Math 6710, Fall 2016 Final Exam Solutions

Transcription

1 Mah 67, Fall 6 Fial Exam Soluios. Firs, a sude poied ou a suble hig: if P (X i p >, he X + + X (X + + X / ( evaluaes o / wih probabiliy p >. This is roublesome because a radom variable is supposed o be defied uder all circumsaces ad here is o aural choice for he /. Foruaely, i does maer. Suppose we make wo differe choices for how o assig he value of ( whe X X, leadig o radom variables U, U such ha P (U U p. Sice σ > we have p <, so P (U U. For all x R, P (U x P (U U + P (U x, P (U x P (U U + P (U x. If U U he P (U x P (U x for all x a which he disribuio fucio of U is coiuous. The iequaliies above imply ha P (U x P (U x for all such x, implyig ha U U also. For our problem, if we make ay choice for he value of /, for example X + + X U (X + + if X X /,..., X are o all zero, if X X, ad show ha U Z N(,, he he same weak covergece would hold for ay oher choice U. Wih ha preamble, we proceed o he argume. Wrie U W Y where W X ( + + X σ σ, Y X + + X ad Y whe X X. The W Z N(, by he classical Ceral Limi Theorem. Wih probabiliy, X + +X will be eveually posiive for all pas a cerai poi, so X + + X σ for sufficiely large. Y The righ side coverges o a.s. by he classical Srog Law of Large Numbers, so he lef side does also. Le g(x / x; he each g(/y Y sice < Y <. Because g is coiuous a x, ( Y g Y g( a.s. We have show ha W Z ad Y a.s., which implies Y. Usig Problem o Homework, we coclude ha U W Y Z. /

2 . (a Le S E be he umber of edges i (V, E for. Number he poeial edges (i.e. pairs of disic verices from o ( ad defie he idicaor variables X,k by X,k if edge k is prese i E ad X,k if edge k is abse from E. For each, he X,k are iid wih P (X,k P (X,k. I addiio, S X, + + X,(. To prove a srog law of large umbers for S, we mimic he proof of Theorem 8. i he oes (Srog Law for bouded fourh momes. Defie he ceered variables X,k X,k, S S ( X, + +X,(. The argume i he proof of Theorem 8. shows ha E[S4 ] C ( where C E[X 4,] 6. Coiuig as i ha proof, Chebyshev/Markov s iequaliy shows ha for all ε >, ( ( ( 4 S P ( > ε P S 4 > ε 4 C( 4 C. As ( (, P ( S ( > ε C ε 4 The firs Borel-Caelli lemma implies ha ( S P ( > ε i.o.. ε 4( 4 ( 4 <. ε 4( Sice his holds for all ε >, i follows ha S ( S ( a.s. ( which is a srog law of large umbers for S. For a ceral limi heorem, oe ha each S has disribuio Biomial( (,, wih mea ( ad variace 4(. We aim o show ha S ( ( Z N(,. ( Le Y, Y,... be iid, P (Y i P (Y i, ad le T k Y + + Y k Biomial(k,. The classical Ceral Limi Theorem says ha T k k Z. ( k Boh ( ad ( are saemes abou he covergece of sequeces of disribuio fucios. The sequece of disribuio fucios i ( is a subsequece of he oe i (: ake k ( (, (, 4,.... Therefore ( direcly implies (, sice weak covergece of a sequece implies weak covergece of every subsequece. This complees he proof.

