Probability Essentials Solutions部分答案

Solutions of Selected Problems from Probability Essentials, Second Edition Solutions to selected problems of Chapter 2 2.1 Let’s first prove by induction that #(2Ωn) = 2n if Ω = {x1, . . . , xn}. For n = 1 it is clear that #(2Ω1) = #({∅, {x1}}) = 2. Suppose #(2Ωn−1) = 2n−1. Observe that 2Ωn = {{xn} ∪ A,A ∈ 2Ωn−1} ∪ 2Ωn−1} hence #(2Ωn) = 2#(2Ωn−1) = 2n. This proves finiteness. To show that 2Ω is a σ-algebra we check: 1. ∅ ⊂ Ω hence ∅ ∈ 2Ω. 2. If A ∈ 2Ω then A ⊂ Ω and Ac ⊂ Ω hence Ac ∈ 2Ω. 3. Let (An)n≥1 be a sequence of subsets of Ω. Then ⋃∞ n=1An is also a subset of Ω hence in 2Ω. Therefore 2Ω is a σ-algebra. 2.2 We check if H = ∩α∈AGα has the three properties of a σ-algebra: 1. ∅ ∈ Gα ∀α ∈ A hence ∅ ∈ ∩α∈AGα. 2. If B ∈ ∩α∈AGα then B ∈ Gα ∀α ∈ A. This implies that Bc ∈ Gα ∀α ∈ A since each Gα is a σ-algebra. So Bc ∈ ∩α∈AGα. 3. Let (An)n≥1 be a sequence in H. Since each An ∈ Gα, ⋃∞ n=1An is in Gα since Gα is a σ-algebra for each α ∈ A. Hence ⋃∞n=1An ∈ ∩α∈AGα. Therefore H = ∩α∈AGα is a σ-algebra. 2.3 a. Let x ∈ (∪∞n=1An)c. Then x ∈ Acn for all n, hence x ∈ ∩∞n=1Acn. So (∪∞n=1An)c ⊂ ∩∞n=1Acn. Similarly if x ∈ ∩∞n=1Acn then x ∈ Acn for any n hence x ∈ (∪∞n=1An)c. So (∪∞n=1An)c = ∩∞n=1Acn. b. By part-a ∩∞n=1An = (∪∞n=1Acn)c, hence (∩∞n=1An)c = ∪∞n=1Acn. 2.4 lim infn→∞An = ∪∞n=1Bn where Bn = ∩m≥nAm ∈ A ∀n since A is closed under taking countable intersections. Therefore lim infn→∞An ∈ A since A is closed under taking countable unions. By De Morgan’s Law it is easy to see that lim supAn = (lim infn→∞Acn) c, hence lim supn→∞An ∈ A since lim infn→∞Acn ∈ A and A is closed under taking complements. Note that x ∈ lim infn→∞An ⇒ ∃n∗ s.t x ∈ ∩m≥n∗Am ⇒ x ∈ ∩m≥nAm∀n ⇒ x ∈ lim supn→∞An. Therefore lim infn→∞An ⊂ lim supn→∞An. 2.8 Let L = {B ⊂ R : f−1(B) ∈ B}. It is easy to check that L is a σ-algebra. Since f is continuous f−1(B) is open (hence Borel) if B is open. Therefore L contains the open sets which implies L ⊃ B since B is generated by the open sets of R. This proves that f−1(B) ∈ B if B ∈ B and that A = {A ⊂ R : ∃B ∈ B with A = f−1(B) ∈ B} ⊂ B. 1 Solutions to selected problems of Chapter 3 3.7 a. Since P (B) > 0 P (.|B) defines a probability measure on A, therefore by Theorem 2.4 limn→∞ P (An|B) = P (A|B). b. We have that A ∩ Bn → A ∩ B since 1A∩Bn(w) = 1A(w)1Bn(w) → 1A(w)1B(w). Hence P (A ∩Bn) → P (A ∩B). Also P (Bn) → P (B). Hence P (A|Bn) = P (A ∩Bn) P (Bn) → P (A ∩B) P (B) = P (A|B). c. P (An|Bn) = P (An ∩Bn) P (Bn) → P (A ∩B) P (B) = P (A|B) since An ∩Bn → A ∩B and Bn → B. 3.11 Let B = {x1, x2, . . . , xb} and R = {y1, y2, . . . , yr} be the sets of b blue balls and r red balls respectively. Let B′ = {xb+1, xb+2, . . . , xb+d} and R′ = {yr+1, yr+2, . . . , yr+d} be the sets of d-new blue balls and d-new red balls