Solution. 1 Solutions of Homework 6. Sangchul Lee. April 28, Problem 1.1 [Dur10, Exercise ]

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1 Soluio Sagchul Lee April 28, 28 Soluios of Homework 6 Problem. [Dur, Exercise 2.3.2] Le A be a sequece of idepede eves wih PA < for all. Show ha P A = implies PA i.o. =. Proof. Noice ha = P A c = P A c = PA c. Sice PA c >, his fac is o affeced by droppig ou firs m erms, i.e., m= PA c m =. This shows ha P m=a m c = ad hece P m=a m =. Therefore akig, we have PA i.o. =. Problem.2 Koche-Soe lemma [Dur, Exercise 2.3.2] Suppose PA k =. Use Exercise.6.6 ad 2.3. o show ha if 2 / lim sup PA k k= he PA i.o. α. The case α = coais Theorem PA j A k j,k = α > Proof. We iroduce some oaios as follows: N = k= Ak, E m,] = k=m+ A k, E = E,] = A k. If m < ad ω is such ha N ω N m ω >, he ω E \ E m E m,]. So we have N N m = N N m Em,] ad hece by he Cauchy-Schwarz iequaliy EN N m 2 EN N m 2 P E m,] E N 2 P Em,]. Now usig he assumpio EN = k= PA k, we have EN EN m 2 P A k = lim PE m,] limsup EN 2 k=m+ k= EN 2 = limsup EN 2 = α. Here, he las lie follows from EN 2 = j,k PA j A k. Therefore leig m proves he desired claim.

2 Problem.3 [Dur, Exercise 2.4.2] Suppose he ih ligh bulb burs for a amou of ime X i ad he remais bured ou for ime Y i before beig replaced. Suppose he X i, Y i are posiive ad idepede wih X s havig disribuio F ad he Y s havig disribuio G, boh of which have fiie mea. Le R be he amou of ime i [,] ha we have a workig ligh bulb. Show ha R / EX i /EX i + EY i almos surely. Proof. We iroduce some quaiies: S = X + + X, T = X +Y + + X +Y, N = sup : T }. Now by he Reewal heory Theorem of [Dur] ad he SLLN, we kow ha he eve } } } N S Ω = EX + EY EX T EX + EY has probabiliy oe. Now o he eve Ω, we have N ad hus S N /N EX ad T N /N EX + EY hold as well. Also, usig he obvious iequaliy S N R S N +, we have S N /N TN N R S N +/N + T N /N N T N +/N + TN + N + N + N N. So akig ogeher wih he squeezig heorem proves ha R / EX /EX + EY o Ω. Therefore he claim follows. Problem.4 [Dur, Exercise 2.4.3] Le X =, ad defie X R 2 iducively by declarig ha X + is chose a radom from he ball of radius X ceered a he origi, i.e., X + / X is uiformly disribued o he ball of radius ad idepede of X,,X. Prove ha log X c a.s ad compue c. Proof. Le U = X / X for =,2,. The U is a uiform disribuio o he ui ball ad is idepede of X,,X. The for x, 2π x r P U x = π drdθ = x2, Elog U = 2ulogudu = 2. The by he SLLN, we have log X k= log U k Elog U = 2 a.s. Problem.5 Prove ha if f : R N R is measurable ad bouded ad X,X 2, is a sequece of i.i.d. radom variables, he k= f X k,x k+, E f X,X 2,, a.s. Hi: Approximae f by a fucio of fiiely may variables. Proof. Defie A f = k= f X k,x k+, ad le H = f : f : R N R measurable ad A f E f X,X 2, a.s.} be he family of bouded measurable fucios f : R N R for which he claim holds. I is immediae ha H is a vecor space. Our aim is o prove ha his coais all bouded measurable fucios. 2

