MATH 507a ASSIGNMENT 4 SOLUTIONS FALL 2018 Prof. Alexander. g (x) dx = g(b) g(0) = g(b),

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1 MATH 57a ASSIGNMENT 4 SOLUTIONS FALL 28 Prof. Alexader (2.3.8)(a) Le g(x) = x/( + x) for x. The g (x) = /( + x) 2 is decreasig, so for a, b, g(a + b) g(a) = a+b a g (x) dx b so g(a + b) g(a) + g(b). Sice g is icreasig we he have g (x) dx = g(b) g() = g(b), g( X Z ) g( X Y + Y Z ) g( X Y ) + g( Y Z ). Takig expecaio here gives d(x, Z) d(x, Y ) + d(y, Z). Sice g, we have d(x, Y ) = = Eg( X Y ) = = X Y = a.s. = X = Y a.s. Clearly d(x, Y ) = d(y, X) so d is a meric. (b) If d(x, X) he for every ɛ >, usig Chebyshev s Iequaliy, P ( X X > ɛ) = P ( g( X X ) > ɛ + ɛ ) + ɛ Eg( X X ) as, ɛ so X X i probabiliy. Coversely if X X i probabiliy, he for ɛ >, Eg( X X ) = g( X X ) dp + g( X X ) dp { X X ɛ} { X X >ɛ} g(ɛ) dp + dp { X X ɛ} g(ɛ) + P ( X X > ɛ), { X X >ɛ} so lim sup Eg( X X ) g(ɛ). Sice ɛ is arbirary, his shows lim sup Eg( X X ) =, ha is, Eg( X X ). (2.3.) Choose ( a sequece ) covergig o, say /. The idea is o fid cosas c such ha P X c > i.o. =, which implies ha X /c a.s. For fixed, sice { X > c/} φ as c, we have P ( X > c/) as c, so here exiss c such ha P ( X > c /) < 2. The ( X P > ) < 2 < c

2 ( ) so by Borel-Caelli, P X c > i.o. =, as desired. (2.3.8)(ii) Cosider a oradom oegaive sequece {x } ad le m = max j x j. If m / log > c i.o. he here exiss a subsequece (k) wih m (k) / log (k) > c. This meas ha for all k, for some j(k) (k) we have x j(k) / log (k) > c. Sice log (k), his requires ha j(k) as k. Sice log (k) log j(k), we have x j(k) / log j(k) > c for all k. Thus x / log > c i.o. Applyig his o he radom variables, i says ha { M log } > c i. o. { X log } > c i. o.. The opposie iclusio is clearly also rue, sice X M, so hese eves are equal. By par (i) ad he Geeral Hi for his assigme, his shows ha lim sup M log = lim sup X log Now sice x e x for oegaive x, we have ( ) M P log < ɛ = P ( k={x k < ( ɛ) log }) = ( e ( ɛ) log ) e ( ɛ) = e ɛ, = a.s. () so ( ) P M < ɛ M < for all ɛ >. By Borel-Caelli, his meas lim if log log a.s. Wih () his shows ha he limi is acually a.s. (2.4.2) Le us defie a cycle o be he ime from oe bulb replaceme o he ex, so he ih cycle has legh X i + Y i. Le N be he umber of cycles compleed by ime. Le S = X + + X, T = Y + + Y. The S N R S N+, ad by Theorem 2.4.6, ad so also S N+ S N = S N N N = S N + N + N + R EX EX EX EX + EY, E(X + Y ) E(X + Y ), a.s. 2

