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t t 1 Y (s) ◦ dX(s) , Y (s) dX(s)+ Y,X , (7) Z Z 2 t 0 0 where the integral on the right hand side is the Itˆointegral. If both X and Y are continuous semimartingales, then
hX, Y it = X, Y t, a.s., and the Stratonovich integral is equivalent to the Fisk-Stratonovich integral. For a continuous semimartingale X and C2 function F defined on the range of X, Itˆo’s rule establishes that t t ′ 1 ′′ F (X(t)) − F (X(0)) = F (X(s)) dX(s)+ F (X(s)) dhXis, a.s., Z0 2 Z0 and with the Fisk-Stratonovich integral, this becomes
t F (X(t)) − F (X(0)) = F ′(X(s)) ◦ dX(s), a.s., (8) Z0 as in ordinary calculus (see Karatzas and Shreve (1991)). The relationship (8) can be extended to a wider class of functions in some cases. For example, for an absolutely continuous function F and Brownian motion W , it was shown in F¨ollmer et al. (1995), Corollary 4.2, that (8) holds, so for the absolute-value function we have t |W (t)| = sgn(W (s)) ◦ dW (s), a.s., (9)
Z0
½ , ½ where sgn(x) {x>0} − {x≤0}. The Russo and Vallois (2007) results allow us to extend this relationship to a class of continuous semimartingales.
FX Definition 1. Let X be a continuous semimartingale defined on [0,T ] under the filtration t . Then X is reversible if the time-reversed process X defined by X(t)= X(T − t) is also a continuous semimartingale on b [0,T ] under the time-reversed filtration FX . b t b
Definition 2. The continuous semimartingales X1,...,Xn have nondegenerate crossings if for any i 6= j the set {t : Xi(t)= Xj (t)} almost surely has measure zero with respect to dhXkit, for 1 ≤ k ≤ n.
Lemma 1. Suppose that the continuous semimartingales X1,...,Xn have nondegenerate crossings. Then the same is true for the rank processes X(1),...,X(n).
Proof. Let pt ∈ Σn be the inverse permutation to the rank function rt. If X(k)(t)= X(ℓ)(t) for k 6= ℓ, then Xi(t)= Xj(t) for i = pt(k) 6= pt(ℓ)= j, so
{t : Xi(t)= Xj (t)} = {t : X(k)(t)= X(ℓ)(t)}. (10) i[6=j k[6=ℓ
2 From Banner and Ghomrasni (2008) we have the representation
n
dX(k)(t)= ½{Xi(t)=X(k) (t)}dXi(t) + finite variation terms, a.s., Xi=1 for k = 1,...,n, so dhX(k)it ≪ dhX1it + ··· + dhX1it, for k = 1,...,n. In the same manner we can show that dhXiit ≪ dhX(1)it + ··· + dhX(n)it, for i = 1,...,n, so sets of the form {t : Xi(t) = Xj(t)}, for i 6= j, or {t : X(k)(t)= X(ℓ)(t)}, for k 6= ℓ, will almost surely have measure zero with respect to dhXiit, for i =1,...,n, and with respect to dhX(k)it, for k =1,...,n, and the same holds for finite unions of such sets, as in (10).
Lemma 2. Let X be a reversible continuous semimartingale defined on [0,T ], and suppose that the set {t : X(t)=0} almost surely has measure zero with respect to dhXit. Then
t |X(t)| − |X(0)| = sgn(X(s)) ◦ dX(s), a.s. (11) Z0
Proof. References in this proof denoted by R&V are from Russo and Vallois (2007). The Tanaka-Meyer formula states that for the Itˆointegral
t sgn(X(s)) dX(s)= |X(t)| − |X(0)| + 2ΛX (t), a.s., (12) Z0 where ΛX is the local time at zero for X (see Karatzas and Shreve (1991)). From R&V, Proposition 1,
t 1 t t sgn(X(s)) ◦ dX(s)= sgn(X(s)) d−X(s)+ sgn(X(s)) d+X(s) , (13) Z0 2 Z0 Z0 with the forward and backward integrals defined by (4) and (5). Since sgn(X) is continuous outside the set {t : X(t)=0}, and this set almost surely has measure zero with respect to dhXit, R&V Proposition 6 implies that
t t sgn(X(s)) d−X(s)= sgn(X(s)) dX(s) Z0 Z0
= |X(t)| − |X(0)| + 2ΛX (t), a.s., (14) by equation (12). By R&V, Proposition 1,
t T sgn(X(s)) d+X(s)= − sgn(X(s)) d−X(s), (15) Z0 ZT −t b b where X is the time-reversed version of X. By hypothesis, X is a continuous semimartingale on [0,T ] with respect to the reverse filtration, so as in (14) we have b b T − sgn(X(s)) d X(s)= |X(T )| − |X(T − t)| +2 Λ b (T ) − Λ b (T − t) Z X X T −t b b b b = |X(0)| − |X(t)| + 2ΛX (t), a.s. (16)
If we combine (13), (14), (15), and (16), then (11) follows.
