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Math 546 Assignment 5 (due November 26) 1. Let Z be a continuous non-negative (Ft )-local martingale such that E(Z0 ) < ∞. Show that: (a) Z is an (Ft )-supermartingale and Zt → Z∞ a.s. (b) Z is an (Ft )-martingale iff E(Zt ) = E(Z0 ) for all t > 0. (c) Z is a uniformly integrable (Ft )-martingale iff E(Z∞ ) = E(Z0 ). Rt 2. (a) Let f : R+ → R be Borel measurable and satisfy 0 f (s)2 ds < ∞ for all t > 0, and Rt B be a standard 1-dimensional (Ft )-Brownian motion. Show that Zt = 0 f (s)dBs is a mean 0 normal r.v. and find its variance. R t Hint. Consider Y (t) = exp[iθZt + (θ2 /2) 0 f (s)2 ds] and use Ito’s Lemma. Rt (b) Consider the Stochastic Differential Equation: Xt = x + σBt − λ 0 Xs ds, where σ, λ > 0, x ∈ R andRB is as above. Show X has a unique solution and show that it is t given by Xt = σe−λt 0 eλs dBs + xe−λt . Do this by considering eλt Xt . (c) Show that Xt converges in distribution as t → ∞ and find the limiting distribution. 3. (Tanaka’s formula and local time).√Let B be a standard 1-dimensional (Ft )-Brownian motion. For every ε > 0 set gε (x) = ε + x2 for x ∈ R. Below you will show that Ito’s lemma remains valid in an appropriate sense for Brownian motion and the non-C 2 function f (x) = |x|. √ (a) Show that gε (Bt ) = ε + Mtε + Aεt , where M ε is a square integrable martingale and Aε is a continuous increasing process. Give explicit formulae for M ε and Aε . (b) Set sgn(x) = 1{x>0} − 1{x<0} . Show that for for every T > 0 sup |Mtε t≤T Z − t sgn(Bs )dBs | → 0 in L2 as ε → 0. 0 Conclude that there is a continuous increasing process L such that Z t |Bt | = sgn(Bs )dBs + Lt for all t ≥ 0, 0 and for all T > 0, sup |Aε (t) − L(t)| → 0 in L2 as ε → 0. t≤T Rt (c) Show that βt = 0 sgn(Bs )dBs is an (Ft )-Brownian motion. (d) Prove that with probability 1, for all 0 ≤ u < v, B 6= 0 on (u, v) implies Lv = Lu . Hence t → Lt only increases on the zero set of B. Hint: Explain why it suffices to show that for fixed rationals p < q, w.p. 1 B 6= 0 on [p, q] implies Lp = Lq . (e) Show that Lt = sups≤t −βs for all t a.s., and conclude that Lt > 0 for all t > 0 a.s. Hint: In order to show Lt ≤ sups≤t −βs consider Lαt , where αt = sup{s ≤ t : Bs = 0}, and recall (d). The reverse inequality is easy to prove.