Explicit construction of a dynamic Bessel bridge of dimension 3
aa r X i v : . [ m a t h . P R ] F e b Explicit construction of a dynamic Bessel bridge of dimension 3
Luciano Campi ∗ Umut C¸ etin † Albina Danilova ‡ October 11, 2018
Abstract
Given a deterministically time-changed Brownian motion Z starting from 1, whose time-change V ( t ) satisfies V ( t ) > t for all t >
0, we perform an explicit construction of a process X which is Brownian motion in its own filtration and that hits zero for the first time at V ( τ ),where τ := inf { t > Z t = 0 } . We also provide the semimartingale decomposition of X under the filtration jointly generated by X and Z . Our construction relies on a combination ofenlargement of filtration and filtering techniques. The resulting process X may be viewed asthe analogue of a 3-dimensional Bessel bridge starting from 1 at time 0 and ending at 0 at therandom time V ( τ ). We call this a dynamic Bessel bridge since V ( τ ) is not known in advance.Our study is motivated by insider trading models with default risk, where the insider observesthe firm’s value continuously on time. The financial application, which uses results proved inthe present paper, has been developed in the companion paper [6]. In this paper, we are interested in constructing a Brownian motion starting from 1 at time t = 0and conditioned to hit the level 0 for the first time at a given random time. More precisely, let Z be the deterministically time-changed Brownian motion Z t = 1 + R t σ ( s ) dW s and let B be anotherstandard Brownian motion independent of W . We denote V ( t ) the associated time-change, i.e. V ( t ) = R t σ ( s ) ds for t ≥
0. Consider the first hitting time of Z of the level 0, denoted by τ .Our aim is to build explicitly a process X of the form dX t = dB t + α t dt , X = 1, where α is anintegrable and adapted process for the filtration jointly generated by the pair ( Z, B ) and satisfyingthe following two properties:1. X hits level 0 for the first time at time V ( τ );2. X is a Brownian motion in its own filtration.This resulting process X can be viewed as an analogue of 3-dimensional Bessel bridge with a randomterminal time. Indeed, the two properties above characterising X can be reformulated as follows: X is a Brownian motion conditioned to hit 0 for the first time at the random time V ( τ ). In orderto emphasise the distinct property that V ( τ ) is not known at time 0, we call this process a dynamicBessel bridge of dimension 3 . The reason that X hits 0 at V ( τ ) rather than τ is simply due to the ∗ LAGA, University Paris 13, and CREST, [email protected]. † Department of Statistics, London School of Economics, [email protected]. ‡ Department of Mathematics, London School of Economics, [email protected]. Z and a standard Brownian motion starting at1. The solution to the above problem consists of two parts with varying difficulties. The easypart is the construction of this process after time τ . Since V is a deterministic function, the firsthitting time of 0 is revealed at time τ . Thus, one can use the well-known relationship between the3-dimensional Bessel bridge and Brownian motion conditioned on its first hitting time to write for t ∈ ( τ, V ( τ )) dX t = dB t + (cid:26) X t − X t V ( τ ) − t (cid:27) dt. The difficult part is the construction of X until time τ . Thus, the challenge is to construct aBrownian motion which is conditioned to stay strictly positive until time τ using a drift termadapted to the filtration generated by B and Z .Our study is motivated by the equilibrium model with insider trading and default as in [5],where a Kyle-Back type model with default is considered. In such a model, three agents act inthe market of a defaultable bond issued by a firm, whose value process is modelled as a Brownianmotion and whose default time is set to be the first time that the firm’s value hits a given constantdefault barrier. It has been shown in [5] that the equilibrium total demand for such a bond, afteran appropriate translation, is a process X ∗ which is a 3-dimensional Bessel bridge in insider’s(enlarged) filtration but is a Brownian motion in its own filtration. These two properties can berephrased as follows: X ∗ is a Brownian motion conditioned to hit 0 for the first time at the defaulttime τ . However, the assumption that the insider knows the default time from the beginning mayseem too strong from the modelling viewpoint. To approach the reality, one might consider a morerealistic situation when the insider doesn’t know the default time but however she can observe theevolution through time of the firm’s value. Equilibrium considerations, akin to the ones employedin [5], lead one to study the existence of processes which we called dynamic Bessel bridges ofdimension 3 at the beginning of this introduction. The financial application announced here hasbeen performed in the companion paper [6], where the tools developed in the present paper areused to solve explicitely the equilibrium model with default risk and dynamic insider information,as outlined above. We refer to that paper for further details.We will observe in the next section that in order to make such a construction possible, one hasto assume that Z evolves faster than its underlying Brownian motion W , i.e. V ( t ) ≥ t for all t ≥ V ( t ) cannot be equal to t in any interval ( a, b ) of R + .We will nevertheless impose a stronger assumption that V ( t ) > t for all t > V ( t ) in a neighbourhood of 0 will be needed.Apart from the financial application, which is our first motivation, such a problem is interestingfrom a probabilistic point of view as well. We have observed above that the difficult part inobtaining the dynamic Bessel bridge is the construction of a Brownian motion which is conditionedto stay strictly positive until time τ using a drift term adapted to the filtration generated by B and Z . Such a construction is related to the conditioning of a Markov process, which has beenthe topic of various works in the literature. The canonical example of this phenomenon is the3-dimensional Bessel process which is obtained when one conditions a standard Brownian motionto stay positive. Chaumont [8] studies the analogous problem for L´evy process whereas Bertoin and2oney [2] are concerned with the situation for random walks and the convergence of their respectiveprobability laws. Bertoin et al. [3] constructs a Brownian path over a fixed time interval with a given minimum by performing transformations on a Brownian bridge. More recently, Chaumontand Doney [9] revisits the L´evy processes conditioned to stay positive and shows a Williams’ typepath decomposition result at the minimum of such processes. However, none of these approachescan be adopted to perform the construction that we are after since i) the time interval in which wecondition the Brownian motion to be positive is random and not known in advance; and ii) we arenot allowed to use transformations that are not adapted to the filtration generated by B and Z .The paper is structured as follows. In Section 2, we formulate precisely our main result (Theorem2.1) and provide a partial justification for its assumptions. Section 3 contains the proof of Theorem2.1, that uses, in particular, a technical result on the density of the signal process Z , whose proofis given in Section 4. Finally, several technical results used along our proofs have been relegated inthe Appendix for reader’s convenience. Let (Ω , H , H = ( H t ) t ≥ , P ) be a filtered probability space satisfying the usual conditions. Wesuppose that H contains only the P -null sets and there exist two independent H -Brownian motions, B and W . We introduce the process Z t := 1 + Z t σ ( s ) dW s , (2.1)for some σ whose properties are given in the assumption below. Assumption 2.1
There exist a measurable function σ : R + (0 , ∞ ) such that:1. V ( t ) := R t σ ( s ) ds ∈ ( t, ∞ ) for every t > ;2. There exists some ε > such that R ε V ( t ) − t ) dt < ∞ . Notice that under this assumptions, Z and W generate the same minimal filtration satisfying theusual conditions. Consider the following first hitting time of Z : τ := inf { t > Z t = 0 } , (2.2)where inf ∅ = ∞ by convention. One can characterize the distribution of τ using the well-knowndistributions of first hitting times of a standard Brownian motion. To this end let H ( t, a ) := P [ T a > t ] = Z ∞ t ℓ ( u, a ) du, (2.3)for a > T a := inf { t > B t = a } , and ℓ ( t, a ) := a √ πt exp (cid:18) − a t (cid:19) . P [ T a > t |H s ] = [ T a >s ] H ( t − s, a − B s ) , s < t. Thus, since V is deterministic and strictly increasing, ( Z V − ( t ) ) t ≥ is a standard Brownian motionin its own filtration starting at 1, and consequently P [ τ > t |H s ] = [ τ>s ] H ( V ( t ) − V ( s ) , Z s ) . (2.4)Hence, P [ V ( τ ) > t ] = H ( t, , for every t ≥
