P ( N m=na c m) (σ-additivity) exp{ P (A m )} (1 x e x for x 0) m=n P (A m ) 0

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1 MA414 STOCHASTIC ANALYSIS: EXAMINATION SOLUTIONS, 211 Q1.(i) Firt Borel-Cantelli Lemma). A = lim up A n = n m=n A m, o A m=na m for each n. So P (A) P ( m=na m ) m=n P (A m ) (n ) (tail of a convergent erie): P (A) =. [5] (ii) (Second Borel-Cantelli lemma). If A n are event, A := lim up A n = {A n i.o.}: if P (A n ) = and the A n are independent, then P (A) = 1. Proof. By the De Morgan law, A c = n m=n A c m. But for each n P ( m=na c m) = lim N P ( N m=na c m) (σ-additivity) = N (1 P (A m )) (independence) m=n N m=n exp{ P (A m )} (1 x e x for x ) = N exp{ P (A m )} (N ), m=n a P (A n ) diverge. So m=na c m i null, o A c = n m=n A c m i null. // [8] (iii) Second Borel-Cantelli Lemma for Pairwie Independence. For A n pairwie independent, if P (A n ) diverge then P (lim up A n ) = P (A n i.o.) = 1. Proof. Put S n := n 1 I(A i ), S := 1 I(A i ), m n := E[S n ] = n 1 P (A i ). var(s n ) = E[(S n m n ) 2 ] = E[( n i=1 (I(A i ) EI(A i ))( n j=1 (I(A j ) EI(A j ))] = j(...)(...)] = i E[(...) 2 ] + i j E(...)(...)] = i E[(...) 2 ] (the um E[ i over i j i, a there by pairwie independence and the Multiplication Theorem E[(...)(...)] = E[(...)]E[(...)] =. = variance of um = um of variance under pairwie independence). A I(A i ) i Bernoulli with parameter P (A i ), it variance i P (A i )[1 P (A i )] P (A i ). So E[(S n m n ) 2 ] n1 P (A i ) = m n, which increae to + a P (A n ) diverge. But P (S m n /2) P (S n m n /2) (S n S) = P (S n m n m n /2) P ( S n m n m n /2) 4 m 2 n var(s n ) (by Tchebycheff Inequality) 4/m n (by above) (n ). But the LHS increae to P (S < ), by continuity (= σ-additivity) of P (.). So P (S < ) = : P ( I(A n ) < ) =, i.e. P ( I(A n ) = ) = 1. Thi ay that P (A n i.o.) = 1: P (lim up A n ) = 1. // [12] (Standard bookwork: (i), (ii) done in lecture, (iii) done on a problem heet.) 1

2 Q2. Take the Lebegue probability pace ([, 1], λ, L) modelling the uniform ditribution U[, 1] on the unit interval (probability = length). For a random variable X U[, 1], take it dyadic expanion X = 1 ϵ n /2 n. Thu ϵ 1 = iff X [, 1/2), 1 iff X [1/2, 1) (or [1/2, 1]: we can omit 1, a it carrie probability). If ϵ 1,..., ϵ n 1 are already defined, on the dyadic interval [k/2 n 1, (k+1)/2 n 1 ), and independent fair coin-toe (Bernoulli B( 1)), 2 plit each interval into two halve: ϵ n = on the left half, 1 on the right half. Then ϵ n i again B( 1), and i independent of ϵ 2 1,..., ϵ n 1. By induction, ϵ n (n = 1, 2,...) are independent B( 1). Converely, given ϵ 2 n independent coin toe, form X := 1 ϵ n /2 n. Then X n := n 1 ϵ k /2 k X a.. The ditribution function F n of X n ha jump 1/2 n at k/2 n, k =, 1,..., 2 n 1. Thi aw-tooth jump function converge to x on [, 1], the ditribution function of U[, 1] (up F n (x) x = 2 n a n ). So X U[, 1]. So if X = 1 ϵ n /2 n, X U[, 1] iff ϵ n are independent coin toe the Lebegue probability pace model both (a) length on the unit interval and (b) infinitely many independent coin toe. (i) From the given U[, 1], we get by dyadic expanion a above a equence of independent coin-toe ϵ n. Rearrange thee into a two-uffix array ϵ jk (a with Cantor proof of 1873 that the rational are countable). The ϵ jk are all independent, o the X j := ϵ jk /2 k are independent, and U[, 1] by above. So from one U(, 1), we get in thi way infinitely many copie. (ii) If F i a ditribution function (right-continuou; increaing from at to 1 at ), define it (left-continuou) invere function by F 1 (t) := inf{f (x) t} for < t < 1. Then if U U[, 1], X := F 1 (U) F. For, {X x} = {F 1 (U) x} = {U F (x)}, which ha probability F (x) a U i uniform. By thi probability integral tranformation we can pa from generating copie from the uniform ditribution (ay by Monte Carlo imulation) to generating copie from the ditribution F, in particular, tandard normal. Hence by (i) above we can then generate infinitely many independent tandard normal. (iii) We can hence imulate a Brownian motion B = (B t ) from B t = λ n Z n n (t), with Z n independent tandard normal, n (t) the Schauder function and λ n uitable normaliing contant. (iv) Similarly, uing (ii) rather than (i), we may imulate infinitely many independent Brownian motion. (Largely tandard book work all covered, in lecture or problem heet.) 2

