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1 Stochastic Integration Itˆ’s Formula: For f ∈ C 2 (R) and X continuous we have: o 1 f (t, Xt ) = f (0, X0 ) + Dt f (·, X) • It + Dx f (·, X) • Xt + Dxx f (·, X) • [X]t . 2 Exercise 1.1 1. Show that the exponential martingale Zt = eWt − 2 , satisﬁes the equation Zt = 1 + Z • Wt . Hint: Use Itˆ’s formula with f (t, x) = ex− 2 and [W ]t = t. o 2. Show that satisﬁes Xt = 1 + a + Xt = eat+bWt , b2 2 t≥0 t t t≥0 X • It + bX • Wt . Hint: Use Itˆ’s formula with f (t, x) = eat+bx o We consider the general stochastic exponential. Theorem 1.1 For every real-valued semimartingale X there exists a unique real-valued semimartingale Z which is the solution to the equation Z = 1 + Z− • X. It is given by 1 Zt = exp Xt − X0 − [X c ]t 2 (1 + ∆Xs )e∆Xs . s≤t (1) Proof. We only consider the case ∆X > −1, in which Z is of the form Zt = exp{Yt } with 1 Yt = Xt − X0 − [X c ]t + 2 (log(1 + ∆Xs ) − ∆Xs ). s≤t Making use of the above deﬁntions and theorems we obtain Y c = X c, ∆Ys = log(1 + ∆Xs ) 1 and s≤t Zs− (eYs − 1 − ∆Ys ) = Z− • ( s≤· (eYs − 1 − ∆Ys ))t . An application of Itˆ’s formula to Z = exp(Y ) gives o 1 Z = Z0 + Z− • Y + Z− • [Y c ] + 2 1 = 1 + Z− • (Y + [Y c ] + 2 (Zs− e∆Ys − Zs− − Zs− ∆Ys ) s≤· (e∆Ys − 1 − ∆Ys )) s≤· 1 = 1 + Z− • (X − X0 − [X c ] + 2 1 + [X c ] + 2 (log(1 + ∆Xs ) − ∆Xs ) s≤· (1 + ∆Xs − 1 − log(1 + ∆Xs )) s≤· = 1 + Z− • X, i.e. Z solves (1). Here is an example of a local martingale which is not a martingale. Let W be a standard Brownian motion. For n ∈ N ﬁxed deﬁne the process K n by Ktn = 1[0,arctan n] (t) 1 + tan2 t. Since K n is predictable and bounded, we have K n ∈ L(W ). Deﬁne stopping times Tn := inf{t ∈ R+ : K n • Wt ≥ 1} and the process H by Ht := Ktn 1[[0,Tn ]] (t) if t < arctan n for n ∈ N 0 if t ≥ π . 2 First, we show that H is well-deﬁned. Therefore, assume that m, n ∈ N with t < arctan m ∧ arctan n. We have K m = K n on [0, t] and hence K m • W = K n • W on [0, t]. Thus {t ≤ Tm } = {K m • W < 1 for all s ∈ [0, t)} = {K n • W < 1 for all s ∈ [0, t)} = {t ≤ Tn } yields Ktm 1[[0,Tm ]] (t) = Ktn 1[[0,Tn]] (t). 2 Furthermore, H is left-continuous and adapted and thus predictable. ∞ Now, we show hat 0 Ht2 dt is a.s. ﬁnite, since then it follows from Theorem 1.65 that H is integrable with respect to W . Obviously, ∞ N := ⊂ {there is no n such that Tn < π } 2 ⊂ ∩n∈N {K n • Wt < 1 for all t ∈ [0, arctan n]}. 0 Ht2 dt = ∞ For ϑ ∈ R+ deﬁne Gϑ := Farctan ϑ , Bϑ := K n • Warctan ϑ , where n > arctan ϑ. As before one can show that Bϑ is well-deﬁned. Moreover, (Gϑ )ϑ∈R+ is a ﬁltration and B = {Bϑ }ϑ∈R+ is a continuous and adapted (with respect to this ﬁltration) process. For ϑ ≥ υ we have from Theorem 1.68 that Bϑ − Bυ = K n • Warctan ϑ − K n • Warctan υ is independent of Farctan υ = Gυ , i.e. B has independent increments. Furthermore, it follows from Theorem 1.68 that Bϑ − Bυ = (K n 1]] arctan υ,∞[[ ) • Warctan ϑ is normally distributed with zero mean and variance arctan ϑ arctan ϑ (Ktn )2 1]] arctan υ,∞[[ 0 dt = arctan υ ϑ (1 + tan2 t) dt = υ (1 + tan2 (arctan ξ)) arctan ξ dξ = ϑ−υ for ϑ ≥ υ. In particular, the above independence yields E[Bϑ − Bυ | Gυ ] = E[Bϑ − Bυ ] = 0, i.e. B is a martingale and thus a semimartingale. Therefore, B is a standard Brownian motion. A Brownian motion reaches a.s. the value 1. Hence, continuity from above yields P (K n • Wt < 1 for all t ∈ [0, arctan n]) = P (Bϑ < 1 for all ϑ ≤ n) → 0 as n → ∞. Consequently N is a null set. Since K n reaches the value 1 for suﬃciently large n we have H • W π = 1. Moreover, it is obvious that H • W = 2 (K n • W )Tn ≤ 1 on [0, arctan n] and therefore H • W ≤ 1. Consequently, for M := H • W : 3 • M is an integral with respect to a continuous martingale. • M is therefore a local martingale. • M is not a martingale, since E[M π ] = 1 = 0 = E[M0 ]. 2 • M is a submartingale, since 1 − M is a non-negative local martingale and thus a supermartingale. 4

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posted: | 12/19/2009 |

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