0, i.e. V ( τ ) = T in distribution. Here we would like to give another formulation forthe function H in terms of the transition density of a Brownian motion killed at 0 . Recall that thistransition density is given by q ( t, x, y ) := 1 √ πt (cid:18) exp (cid:18) − ( x − y ) t (cid:19) − exp (cid:18) − ( x + y ) t (cid:19)(cid:19) , (2.5)for x > y > H ( t, a ) = Z ∞ q ( t, a, y ) dy. (2.6)In the sequel, for any process Y , F Y is going to denote the minimal filtration satisfying theusual conditions and with respect to which Y is adapted. The following is the main result of thispaper. Theorem 2.1
There exists a unique strong solution to X t = 1 + B t + Z τ ∧ t q x ( V ( s ) − s, X s , Z s ) q ( V ( s ) − s, X s , Z s ) ds + Z V ( τ ) ∧ tτ ∧ t ℓ a ( V ( τ ) − s, X s ) ℓ ( V ( τ ) − s, X s ) ds. (2.7) Moreover,i) Let F Xt = N W σ ( X s ; s ≤ t ) , where N is the set of P -null sets. Then, X is a standardBrownian motion with respect to F X := ( F Xt ) t ≥ ;ii) V ( τ ) = inf { t > X t = 0 } . The proof of this result is postponed to the next section. We conclude this section by providing ajustification for our assumption V ( t ) > t for all t > V ( t ) ≥ t for any t ≥
0. This follows from the fact thatif the construction in Theorem 2.1 is possible, then V ( τ ) is an F B,Z -stopping time since it is anexit time from the positive real line of the process X . Indeed, if V ( t ) < t for some t > V − ( t ) > t , then [ V ( τ ) < t ] cannot belong to F B,Zt since [ V ( τ ) < t ] ∩ [ τ > t ] = [ τ < V − ( t )] ∩ [ τ >t ] / ∈ F Zt , and that τ is not F B ∞ -measurable.We will next see that when V ( t ) ≡ t construction of a dynamic Bessel bridge is not possible.Similar arguments will also show that V ( t ) cannot be equal to t in an interval. We are going toadapt to our setting the arguments used in [11], Proposition 5.1.4o this end consider any process X t = 1+ B t + R t α s ds for some H -adapted and integrable process α . Assume that X is a Brownian motion in its own filtration an that τ = inf { t : X t = 0 } a.s. andfix an arbitrary time t ≥
0. The two processes M Zs := P [ τ > t |F Zs ] and M Xs := P [ τ > t |F Xs ], for s ≥
0, are uniformly integrable continuous martingales, the former for the filtration F Z,B and thelatter for the filtration F X . In this case, Doob’s optional sampling theorem can be applied to anypair of finite stopping times, e.g. τ ∧ s and τ , to get the following: M Xτ ∧ s = E [ M Xτ |F Xτ ∧ s ] = E [ τ>t |F Xτ ∧ s ]= E [ M Zτ |F Xτ ∧ s ] = E [ M Zτ ∧ s |F Xτ ∧ s ] , where the last equality is just an application of the tower property of conditional expectations andthe fact that M Z is martingale for the filtration F Z,B which is bigger than F X . We also obtain E [( M Xτ ∧ s − M Zτ ∧ s ) ] = E [( M Xτ ∧ s ) ] + E [( M Zτ ∧ s ) ] − E [ M Xτ ∧ s M Zτ ∧ s ] . Notice that, since the pairs (
X, τ ) and (
Z, τ ) have the same law by assumption, the randomvariables M Xτ ∧ s and M Zτ ∧ s have the same law too. This implies E [( M Xτ ∧ s − M Zτ ∧ s ) ] = 2 E [( M Xτ ∧ s ) ] − E [ M Xτ ∧ s M Zτ ∧ s ] . On the other hand we can obtain E [ M Xτ ∧ s M Zτ ∧ s ] = E [ M Xτ ∧ s E [ M Zτ ∧ s |F Xτ ∧ s ]] = E [( M Xτ ∧ s ) ] , which implies that M Xτ ∧ s = M Zτ ∧ s for all s ≥
0. Using the fact that M Zs = τ>s H ( t − s, Z s ) , M Xs = τ>s H ( t − s, X s ) , s < t, one has H ( t − s, X s ) = H ( t − s, Z s ) on [ τ > s ] . Since the function a H ( u, a ) is strictly monotone in a whenever u >
0, the last equality aboveimplies that X s = Z s for all s < t on the set [ τ > s ]. t being arbitrary, we have that that X τs = Z τs for all s ≥ τ , X and Z coincide, which contradicts the fact that B and Z are independent, so that the construction of a Brownian motion conditioned to hit 0 for the firsttime at τ is impossible. A possible way out is to assume that the signal process Z evolves fasterthan its underlying Brownian motion W , i.e. V ( t ) ∈ ( t, ∞ ) for all t ≥ σ . We prove our main result in the following section. Note first that in order to show the existence and the uniqueness of the strong solution to the SDEin (2.7) it suffices to show these properties for the following SDE Y t = y + B t + Z τ ∧ t q x ( V ( s ) − s, Y s , Z s ) q ( V ( s ) − s, Y s , Z s ) ds, y > , (3.8)5nd that Y τ >
0. Indeed, the drift term after τ is the same as that of a 3-dimensional Bessel bridgefrom X τ to 0 over the interval [ τ, V ( τ )]. Note that V ( τ ) = T in distribution implies that τ hasthe same law as V − ( T ) which is finite since T is finite and the function V ( t ) is increasing toinfinity as t tends to infinity. Thus τ is a.s. finite.By Corollary 5.3.23 in [14] the existence and uniqueness of the strong solution of (3.8) is equiv-alent to the existence of a weak solution and pathwise uniqueness of strong solution when thelatter exists. More precisely, after proving pathwise uniqueness for the SDE (3.8), and thus estab-lishing the uniqueness of the system of (2.1) and (3.8), in Lemma 3.1, we will construct a weaksolution, ( Y, Z ) , to this system. The weak existence and pathwise uniqueness will then imply(
Y, Z ) = h (1 , y, β, W ) for some measurable h and some Brownian motion β in view of Corollary5.3.23 in [14]. Moreover, the second part of Corollary 5.3.23 in [14] will finally give us h (1 , y, B, W )as the strong solution of the system described by (2.1) and (3.8).In the sequel we will often work with a pair of SDEs defining ( A, Z ) where A is a semimartingalegiven by an SDE whose drift coefficient depends on Z . In order to simplify the statements of thefollowing results, we will shortly write existence and/or uniqueness of the SDE for A , when weactually mean the corresponding property for the whole system.We start with demonstrating the pathwise uniqueness property. Lemma 3.1
Pathwise uniqueness holds for the SDE in (3.8).
Proof.
It follows from direct calculations that q x ( t, x, z ) q ( t, x, z ) = z − xt + exp (cid:0) − xzt (cid:1) − exp (cid:0) − xzt (cid:1) zt . (3.9)Moreover, q x ( t,x,z ) q ( t,x,z ) is decreasing in x for fixed z and t . Now, suppose there exist two strong solutions, Y and Y . Then( Y t ∧ τ − Y t ∧ τ ) = 2 Z τ ∧ t ( Y s − Y s ) (cid:26) q x ( V ( s ) − s, Y s , Z s ) q ( V ( s ) − s, Y s , Z s ) − q x ( V ( s ) − s, Y s , Z s ) q ( V ( s ) − s, Y s , Z s ) (cid:27) ds ≤ . (cid:4) The existence of a weak solution will be obtained in several steps. First we show the existence ofa weak solution to the SDE in the following proposition and then conclude via Girsanov’s theorem.
Proposition 3.1
There exists a unique strong solution to Y t = y + B t + Z τ ∧ t f ( V ( s ) − s, Y s , Z s ) ds y > , (3.10) where f ( t, x, z ) := exp (cid:0) − xzt (cid:1) − exp (cid:0) − xzt (cid:1) zt . Moreover, P [ Y τ > and Y t ∧ τ > , ∀ t >
0] = 1 . roof. Pathwise uniqueness can be shown as in Lemma 3.1; thus, its proof is omitted. Observethat if Y is a solution to (3.10), then dY t = 2 Y t dB t + (cid:0) [ τ>t ] Y t f ( V ( t ) − t, Y t , Z t ) + 1 (cid:1) dt. Inspired by this formulation we consider the following SDE: dU t = 2 p | U t | dB t + (cid:16) [ τ>t ] p | U t | f ( V ( t ) − t, p | U t | , Z t ) + 1 (cid:17) dt, (3.11)with U = y . In Lemma 3.2 it is shown that there exists a weak solution to this SDE whichis strictly positive in the interval [0 , τ ]. This yields in particular that the absolute values can beremoved from the SDE (3.11) considered over the interval [0 , τ ]. Thus, it follows from an applicationof Itˆo’s formula that √ U is a weak, therefore strong, solution to (3.10) in [0 , τ ] due to pathwiseuniqueness and Corollary 5.3.23 in [14]. The global solution can now be easily constructed by theaddition of B t − B τ after τ . This further implies that Y is strictly positive in [0 , τ ] since √ U isclearly strictly positive. (cid:4) Lemma 3.2
There exists a weak solution to dU t = 2 p | U t | dB t + (cid:16) p | U t | f ( V ( t ) − t, p | U t | , Z t ) + 1 (cid:17) dt, (3.12) with U = y upto and including τ . Moreover, the solution is strictly positive in [0 , τ ] . Proof.
