Solutions of Homework 2.

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1 1 Solutios of Homework Suppose X Y with E(Y ) <, ad X X i probability. Show that E X X 0 (i.e., X X i L 1 ). Solutio. Method 1. Suppose X X i L 1. The, there exists a subsequece { k } such that E X k X > ǫ for some ǫ > 0 ad all { k }. Sice X k X i probability, there exists a further subsequece kj such that X kj X a.e.. The the domiated covergece theorem implies that E X kj X 0. This cotradicts with E X k X > ǫ for some ǫ > 0 ad all { k }. Hece, X X i L 1. Method 2. Let Z X X. The Z 2Y. E(Z ) E(Z 1 {Zǫ}) + E(Z 1 {Z>ǫ}) ǫ + 2E(Y 1 {Z>ǫ}) ǫ + 2E(Y 1 {Y >1/ǫ,Z>ǫ}) + 2E(Y 1 {Y 1/ǫ,Z>ǫ}) ǫ + 2E(Y 1 {Y >1/ǫ} ) + 2/ǫE(1 {Z>ǫ}) ǫ + 2E(Y 1 {Y >1/ǫ} ) + 2/ǫP(Z > ǫ) ǫ + E(Y 1 {Y >1/ǫ} ) as which ca be arbitrarily small sice ǫ ca be chose arbitrarily small. Therefore lim E(Z ) X X i probability if ad oly if, for ay sub-sequece of X, 1, there always exists a further sub-sequece, which may be called sub-sub-sequece, that coverges a.s. to X. Solutio. Suppose X does ot coverge to X i probability. There exists a subsequece k ad a ǫ > 0 such that P(X k X > ǫ) > ǫ. The, o further subsequece of X k ca coverge to X, a.s. or i probability, which leads to a cotradictio. Ay subsequece of {X } still coverges to X i probability. Ad there is always of further subsequece of this subsequece that coverges to X a.s.. 3. Two r.v.s X, Y are called coditioally idepedet give r.v. Z if P(X t, Y s Z z) P(X t Z z)p(y s Z z) for all t, s, z. Let X be the total umber of heads of the first tosses of a fair coi. Set X 0 0. Show that X 1 ad X +1 are coditioally idepedet give X. Solutio. Let ξ be 1 or 0 whe the -th toss is a head or tail. Set ξ 0 0. The ξ i, i 0, 1, 2,... are idepedet. X ξ j. Ad ξ +1 is idepedet of X ad ξ P(X +1 k, X 1 j X l) P(ξ +1 k l, ξ l j X l) P(ξ +1 k l, ξ l j, X l) P(X l) P(ξ +1 k l) P(ξ l j, X l) P(X l) P(ξ +1 k l X l)p(ξ l j X l) P(X k X l)p(x 1 j X l) 4. Suppose X 1, X 2,... are i.i.d. radom variables followig expoetial distributio with mea 1. Show that Y a.e., where Y max 1i X i. (You might cosider first show the covergece i probability.)

2 2 Solutio. For ay costat C > 0, P(Y > C) P( max 1i X i > C) 1 P( max 1i X i C) 1 (1 e C ) 1 as. Therefore Y i probability. Sice Y is odecreasig, Y a.e.. (why?) 5. Raise a example to show that Fatou s lemma does ot hold if the coditio of X 0 is dropped. Solutio. Suppose ξ Uif[0, 1]. Let X 1/ξ1 {ξ1/}. The X 0 a.e., but E(X ) for all. Therefore E(limif X ) 0 > limif E(X ). DIY Exercises 1. (Extesio of Fatou s lemma). Suppose X Y ad E(Y ) <. The E(limif X ) limif E(X ). Solutio. X Y X Y 0 X (Y + Y ) 0 X + Y 0. By Fatou s lemma, E(lim if lim if X ) E(limif(X + Y ) E(Y ) E(X + Y ) E(Y ) limif E(X ). 2. Suppose X Y, E( Y Y ) 0 with E(Y ) <, ad X X i probability. The E(X ) E(X). Solutio. (This problem is a further extesio of Problem 1) Method 1: X X Y + Y. Let ξ X X. The ξ 0 i probability. Choose ay 0 < ǫ < c <. The, E(ξ ) E(ξ 1 {ξc}) + E(ξ 1 {ξ>c}) E(ξ 1 {ξc}) + E(ξ 1 {ξ>c}) + E(ξ 1 {ξ>c}) ǫp(ξ ǫ) + E(ξ 1 {ǫ<ξc}) + E[(Y + Y )1 {Y+Y >c} ǫ + cp(ξ > ǫ) + E( Y Y ) + 2E(Y 1 {Y+Y >c}) ǫ E(Y 1 {2Y >c} ). Lettig ǫ 0 ad c, we have E(ξ ) 0. Method 2: Assume first X X a.e. ad Y Y a.e.. The Fatou s lemma implies E(Y X) E(limif E(Y + X) E(limif (Y X )] limif E(Y X ) E(Y ) limsup E(X ) (Y + X )] limif E(Y + X ) E(Y ) + limif E(X ) So, the limit of E(X ) exists ad equals to E(X). Now suppose E(X ) E(X). There exists a subsequece { k } such that E(X k E(X) > ǫ for some ǫ > 0 ad all { k }. But for this subsequece, sice X k X ad Y Y i probability, there exists a further subsequece {X kj } X a.e. ad {Y kj } Y a.e.. The the above proof implies E(X kj ) E(X), which cotracts with E(X k ) E(X) > ǫ. 3. Show that Bi(, p ) P(λ) if ad p λ > 0. (This problem ad the ext oe are for your kowledge about the basic facts of commoly used distributios.)