3 Sep. We firs show he claim whe f depeds oly o fiiely may variables, i.e., here exiss a measurable bouded fucio f : R d R such ha f x,x 2, = f x,,x d. The we have f X k,x k+, = f X k,,x k+d. Now wrie A f = d r= r/d q= f X qd+r,,x q+d+r ad oice ha f X qd+r,,x q+d+r : q is a sequece of i.i.d. radom variables for each give r,,d}. So we ca apply SLLN o each summad o obai A f SLLN d r= d E[ f X r,,x r+d ] = E[ f X,X 2, ], Here, he las equaliy follows sice X r,,x r+d law = X,,X d. Therefore f H. Sep 2. Nex we prove he claim whe f = B is a idicaor fucio of a se B BR N. Recall ha BR N is geeraed by he family F = i= A i : A i BR ad A i = R for all bu fiiely may i}. Also we deoe by µ he law of X,X 2,, i.e., µ is he measure defied as µb = PX,X 2, B for each B BR N. Now we fix ε >. The by Lemma A.2. of [Dur], here exiss A F such ha µb A < ε ad N F such ha B A N ad µn 2ε. Also oice ha A B µb A B A + A A µa + µa µb A N + A A µa + µn. So akig limsup as ad uilizig he fac ha A, N H, i follows ha lim sup A B µb 2µN < 4ε, P-a.s. So by leig ε alog a subsequece, we fid ha A B µb ad hece f H. Sep 3. The previous sep esseially resolves he problem. Sice H is a vecor space, ay bouded simple fucios o R N is i H. The by choosig simple fucios l j ad u j such ha l j f ad u j f as j, we obai El j X,X 2, = limif A l j limif A f limsup P-a.s. A f limsupa u j = Eu j X,X 2,, Leig j ogeher wih he domiaed covergece ells ha f H. Therefore his proves ha H coais all bouded measurable fucios R N R ad he complees he proof. Problem.6 P-a.s. Le T,T 2, be radom variables o he probabiliy space Ω,F,P which are i.i.d., o-egaive ad obey ET i,. Le N := supm : T + + T m }. Prove ha here is Ω F wih PΩ = such ha o Ω we have ET T x} x R : TNs+ x} ds ET This correcs a fauly calculaio from Wedesday s class. Remark. Provig ha he covergece occurs a.s. for a fixed x R is o so hard. The rue echical challege is o prove ha he covergece occurs simulaeously for all x R o a eve of probabiliy oe. To see wha may go wrog, oice ha Ω = x R lim TNs+ x} ds = ET } T x} ET 3

4 is a ucouable iersecio of eves so ha Ω eve eed o defie a eve, much less wheher is probabiliy is or o. This is why he problem is carefully formulaed. Alhough i may happe ha Ω fails o be a eve, you ca sill exploi some srucure of he problem o show ha here is a eve Ω of probabiliy oe ha saisfies Ω Ω. Proof. Noice ha we always have S N S N + ad SN TNs+ x} ds = N k= Sk S k TNs+ x} ds = N k= Sk These ogeher provides he followig bouds o he iegral i quesio. TNs+ x} ds S N TNs+ x} ds S N + I order o uilize hese bouds, we observe ha SN + SN Ω := S / ET } has probabiliy oe by SLLN, Ω 2 := N } has probabiliy oe, ad S k Tk x} ds = TNs+ x} ds = N + S N k= TNs+ x} ds = S N + N k= N k= T k Tk x}, T k Tk x}. T k Tk x}. Ω 3,x := k= T k Tk x} E[T T x}]} has probabiliy oe by SLLN for each x R. Now choose a couable dese subse D R which coais all he aoms of he law of T. Tha is, we have PT = x = for all x R \ D. Ad fially, we defie Ω := Ω Ω 2 Ω 3,r. Sice Ω is a couable iersecio of eves of probabiliy, i also has probabiliy oe. Moreover, for ay x R ad r D wih x r, we have he followig compuaio o he eve Ω. lim sup TNs+ x} ds limsup limsup N r D TNs+ r} ds N + S N N N + N + k= T k Tk r} = E[T T r}] ET Leig r x ad applyig he domiaed covergece heorem, we obai he required upper boud. The machig lower boud ca be obaied aalogously. If x R ad r D are such ha x r, he o he eve Ω we have lim if TNs+ x} ds limif N + limif S N + TNs+ r} ds N N + N + N k= T k Tk r} = E[T T r}] ET If x D, he his is already good. Oherwise, x is a coiuiy poi of he law of T so ha PT = x =. The leig r x, by he moooe covergece heorem we obai lim if TNs+ x} ds E[T T <x}] = E[T T x}]. ET ET Therefore wo iequaliies are sauraed ad hece we have he desired saeme. 4

5 Problem.7 Buildig o he previous problem, le Z be a radom variable wih CDF PZ x := ET T x} ET, x R. Prove ha EZ = ET 2 /ET ad hus EZ > ET. Prove ha, i fac, These are demosraios of he Ispecor s Paradox. x R : PZ > x PT > x Proof. Le µ be he law of T. Sice Z is o-egaive, E[T T >z}] EZ = PZ > zdz = dz = ET ET >z} µd dz. Applyig he Fubii s heorem o swich he order of iegraio ad proceedig, EZ = ET >z} dzµd = 2 µd = ET 2. ET ET Now by he Cauchy-Schwarz iequaliy, we have ET 2 = E[T ] 2 ET 2 E = ET 2 ad he iequaliy is sric uless T is a.s. cosa. This proves he secod claim modulo he modificaio of he saeme ha he equaliy is achieved whe T is a.s. cosa. x Fially, oice ha he fucio a+x is icreasig i x [, ad decreasig i a [,. So by uilizig he iequaliies ET T >x} xpt > x ad ET T x} xpt x, we obai wrie PZ > x = This complee he proof as desired. ET T >x} ET T x} + ET T >x} xpt > x xpt x + xpt > x = PT > x. Problem.8 [Dur, Exercise 2.5.5] Le X be idepede for. The followig are equivale: i = X < a.s. ii = [PX > + EX X }] < iii = EX / + X < Proof. i = ii : By he Kolmogorov 3-series Theorem, boh PX > < ad EX X } < hold. This implies ii. ii iii : Observe ha PX > +EX X } = EX. Sice x/+x x 2x/+x, his observaio esablishes he equivalece bewee ii ad iii. ii = i : By he Fubii s Theorem, we have E = X = = EX < ucodiioally. This i paricular implies ha = X < a.s. The X shows ha X = X eveually, hece i follows ha = X < a.s. as required. Problem.9 [Dur, Exercise 2.5.8] Le X,X 2, be i.i.d. ad o. The he radius of covergece of he power series X ωz is a.s. or a.s., accordig as Elog + X < or = where log + x = maxlogx,. 5