3 Q3. A ditribution i infinitely diviible (id) iff, for each n = 1, 2,..., it i the n-fold convolution of a probability ditribution equivalently, if it CF i the nth power of the CF of a probability ditribution. The Lévy-Khintchine formula tate that a probability ditribution i id iff it CF ha the form exp{ Ψ(u)}, where Ψ(u) = iau σ2 u 2 + (e iux 1 iuxi( x < 1)µ(dx), (a i real, σ and the Lévy meaure µ atifie min(1, x 2 )µ(dx) < ). (i) ϕ(t) = eitx /(π(1 + x 2 ))dx. Take γ the emicircle in the upper halfplane on bae [ R, R], t >, and conider f(z) := e itz /(π(1 + z 2 )). The only ingularity inide γ i at y = i, a imple pole. e itz By Cauchy Reidue Theorem: f = 2πi. But γ Re i π(z 1)(z + 1) = e t π.2i = ie t 2π. f = f + γ 1 f γ 2 γ ( ) ie t = e t. 2π e iλt + (Jordan Lemma). π(1 + x 2 ) Thi give the reult for t >. For t =, it i an arctan (or tan 1 ) integral. For t < : replace t by t. // Thu the CF of the ymmetric Cauchy denity 1/(π(1 + x 2 )) i e t. (ii) Thi i id, a e t = [e t /n ] n for each n, and each [.] i a CF. (iii) Subtituting µ(dx) = 1/(π x 2 )dx above give Ψ(u) a the um of two integral, I 1 over ( 1, 1) and I 2 over it complement. In I 1, the ±iux term over ( 1, ) and (, 1) cancel; we can then combine I 1 and I 2 to get Ψ(u) = 2 (co ux 1)dx/x 2. π Thi give Ψ (u) = (2/π) in ux dx/x = (2/π) in t dt/t = (2/π).π/2 = 1. So Ψ(u) = u for u >. So Ψ(u) = u. // For X i independent Cauchy, (X X n )/n ha CF [e t /n ] n = e t, the CF of X 1. So (X X n )/n = d X 1. Thi doe not contradict the SLLN: it doe not apply, a the mean of X i i undefined. ((i), (ii): tandard bookwork, covered in lecture; (iii): uneen (but obviou) example; lat part: imilar een.) 3