Consider the measurable function g : R + × R [0 ,
1] defined by g ( t, x, z ) = p | x | f ( t, p | x | , z ) , for ( t, x, z ) ∈ (0 , ∞ ) × R × R + , for ( t, x, z ) ∈ (0 , ∞ ) × R × ( −∞ , , for ( t, x, z ) ∈ { } × R , and the following SDE: d ˜ U t = 2 q | ˜ U t | dB t + (cid:18) g ( V ( t ) − t, q | ˜ U t | , Z t ) + 1 (cid:19) dt. (3.13)Observe that if we can show the existence of a positive weak solution to (3.13), then U =( ˜ U t ∧ τ ) t ≥ is a positive weak solution to (3.12) upto time τ .It follows from Corollary 10.1.2 and Theorem 6.1.7 in [19] that the martingale problem definedby the stochastic differential equations for ( ˜ U , Z ) with the state space R is well-posed upto anexplosion time, i.e. there exists a weak solution to (3.13), along with (2.1), valid upto the explosiontime by Theorem 5.4.11 in [14]. Fix one of these solutions and call it ( ˜ U , Z ). Then, since the rangeof g is [0 , U is nonnegative and there is no explosion.Next it remains to show the strict positivity of U in [0 , τ ]. First, let a and b be strictly positivenumbers such that ae − a − e − a = 34 and be − b − e − b = 12 . As xe − x − e − x is strictly decreasing for positive values of x , one has 0 < a < b . Now define the stoppingtime I := inf { < t ≤ τ : p U t Z t ≤ V ( t ) − t a } , ∅ = τ by convention. As √ U τ Z τ = 0, √ U Z = y , and V ( t ) − t > t >
0, wehave that 0 < I < τ , ν y -a.s. by continuity of ( U, Z ) and V , where ν y is the probability measureassociated to the fixed weak solution. Moreover, U t > t ≤ I ].Note that C t := √ U t Z t V ( t ) − t is continuous on (0 , ∞ ) and C I = a . Thus, ¯ τ := inf { t > I : C t = 0 } >I . Consider the following sequence of stopping times: J n := inf { I n ≤ t ≤ ¯ τ : C t / ∈ (0 , b ) } I n +1 := inf { J n ≤ t ≤ ¯ τ : C t = a } for n ∈ N ∪ { } , where inf ∅ = ¯ τ by convention.Our aim is to show that τ = ¯ τ = lim n →∞ J n , a.s.. We start with establishing the secondequality. As J n s are increasing and bounded by ¯ τ , the limit exists and is bounded by ¯ τ . Supposethat J := lim n →∞ J n < ¯ τ with positive probability. Note that by construction we have I n ≤ J n ≤ I n +1 and, therefore, lim n →∞ I n = J . Since C is continuous, one has lim n →∞ C I n = lim n →∞ C J n .However, as on the set [ J < ¯ τ ] we have C I n = a and C J n = b for all n , we arrive at a contradiction.Therefore, ¯ τ = J .Next, we will demonstrate that ¯ τ = τ . Observe that since τ is finite, a.s., and U does notexplode until τ , one has that C τ = 0. Therefore, ¯ τ ≤ τ and thus C ¯ τ = 0. Suppose that ¯ τ < τ with positive probability. Then, we claim that on this set C J n = b for all n , which will lead to acontradiction since then b = lim n →∞ C J n = C ¯ τ = 0. We will show our claim by induction.1. For n = 0, recall that I < ¯ τ by construction. Also note that on ( I , J ] the drift term in(3.12) is greater than 2 as xe − x − e − x is strictly decreasing for positive values of x and due to thechoice of a and b . Therefore the solution to (3.12) is strictly positive in ( I , J ] in view ofLemma A.2 since a 2-dimensional Bessel process is always strictly positive. Thus, C J = b .2. Suppose we have C J n − = b . Then, due to continuity of C , I n < ¯ τ . For the same reasons asbefore, the solution to (3.12) is strictly positive in ( I n , J n ]. Thus, C J n = b .Thus, we have shown that for all t > U τ ∧ t >
0, a.s.. In order to show that U τ > I := sup { I n : I n < τ } . Then, we must have that I < τ a.s. since otherwise a = C I = C τ = 0, another contradiction. Similar to the earlier cases the drift term in ( I, τ ] islarger than 2, thus, U τ > (cid:4) Proposition 3.2
There exists a unique strong solution to (3.8) which is strictly positive on [0 , τ ] . Proof.
Due to Proposition 3.1 there exists a unique strong solution, Y , of (3.10). Define ( L t ) t ≥ by L = 1 and dL t = [ τ>t ] L t Y t − Z t V ( t ) − t dB t . Observe that there exists a solution to the above equation since Z t [ τ>s ] (cid:18) Y s − Z s V ( s ) − s (cid:19) ds < ∞ , a.s. ∀ t ≥ . Indeed, since Y and Z are well-defined and continuous upto τ , we have sup s ≤ τ | Y s − Z s | < ∞ , a.s.and thus the above expression is finite in view of Assumption 2.1.2.8f ( L t ) t ≥ is a true martingale, then for any T > Q T on H T defined by d Q T d P T = L T , where P T is the restriction of P to H T , is a probability measure on H T equivalent to P T . Then, byGirsanov Theorem (see, e.g., Theorem 3.5.1 in [14]) under Q T Y t = y + β Tt + Z τ ∧ t q x ( V ( s ) − s, Y s , Z s ) q ( V ( s ) − s, Y s , Z s ) ds, for t ≤ T where β T is a Q T -Brownian motion. Thus, Y is a weak solution to (3.8) on [0 , T ].Therefore, due to Lemma 3.1 and Corollary 5.3.23 in [14], there exists a unique strong solution to(3.8) on [0 , T ], and it is strictly positive on [0 , τ ] since Y has this property. Since T is arbitrary,this yields a unique strong solution on [0 , ∞ ) which is strictly positive on [0 , τ ].Thus, it remains to show that L is a true martingale. Fix T > ≤ t n − < t n ≤ T consider E " exp Z t n ∧ τt n − ∧ τ (cid:18) Y t − Z t V ( t ) − t (cid:19) dt ! . (3.14)As both Y and Z are positive until τ , ( Y t − Z t ) ≤ Y t + Z t ≤ R t + Z t by comparison where R satisfies R t = y + 2 Z t p R s dB s + 3 t. Therefore, since R and Z are independent, the expression in (3.14) is bounded by E " exp Z t n t n − R t υ ( t ) dt ! E " exp Z t n t n − Z t υ ( t ) dt ! (3.15) ≤ E " exp R ∗ T Z t n t n − υ ( t ) dt ! E " exp
12 ( Z ∗ T ) Z t n t n − υ ( t ) dt ! , where Y ∗ t := sup s ≤ t | Y s | for any c`adl`ag process Y and υ ( t ) := (cid:16) V ( t ) − t (cid:17) . Recall that Z is only atime-changed Brownian motion where the time change is deterministic and R t is the square of theEuclidian norm of a 3-dimensional standard Brownian motion with initial value ( y , , V ( T ) > T , the above expression is going to be finite if E y ∨ " exp
12 ( β ∗ V ( T ) ) Z t n t n − υ ( t ) dt ! < ∞ , (3.16)where β is a standard Brownian motion and E x is the expectation with respect to the law of astandard Brownian motion starting at x . Indeed, it is clear that, by time change, (3.16) impliesthat the second expectation in the RHS of (3.15) is finite. Moreover, since R ∗ T is the supremumover [0 , T ] of a 3-dimensional Bessel square process, it can be bounded above by the sum of threesupremums of squared Brownian motions over [0 , V ( T )] (remember that V ( T ) > T ), which givesthat (3.16) is an upper bound for the first expectation in the RHS of (3.15) as well.9n view of the reflection principle for standard Brownian motion (see, e.g. Proposition 3.7 inChap. 3 of [17]) the above expectation is going to be finite if Z t n t n − υ ( t ) dt < V ( T ) . (3.17)However, Assumption 2.1 yields that R T υ ( t ) dt < ∞ . Therefore, we can find a finite sequenceof real numbers 0 = t < t < . . . < t n ( T ) = T that satisfy (3.17). Since T was arbitrary, this meansthat we can find a sequence ( t n ) n ≥ with lim n →∞ t n = ∞ such that (3.14) is finite for all n . Then,it follows from Corollary 3.5.14 in [14] that L is a martingale. (cid:4) The above proposition establishes 0 as a lower bound to the solution of (3.8) over the interval[0 , τ ], however, one can obtain a tighter bound. Indeed, observe that q x q ( t, x, z ) is strictly increasingin z on [0 , ∞ ) for fixed ( t, x ) ∈ R . Moreover, q x q ( t, x,
0) := lim z ↓ q x q ( t, x, z ) = 1 x − xt . Therefore, q x q ( V ( t ) − t, Y t , Z t ) > q x q ( V ( t ) − t, Y t ,
0) = Y t − Y t V ( t ) − t for t ∈ (0 , τ ]. Although q x q ( t, x, z ) isnot Lipschitz in x (thus, standard comparison results don’t apply), if Y < Z then the comparisonresult of Exercise 5.2.19 in [14] can be applied to obtain P [ Y t ≥ R t ; 0 ≤ t < τ ] = 1 where R is givenby(3.18).However, this strict inequality may break down at t = 0 when Y ≥ Z , and, thus, renderingthe results of Exercise 5.2.19 is inapplicable. Nevertheless, we will show in Proposition 3.4 that P [ Y t ≥ R t ; 0 ≤ t < τ ] = 1 where R is the solution of R t = y + B t + Z t (cid:26) R s − R s V ( s ) − s (cid:27) ds, y > . (3.18)Before proving the comparison result we first establish that there exists a unique strong solutionto the SDE above and it equals in law to a scaled, time-changed 3-dimensional Bessel process. Weincidentally observe that the existence of a weak solution to an SDE similar to that in (3.18) isproved in Proposition 5.1 in [7] along with its distributional properties. Unfortunately, our SDE(3.18) cannot be reduced to theirs and moreover, in our setting, existence of a weak solution is notenough. Proposition 3.3
There exists a unique strong solution to (3.18). Moreover, the law of R is equalto the law of ˜ R = ( ˜ R t ) t ≥ , where ˜ R t = λ t ρ Λ t where ρ is a 3-dimensional Bessel process starting at y and λ t := exp (cid:18) − Z t V ( s ) − s ds (cid:19) , Λ t := Z t λ s ds. Proof.