3 Aleraively, we could use he Lideberg-Feller Ceral Limi Theorem o prove ( direcly. Se ( U,k X,k (, W U, + + U,( S (. The h row of he riagular array has ( eries raher ha eries as i he saeme of Lideberg-Feller i he oes. However, he proof exeds wihou chage o he siuaio where each row has a arbirary umber of eries. Thus, o show ha W Z i suffices o check ha: (i ( ( lim E[U,k], (ii lim E[U,k; U,k > ε] for all ε >. k Boh codiios are easily verified. For (i, k ( Var(X, + + Var(X,( E[U,k]. k ( For (ii, sice each U,k /, if ε > is fixed he for large eough, each E[U,k ; U,k > ε]. Thus W Z, as desired. (b For each, umber he riples of disic verices from o (. Le X,k if riagle k is prese i he graph ad X,k if riagle k is abse. Thus, P (X,k 8 ad P (X,k 7 8. Sice T X, + + X,(, we have E[T ] 8(. Nex we compue Var(T E[T] E[T ] : E[T] E[X,j X,k ], E[T ] E[X,j ]E[X,k ]. j,k ( j,k ( 4( j,k ( (Noe ha each erm i he laer sum is 64, which is cosise wih E[T ] 64(. So, Var(T ( E[X,j X,k ] E[X,j ]E[X,k ] ( E[X,j X,k ]. 64 j,k ( The radom variable X,j X,k is if boh riagle j ad riagle k are prese i he graph, ad oherwise. There are hree cases: ( j k, so boh riagles are he same. The E[X,j X,k ] There are ( erms of his ype. ( The wo riagles share a sigle edge. There are five disic edges which mus be prese i he graph i order for X,j X,k o equal, so E[X,j X,k ] To cou he umber of erms of his ype, suppose riagle j has verices u, v, w. Triagle k could have verices u, v, x or u, w, x or v, w, x for ay x amog he remaiig verices. Therefore, for each j here are ( choices for k, ad he oal umber of erms is ( (.

4 ( The wo riagles share o edges. The X,j ad X,k are idepede, so E[X,j X,k ] 64 E[X,j ]E[X,k ] 64. I does o maer how may of hese erms here are, sice hey coribue zero o Var(T, bu we could sill cou hem if we wa o. Give riagle j wih verices u, v, w, riagle k could have verices u, x, y or v, x, y or w, x, y or x, y, z. This gives ( ( + possibiliies, so he oal umber of erms is [ ( ( + ] (. Addig up he oal umber of erms from cases (,(,( gives [ + ( + ( + ( ] ( ( which is he correc umber of erms i he sum. Agai, his is uecessary bu i is a good way o check ha he umbers i cases ( ad ( are righ. From he couig procedure above, Var(T 7 ( + ( ( C where we ca ake C 8. To prove a srog law of large umbers for T, use Chebyshev s iequaliy. For all ε >, ( P T 8( ( > ε C4 ε (. The deomiaor has order 6, so he sum over of hese probabiliies should be fiie. Ideed, sice ( 6 as ad C 4 ε ( /6 6C ε <, he limi compariso es implies ha ( P T 8( ( > ε <. By he firs Borel-Caelli lemma, ( P T 8( ( > ε i.o.. Sice his holds for all ε >, i follows ha T 8( ( a.s. which is he desired srog law of large umbers for T. 4

5 . (a Defie X X, X X if X > X, if X X, if X < X, X, X X,. Because X X X, X + X X, X, X + X,, i suffices o prove ha E X, X ad E X,. By cosrucio, { X X if X > X, X, if X X. Sice X X a.s., also X X a.s. ad herefore X, a.s. I follows ha X, X X, X a.s., so ha X, X a.s. Also, X, X ad X, X X. Thus E X, E X ad E X, X by domiaed covergece. Nex, observe ha X, X X,. Therefore, E X, E X E X, E X E X. This complees he proof. (b Saeme: For p >, if X X a.s. ad E[ X p ] E[ X p ], he E[ X X p ]. Proof: Defie X,, X, as i par (a. If we ca show ha E[ X, X p ] ad E[ X, p ], he X X p X, X p + X, p, which implies ha E[ X X p ]. By he argume i par (a, X, X a.s., which implies ha X, p X p a.s. ad X, X p a.s. The bouds X, p X p ad X, X p ( X p, alog wih E[ X p ] <, imply ha E[ X, p ] E[ X p ] ad E[ X, X p ] by domiaed covergece. Nex, sice X, + X, X, we have X, p + X, p X p. This is a geeral fac: If p > ad a, b, he a p + b p (a + b p. Quick proof: The saeme is rivial if a + b. If a + b > he a/(a + b, b/(a + b [, ] ad so ( a a + b p + ( b a + b p a a + b + b a + b ap + b p (a + b p. (The fac also follows from covexiy of x x p. We obai E[ X, p ] E[ X p ] E[ X, p ], so lim sup E[ X, p ] lim sup E[ X p ] lim if E[ X, p ] E[ X p ] E[ X p ], meaig ha E[ X, p ]. This complees he proof. 4. Our overall sraegy is o defie radom variables X,..., X ad Y,..., Y o he same probabiliy space such ha he X i are iid wih disribuio fucio F ad he Y i are iid wih disribuio fucio U (i.e. Y i Uiform[, ]. Le F (x #{ i : X i x}, U (y #{ i : Y i y} ad D (F sup F (x F (x, D (U sup U (y U(y. y R 5