4 Q4. (i) For t, X i Gauian with zero mean (a B i), and continuou (again, a B i). The covariance of B i min(, t). The covariance of X i cov(x, X t ) = cov(b(1/), tb(1/t)) = E[B(1/).tB(1/t)] = t.e[b(1/)b(1/t)] = t.cov(b(1/), B(1/t)) = t. min(1/, 1/t) = min(t, ) = min(, t). Thi i the ame covariance a Brownian motion. So, away from the origin, X i Brownian motion, a a Gauian proce i uniquely characterized by it mean and covariance (from the propertie of the multivariate normal ditribution). So X i continuou. So we can define it at the origin by continuity. So X i Brownian motion everywhere X i BM. (ii) Since Brownian motion i at the origin, X() =. Since Brownian motion i continuou at the origin, X(t) a t. Thi ay that which i tb(1/t) (t ), B(t)/t (t ), a required. By contruction, Brownian motion i given by it expanion B t = λ n Z n n (t), n= where the Z n are independent tandard normal random variable, the n (t) are the Schauder function and the λ n are normaliing contant. Now n () = for n 1, while (t) = t, o (1) = 1. Alo λ = 1. Putting t = 1, B 1 = Z. So Brownian bridge i B (t) := B(t) tb(1) = B(t) tz : the expanion of Brownian bridge in the Schauder function i B (t) = λ n Z n n (t). n=1 (Standard bookwork: covered in lecture or problem heet.) 4

5 Q5. A function ϕ i convex if ϕ(λx + (1 λ)y) λϕ(x) + (1 λ)ϕ(y) λ [, 1], x, y. Jenen inequality tate that ϕ(e[x]) E[ϕ(X)] for convex ϕ and random variable X with X, ϕ(x) both integrable. The conditional Jenen inequality tate that for C a σ-field, ϕ, X a above, ϕ(e[x C]) E[ϕ(X) C]. (i) For < t, M = E[M t F ] a M i a martingale. So by the conditional Jenen inequality, ϕ(m ) = ϕ(e[m t F ]) E[ϕ(M t ) F ], which ay that ϕ(m) i a ubmartingale. (ii) If M i a ubmartingale, M E[M t F ]. A ϕ i non-decreaing on the range of M, ϕ(m ) ϕ(e[m t F ]), E[ϕ(M t ) F ] by the conditional Jenen inequality again, and again ϕ(m) i a ubmartingale. (iii) A Brownian motion B i a martingale (lecture), and x 2 i convex (it econd derivative i 1 ), B 2 i a ubmartingale by (i). (iv) A B 2 t t i a martingale (which you may quote here a it i not aked but i eay to prove, a in lecture) B 2 t = [B 2 t t] + t (ubmg = mg + incr) i the Doob-Meyer decompoition of B 2 t. The increaing proce here i t, which i thu the quadratic variation of Brownian motion B. (Standard bookwork covered in lecture or problem heet.) 5

6 Q6. (i) The Itô iometry tate that for f H 2 := H 2 (, T ), the cla of meaurable f with {f : E[ T f 2 (ω, t)dt] < }, E[( f 2 (ω, u)db u ) 2 ] = E[ f 2 (ω, t)dt]. [2] (ii) Conditional Itô iometry. For t T, E[( f 2 (ω, u)db u ) 2 F ] = E[ f 2 (ω, t)dt F ]. Proof. It uffice to how that for all A F, E[I(A)( f 2 (ω, u)db u ) 2 ] = E[I(A) f 2 (ω, t)dt]. Thi follow from the unconditional Itô iometry, applied to the integrand g(ω, u) := fi A (ω)i (,t] (u). // [6] (iii) For t, E[M t F ] = E[{( + )f u db u } 2 F ] fudu 2 E[ fudu F 2 ] = E[( f u db u ) 2 ]+2( b u db u )E[ b u db u F ]+E[( f u db u ) 2 F ] fudu E[ 2 fudu F 2 ]. The firt and fourth term give M. The third and fifth term cancel, by the conditional Itô iometry (ii). The econd factor in the econd term involve an Itô integral, which (for an integral f H 2 ) i a martingale, o ha contant expectation, which i on taking t =, o the econd term i. Combining, the RHS i M, which prove that M i a martingale. [13] (iv) Taking f 1 give M t := Bt 2 t i a martingale. [4] ((i), (ii): tandard book work, covered in lecture; (iii), (iv): imilar problem een.) N. H. Bingham 6