Note that x − xt is decreasing in x and, thus, pathwise uniqueness holds for (3.18).Thus, it suffices to find a weak solution for the existence and the uniqueness of strong solution.10onsider the 3-dimensional Bessel process ρ which is the unique strong solution (see Proposition3.3 in Chap. VI in [17]) to ρ t = y + B t + Z t ρ s ds. Therefore, ρ Λ t = y + B Λ t + R Λ t ρ s ds. Now, M t = B Λ t is a martingale with respect to the time-changed filtration ( H Λ t ) with quadratic variation given by Λ. By integration by parts we see that d ( λ t ρ Λ t ) = λ t dM t + (cid:26) λ t ρ Λ t − λ t ρ Λ t V ( t ) − t (cid:27) dt. Since λ ρ Λ = y and R t λ s d [ M, M ] s = t , we see that λ t ρ Λ t is a weak solution to (3.18). Thisobviously implies the equivalence in law. (cid:4) Proposition 3.4
Let R be the unique strong solution to (3.18). Then, P [ Y t ≥ R t ; 0 ≤ t < τ ] = 1 where Y is the unique strong solution of (3.8). Proof.
Note that R t − Y t = Z t (cid:26) q x q ( V ( s ) − s, R s , − q x q ( V ( s ) − s, Y s , Z s ) (cid:27) ds, so that by Tanaka’s formula (see Theorem 1.2 in Chap. VI of [17])( R t − Y t ) + = Z t [ R s >Y s ] (cid:26) q x q ( V ( s ) − s, R s , − q x q ( V ( s ) − s, Y s , Z s ) (cid:27) ds = Z t [ R s >Y s ] (cid:26) q x q ( V ( s ) − s, R s , − q x q ( V ( s ) − s, Y s , (cid:27) ds + Z t [ R s >Y s ] (cid:26) q x q ( V ( s ) − s, Y s , − q x q ( V ( s ) − s, Y s , Z s ) (cid:27) ds ≤ Z t [ R s >Y s ] (cid:26) q x q ( V ( s ) − s, R s , − q x q ( V ( s ) − s, Y s , (cid:27) ds, since the local time of R − Y at 0 is identically 0 (see Corollary 1.9 n Chap. VI of [17]). Let τ n := inf { t > R t ∧ Y t = n } . Note that as R is strictly positive and Y is strictly positive on [0 , τ ],lim n →∞ τ n > τ . Since for each t ≥ (cid:12)(cid:12)(cid:12)(cid:12) q x q ( t, x, − q x q ( t, y, (cid:12)(cid:12)(cid:12)(cid:12) ≤ (cid:18) t + 1 n (cid:19) | x − y | for all x, y ∈ [1 /n, ∞ ), we have( R t ∧ τ n − Y t ∧ τ n ) + ≤ Z t ( R s ∧ τ n − Y s ∧ τ n ) + (cid:18) V ( s ) − s + 1 n (cid:19) ds. Thus, by Gronwall’s inequality (see Exercise 14 in Chap. V of [18]), we have ( R t ∧ τ n − Y t ∧ τ n ) + = 0since Z t (cid:18) V ( s ) − s + 1 n (cid:19) ds < ∞ by Assumption 2.1. Thus, the claim follows from the continuity of Y and R and the fact thatlim n →∞ τ n > τ . (cid:4) emark 3.1 Note that the above proof does not use the particular SDE satisfied by Z . The resultof the above proposition will remain valid as long as Z is nonnegative and Y is the unique strongsolution of (3.8), strictly positive on [0 , τ ] . Since the solution to (3.8) is strictly positive on [0 , τ ] and the drift term in (2.7) after τ is the sameas that of a 3-dimensional Bessel bridge from X τ to 0 over [ τ, V ( τ )], we have proved Proposition 3.5
There exists a unique strong solution to (2.7). Moreover, the solution is strictlypositive in [0 , τ ] . Using the well-known properties of a 3-dimensional Bessel bridge (see, e.g., Section 12.1.3, inparticular expression (12.9) in [20]), we also have the following
Corollary 3.1
Let X be the unique strong solution of (2.7). Then, V ( τ ) = inf { t > X t = 0 } . Thus, in order to finish the proof of Theorem 2.1 it remains to show that X is a standardBrownian motion in its own filtration. We will achieve this result in several steps. First, we willobtain the canonical decomposition of X with respect to the minimal filtration, G , satisfying theusual conditions such that X is G -adapted and τ is a G -stopping time. More precisely, G = ( G t ) t ≥ where G t = ∩ u>t ˜ G u , with ˜ G t := N W σ ( { X s , s ≤ t } , τ ∧ t ) and N being the set of P -null sets. Then,we will initially enlarge this filtration with τ to show that the canonical decomposition of X in thisfiltration is the same as that of a Brownian motion starting at 1 in its own filtration enlarged withits first hitting time of 0. This observation will allow us to conclude that the law of X is the lawof a Brownian motion.In order to carry out this procedure we will use the following key result, the proof of which isdeferred until the next section for the clarity of the exposition. We recall that H ( t, a ) = Z ∞ q ( t, a, y ) dy, where q ( t, a, y ) is the transition density of a Brownian motion killed at 0. Proposition 3.6
Let X be the unique strong solution of (2.7) and f : R + R be a boundedmeasurable function with a compact support contained in (0 , ∞ ) . Then E [ [ τ>t ] f ( Z t ) |G t ] = [ τ>t ] Z ∞ f ( z ) q ( V ( t ) − t, X t , z ) H ( V ( t ) − t, X t ) dz. Using the above proposition we can easily obtain the G -canonical decomposition of X . Corollary 3.2
Let X be the unique strong solution of (2.7). Then, M t := X t − − Z τ ∧ t H x ( V ( s ) − s, X s ) H ( V ( s ) − s, X s ) ds − Z V ( τ ) ∧ tτ ∧ t ℓ a ( V ( τ ) − s, X s ) ℓ ( V ( τ ) − s, X s ) ds is a standard G -Brownian motion starting at . roof. It follows from Theorem 8.1.5 in [13] and Lemma A.4 that X t − − Z t E (cid:20) [ τ>s ] q x ( V ( s ) − s, X s , Z s ) q ( V ( s ) − s, X s , Z s ) (cid:12)(cid:12)(cid:12)(cid:12) G s (cid:21) ds − Z V ( τ ) ∧ tτ ∧ t ℓ a ( V ( τ ) − s, X s ) ℓ ( V ( τ ) − s, X s ) ds is a G -Brownian motion. However, E (cid:20) [ τ>s ] q x ( V ( s ) − s, X s , Z s ) q ( V ( s ) − s, X s , Z s ) (cid:12)(cid:12)(cid:12)(cid:12) G s (cid:21) = [ τ>s ] Z ∞ q x ( V ( s ) − s, X s , z ) q ( V ( s ) − s, X s , z ) q ( V ( s ) − s, X s , z ) H ( V ( s ) − s, X s ) dz = [ τ>s ] H ( V ( s ) − s, X s ) Z ∞ q x ( V ( s ) − s, X s , z ) dz = [ τ>s ] H ( V ( s ) − s, X s ) ∂∂x Z ∞ q ( V ( s ) − s, x, z ) dz (cid:12)(cid:12)(cid:12)(cid:12) x = X s = [ τ>s ] H x ( V ( s ) − s, X s ) H ( V ( s ) − s, X s ) . (cid:4) A naive way to show that X as a solution of (2.7) is a Brownian motion is to calculate theconditional distribution of τ given the minimal filtration generated by X satisfying the usual con-ditions. Although, as we will see later, the conditional distribution of V ( τ ) given an observationof X is defined by the function H as defined in (2.3), verification of this fact leads to a highlynon-standard filtering problem. For this reason we use an alternative approach which utilizes thewell-known decomposition of Brownian motion conditioned on its first hitting time as in [5].We shall next find the canonical decomposition of X under G τ := ( G τt ) t ≥ where G τt = G t W σ ( τ ).Note that G τt = F Xt + W σ ( τ ). Therefore, the canonical decomposition of X under G τ would be itscanonical decomposition with respect to its own filtration initially enlarged with τ . As we shall seein the next proposition it will be the same as the canonical decomposition of a Brownian motionin its own filtration initially enlarged with its first hitting time of 0. Proposition 3.7
Let X be the unique strong solution of (2.7). Then, X t − − Z V ( τ ) ∧ t ℓ a ( V ( τ ) − s, X s ) ℓ ( V ( τ ) − s, X s ) ds is a standard G τ -Brownian motion starting at . Proof.