3 (I) Le Y = i= A k, so A occurs i.o. Y Y > a eveually for all a >. Sice he eves {Y > a} are icreasig i ad decreasig i a, we have so P (Y > a eveually) = lim P (Y > a) P (A i.o.) = P (Y ) = P (Y > a eveually for all a) = lim a lim P (Y > a). Le ɛ >. Give a >, choose N so EY N = N i= P (A i) > a/ɛ. This he also holds for all N, meaig EY a ( ɛ)ey. By (II) i Assigme 2, for N, P (Y > a) (EY a) 2 E(Y 2 ) ( ɛ)2 (EY ) 2 E(Y 2 ) = ( ɛ) 2 ( k= P (A k)) 2 k= P (A j A k ) j= so compaig lim sups we ge lim P (Y > a) ( ɛ) 2 α for all a, ɛ Hece P (A i.o.) ( ɛ) 2 α for all ɛ >, so P (A i.o.) α. (II)(a) EX = 2 ( 2 ) + ( 2 )( ) =. Also P (X = 2 i.o.) = sice P (X = 2 ) <, so X = for all sufficiely large a.s., so X a.s. Bu his meas also S / a.s. (b) To ge a coradicio, suppose S µ a.s. The he followig also coverges o a.s.: S µ S µ + = S ( + ) X µ + µ = ( ) S + µ X µ + Sice he firs erm o he righ approaches, his meas we have + µ. X µ + + µ a.s. (2) Sice P (X + = ) =, we have P (X + = i.o.) =. This ad (2) show here mus exis a (oradom) subsequece wih The (2) says k k + + µ k + k k + µ k so µ k + k k + µ k. X k + k+ a.s. 3

4 Bu alog ay oradom subsequece we have P (X k+ = i.o.) = P (lim sup{x k + = }) lim sup P (X k + = ) =, k k so his is a coradicio. (III) If such ɛ exis, he for a.e. ω here exiss (ω) for which (ω) = X (ω) X(ω) ɛ, which meas X (ω) X(ω). I he oher direcio, suppose X X a.s. Le p N (ɛ) = P ( X X ɛ for some N). These eves are decreasig i N so p N (ɛ) as N for all ɛ >. Pick a summable sequece, say /k 2. Sice for each k we have p N (/k) as N, here exiss N k such ha p Nk (/k) /k 2. This propery is preserved if we icrease ay N k, so we may sequeially choose N =, he N 2 > N, N 3 > N 2, ec. Sice k /k2 <, by Borel-Caelli we have ( P X X ) k for some N k, for ifiiely may k =. If we defie ɛ o be /k for all N k < N k+ (so ɛ ), he his eve is he same as { X X ɛ i.o.}, so we have P ( X X ɛ i.o.) =. (IV)(a) We have log X = i= log U i c = E(log U ) a.s.; he las expecaio exiss because log U is a bouded r.v. This meas ha for some (depedig o ɛ, ω) we have c ɛ < log X < c + ɛ, equivalely e (c ɛ) < X < e (c+ɛ). (b) The desiy is /2 o [,2], so E(log U ) = 2 2 (log x) dx = 2 (x log x x) 2 = log 2 Sice his expecaio exiss, we have c = log 2 as i (a). (c) Here log U = log /U is oegaive, so ) E (log U = P (log U ) d = P (U < e ) d = + d =. Therefore log X a.s., so o such c exiss; we have X faser ha expoeially. 4

5 (V) For c > ad large, we have from Theorem.2.3 P (ξ > (2c log ) /2 ) 2π (2c log ) /2 e c log c, so P (ξ > (2c log ) /2 ) <, so by Borel-Caelli, P (ξ /(2 log ) /2 > c /2 i.o.) =. Therefore lim sup ξ /(2 log ) /2 c /2 a.s. Similarly, for c < ad A >, akig δ < c ad large, we have A(log ) /2 > δ, ad he also P (ξ > (2c log ) /2 ) ( ) 2π (2c log ) e c log c /2 (2c log ) 3/2 2π 2(2c log ) /2 (c+δ), so P (ξ > (2c log ) /2 ) =, so by idepedece ad Borel-Caelli, P (ξ /(2 log ) /2 > c /2 i.o.) =. Therefore lim sup ξ /(2 log ) /2 c /2 a.s. Combiig he resuls for c < ad c > we obai lim sup ξ /(2 log ) /2 = a.s. (VI) Le a = sup{x : F (x) = } (wih a = if here is o such x) ad b = if{x : F (x) = } (wih b = if here is o such x.) The F (a ) =, F (b) =, so we mus have a < b (oherwise P (X = a) =.) Hece we ca choose a < c < d < b ad we have F (c) >, F (d) >. The P (X c) = ad P (X > d) = so X c i.o. ad X > d i.o., a.s. Sice c < d, his meas P ({X } coverges) =. 5