respectively. Then we can write down the sample space Ω as Ω = {(a, b) : (a ∈ B and b ∈ B ∪B′ ∪R) or (a ∈ R and b ∈ R ∪R′ ∪B)}. Clearly card(Ω) = b(b + d + r) + r(b + d + r) = (b + r)(b + d + r). Now we can define a probability measure P on 2Ω by P (A) = card(A) card(Ω) . a. Let A = { second ball drawn is blue} = {(a, b) : a ∈ B, b ∈ B ∪B′} ∪ {(a, b) : a ∈ R, b ∈ B} card(A) = b(b + d) + rb = b(b + d + r), hence P (A) = b b+r . b. Let B = { first ball drawn is blue} = {(a, b) ∈ Ω : a ∈ B} Observe A ∩B = {(a, b) : a ∈ B, b ∈ B ∪B′} and card(A ∩B) = b(b + d). Hence P (B|A) = P (A ∩B) P (A) = card(A ∩B) card(A) = b + d b + d + r . 2 3.17 We will use the inequality 1−x > e−x for x > 0, which is obtained by taking Taylor’s expansion of e−x around 0. P ((A1 ∪ . . . ∪ An)c) = P (Ac1 ∩ . . . ∩ Acn) = (1− P (A1)) . . . (1− P (An)) ≤ exp(−P (A1)) . . . exp(−P (An)) = exp(− n∑ i=1 P (Ai)) 3 Solutions to selected problems of Chapter 4 4.1 Observe that P (k successes) = ( n 2 ) λ n k ( 1− λ n )n−k = Canb1,n . . . bk,ndn where C = λk k! an = (1− λ n )n bj,n = n− j + 1 n dn = (1− λ n )−k It is clear that bj,n → 1 ∀j and dn → 1 as n→∞. Observe that log((1− λ n )n) = n( λ n − λ 2 n2 1 ξ2 ) for some ξ ∈ (1− λ n , 1) by Taylor series expansion of log(x) around 1. It follows that an → e−λ as n → ∞ and that |Error| = |en log(1−λn ) − e−λ| ≥ |n log(1− λ n )− λ| = nλ 2 n2 1 ξ2 ≥ λp Hence in order to have a good approximation we need n large and p small as well as λ to be of moderate size. 4 Solutions to selected problems of Chapter 5 5.7 We put xn = P (X is even) for X ∼ B(p, n). Let us prove by induction that xn = 1 2 (1 + (1− 2p)n). For n = 1, x1 = 1− p = 12(1 + (1− 2p)1). Assume the formula is true for n− 1. If we condition on the outcome of the first trial we can write xn = p(1− xn−1) + (1− p)xn = p(1− 1 2 (1 + (1− 2p)n−1)) + (1− p)(1 2 (1 + (1− 2p)n−1)) = 1 2 (1 + (1− 2p)n) hence we have the result. 5.11 Observe that E(|X − λ|) = ∑i<λ(λ − i)pi +∑i≥λ(i − λ)pi. Since ∑i≥λ(i − λ)pi =∑∞ i=0(i− λ)pi − ∑ i<λ(i− λ)pi we have that E(|X − λ|) = 2 ∑ i<λ(λ− i)pi. So E(|X − λ|) = 2 ∑ i<λ (λ− i)pi = 2 λ−1∑ i=1 (λ− i)e −λλk k! = 2e−λ λ−1∑ i=0 ( λk+1 k! − λ k (k − 1)!) = 2e−λ λλ (k − 1)! . 5 Solutions to selected problems of Chapter 7 7.1 Suppose limn→∞ P (An) 6= 0. Then there exists � > 0 such that there are dis- tinct An1 , An2 , . . . with P (Ank) > 0 for every k ≤ 1. This gives ∑∞ k=1 P (Ank) = ∞ which is a contradiction since by the hypothesis that the An are disjoint we have that∑∞ k=1 P (Ank) = P (∪∞n=1Ank) ≤ 1 . 