First, we will determine the law of τ conditional on G t for each t . Let f be a test13unction. Then E (cid:2) [ τ>t ] f ( τ ) |G t (cid:3) = E (cid:20) E (cid:2) [ τ>t ] f ( τ ) |H t (cid:3) (cid:12)(cid:12)(cid:12)(cid:12) G t (cid:21) = E (cid:20) [ τ>t ] Z ∞ t f ( u ) σ ( u ) ℓ ( V ( u ) − V ( t ) , Z t ) du (cid:12)(cid:12)(cid:12)(cid:12) G t (cid:21) = [ τ>t ] Z ∞ t f ( u ) σ ( u ) Z ∞ ℓ ( V ( u ) − V ( t ) , z ) q ( V ( t ) − t, X t , z ) H ( V ( t ) − t, X t ) dz du = − [ τ>t ] Z ∞ t f ( u ) σ ( u ) Z ∞ H t ( V ( u ) − V ( t ) , z ) q ( V ( t ) − t, X t , z ) H ( V ( t ) − t, X t ) dz du = − [ τ>t ] Z ∞ t f ( u ) σ ( u ) ∂∂s Z ∞ Z ∞ q ( s, z, y ) dy q ( V ( t ) − t, X t , z ) H ( V ( t ) − t, X t ) dz (cid:12)(cid:12)(cid:12)(cid:12) s = V ( u ) − V ( t ) du = − [ τ>t ] Z ∞ t f ( u ) σ ( u ) ∂∂s Z ∞ Z ∞ q ( V ( t ) − t, X t , z ) H ( V ( t ) − t, X t ) q ( s, z, y ) dz dy (cid:12)(cid:12)(cid:12)(cid:12) s = V ( u ) − V ( t ) du = − [ τ>t ] Z ∞ t f ( u ) σ ( u ) ∂∂s Z ∞ q ( V ( t ) − t + s, X t , y ) H ( V ( t ) − t, X t ) dy (cid:12)(cid:12)(cid:12)(cid:12) s = V ( u ) − V ( t ) du = − [ τ>t ] Z ∞ t f ( u ) σ ( u ) H t ( V ( u ) − t, X t ) H ( V ( t ) − t, X t ) du = [ τ>t ] Z ∞ t f ( u ) σ ( u ) ℓ ( V ( u ) − t, X t ) H ( V ( t ) − t, X t ) du. Thus, P [ τ ∈ du, τ > t |G t ] = [ τ>t ] σ ( u ) ℓ ( V ( u ) − t,X t ) H ( V ( t ) − t,X t ) du. Then, it follows from Theorem 1.6 in [16] that M t − Z τ ∧ t (cid:18) ℓ a ( V ( τ ) − s, X s ) ℓ ( V ( τ ) − s, X s ) − H x ( V ( s ) − s, X s ) H ( V ( s ) − s, X s ) (cid:19) ds is a G τ -Brownian motion as in Example 1.6 in [16]. This completes the proof. (cid:4) Corollary 3.3
Let X be the unique strong solution of (2.7). Then, X is a Brownian motion withrespect to F X . Proof.
It follows from Proposition 3.7 that G τ - decomposition of X is given by X t = 1 + µ t + Z V ( τ ) ∧ t (cid:26) X s − X s V ( τ ) − s (cid:27) ds, where µ is a standard G τ -Brownian motion vanishing at 0. Thus, X is a 3-dimensional Besselbridge from 1 to 0 of length V ( τ ). As V ( τ ) is the first hitting time of 0 for X and V ( τ ) = T indistribution, the result follows using the same argument as in Theorem 3.6 in [5]. (cid:4) Next section is devoted to the proof of Proposition 3.6. Z Recall from Proposition 3.6 that we are interested in the conditional distribution of Z t on the set[ τ > t ]. To this end we introduce the following change of measure on H t . Let P t be the restriction14f P to H t and define P τ,t on H t by d P τ,t d P t = [ τ>t ] P [ τ > t ] . Note that this measure change is equivalent to an h-transform on the paths of Z until time t where the h-transform is defined by the function H ( V ( t ) − V ( · ) , · ) and H is the function defined in(2.3) (see Part 2, Sect. VI.13 of [10] for the definition and properties of h-transforms). Note alsothat ( [ τ>s ] H ( V ( t ) − V ( s ) , Z s )) s ∈ [0 ,t ] is a ( P , H )-martingale as a consequence of (2.4). Therefore,an application of Girsanov’s theorem yields that under P τ,t ( X, Z ) satisfy dZ s = σ ( s ) dβ ts + σ ( s ) H x ( V ( t ) − V ( s ) , Z s ) H ( V ( t ) − V ( s ) , Z s ) ds (4.19) dX s = dB s + q x ( V ( s ) − s, X s , Z s ) q ( V ( s ) − s, X s , Z s ) ds, (4.20)with X = Z = 1 and β t being a P τ,t -Brownian motion. Moreover, due to the property of h-transforms, transition density of Z under P τ,t is given by P τ,t [ Z s ∈ dz | Z r = x ] = q ( V ( s ) − V ( r ) , x, z ) H ( V ( t ) − V ( s ) , z ) H ( V ( t ) − V ( r ) , x ) . (4.21)Thus, P τ,t [ Z s ∈ dz | Z r = x ] = p ( V ( t ); V ( r ) , V ( s ) , x, z ) where p ( t ; r, s, x, z ) = q ( s − r, x, z ) H ( t − s, z ) H ( t − r, x ) . (4.22)Note that p is the transition density of the Brownian motion killed at 0 after the analogous h-transform where the h-function is given by H ( t − s, x ). Lemma 4.1
Let F τ,t,Xs = σ ( X r ; r ≤ s ) ∨N τ,t where X is the process defined by (4.20) with X = 1 ,and N τ,t is the collection of P τ,t -null sets. Then the filtration ( F τ,t,Xs ) s ∈ [0 ,t ] is right-continuous. The proof of the above lemma is trivial once we observe that ( F τ,t,Xτ n ∧ s ) s ∈ [0 ,t ] , where τ n := inf { s > X s = n } , is right continuous. This follows from the observation that X τ n is a Brownian motionunder an equivalent probability measure, which can be shown using the arguments of Proposition3.2 along with the identity (3.9) and the fact that X is bounded upto τ n . Thus, for each n one has F τ,t,Xτ n ∩ F τ,t,Xu = F τ,t,Xτ n ∧ u = \ s>u F τ,t,Xτ n ∧ s = \ s>u (cid:0) F τ,t,Xτ n ∩ F τ,t,Xs (cid:1) = \ s>u F τ,t,Xs ! ∩ F τ,t,Xτ n Indeed, since ∪ n F τ,t,Xτ n = F τ,t,Xτ , letting n tend to infinity yields the conclusion.The reason for the introduction of the probability measure P τ,t and the filtration ( F τ,t,Xs ) s ∈ [0 ,t ] isthat ( P τ,t , ( F τ,t,Xs ) s ∈ [0 ,t ] )-conditional distribution of Z can be characterised by a Kushner-Stratonovichequation which is well-defined. Moreover, it gives us ( P , G )-conditional distribution of Z . Indeed,observe that P τ,t [ τ > t ] = 1 and for any set E ∈ G t , [ τ>t ] E = [ τ>t ] F for some set F ∈ F τ,t,Xt (see15emma 5.1.1 in [4] and the remarks that follow). Then, it follows from the definition of conditionalexpectation that E (cid:2) f ( Z t ) [ τ>t ] |G t (cid:3) = [ τ>t ] E τ,t h f ( Z t ) (cid:12)(cid:12) F τ,t,Xt i , P − a.s.. (4.23)Thus, it is enough to compute the conditional distribution of Z under P τ,t with respect to ( F τ,t,Xs ) s ∈ [0 ,t ] .In order to achieve this goal we will use the characterization of the conditional distributions obtainedby Kurtz and Ocone [15]. We refer the reader to [15] for all unexplained details and terminology.Let P be the set of probability measures on the Borel sets of R + topologized by weak conver-gence. Given m ∈ P and m − integrable f we write mf := R R f ( z ) m ( dz ). The next result is directconsequence of Lemma 1.1 and subsequent remarks in [15]: Lemma 4.2
There is a P -valued F τ,t,X -optional process π t ( ω, dx ) such that π ts f = E τ,t [ f ( Z s ) |F τ,t,Xs ] for all bounded measurable f . Moreover, ( π ts ) s ∈ [0 ,t ] has a c`adl`ag version. Let’s recall the innovation process I s = X s − Z s π tr κ r dr where κ r ( z ) := q x ( V ( r ) − r,X r ,z ) q ( V ( r ) − r,X r ,z ) . Although it is clear that I depends on t , we don’t emphasize it inthe notation for convenience. Due to Lemma A.4 π ts κ s exists for all s ≤ t .In order to be able to use the results of [15] we first need to establish the Kushner-Stratonovichequation satisfied by ( π ts ) s ∈ [0 ,t ) . To this end, let B ( A ) denote the set of bounded Borel measurablereal valued functions on A , where A will be alternatively a measurable subset of R or a measurablesubset of R + . Consider the operator A : B ([0 , t ] × R + ) B ([0 , t ] × R + ) defined by A φ ( s, x ) = ∂φ∂s ( s, x ) + 12 σ ( s ) ∂ φ∂x ( s, x ) + σ ( s ) H x H ( V ( t ) − V ( s ) , x ) ∂φ∂x ( s, x ) , (4.24)with the domain D ( A ) = C ∞ c ([0 , t ] × R + ), where C ∞ c is the class of infinitely differentiable functionswith compact support. By Lemma A.3 the martingale problem for A is well-posed over the timeinterval [0 , t − ε ] for any ε >
0. Therefore, it is well-posed on [0 , t ) and its unique solution is givenby ( s, Z s ) s ∈ [0 ,t ) where Z is defined by (4.19). Moreover, the Kushner-Stratonovich equation for theconditional distribution of Z is given by the following: π ts f = π t f + Z s π tr ( A f ) dr + Z s (cid:2) π tr ( κ r f ) − π tr κ r π tr f (cid:3) dI r , (4.25)for all f ∈ C ∞ c ( R + )(see Theorem 8.4.3 in [13] and note that the condition therein is satisfied dueto Lemma A.4). Note that f can be easily made an element of D ( A ) by redefining it as f n where n ∈ C ∞ c ( R + ) is such that n ( s ) = 1 for all s ∈ [0 , t ). Thus, the above expression is rigorous. Thefollowing theorem is a corollary to Theorem 4.1 in [15]. Theorem 4.1
Let m t be an F τ,t,X -adapted c`adl`ag P -valued process such that m ts f = π t f + Z s m tr ( A f ) dr + Z s (cid:2) m tr ( κ r f ) − m tr κ r m tr f (cid:3) dI mr , (4.26) for all f ∈ C ∞ c ( R + ) , where I ms = X s − R s m tr κ r dr. Then, m ts = π ts for all s < t , a.s.. roof. Proof follows along the same lines as the proof of Theorem 4.1 in [15], even though,differently from [15], we allow the drift of X to depend on s and X s , too. This is due to the factthat [15] used the assumption that the drift depends only on the signal process, Z , in order toensure that the joint martingale problem ( X, Z ) is well-posed, i.e. conditions of Proposition 2.2 in[15] are satisfied. Note that the relevant martingale problem is well posed in our case by PropositionA.1. (cid:4)
Now, we can state and prove the following corollary.