6 I our cosrucio, i will hold ha D (F D (U, wih equaliy whe F is coiuous. This will prove boh pars (a ad (b. The probabiliy space is Ω (, wih Lebesgue measure. (Tha is, pu Lebesgue measure o each ierval (, ad ake he produc measure. Give ω (ω,..., ω, le Y i (ω ω i for i, so he Y i are iid Uiform[, ] radom variables. For < y <, defie F (y if{x R : F (x y}. For each i, le X i F (Y i. The X i are idepede, ad each X i has disribuio fucio F by he proof of Theorem. i he oes. I ha proof, i is show ha for all x R, X i x if ad oly if Y i F (x. Hece F (x #{i : X i x} #{i : Y i F (x} U (F (x ad so F (x F (x U (F (x F (x U (F (x U(F (x. (The las equaliy is because U(y y for all y. I follows ha sup I oher words, D (F F (x F (x sup D (U, provig par (b. U (F (x U(F (x sup U (y U(y. y R If F is coiuous, he rage of F icludes every < y < by he Iermediae Value Theorem. Thus sup F (x F (x sup U (F (x U(F (x sup U (y U(y. <y< I fac, U (y U(y for all y / (,. If y he U(y, ad also U (y sice each Y i >. If y he U(y, ad also U (y sice each Y i <. Therefore, sup <y< so we have show he reverse iequaliy D (F U (y U(y sup U (y U(y y R D (U, provig par (a. 5. (a From Theorems 8. ad 8. i he oes, {S } is recurre if ad oly if P (S. Sice P (R P (N( P (R N( P (N( P (S, we use Fubii s Theorem o compue For each, P (R d P (N( P (S d P (S P (N( d. P (N( d is he expeced amou of ime ha he Poisso process says a level, which is he expeced amou of ime bewee he h ad ( + s arrivals, which 6

7 equals. This argume ca be made rigorous, bu we ca also evaluae he iegral direcly. Sice N( Poisso(, ( P (N( d e d.! We prove by iducio ha his iegral equals for every. Whe, e d. Whe, iegraio by pars gives ( ( e d e!! e ( which equals by he iducive hypohesis. I coclusio, P (R d which is ifiie if ad oly if {S } is recurre. (! d P (S, e ( (! d, (b If x (x,..., x d R d, defie he orm x max{ x i : i d}. This is jus for cosisecy wih he oes; he saemes ad argumes work for ay orm o R d. Saeme: {S } is recurre if ad oly if Proof: There are wo seps: ( {S } is recurre if ad oly if ( For fixed ε >, P ( R < ε d for all ε >. P ( S < ε for all ε >. P ( S < ε Sep is proved exacly as i par (a: P ( R < ε d P ( R < ε d. P (N( P ( S < ε d P ( S < ε P (N( d P ( S < ε. Therefore i suffices o prove sep. Oe direcio is quick. Suppose he sum i sep is fiie for some ε >. The by he firs Borel-Caelli lemma, P ( S < ε i.o., so is o a recurre value of {S }. I follows from Theorem 8. ha {S } is o recurre. Coversely, suppose {S } is o recurre. By Theorem 8., is o a recurre value, so here exiss ε > such ha P ( S < ε i.o.. I follows ha here exiss δ > such ha P ( S 7