7.2 Let An = {Aβ : P (Aβ) > 1/n}. An is a finite set otherwise we can pick disjoint Aβ1 , Aβ2 , . . . in An. This would give us P ∪∞m=1 Aβm = ∑∞ m=1 P (Aβm) = ∞ which is a contradiction. Now {Aβ : β ∈ B} = ∪∞n=1An hence (Aβ)β∈B is countable since it is a countable union of finite sets. 7.11 Note that {x0} = ∩∞n=1[x0 − 1/n, x0] therefore {x0} is a Borel set. P ({x0}) = limn→∞ P ([x0 − 1/n, x0]). Assuming that f is continuous we have that f is bounded by some M on the interval [x0 − 1/n, x0] hence P ({x0}) = limn→∞M(1/n) = 0. Remark: In order this result to be true we don’t need f to be continuous. When we define the Lebesgue integral (or more generally integral with respect to a measure) and study its properties we will see that this result is true for all Borel measurable non-negative f . 7.16 First observe that F (x) − F (x−) > 0 iff P ({x}) > 0. The family of events {{x} : P ({x}) > 0} can be at most countable as we have proven in problem 7.2 since these events are disjoint and have positive probability. Hence F can have at most countable discon- tinuities. For an example with infinitely many jump discontinuities consider the Poisson distribution. 7.18 Let F be as given. It is clear that F is a nondecreasing function. For x < 0 and x ≥ 1 right continuity of F is clear. For any 0 < x < 1 let i∗ be such that 1 i∗+1 ≤ x < 1i∗ . If xn ↓ x then there exists N such that 1i∗+1 ≤ xn < 1i∗ for every n ≥ N . Hence F (xn) = F (x) for every n ≥ N which implies that F is right continuous at x. For x = 0 we have that F (0) = 0. Note that for any � there exists N such that ∑∞ i=N 1 2i < �. So for all x s.t. |x| ≤ 1 N we have that F (x) ≤ �. Hence F (0+) = 0. This proves the right continuity of F for all x. We also have that F (∞) = ∑∞i=1 12i = 1 and F (−∞) = 0 so F is a distribution function of a probability on R. a. P ([1,∞)) = F (∞)− F (1−) = 1−∑∞n=2 = 1− 12 = 12 . b. P ([ 1 10 ,∞)) = F (∞)− F ( 1 10 −) = 1−∑∞n=11 12i = 1− 2−10. c P ({0}) = F (0)− F (0−) = 0. d. P ([0, 1 2 )) = F (1 2 −)− F (0−) = ∑∞n=3 12i − 0 = 14 . e. P ((−∞, 0)) = F (0−) = 0. f. P ((0,∞)) = 1− F (0) = 1. 6 7 Solutions to selected problems of Chapter 9 9.1 It is clear by the definition of F that X−1(B) ∈ F for every B ∈ B. So X is measurable from (Ω,F) to (R,B). 9.2 Since X is both F and G measurable for any B ∈ B, P (X ∈ B) = P (X ∈ B)P (X ∈ B) = 0 or 1. Without loss of generality we can assume that there exists a closed interval I such that P (I) = 1. Let Λn = {tn0 , . . . tnln} be a partition of I such that Λn ⊂ Λn+1 and supk t n k − tnk−1 → 0. For each n there exists k∗(n) such that P (X ∈ [tnk∗ , tnk∗+1]) = 1 and [tnk∗(n+1, t n k∗(n+1)+1] ⊂ [tnk∗(n), tnk∗(n)+1]. Now an = tnk∗(n) and bn = tnk∗(n) + 1 are both Cauchy sequences with a common limit c. So 1 = limn→∞ P (X ∈ (tnk∗ , tnk∗+1]) = P (X = c). 