Corollary 4.1
Let f ∈ B ( R + ) . Then, π ts f = Z R + f ( z ) p ( V ( t ); s, V ( s ) , X s , z ) dz, for s < t where p is as defined in (4.22). Proof.
Let ρ ( t ; s, x, z ) := p ( V ( t ); s, V ( s ) , x, z ). Direct computations lead to ρ s + H x ( V ( t ) − s, x ) H ( V ( t ) − s, x ) ρ x + 12 ρ xx (4.27)= − σ ( s ) (cid:18) H x ( V ( t ) − V ( s ) , z ) H ( V ( t ) − V ( s ) , z ) ρ (cid:19) z + 12 σ ( s ) ρ zz . Define m t ∈ P by m ts f := R R + f ( z ) ρ ( t ; s, X s , z ) dz . Then, using the above pde and Ito’s formula onecan directly verify that m t solves (4.26). Finally, Theorem 4.1 gives the statement of the corollary. (cid:4) Now, we have all necessary results to prove Proposition 3.6.
Proof of Proposition 3.6.
Note that as X is continuous, F τ,t,Xt = W s
0. This yields that H ( V ( t ) − s,X s ) is bounded ( ω -by- ω ) for s ≤ t . Moreover, q ( V ( s ) − s, X s , · ) is bounded by √ π ( V ( s ) − s ) . Thus, inview of (4.22), p ( V ( t ); V ( s ) − s, X s , z ) ≤ K ( ω ) p V ( s ) − s H ( V ( t ) − V ( s ) , z ) , K is a constant. Since ( V ( s ) − s ) − can be bounded when s is away from 0, H is boundedby 1, and f has a compact support, it follows from the Dominated Convergence Theorem thatlim s ↑ t Z R + f ( z ) p ( V ( t ); V ( s ) − s, X s , z ) dz = Z R + f ( z ) q ( V ( t ) − t, X t , z ) H ( V ( t ) − t, X t ) dz, P τ,t − a.s..This in turn shows, E τ,t [ f ( Z t ) E ] = E τ,t [lim s ↑ t f ( Z s ) E ] = E τ,t (cid:20)Z R + f ( z ) q ( V ( t ) − t, X t , z ) H ( V ( t ) − t, X t ) dz E (cid:21) . The claim now follows from (4.23). (cid:4)
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A Auxiliary results and their proofs
A.1 Comparison results
Lemma A.1
Suppose that d : R + × R [0 , M ] for some constant M > is a measurable functionand Y is a strong solution to Y t = y + 2 Z t p | Y s | dB s + Z t d ( s, Y s , Z s ) ds for some y ≥ upto an explosion time τ . Then, P [ τ = ∞ ] = 1 and P [0 ≤ Y t ≤ Y Mt , ∀ t ] = 1 , where Y Mt = y + 2 Z t q | Y Ms | dB s + Z t M ds.
Proof.
Let τ n := inf { t > | Y t | ≥ n } . By Tanaka’s formula,( Y t ∧ τ n − Y Mt ∧ τ n ) + = 2 Z t ∧ τ n ( p | Y s | − q Y Ms ) [ Y s >Y Ms ] dB s − Z t ∧ τ n ( M − d ( s, Y s , Z s )) [ Y s >Y Ms ] ds + L ( Y − Y M ) t ∧ τ n ,Y − t ∧ τ n = − Z t ∧ τ n p | Y s | [ Y s < dB s − Z t ∧ τ n d ( s, Y s , Z s ) [ Y s < ds + L ( Y ) t ∧ τ n L ( Y − Y M ) and L ( Y ) are the local times of Y − Y M and Y at 0, respectively. We willfirst show that Y is nonnegative upto τ n . Since Z t ∧ τ n [0 < − Y s < | Y s || Y s | ds ≤ t and R ∞ x dx = ∞ , it follows from Lemma 3.3 in Chap. IX of [17] that L ( Y t ∧ τ n ) = 0 for all t ≥ E (cid:2) Y − t ∧ τ n (cid:3) = E (cid:20) − Z t ∧ τ n p Y s [ Y s < dB s − Z t ∧ τ n d ( s, Y s , Z s ) [ Y s < ds (cid:21) ≤ , since the stochastic integral is a martingale having a bounded integrand. Thus, Y t ∧ τ n ≥
0, a.s. forevery t ≥ n .Similarly, Z t ∧ τ n [0
0] = 1 . By taking the limit as n → ∞ , we obtain P [0 ≤ Y t ∧ τ ≤ Y Mt ∧ τ , ∀ t ≥
0] = 1 . Since Y M is non-explosive, this implies that τ = ∞ , a.s.. (cid:4) In view of the above lemma, the hypothesis of the next lemma is not vacuous.
Lemma A.2
Suppose that d : R + × R [0 , M ] for some constant M > is a measurable functionand Y is the nonnegative strong solution to Y t = y + 2 Z t p Y s dB s + Z t d ( s, Y s , Z s ) ds, for some y ≥ . Moreover, suppose that there exists two stopping times S ≤ T such that d (( t ∨ S ) ∧ T, Y ( t ∨ S ) ∧ T , Z ( t ∨ S ) ∧ T ) ∈ [ a, b ] ⊆ [0 , M ] for some constants a and b . Then, P [ Y at ∧ T ≤ Y t ∧ T ≤ Y bt ∧ T , ∀ t ] = 1 , where Y at = Y t ∧ S + Z t ∨ SS n p Y as dB s + ads o Y bt = Y t ∧ S + Z t ∨ SS (cid:26) q Y bs dB s + bds (cid:27) . roof. Observe that using the similar arguments as in the previous lemma, one obtains that L ( Y − Y a ) = L ( Y − Y b ) = 0. Thus, by Tanaka’s formula,( Y t ∧ T − Y bt ∧ T ) + = 2 Z t ∧ Tt ∨ S ( p Y s − q Y bs ) [ Y s >Y bs ] dB s − Z t ∧ Tt ∨ S ( b − d ( s, Y s , Z s )) [ Y s >Y bs ] ds ( Y at ∧ T − Y t ∧ T ) + = 2 Z t ∧ Tt ∨ S ( p Y as − p Y s ) [ Y as >Y s ] dB s − Z t ∧ Tt ∨ S ( d ( s, Y s , Z s ) − a ) [ Y as >Y s ] ds. Observe that the stochastic integrals above are nonnegative local martingales, therefore they aresupermartingales. Thus, by taking the expectations we obtain E h ( Y t ∧ T − Y bt ∧ T ) + i ≤ E (cid:2) ( Y at ∧ T − Y t ∧ T ) + (cid:3) ≤ . Hence, the conclusion follows. (cid:4)
A.2 Martingale problems and some L estimates In the next lemma we show that the martingale problem related to Z as defined in (4.19) is wellposed. Recall that A is the associated infinitesimal generator defined in (4.24). We will denotethe restriction of A to B ([0 , t − ε ] × R + ) by A ε . Lemma A.3
Fix ε > and let µ ∈ P . Then, the martingale problem ( A ε , µ ) is well-posed.Moreover, the SDE (4.19) has a unique weak solution for any nonnegative initial condition and thesolution is strictly positive on ( s, t − ε ] for any s ∈ [0 , t − ε ] . Proof.