3 Lemma 3. Let X and Y be reversible continuous semimartingales defined on [0,T ] under under a common filtration and suppose that they have nondegenerate crossings. Then
t X(t) − Y (t) − X(0) − Y (0) = sgn X(s) − Y (s) ◦ dX(s) Z 0 t − sgn X(s) − Y (s) ◦ dY (s), a.s. (17) Z 0
Proof. References in this proof denoted by R&V are from Russo and Vallois (2007). Since X − Y is a reversible continuous semimartingale, it follows from Lemma 2 that
t X(t) − Y (t) − X(0) − Y (0) = sgn X(s) − Y (s) ◦ d X(s) − Y (s) , a.s., Z 0
so, due to linearity of the integral with respect to the differentials, it suffices to show that the integrals in (17) are defined. Let us first consider the integral with respect to dX. By the definition of the Stratonovich integral in (7),
t t sgn X(s) − Y (s) ◦ dX(s)= sgn X(s) − Y (s) dX(s)+ sgn X − Y ,X , (18) Z Z t 0 0 if the terms on the right hand side are defined. The Itˆointegral in (18) is defined, so we need only consider the covariation term. From R&V Proposition 1,
t t sgn X − Y ,X = sgn X(s) − Y (s) d+X(s) − sgn X(s) − Y (s) d−X(s), (19) t Z Z 0 0 and will be defined if the two integrals are. Since sgn(X − Y ) is continuous outside {t : X(t)= Y (t)}, which almost surely has measure zero with respect to dhXit, R&V Proposition 6 implies that
t t sgn X(s) − Y (s) d−X(s)= sgn X(s) − Y (s) dX(s), (20) Z Z 0 0 and since this Itˆointegral is defined, so is the forward integral. By R&V Proposition 1,
t T sgn X(s) − Y (s) d+X(s)= − sgn X(s) − Y (s) d−X(s), (21) Z Z 0 T −t b b b where X and Y are the time-reversed versions of X and Y on [0,T ]. By hypothesis, the time-reversed process X is a continuous semimartingale, so as in (20) its forward integral is defined, and this defines the backward b b integral in (21). Hence, the covariation in (19) is defined, so both terms on the right hand side of (18) are b defined, and this defines the Stratonovich integral with respect to dX in (17). The same reasoning holds for the integral with respect to dY .
A Stratonovich representation for rank processes
We would like to prove (1), and we shall start with a lemma that establishes this result for n = 2 and then apply the lemma to prove the general case with n ≥ 2.
Lemma 4. Let X1 and X2 be reversible continuous semimartingales defined on [0,T ] under a common filtration, and suppose that they have nondegenerate crossings. Then
t t ½ X1(t) ∨ X2(t) − X1(0) ∨ X2(0) = ½{X1(s)≥X2(s)} ◦ dX1(s)+ {X1(s) 4 and t t ½ X1(t) ∧ X2(t) − X1(0) ∧ X2(0) = ½{X1(s) Proof. For t ∈ [0,T ] we have 1 X (t) ∨ X (t)= X (t)+ X (t)+ X (t) − X (t) , a.s., 1 2 2 1 2 2 1 so by Lemma 3, 1 t X1(t) ∨ X2(t) − X1(0) ∨ X2(0) = X1(t)+ X2(t) − X1(0) − X2(0) + sgn X2(s) − X1(s) ◦ dX2(s) 2 Z0 t − sgn X2(s) − X1(s) ◦ dX1(s) Z 0 1 t = 1+sgn X2(s) − X1(s) ◦ dX2(s) 2 Z 0 1 t + 1 − sgn X2(s) − X1(s) ◦ dX1(s) 2 Z0 t t ½ = ½{X1(s) X1(t) ∧ X2(t)= X1(t)+ X2(t) − X1(t) ∨ X2(t), a.s. Proposition 1. Let X1,...,Xn be continuous semimartingales defined on [0,T ] that are reversible and have nondegenerate crossings. Then the rank processes X(1),...,X(n) satisfy n dX(k)(t)= ½{Xi(t)=X(k) (t)} ◦ dXi(t), a.s. (24) Xi=1 Proof. It follows from Lemma 4 that (24) holds for n = 2, so let us assume that it holds for X1,...,Xn−1, and prove that it then holds for X1,...,Xn. Let X(1),..., X(n−1) be the ranked processes X1,...,Xn−1, so by our inductive hypothesis we have e e n−1 dX k (t)= ½ e ◦ dXi(t), a.s., (25) ( ) {Xi(t)=X(k) (t)} Xi=1 e for k = 1,...,n − 1. By Lemma 1, the processes X(1),..., X(n−1) have nondegenerate crossings as do X1,...,Xn, so the same holds for holds for X(1),..., Xe(n−1),Xne. Now, e e X(1)(t)= X(1)(t) ∨ Xn(t), a.s., and we can apply Lemma 4, so by our inductivee hypotheses, ½ dX = ½ e ◦ dX (t)+ e ◦ dXn(t) (1) {X(1)(t)≥Xn(t)} (1) {X(1) (t) n−1e ½ ½ = ½ e e ◦ dXi(t)+ {Xn(t)=X (t)} ◦ dXn(t) {X(1)(t)≥Xn(t)} {Xi(t)=X(1) (t)} (1) Xi=1 n−1 ½ = ½{Xi(t)=X(1) (t)} ◦ dXi(t)+ {Xn(t)=X(1) (t)} ◦ dXn(t) Xi=1 5 n = ½{Xi(t)=X(1) (t)} ◦ dXi(t), a.s. Xi=1 For 2 ≤ k ≤ n − 1, we have X(k)(t)= X(k−1)(t) ∧ X(k)(t) ∨ Xn(t) , a.s. Since e e {t : X(k−1)(t)= X(k)(t) ∨ Xn(t) }⊂{t : X(k−1)(t)= X(k)(t)}∪{t : X(k−1)(t)= Xn(t)} e e e e e and dhX(k) ∨ Xnit ≪ dhX(k)it + dhXnit, it follows that X(k−1) and X(k) ∨ Xn have nondegenerate crossings. Hence, by Lemma 4, e e e e ½ dX k (t)= ½ e e ◦ dX k (t)+ e e ◦ d X k (t) ∨ Xn(t) ( ) {X(k−1) (t)