9.3 X−1(A) = (Y −1(A) ∩ (Y −1(A) ∩X−1(A)c)c)∪(X−1(A) ∩ Y −1(A)c). Observe that both Y −1(A)∩ (X−1(A))c and X−1(A)∩Y −1(A)c are null sets and therefore measurable. Hence if Y −1(A) ∈ A′ then X−1(A) ∈ A′. In other words if Y is A′ measurable so is X. 9.4 Since X is integrable, for any � > 0 there exists M such that ∫ |X|1{X>M}dP < � by the dominated convergence theorem. Note that E[X1An ] = E[X1An1{X>M}] + E[X1An1{X≤M}] ≤ E[|X|1{X≤M}] + MP (An) Since P (An) → 0, there exists N such that P (An) ≤ �M for every n ≥ N . Therefore E[X1An ] ≤ � + � ∀n ≥ N , i.e. limn→∞E[X1An ] = 0. 9.5 It is clear that 0 ≤ Q(A) ≤ 1 and Q(Ω) = 1 since X is nonnegative and E[X] = 1. Let A1, A2, . . . be disjoint. Then Q(∪∞n=1An) = E[X1∪∞n=1An ] = E[ ∑ n=1 X1An ] = ∞∑ n=1 E[X1An ] where the last equality follows from the monotone convergence theorem. Hence Q(∪∞n=1An) =∑∞ n=1Q(An). Therefore Q is a probability measure. 9.6 If P (A) = 0 then X1A = 0 a.s. Hence Q(A) = E[X1A] = 0. Now assume P is the uniform distribution on [0, 1]. Let X(x) = 21[0,1/2](x). Corresponding measure Q assigns zero measure to (1/2, 1], however P ((1/2, 1]) = 1/2 6= 0. 9.7 Let’s prove this first for simple functions, i.e. let Y be of the form Y = n∑ i=1 ci1Ai 8 for disjoint A1, . . . , An. Then EQ[Y ] = n∑ i=1 ciQ(Ai) = n∑ i=1 ciE[X1Ai ] = EP [XY ] For non-negative Y we take a sequence of simple functions Yn ↑ Y . Then EQ[Y ] = lim n→∞ EQ[Yn] = lim n→∞ EP [XYn] = EP [XY ] where the last equality follows from the monotone convergence theorem. For general Y ∈ L1(Q) we have that EQ[Y ] = EQ[Y +]− EQ[Y −] = EP [(XY )+]− EQ[(XY )−] = EP [XY ]. 9.8 a. Note that 1 X X = 1 a.s. since P (X > 0) = 1. By problem 9.7 EQ[ 1 X ] = EP [ 1 X X] = 1. So 1 X is Q-integrable. b. R : A → R, R(A) = EQ[ 1X1A] is a probability measure since 1X is non-negative and EQ[ 1 X ] = 1. Also R(A) = EQ[ 1 X 1A] = EP [ 1 X X1A] = P (A). So R = P . 9.9 Since P (A) = EQ[ 1 X 1A] we have that Q(A) = 0 ⇒ P (A) = 0. Now combining the results of the previous problems we can easily observe that Q(A) = 0 ⇔ P (A) = 0 iff P (X > 0) = 1. 9.17. Let g(x) = ((x− µ)b + σ)2 σ2(1 + b2)2 . Observe that {X ≥ µ + bσ} ∈ {g(X) ≥ 1}. So P ({X ≥ µ + bσ}) ≤ P ({g(X) ≥ 1}) ≤ E[g(X)] 1 where the last inequality follows from Markov’s inequality. Since E[g(X)] = σ 2(1+b2) σ2(1+b2)2 we get that P ({X ≥ µ + bσ}) ≤ 1 1 + b2 . 9.19 xP ({X > x}) ≤ E[X1{X > x}] = ∫ ∞ x z√ 2pi e− z2 2 dz = e− x2 2√ 2pi Hence P ({X > x}) ≤ e −x2 2 x √ 2pi 9 . 9.21 h(t+s) = P ({X > t+s}) = P ({X > t+s,X > s}) = P ({X > t+s|X > s})P ({X > s}) = h(t)h(s) for all t, s > 0. Note that this gives h( 1 n ) = h(1) 1 n and h(m n ) = h(1) m n . So for all rational r we have that h(r) = exp (log(h(1))r). Since h is right continuous this gives h(x) = exp(log(h(1))x) for all x > 0. Hence X has exponential distribution with parameter − log h(1). 10 Solutions to selected problems of Chapter 10 10.5 Let P be the uniform distribution on [−1/2, 1/2]. Let X(x) = 1[−1/4,1/4] and Y (x) = 1[−1/4,1/4]c . It is clear that XY = 0 hence E[XY ] = 0. It is also true that E[X] = 0. So E[XY ] = E[X]E[Y ] however it is clear that X and Y are not independent. 