Let s ∈ [0 , t − ε ] and z ∈ R + . Then, direct calculations yield dZ r = σ ( r ) dβ r + σ ( r ) (cid:26) Z r − Z r η t ( r, Z r ) (cid:27) dr, for r ∈ [ s, t − ε ] , (A.28)with Z s = z , where η t ( r, y ) := R ∞ V ( t ) − V ( r ) 1 √ πu exp (cid:16) − y u (cid:17) du R ∞ V ( t ) − V ( r ) 1 √ πu exp (cid:16) − y u (cid:17) du , (A.29)thus, η t ( r, y ) ∈ [0 , V ( t ) − V ( t − ε ) ] for any r ∈ [0 , t − ε ] and y ∈ R + .First, we show the uniqueness of the solutions to the martingale problem. Suppose there existsa weak solution taking values in R + to the SDE above. Thus, there exists ( ˜ Z, ˜ β ) on some filteredprobability space ( ˜Ω , ˜ F , ( ˜ F r ) r ∈ [0 ,t − ε ] , ˜ P ) such that d ˜ Z r = σ ( r ) d ˜ β r + σ ( r ) (cid:26) Z r − ˜ Z r η t ( r, ˜ Z r ) (cid:27) dr, for r ∈ [ s, t − ε ] , Z s = z . Consider ˜ R which solves d ˜ R r = σ ( r ) d ˜ β r + σ ( r ) 1˜ R r dr, (A.30)with ˜ R s = z . Note that this equation is the SDE for a time-changed 3-dimensional Bessel processwith a deterministic time change and the initial condition ˜ R s = z . Therefore, it has a unique strongsolution which is strictly positive on ( s, t − ε ] (see 9. 446 in Chap. XI of [17]). Then, from Tanaka’sformula (see Theorem 1.2 in Chap. VI of [17]), since the local time of ˜ R − ˜ Z at 0 is identically 0(see Corollary 1.9 in Chap. VI of [17]), we have( ˜ Z t − ˜ R t ) + = Z t [ ˜ Z r > ˜ R r ] σ ( r ) (cid:26) Z r − ˜ Z r η t ( r, ˜ Z r ) − R r (cid:27) dr ≤ , where the last inequality is due to η t ≥
0, and a < b whenever a > b >
0. Thus, ˜ Z r ≤ ˜ R r for r ∈ [ s, t − ε ].Define ( L r ) r ∈ [0 ,t − ε ] by L = 1 and dL r = − L r ˜ Z r η t ( r, ˜ Z r ) d ˜ β r . If ( L r ) r ∈ [0 ,t − ε ] is a true martingale, then Q on ˜ F t − ε defined by dQd ˜ P = L t − ε , is a probability measure on ˜ F t − ε equivalent to ˜ P . Then, by Girsanov Theorem (see, e.g., Theorem3.5.1 in [14]) under Q d ˜ Z r = σ ( r ) d ˜ β Qr + σ ( r ) 1˜ Z r dr, for r ∈ [ s, t − ε ] , with ˜ Z s = z , where ˜ β Q is a Q -Brownian motion. This shows that ( ˜ Z, ˜ β Q ) is a weak solution to(A.30). As (A.30) has a unique strong solution which is strictly positive on ( s, t − ε ], any weaksolution to (4.19) is strictly positive on ( s, t − ε ]. Thus, due to Theorem 6.4.2 in [19], the martingaleproblem for ( δ z , A ε ) has a unique solution. Note that although the drift coefficient is not bounded,Theorem 6.4.2 in [19] is still applicable when L is a martingale.Thus, it remains to show that L is a true martingale when ˜ Z is a positive solution to (A.28).For some 0 ≤ t n − < t n ≤ t − ε consider E " exp Z t n t n − ( ˜ Z r η t ( r, ˜ Z r )) dr ! . (A.31)The expression in (A.31) is bounded by E " exp Z t n t n − ˜ R r (cid:18) V ( t ) − V ( t − ε ) (cid:19) dr ! ≤ E (cid:20) exp (cid:18)
12 ( ˜ R ∗ r ) t n − t n − ( V ( t ) − V ( t − ε )) (cid:19)(cid:21) where Y ∗ t := sup s ≤ t | Y s | for any c`adl`ag process Y . Recall that ˜ R is only a time-changed Besselprocess where the time change is deterministic and, therefore, ˜ R r is the square of the Euclidian22orm at time V ( r ) of a 3-dimensional standard Brownian motion, starting at ( z, ,
0) at time V ( s ).Thus, by using the same arguments as in Proposition 3.2, we get that the above expression is goingto be finite if E zV ( s ) (cid:20) exp (cid:18)
12 ( β ∗ V ( t − ε ) ) t n − t n − ( V ( t ) − V ( t − ε )) (cid:19)(cid:21) < ∞ , where β is a standard Brownian motion and E xs is the expectation with respect to the law of astandard Brownian motion starting at x at time s . In view of the reflection principle for standardBrownian motion (see, e.g. Proposition 3.7 in Chap. 3 of [17]) the above expectation is going to befinite if t n − t n − ( V ( t ) − V ( t − ε )) < V ( t − ε ) . Clearly, we can find a finite sequence of real numbers 0 = t < t < . . . < t n ( T ) = T that satisfyabove. Now, it follows from Corollary 3.5.14 in [14] that L is a martingale.In order to show the existence of a nonnegative solution, consider the solution, ˜ R , to (A.30),which is a time-changed 3-dimensional Bessel process, thus, nonnegative. Then, define ( L − r ) r ∈ [0 ,t − ε ] by L − = 1 and dL − r = L − r ˜ R r η t ( r, ˜ R r ) d ˜ β r . Applying the same estimation to L − as we did for L yields that L − is a true martingale. Then, Q on ˜ F t − ε defined by dQd ˜ P = L − t − ε , is a probability measure on ˜ F t − ε under which ˜ R solves d ˜ Z r = σ ( r ) d ˜ β Qr + σ ( r ) (cid:26) Z r − ˜ Z r η t ( r, ˜ Z r ) (cid:27) dr, for r ∈ [ s, t − ε ] , with ˜ Z s = z and ˜ β Q is a Q -Brownian motion. This means that the nonnegative process ˜ R is aweak solution of (A.28). Therefore, the martingale problem ( δ z , A ε ) has a solution by Proposition5.4.11 and Corollary 5.4.8 in [14] since σ is locally bounded. Thus, the martingale problem ( δ z , A ε )is well-posed for any z ∈ R + .The well-posedness of the martingale problem for ( µ, A ε ) follows from Theorem 21.10 in [12]since P z is the unique solution of the martingale problem for ( δ z , A ε ) for any z ∈ R + . (cid:4) We are now ready to show that the joint martingale problem for (
X, Z ) defined by the operator A : B ([0 , t ) × R ) B ([0 , t ) × R ) which is given by A φ ( s, x, z ) = ∂φ∂s ( s, x, z ) + 12 ∂ φ∂x ( s, x, z ) + 12 σ ( s ) ∂ φ∂z ( s, x, z )+ q x q ( V ( t ) − V ( s ) , x, z ) ∂φ∂x ( s, x, z ) + σ ( s ) H z H ( V ( t ) − V ( s ) , z ) ∂φ∂z ( s, x, z ) , with the domain D ( A ) = C ∞ c ([0 , t ) × R ). Proposition A.1
Let µ ∈ P where P is the set of probability measures on the Borel sets of R topologized by weak convergence. Then, the martingale problem ( µ, A ) is well-posed. roof. Clearly, if ( µ, A ε ) is well-posed for any ε >
0, where A ε is the restriction of A to B ([0 , t − ε ] , R + ), then ( µ, A ) is well-posed. As in the proof of Lemma A.3, the problem of well-posedness of ( µ, A ε ) can be reduced to that of ( δ x,z , A ε ) for any fixed ( x, z ) ∈ R due to Theorem21.11 in [12] and Proposition 1.6 in Chap. III of [17]. To this end, in view of Proposition 5.4.11 andCorollary 5.4.8 in [14], it suffices to show the existence and the uniqueness of weak solutions to thesystem of SDEs defined by (4.19) and (4.20) with the initial condition that X s = x and Z s = z fora fixed s ∈ [0 , t − ε ]. We will consider the following three cases to finish the proof. Case 1: x > , z >