8 δ for all >. To see why, assume for coradicio ha for every δ >, P ( S δ for all. Le τ ad for k, τ k mi{ > τ k : S S τk < ε/ k }. We show by iducio ha each τ k is a.s. fiie. Sice τ k is a a.s. fiie soppig ime, he sequece {S τk +m S τk } m has he same law as {S m } m eve whe we codiio o he value of τ k. Usig he assumpio, wih probabiliy here exiss > τ k such ha S S τk < ε/ k, so τ k is a.s. fiie. For each k, S τk S τk S τ Hece P ( S < ε i.o., a coradicio. k S τj S τj < j k j ε k < ε. We have show ha if {S } is o recurre, he here exiss δ > such ha p : P ( S δ for all >. Le V #{ : S < δ/}. The E[V ] P ( S < δ/ ad also E[V ] k P (V > k. Le α ad for k, α k mi{ > α k : S < δ/}. Each α k is a soppig ime, ad P (V > k P (α k <. Give ha α k <, ad give he hisory of he radom walk up o ime α k, we kow ha S αk < δ/. Also, wih probabiliy p >, S S αk δ for every > α k. This implies ha S S S αk S αk δ/ for every > α k, so α k. We coclude ha P (α k < α k < p, ad so P (α k < ( p k by iducio. Hece P ( S < δ/ E[V ] P (α k < ( p k <. k This fiishes he proof of he coverse direcio i sep. (c We cosruc he simple radom walk i he followig maer. Le e,..., e d be he sadard basis vecors i R d (e.g. e (,,,...,. Le V, V,... be iid, P (V i j /d for j d. Le W, W,... be iid ad idepede of he V i, P (W i P (W i /. The we ca se X i W i e Vi ad S X + + X. The jh coordiae of he radom walk afer seps is S (j {V i j}w i. Passig o he coiuous ime versio R S N(, he jh coordiae a ime is i N( R (j {V i j}w i. i 8 k

9 Fix. For each j d, le N j ( #{ i N( : V i j}. Problem 5 o Homework (hiig of Poisso variables shows ha each N j ( Poisso(/d ad ha he N j ( are idepede. Give he values of he N j (, he R (j are codiioally idepede wih P (R (j x N (,..., N d ( d P (T j x where {T } is a discree ime simple radom walk o Z. Therefore, P (R ( x,..., R (d x d P (N (,..., N d ( d P (R (x,..., x d N (,..., N d ( d,..., d,..., d d d P (N j ( j P (T j x j j j ( ( P (N ( P (T x P (N d ( d P ( x d d ( ( P (N(/d P (T x P (N(/d d P ( x d P (Y /d x P (Y /d x d, d where {Y } is a coiuous ime simple radom walk o Z. Summig over all possible values of x,..., x j, x j+,..., x d gives P (R (j x j P (Y /d x j. Thus he R (j are iid ad have he same law as Y /d. I follows ha P (R P (R (,..., R (d P (Y /d P (Y /d P (Y /d d. (d Le {S } ad {R } be discree ad coiuous ime simple radom walks o Z d. Assume ha c P (Y C for all T. Sice P (R d, by par (a, {S } is recurre if ad oly if P (R d <. Par (c shows ha P (R P (Y /d d. We compue P (Y /d d d P (Y /d d d ( d C d C d d d/ d, /d d/ ( d c d c d d d/ d. /d d/ Whe d or d, (/d/ d diverges ad so {S } is recurre. Whe d, he iegral coverges ad so {S } is rasie. 9