10.6 a. P (min(X, Y ) > i) = P (X > i)P (Y > i) = 1 2i 1 2i = 1 4i . So P (min(X, Y ) ≤ i) = 1− P (min(X,Y ) > i) = 1− 1 4i . b. P (X = Y ) = ∑∞ i=1 P (X = i)P (Y = i) = ∑∞ i=1 1 2i 1 2i = 1 1− 1 4i − 1 = 1 3 . c. P (Y > X) = ∑∞ i=1 P (Y > i)P (X = i) = ∑∞ i=1 1 2i 1 2i = 1 3 . d. P (X divides Y ) = ∑∞ i=1 ∑∞ k=1 1 2i 1 2ki = ∑∞ i=1 1 2i 1 2i−1 . e. P (X ≥ kY ) = ∑∞i=1 P (X ≥ ki)P (Y = i) = ∑∞i=1 12i 12ki−1 = 22k+1−1 . 11 Solutions to selected problems of Chapter 11 11.11. Since P{X > 0} = 1 we have that P{Y < 1} = 1. So FY (y) = 1 for y ≥ 1. Also P{Y ≤ 0} = 0 hence FY (y) = 0 for y ≤ 0. For 0 < y < 1 P{Y > y} = P{X < 1−yy } = FX( 1−y y ). So FY (y) = 1− ∫ 1−y y 0 fX(x)dx = 1− ∫ y 0 −1 z2 fX( 1− z z )dz by change of variables. Hence fY (y) =  0 −∞ < y ≤ 0 1 y2 fX( 1−y y ) 0 < y ≤ 1 0 1 ≤ y <∞ 11.15 Let G(u) = inf{x : F (x) ≥ u}. We would like to show {u : G(u) > y} = {u : F (Y ) < u}. Let u be such that G(u) > y. Then F (y) < u by definition of G. Hence {u : G(u) > y} ⊂ {u : F (Y ) < u}. Now let u be such that F (y) < u. Then y < x for any x such that F (x) ≥ u by monotonicity of F . Now by right continuity and the monotonicity of F we have that F (G(u)) = infF (x)≥u F (x) ≥ u. Then by the previous statement y < G(u). So {u : G(u) > y} = {u : F (Y ) < u}. Now P{G(U) > y} = P{U > F (y)} = 1− F (y) so G(U) has the desired distribution. Remark:We only assumed the right continuity of F . 12 Solutions to selected problems of Chapter 12 12.6 Let Z = ( 1 σY )Y − (ρXY σX )X. Then σ2Z = ( 1 σ2Y )σ2Y − (ρ 2 XY σ2X )σ2X − 2( ρXYσXσY )Cov(X, Y ) = 1− ρ2XY . Note that ρXY = ∓1 implies σ2Z = 0 which implies Z = c a.s. for some constant c. In this case X = σX σY ρXY (Y − c) hence X is an affine function of Y . 12.11 Consider the mapping g(x, y) = ( √ x2 + y2, arctan(x y )). Let S0 = {(x, y) : y = 0}, S1 = {(x, y) : y > 0}, S2 = {(x, y) : y < 0}. Note that ∪2i=0Si = R2 and m2(S0) = 0. Also for i = 1, 2 g : Si → R2 is injective and continuously differentiable. Corresponding inverses are given by g−11 (z, w) = (z sinw, z cosw) and g −1 2 (z, w) = (z sinw,−z cosw). In both cases we have that |Jg−1i (z, w)| = z hence by Corollary 12.1 the density of (Z,W ) is given by fZ,W (z, w) = ( 1 2piσ2 e −z2 2σ z + 1 2piσ2 e −z2 2σ z)1(−pi 2 ,pi 2 )(w)1(0,∞)(z) = 1 pi 1(−pi 2 ,pi 2 )(w) ∗ z σ2 e −z2 2σ 1(0,∞)(z) as desired. 