6 Before he acual proof, we recall he followig simple observaio from Borel-Caelli s lemmas. Lemma.. Le Y be a sequece of i.i.d. radom variables. The Y lim sup =, if E Y =, if E Y < a.s. Proof of Lemma.. This is a direc cosequece of he followig equivalece. Y = a.s. lim sup = a.s. E Y = A > : P Y > A i.o. = = A > : P Y > A < = < Here, he secod equivalece follows from boh he s ad 2 d Borel-Caelli lemma. //// Lemma.2. Le X be i.i.d. ad o ideically zero. The we have log X lim sup Cosequely, we have limsup log + X / = limsup log X / a.s. Proof of Lemma.2. Le ε >. Sice X is o ideically zero, here exiss N such ha P X e εn >. The P X e ε = P X e ε =. N The by he 2 d Borel-Caelli s lemma, we have P X e ε i.o. = ad hece limsup log X / ε a.s. Now leig ε alog a subsequece proves he firs claim. The log + X log X log X lim sup = lim sup = limsup ad he secod claim follows. //// Proof. Le R = Rω deoe he radius of covergece. The by he Cauchy-Hadamard formula, we have } log X log + } X R = limif = exp limsup = exp limsup a.s. X / where he las equaliy follows from Lemma.2. So If Elog + X <, he by Lemma. we have limsup log + X = ad hece we have R = a.s. If Elog + X =, he ow limsup log + X = ad i follows ha R = a.s. This complees he proof. a.s. 6

7 Problem. [Dur, Exercise 2.5.9] Le X,X 2, be idepede ad le S m, = X m+ + + X. The P max S m, j > 2a mi P S k, a P S m, > a m< j m<k Proof. We wrie S m,] = S m, = X m+ + + X o clarify he meaig of he subscrip. Fix a > ad se T ad S by S := if j > m : S m, j] > 2a ad S j,] a}, T := if j > m : S m, j] > 2a} We use he coveio if = simply o guaraee he well-defiedess of he quaiies above. The by he reverse riagle iequaliy, We also fid ha S } = S m, j] > 2a ad S j,] a for some j m,]} S m,] > a}. T = k} S k,] a} S = k}. Here, he crucial observaio is ha T = k} depeds oly o X m+,,x k while S k,] a} depeds oly o X k+,,x. So hese eves are idepede. The P S m,] > a PS = k=m+ k=m+ PS = k PT = kp S k,] a PT mi m<k P S k,] a. Therefore he claim follows from he equaliy PT = P max m< j S m, j] > 2a. Problem. [Dur, Exercise 2.5.] Use o prove a heorem of P. Lévy: Le X,X 2, be idepede ad le S = X + + X. If lim S exiss i probabiliy he i also exiss a.s. Proof. I suffices o show ha S = X + + X is Cauchy P-a.s. Noice also ha S coverges i probabiliy if ad oly if S is Cauchy i probabiliy. So for each l, here exiss N l such ha N l ad N l k : P S k S > 2 l < 2 l. By, we have So i follows ha P max S Nl,k] N l <k P > 2 l+ P SNl,] > 2 l mi P S k,] 2 l N l <k sup S j S k > 2 l+2 j,k>n l = 2P 2 l 2 l 2 l+. sup S Nl,k] > 2 l+ 2 l+2. k>n l 7

8 So by he Borel-Caelli s Lemma, P sup S j S k 2 l+2 for ifiiely may l j,k>n l This is eough o coclude ha S is Cauchy P-.a.s. ad hece coverge P-a.s. =. Problem.2 [Dur, Exercise 2.5.] Le X,X 2, be i.i.d. ad S = X + + X. Use o coclude ha if S / i probabiliy he max m S m / i probabiliy. Proof. Le ε >. Pick N such ha P S / > ε < 4 for all N. The So by, we have N k <, P S S k > 2ε P S > ε + P S k > ε 2. P max S k S N > 4ε 2P S S N > 2ε. N<k This proves ha max N<k S k S N / i probabiliy ad hece max N<k S k / i probabiliy. Refereces [Dur] R. Durre. Probabiliy: Theory ad Examples. Cambridge Series i Saisical ad Probabilisic Mahemaics. Cambridge Uiversiy Press, 2. 8