3 3 Solutio. For ay fixed iteger k 0, ( P(Bi(, p ) k) k ( 1) ( k + 1) k! k ) p k (1 p ) k ( p 1 p ) k(1 p ) 1 k! λk e λ.! ( p ) k(1 p ) k!( k)! 1 p 4. Suppose M ad N are two idepedet Poisso radom variables with mea λ ad θ. Show that M + N is still Poisso radom variable, ad, moreover, the coditioal distributio of M give M + N k is Bi(k, p), where p λ/(λ + θ). Solutio. P(M + N k) P(M j, N k j) 1 j! λj e λ 1 (k j)! θk j e θ 1 k! (λ + θ)k e λ θ P(M j)p(n k j) ( ) k ( λ ) j ( θ ) k j j λ + θ λ + θ 1 k! (λ + θ)k e λ θ. Ad P(M j, N k j) P(M j M + N k) P(M + N k) k! λ j e λ θ k j e θ ( ) k j!(k j)! (λ + θ) k e λ θ p j (1 p) k j j P(M j)p(n k j) P(M + N k) where p λ/(λ + θ). 5. Suppose X F (meaig that X is measurable to a σ-algebra F). Show that E(X F) X, a.e. Solutio. Sice X satisfies coditio X F ad the coditio E(X1 A ) E(X1 A ) for every A F, which is actually a idetity, so E(X F) X, a.s.. 6. Suppose X 0 ad X X a.e., the E(X F) E(X F), a.e.. (This is the mootoe covergece theorem for coditioal expectatio. Fatou s lemma ad the domiated covergece theorem also hold for coditioal expectatio.) Solutio. Sice X, X +1 X 0. Therefore E(X +1 X F) 0. Hece, E(X F). Let the limit be ξ. Sice E(X F) F, the limit ξ is also F-measurable. For ay set A F, E(X F)1 A ξ1 A. by mootoe covergece theorem, E(ξ1 A ) lim E[E(X F)1 A ] by mootoe covergece theorem lim E(X 1 A ) by the defiitio of coditioal expectatio w.r.t. σ-algebra E(X1 A ) by mootoe covergece theorem. Therefore by the defiitio of coditioal expectatio with respect to a σ-algebra, ξ E(X F). 7. Suppose X c i distributio where c is a costat. The X c i probability. Solutio. For costat c as a r.v., its c.d.f. F(t) 1 for all t c ad F(t) 0 for all t < c. Therefore P(X t) 1 for ay t > c ad P(X t) 0 for ay t < c. Hece, P(X > t) 0 for ay t > c. So X c i probability.

4 4 8. Let X 1, X 2,... be i.i.d. r.v.s. with limsup t tp(x 1 > t) 0. Show that Y / 0 i probability, where Y max 1i X i. (Hit: use Chebyshev iequality). Solutio. Let ǫ > 0. P(Y / > ǫ) P( max 1i X i > ǫ) 1 P( max 1i X i ǫ) 1 P(X 1 ǫ) 1 (1 P(X 1 > ǫ)) 1 e log(1 P(X1>ǫ)) 1 e P(X1>ǫ) 0 Next, So, Y / 0 i probability. P(Y / < ǫ) P( max 1i X i < ǫ) P(X 1 < ǫ) Let f ad f be bouded cotiuous fuctios such that lim sup t f (t) f(t) 0. Suppose X X i distributio. The, E(f (X )) E(f(X)). Solutio. E(f (X )) E(f(X)) E f (X ) f(x ) + E(f(X )) E(f(X)) sup f (t) f(t) + E(f(X )) E(f(X)) 0. t 10. For ay sequece of r.v.s. X, there exists a sequece of costats A such that X /A 0 a.e.. (Hit: use Borel-Cotelli Lemma). Solutio. Choose a such that P( X > a ) 1/2. Let A a. The, X 1 { X a }/A a /A 1/ 0. Ad P( X > a, i.o.) 0 as 1 P( X > a ) < by the Borel-Catelli lemma. Therefore, X 1 { X >a }/A 0, a.e. Cosequetly, X /A 0 a.e Suppose F A. Show that E(E(X F) A) E(X F). Solutio. Let Y E(X F). Y is F-measurable, so Y must be A-measurable sice F A. Therefore E(Y A Y. 12. (crude versio of martigale covergece theorem) Suppose F F +1 for 1. Let F σ( 1 F ). For ay radom variable X with X c > 0, a.s., assume E(X F ) has a a.s. limit. show that E(X F ) E(X F), a.s. Solutio. Perhaps for a better uderstadig, deote Y E(X + c F ) ad Y E(X + c F), which are oegative r.v.s. For ay A F m, E(Y 1 A ) E(Y 1 A ) E(X + 1 A ) for all m. The, Fatou s lemma esures i.e., E(lim if Y 1 A ) limif E(Y 1 A ) E(Y 1 A ). E((Y limif Y )1 A ) 0, for all A F m, m 1 It implies E((Y limif Y )1 A ) 0, for all A F, which ca be proved by showig {A F : E((Y limif Y )1 A ) 0} is a σ-algebra which cotais F m, m 1, ad therefore must be the

5 5 same as F. The, Y limif Y, beig F-measurable, must be oegative a.s.. As a result, we have show lim if E(X F ) E(X F), a.s. Next, by cosiderig Y E(c X F ) ad Y E(c X F), oe ca likewise show lim sup E(X F ) E(X F). a.s. Cosequetly, lim E(X F ) E(X F) a.s..