12.12 Let P be the set of all permutations of {1, . . . , n}. For any pi ∈ P let Xpi be the corresponding permutation of X, i.e. Xpik = Xpik . Observe that P (Xpi1 ≤ x1, . . . , Xpin ≤ xn) = F (x1) . . . F (Xn) hence the law of Xpi and X coincide on a pisystem generating Bn therefore they are equal. Now let Ω0 = {(x1, . . . , xn) ∈ Rn : x1 < x2 < . . . < xn}. Since Xi are i.i.d and have continuous distribution PX(Ω0) = 1. Observe that P{Y1 ≤ y1, . . . , Yn ≤ yn} = P (∪pi∈P{Xpi1 ≤ y1, . . . , Xpin ≤ yn} ∩ Ω0) Note that {Xpi1 ≤ y1, . . . , Xpin ≤ yn} ∩ Ω0, pi ∈ P are disjoint and P (Ω0 = 1) hence P{Y1 ≤ y1, . . . , Yn ≤ yn} = ∑ pi∈P P{Xpi1 ≤ y1, . . . , Xpin ≤ yn} = n!F (y1) . . . F (yn) for y1 ≤ . . . ≤ yn. Hence fY (y1, . . . , yn) = { n!f(y1) . . . f(yn) y1 ≤ . . . ≤ yn 0 otherwise 13 Solutions to selected problems of Chapter 14 14.7 ϕX(u) is real valued iff ϕX(u) = ϕX(u) = ϕ−X(u). By uniqueness theorem ϕX(u) = ϕ−X(u) iff FX = F−X . Hence ϕX(u) is real valued iff FX = F−X . 14.9 We use induction. It is clear that the statement is true for n = 1. Put Yn =∑n i=1Xi and assume that E[(Yn) 3] = ∑n i=1E[(Xi) 3]. Note that this implies d 3 dx3 ϕYn(0) = −i∑ni=1E[(Xi)3]. Now E[(Yn+1)3] = E[(Xn+1 + Yn)3] = −i d3dx3 (ϕXn+1ϕYn)(0) by indepen- dence of Xn+1 and Yn. Note that d3 dx3 ϕXn+1ϕYn(0) = d3 dx3 ϕXn+1(0)ϕYn(0) + 3 d2 dx2 ϕXn+1(0) d dx ϕYn(0) + 3 d dx ϕXn+1(0) d2 dx2 ϕYn(0) + ϕXn+1(0) d3 dx3 ϕYn(0) = d3 dx3 ϕXn+1(0) + d3 dx3 ϕYn(0) = −i ( E[(Xn+1) 3] + n∑ i=1 E[(Xi) 3] ) where we used the fact that d dx ϕXn+1(0) = iE(Xn+1) = 0 and d dx ϕYn(0) = iE(Yn) = 0. So E[(Yn+1) 3] = ∑n+1 i=1 E[(Xi) 3] hence the induction is complete. 14.10 It is clear that 0 ≤ ν(A) ≤ 1 since 0 ≤ n∑ j=1 λjµj(A) ≤ n∑ j=1 λj = 1. Also for Ai disjoint ν(∪∞i=1Ai) = n∑ j=1 λjµj(∪∞i=1Ai) = n∑ j=1 λj ∞∑ i=1 µj(Ai) = ∞∑ i=1 n∑ j=1 λjµj(Ai) = ∞∑ i=1 ν(Ai) 14 Hence ν is countably additive therefore it is a probability mesure. Note that ∫ 1Adν(dx) =∑n j=1 λj ∫ 1A(x)dµj(dx) by definition of ν. Now by linearity and monotone convergence theorem for a non-negative Borel function f we have that ∫ f(x)ν(dx) = ∑n j=1 λj ∫ f(x)dµj(dx). Extending this to integrable f we have that νˆ(u) = ∫ eiuxν(dx) = ∑n j=1 λj ∫ eiuxdµj(dx) =∑n j=1 λjµˆj(u). 14.11 Let ν be the double exponential distribution, µ1 be the distribution of Y and µ2 be the distribution of −Y where Y is an exponential r.v. with parameter λ = 1. Then we have that ν(A) = 1 2 ∫ A∩(0,∞) e −xdx+ 1 2 ∫ A∩(−∞,0) e xdx = 1 2 µ1(A) + 1 2 µ2(A). By the previous exercise we have that νˆ(u) = 1 2 µˆ1(u) + 1 2 µˆ2(u) = 1 2 ( 1 1−iu + 1 1+iu ) = 1 1+u2 . 14.15. Note that E{Xn} = (−i)n dn dxn ϕX(0). Since X ∼ N(0, 1) ϕX(s) = e−s2/2. Note that we can get the derivatives of any order of e−s 2/2 at 0 simply by taking Taylor’s expansion of ex: e−s 2/2 = ∞∑ i=0 (−s2/2)n n! = ∞∑ i=0 1 2n! (−i)2n(2n)! 2nn! s2n hence E{Xn} = (−i)n dn dxn ϕX(0) = 0 for n odd. For n = 2k E{X2k} = (−i)2k d2kdx2kϕX(0) = (−i)2k (−i)2k(2k)! 2kk! = (2k)! 2kk! as desired. 15 Solutions to selected problems of Chapter 15 15.1 a. E{x} = 1 n ∑n i=1E{Xi} = µ. b. Since X1, . . . , Xn are independent Var(x) = 1 n2 ∑n i=1 Var{Xi} = σ 2 n . c. Note that S2 = 1 n ∑n i=1(Xi) 2 − x2. Hence E(S2) = 1 n ∑n i=1(σ 2 + µ2)− (σ2 n + µ2) = n−1 n σ2. 