞ . g ( y ) = 1 π 1 1 + y 2 , −∞ < y < ∞ . 1.7.23 G ( y ) = P ( − 2 log X 4 ≤ y ) = P ( X ≥ e − y/ 8 ) = ∫ 1 e − y/ 8 4 x 3 dx = 1 − e − y/ 2 , 0 < y < ∞ g ( y ) = G ′ ( y ) = { e − y/ 2 0 < y < ∞ 0 elsewhere . 1.7.24 G ( y ) = P ( X 2 ≤ y ) = P ( − √ y ≤ X ≤ √ y ) =    ∫ √ y − √ y 1 3 dx = 2 √ y 3 0 ≤ y < 1 ∫ √ y − 1 1 3 dx = √ y 3 + 1 3 1 ≤ y < 4 g ( y ) =          1 3 √ y 0 ≤ y < 1 1 6 √ y 1 ≤ y < 4 0 elsewhere . 1.8.4 E (1 /X ) = 100 ∑ x =51 1 x 1 50 . The latter sum is bounded by the two integrals ∫ 101 51 1 x dx and ∫ 100 50 1 x dx. An appropriate approximation might be 1 50 ∫ 101 . 5 50 . 5 1 x dx = 1 50 (log 100 . 5 − log 50 . 5) .

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    7 1.8.6 E [ X (1 − X )] = ∫ 1 0 x (1 − x )3 x 2 dx. 1.8.8 When 1 < y < ∞ G ( y ) = P (1 /X ≤ y ) = P ( X ≥ 1 /y ) = ∫ 1 1 /y 2 x dx = 1 − 1 y 2 g ( y ) = 2 y 3 E ( Y ) = ∫ ∞ 1 y 2 y 3 dy = 2 , which equals ∫ 1 0 (1 /x )2 x dx . 1.8.10 The expectation of X does not exist because E ( | X | ) = 2 π ∫ ∞ 0 x 1 + x 2 dx = 1 π ∫ ∞ 1 1 u du = ∞ , where the change of variable u = 1 + x 2 was used. 1.9.2 M ( t ) = ∞ ∑ x =1 ( e t 2 ) x = e t / 2 1 − ( e t / 2) , e t / 2 < 1 . Find E ( X ) = M ′ (0) and Var( X ) = M ′′ (0) − [ M ′ (0)] 2 . 1.9.4 0 ≤ var( X ) = E ( X 2 ) − [ E ( X )] 2 . 1.9.6 E [ ( X − μ σ ) 2 ] = 1 σ 2 σ 2 = 1 . 1.9.8 K ( b ) = E [( X − b ) 2 ] = E ( X 2 ) − 2 bE ( X ) + b 2 K ′ ( b ) = − 2 E ( X ) + 2 b = 0 ⇒ b = E ( X ) . 1.9.11 For a continuous type random variable, K ( t ) = ∫ ∞ −∞ t x f ( x ) dx. K ′ ( t ) = ∫ ∞ −∞ xt x − 1 f ( x ) dx ⇒ K ′ (1) = E ( X ) . K ′′ ( t ) = ∫ ∞ −∞ x ( x − 1) t x − 2 f ( x ) dx ⇒ K ′′ (1) = E [ X ( X 1 )]; and so forth.