15.17 Note that ϕY (u) = ∏α i=1 ϕXi(u) = ( β β−iu) α which is the characteristic function of Gamma(α,β) random variable. Hence by uniqueness of characteristic function Y is Gamma(α,β). 16 Solutions to selected problems of Chapter 16 16.3 P ({Y ≤ y}) = P ({X ≤ y} ∩ {Z = 1}) + P ({−X ≤ y} ∩ {Z = −1}) = 1 2 Φ(y) + 1 2 Φ(−y) = Φ(y) since Z and X are independent and Φ(y) is symmetric. So Y is normal. Note that P (X + Y = 0) = 1 2 hence X + Y can not be normal. So (X, Y ) is not Gaussian even though both X and Y are normal. 16.4 Observe that Q = σXσY [ σX σY ρ ρ σY σX ] So det(Q) = σXσY (1− ρ2). So det(Q) = 0 iff ρ = ∓1. By Corollary 16.2 the joint density of (X,Y ) exists iff −1 < ρ < 1. (By Cauchy-Schwartz we know that −1 ≤ ρ ≤ 1). Note that Q−1 = 1 σXσY (1− ρ2) σY σX −ρ −ρ σX σY Substituting this in formula 16.5 we get that f(X,Y )(x, y) = 1 2piσXσY (1− ρ2) exp { −1 2(1− ρ2) (( x− µX σX )2 − 2ρ(x− µX)(y − µY ) σXσY + ( y − µY σY )2)} . 16.6 By Theorem 16.2 there exists a multivariate normal r.v. Y with E(Y ) = 0 and a diagonal covariance matrix Λ s.t. X − µ = AY where A is an orthogonal matrix. Since Q = AΛA∗ and det(Q) > 0 the diagonal entries of Λ are strictly positive hence we can define B = Λ−1/2A∗. Now the covariance matrix Q˜ of B(X − µ) is given by Q˜ = Λ−1/2A∗AΛA∗AΛ−1/2 = I So B(X − µ) is standard normal. 16.17 We know that as in Exercise 16.6 if B = Λ−1/2A∗ where A is the orthogonal matrix s.t. Q = AΛA∗ then B(X−µ) is standard normal. Note that this gives (X−µ)∗Q−1(X−µ) = (X − µ)∗B∗B(X − µ) which has chi-square distribution with n degrees of freedom. 17 Solutions to selected problems of Chapter 17 17.1 Let n(m) and j(m) be such that Ym = n(m) 1/pZn(m),j(m). This gives that P (|Ym| > 0) = 1 n(m) → 0 as m → ∞. So Ym converges to 0 in probability. However E[|Ym|p] = E[n(m)Zn(m),j(m)] = 1 for all m. So Ym does not converge to 0 in L p. 17.2 Let Xn = 1/n. It is clear that Xn converge to 0 in probability. If f(x) = 1{0}(x) then we have that P (|f(Xn) − f(0)| > �) = 1 for every � ≥ 1, so f(Xn) does not converge to f(0) in probability. 17.3 First observe that E(Sn) = ∑n i=1E(Xn) = 0 and that Var(Sn) = ∑n i=1 Var(Xn) = n since E(Xn) = 0 and Var(Xn) = E(X 2 n) = 1. By Chebyshev’s inequality P (|Snn | ≥ �) = P (|Sn| ≥ n�) ≤ Var(Sn)n2�2 = nn2�2 → 0 as n→∞. Hence Snn converges to 0 in probability. 17.4 Note that Chebyshev’s inequality gives P (|Sn2 n2 | ≥ �) ≤ 1 n2�2 . Since ∑∞ i=1 1 n2�2 <∞ by Borel Cantelli Theorem P (lim supn{|Sn2n2 | ≥ �}) = 0. Let Ω0 = ( ∪∞m=1 lim supn{|Sn2n2 | ≥ 1m} )c . Then P (Ω0) = 1. Now let’s pick w ∈ Ω0. For any � there exists m s.t. 1m ≤ � and w ∈ (lim supn{|Sn2n2 | ≥ 1m})c. Hence there are finitely many n s.t. | Sn2 n2 |