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    8 Probability and Distributions 1.9.12 3 = E ( X − 7) ⇒ E ( X ) = 10 = μ. 11 = E [( X − 7) 2 ] = E ( X 2 ) − 14 E ( X ) + 49 = E ( X 2 ) − 91 ⇒ E ( X 2 ) = 102 and var( X ) = 102 − 100 = 2 . 15 = E [( X − 7) 3 ] . Expand ( X − 7) 3 and continue. 1.9.16 E ( X ) = 0 ⇒ var( X ) = E ( X 2 ) = 2 p. E ( X 4 ) = 2 p ⇒ kurtosis = 2 p/ 4 p 2 = 1 / 2 p. 1.9.17 ψ ′ ( t ) = M ′ ( t ) /M ( t ) ⇒ ψ ′ (0) = M ′ (0) /M (0) = E ( X ) . ψ ′′ ( t ) = M ( t ) M ′′ ( t ) − M ′ ( t ) M ′ ( t ) [ M ( t ) 2 ] ⇒ ψ ′′ (0) = M (0) M ′′ (0) − M ′ (0) M ′ (0) [ M (0) 2 ] = M ′′ (0) − [ M ′ (0)] 2 = var( X ) . 1.9.19 M ( t ) = (1 − t ) − 3 = 1 + 3 t + 3 · 4 t 2 2! + 3 · 4 · 5 t 3 3! + · · · Considering the coefficient of t r /r !, we have E ( X r ) = 3 · 4 · 5 · · · ( r + 2) , r = 1 , 2 , 3 . . . . 1.9.20 Integrating the parts with u = 1 − F ( x ), dv = dx , we get { [1 − F ( x )] x } b 0 − ∫ b 0 x [ − f ( x )] dx = ∫ b 0 xf ( x ) dx = E ( X ) . 1.9.23 E ( X ) = ∫ 1 0 x 1 4 dx + 0 · 1 4 + 1 · 1 2 = 5 8 . E ( X 2 ) = ∫ 1 0 x 2 1 4 dx + 0 · 1 4 + 1 · 1 2 = 7 12 . var( X ) = 7 12 − ( 5 8 ) 2 = 37 192 . 1.9.24 E ( X ) = ∫ ∞ −∞ x [ c 1 f 1 ( x ) + · · · + c k f k ( x )] dx = k ∑ i =1 c i μ i = μ.

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    9 Because ∫ ∞ −∞ ( x − μ ) 2 f i ( x ) dx = σ 2 i + ( μ i − μ ) 2 , we have E [( X − μ ) 2 ] = k ∑ i =1 c i [ σ 2 i + ( μ i − μ ) 2 ] . 1.10.2 μ = ∫ ∞ 0 xf ( x ) dx ≥ ∫ ∞ 2 μ 2 μf ( x ) dx = 2 μP ( X > 2 μ ) . Thus 1 2 ≥ P ( X > 2 μ ) . 1.10.4 If, in Theorem 1.10.2, we take u ( X ) = exp { tX } and c = exp { ta } , we have P (exp { tX } ≥ exp { ta } ] ≤ M ( t ) exp {− ta } . If t > 0, the events exp { tX } ≥ exp { ta } and X ≥ a are equivalent. If t < 0, the events exp { tX } ≥ exp { ta } and X ≤ a are equivalent. 1.10.5 We have P ( X ≥ 1) ≤ [1 − exp {− 2 t } ] / 2 t for all 0 < t < ∞ , and P ( X ≤ − 1) ≤ [exp { 2 t } − 1] / 2 t for all −∞ < t < 0. Each of these bounds has the limit 0 as t → ∞ and t → −∞ , respectively.

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    Chapter 2 Multivariate Distributions 2.1.2 P ( A 5 ) = 7 8 − 4 8 − 3 8 + 2 8 = 2 8 . 2.1.5 ∫ ∞ 0 ∫ ∞ 0 [ 2 g ( √ x 2 1 + x 2 2 ) /π √ x 2 1 + x 2 2 ] dx 1 dx 2 = ∫ ∞ 0 ∫ π/ 2 0 [2 g ( ρ ) /πρ ] ρ dθdρ = ∫ ∞ 0 g ( ρ ) dρ = 1 . 2.1.6 G ( z ) = P ( X + Y ≤ z ) = ∫ z 0 ∫ z − x 0 e − x − y dydx = ∫ z 0 [1 − e − ( z − x ) ] e − x dx = 1 − e − z − ze − z . g ( z ) = G ′ ( z ) = { ze − z 0 < z < ∞ 0 elsewhere . 2.1.7 G ( z ) = P ( XY ≤ z ) = 1 − ∫ 1 z ∫ 1 z/x dydx = 1 − ∫ 1 z ( 1 − z x ) dx = z − z log z g ( z ) = G ′ ( z ) = { − log z 0 < z < 1 0 elsewhere . Why is − log z > 0? 11

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    12 Multivariate Distributions 2.1.8 f ( x, y ) =    ( 13 x )( 13 y )( 26 13 − x − y ) ( 52 13 ) x ≥ 0 , y ≥ 0 , x + y ≤ 13 , x and y integers 0 elsewhere . 2.1.10 P ( X 1 + X 2 ≤ 1) = 15 ∫ 1 / 2 0 x 2 1 [∫ 1 − x 1 x 1 x 2 dx 2 ] dx 1 . 2.1.14 E [ e t 1 X 1 + t 2 X 2 ] = ∞ ∑ i =1 ∞ ∑ j =1 e t 1 i + t 2 j ( 1 2 ) i + j = ∞ ∑ i =1 ( e t 1 1 2 ) i ∞ ∑ j =1 ( e t 1 1 2 ) j = [ 1 1 − 2 − 1 e t 1 − 1 ] [ 1 1 − 2 − 1 e t 2 − 1 ] , provided t i < log 2, i = 1 , 2. 2.2.1 p ( y 1 , y 2 ) = { ( 2 3 ) y 2 ( 1 3 ) 2 − y 2 ( y 1 , y 2 ) = (0 , 0) , ( − 1 , 1) , (1 , 1) , (0 , 2) 0 elsewhere . 2.2.2 p ( y 1 , y 2 ) = { y 1 / 36 y 1 = y 2 , 2 y 2 , 3 y 2 ; y 2 = 1 , 2 , 3 0 elsewhere . y 1 1 2 3 4 6 9 p ( y 1 ) 1/36 4/36 6/36 4/36 12/36 9/36 2.2.4 The inverse transformation is given by x 1 = y 1 y 2 and x 2 = y 2 with Jacobian J = y 2 . By noting what the boundaries of the space S ( X 1 , X 2 ) map into, it follows that the space T ( Y 1 , Y 2 ) = { ( y 1 , y 2 ) : 0 < y i < 1 , i = 1 , 2 } . The pdf of ( Y 1 , Y 2 ) is f Y 1 ,Y 2 ( y 1 , y 2 ) = 8 y 1 y 3 2 . 2.2.5 The inverse transformation is x 1 = y 1 − y 2 and x 2 = y 2 with Jacobian J = 1. The space of ( Y 1 , Y 2 ) is T = { ( y 1 , y 2 ) : −∞ < y i < ∞ , i = 1 , 2 } . Thus the joint pdf of ( Y 1 , Y 2 ) is f Y 1 ,Y 2 ( y 1 , y 2 ) = f X 1 ,X 2 ( y 1 − y 2 , y 2 ) , which leads to formula (2.2.1).

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    13 2.3.2 ( a ) c 1 ∫ x 2 0 x 1 /x 2 2 dx 1 = c 1 2 = 1 ⇒ c 1 = 2 and c 2 = 5 . ( b ) 10 x 1 x 2 2 , 0 < x 1 < x 2 < 1; zero elsewhere ( c ) ∫ 1 / 2 1 / 4 2 x 1 / (5 / 8) 2 dx = 64 25 ( 1 4 − 1 16 ) = 12 25 . ( d ) ∫ 1 / 2 1 / 4 ∫ 1 x 1 10 x 1 x 2 2 dx 2 dx 1 = ∫ 1 / 2 1 / 4 10 3 x 1 (1 − x 3 1 ) dx 1 = 135 512 . 2.3.3 f 2 ( x 2 ) = ∫ x 2 0 21 x 2 1 x 3 2 dx 1 = 7 x 6 2 , 0 < x 2 < 1 . f 1 | 2 ( x 1 | x 2 ) = 21 x 2 1 x 3 2 / 7 x 6 2 = 3 x 2 1 /x 3 2 , 0 < x 1 < x 2 . E ( X 1 | x 2 ) = ∫ x 2 0 x 1 (3 x 2 1 /x 3 2 ) dx 1 = 3 4 x 2 . G ( y ) = P ( 3 4 X 2 ≤ y ) = ∫ 4 y/ 3 0 7 x 6 2 dx 2 = ( 4 y 3 ) 7 , 0 < y < 3 4 g ( y ) = { 7 ( 4 3 ) 7 y 6 0 < y < 3 4 0 elsewhere . E ( Y ) = 7 8 3 4 = 21 32 . Var( Y ) = 7 1024 . E ( X 1 ) = 21 32 . Var( X 1 ) = 553 15360 > 7 1024 . 2.3.8 The marginal pdf of X is f X ( x ) = 2 ∫ ∞ x e − x e − y dy = 2 e − 2 x , 0 < x < ∞ . Hence, the conditional pdf of Y given X = x is f Y | X ( y | x ) = 2 e − x e − y 2 e − 2 x = e − ( y − x ) , 0 < x < y < ∞ , with conditional mean E ( Y | X = x ) = ∫ ∞ x ye − ( y − x ) dy = x + 1 , x > 0 .

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