0. In Lemma A.3 we have proved the existence and the uniqueness of a weaksolution to the SDE (4.19) for any initial condition Z s = z for s ∈ [0 , t − ε ] and z ≥ Z, ˜ β ) on some filtered probability space ( ˜Ω , ˜ F , ( ˜ F r ) r ∈ [0 ,t − ε ] , ˜ P ) such that( ˜ Z, ˜ β ) solves the SDE (4.19) with the initial condition Z s = z . Without loss of generalitywe can assume that the space ( ˜Ω , ˜ F , ( ˜ F r ) r ∈ [0 ,t − ε ] , ˜ P ) supports another Brownian motion, ˜ B ,independent of ˜ β . Then, Proposition 3.2 yields that there exists a unique strong and strictlypositive solution to (4.20) on ( ˜Ω , ˜ F , ( ˜ F r ) r ∈ [0 ,t − ε ] , ˜ P ). Indeed, the proof of Proposition 3.2would remain the same as long as the initial condition for Z is strictly positive and oneobserves that although Z is not a Brownian motion, the finiteness of (3.15) still follows from(3.16) since Z is strictly positive and bounded from above by a time-changed 3-dimensionalBessel process and the time change is given by V ( t ). This demonstrates that there existsa weak solution to the system of SDEs. Moreover, the solution is unique in law since X ispathwise uniquely determined by Z , which is unique in law. Case 2: x = 0 , z ≥
0. We can use the same arguments as in the previous case once we establish Lemma3.2 over the time interval [ s, t − ε ]. Note that we only need to show the strict positivity ofthe solution as the existence of a nonnegative weak solution follows along the same lines.Consider the sequence of stopping times ( τ n ) n ≥ τ n := inf { r ∈ [ s, t − ε ] : U r = 1 n } , where inf ∅ = t − ε . On ( τ n , t − ε ] the solution exists and is strictly positive as in Case 1since Z τ n > U τ n = n when τ n < t − ε . Consider τ := inf n τ n . If τ = s , we are done.Suppose τ > s with some positive probability. Then, on this set U t = 0 for t ≤ τ . However,this contradicts the fact that U solves (3.12) on [ s, t − ε ]. Case 3: x > , z = 0. As in the previous case it only remains to establish the strict positivity of thesolution of (3.12), which exists by the same arguments. Again consider the following sequenceof stopping times: τ n := inf { r ∈ [ s, t − ε ] : Z r = 1 n } , where inf ∅ = t − ε . That the weak solution to (3.12) is strictly positive on ( τ n , t − ε ] followsfrom Case 1 if X τ n >
0, and from Case 2 if X τ n = 0. Since inf n τ n = s by Lemma A.3, wehave the strict positivity on [ s, t − ε ]. (cid:4) Lemma A.4
Let ( Z, X ) be the unique strong solutions to (2.1) and (2.7). Then they solve themartingale problem on the interval [0 , t ) defined by (4.19) and (4.20) with the initial condition X = Z = 1 . Moreover, under Assumption 2.1 we have ) E (cid:20)R t [ τ>s ] (cid:16) q x q ( V ( s ) − s, X s , Z s ) (cid:17) ds (cid:21) < ∞ . ii) E τ,t (cid:20)R t (cid:16) q x q ( V ( s ) − s, X s , Z s ) (cid:17) ds (cid:21) < ∞ . iii) E τ,t hR t − ε σ ( s ) (cid:12)(cid:12) H x H ( V ( t ) − V ( s ) , Z s ) (cid:12)(cid:12) ds i < ∞ , for any ε > . Proof.
Recall that d P τ,t d P t = [ τ>t ] P [ τ>t ] and that E τ,t denotes the expectation operator with respectto P τ,t . Hence, under P τ,t , ( Z, X ) satisfy (4.19) and (4.20) with the initial condition X = Z = 1,which implies that they solve the corresponding martingale problem. i) & ii) Note that P [ τ > t ] E τ,t "Z t (cid:18) q x q ( V ( s ) − s, X s , Z s ) (cid:19) ds = E " [ τ>t ] Z t (cid:18) q x q ( V ( s ) − s, X s , Z s ) (cid:19) ds ≤ E "Z t [ τ>s ] (cid:18) q x q ( V ( s ) − s, X s , Z s ) (cid:19) ds . Thus, it suffices to prove the first assertion since P [ τ > t ] > t ≥
0. Recall from (3.9)that q x ( t, x, z ) q ( t, x, z ) = z − xt + exp (cid:0) − xzt (cid:1) − exp (cid:0) − xzt (cid:1) zt = z − xt + f (cid:18) xzt (cid:19) x , where f ( y ) = e − y − e − y y is bounded by 1 on [0 , ∞ ). As R t V ( s ) − s ) ds < ∞ and sup s ∈ [0 ,t ] E [ Z s ] ≤ V ( t ) + 1, the result will follow once we obtain1. sup s ∈ [0 ,t ] E [ X s [ τ>s ] ] < ∞ , and2. E (cid:16)R t [ τ>s ] 1 X s ds (cid:17) < ∞ ,demonstrated below.1. By Ito formula, [ τ>t ] X t = [ τ>t ] (cid:18) Z t X s dB s + 2 Z t (cid:26) Z s X s − X s V ( s ) − s + f (cid:18) Z s X s V ( s ) − s (cid:19) + 12 (cid:27) ds (cid:19) . (A.32)Observe that the elementary inequality 2 ab ≤ a + b implies2 [ τ>t ] Z t X s dB s ≤ (cid:18) [ τ>t ] Z t X s dB s (cid:19) ≤ (cid:18)Z τ ∧ t X s dB s (cid:19) , and2 Z t Z s X s − X s V ( s ) − s ds ≤ Z t Z s − X s V ( s ) − s ds ≤ Z t Z s V ( s ) − s ds. f is bounded by 1, using the above inequalities and taking expectations of both sidesof (A.32) yield E [ [ τ>t ] X t ] ≤ E (cid:18)Z t [ τ>s ] X s dB s (cid:19) + Z t E [ Z s ] V ( s ) − s ds + 3 t ≤ t + ( V ( t ) + 1) Z t V ( s ) − s ds + Z t E (cid:0) [ τ>s ] X s (cid:1) ds. The last inequality obviously holds when R t E (cid:0) [ τ>s ] X s (cid:1) ds = ∞ , otherwise, it is aconsequence of Ito isometry. Let T > t ∈ [0 , T ] it followsfrom Gronwall’s inequality that E [ [ τ>t ] X t ] ≤ (cid:18) T + ( V ( T ) + 1) Z T V ( s ) − s ds (cid:19) e T .
2. In view of Proposition 3.4 we have [ τ>s ] 1 X s ≤ R s where R is the unique strong solutionof (3.18). Thus, it is enough to show that R t E h R s i ds < ∞ . Recall from Proposition3.3 that the law of R s is that of λ s ρ Λ s where ρ is a 3-dimensional Bessel process startingat 1 and λ t = exp (cid:18) − Z t V ( s ) − s ds (cid:19) , Λ t = Z t λ s ds. Therefore, using the explicit form of the probability density of 3-dimensional Besselprocess (see Proposition 3.1 in Chap. VI of [17]) one has Z t E (cid:20) R s (cid:21) ds ≤ Z t E (cid:20) R s [ R s ≤ √ Λ s ] + Λ − s (cid:21) ds ≤ Z t λ − s Z √ Λ s λ − s y q (Λ s , , y ) dy ds + 3 p Λ t = Z t λ − s Z √ Λ s λ − s q y (Λ s , , y ∗ ) dy ds + 3 p Λ t where the last equality is due to the Mean Value Theorem and y ∗ ∈ [0 , y ]. It followsfrom direct computations that | q y ( t, , y ) | ≤ q πe t for all y ∈ R and t ∈ R + . Therefore,we have Z t E (cid:20) R s (cid:21) ds ≤ r πe Z t λ − s Z √ Λ s λ − s s dy ds + 3 p Λ t = r πe Z t λ − s Λ − s ds + 3 p Λ t ≤ r πe λ − t + 1 ! p Λ t . ii) Recall that H x H ( V ( t ) − V ( s ) , Z s ) = 1 Z s − Z s η t ( s, Z s ) , where η t is as defined in (A.29). Fix an ε >
0. Then, Z t − ε σ ( s ) (cid:12)(cid:12)(cid:12)(cid:12) H x H ( V ( t ) − V ( s ) , Z s ) (cid:12)(cid:12)(cid:12)(cid:12) ds = Z V ( t − ε )0 (cid:12)(cid:12)(cid:12)(cid:12) H x H ( V ( t ) − s, Z V − ( s ) ) (cid:12)(cid:12)(cid:12)(cid:12) ds. Consider the process S r := Z V − ( r ) for r ∈ [0 , V ( t )). Then, E τ,t (cid:20)Z t − ε σ ( s ) (cid:12)(cid:12)(cid:12)(cid:12) H x H ( V ( t ) − V ( s ) , Z s ) (cid:12)(cid:12)(cid:12)(cid:12) ds (cid:21) = E τ,t "Z V ( t − ε )0 (cid:12)(cid:12)(cid:12)(cid:12) S s − S s η t ( V − ( s ) , S s ) (cid:12)(cid:12)(cid:12)(cid:12) ds ≤ E τ,t "Z V ( t − ε )0 S s ds + V ( t − ε )( V ( t ) − V ( t − ε )) Z V ( t − ε )0 E τ,t [ S s ] ds . (A.33)Moreover, under P τ,t dS s = (3 − S s η t ( V − ( s ) , S s )) ds + 2 S s dW ts for all s < V ( t ) for the Brownian motion W t defined by W ts := R V − ( s )0 σ ( r ) dβ tr . Thus, E τ,t [ S s ] ≤ s + 1 + Z s E τ,t [ S r ] dr. Hence, by Gronwall’s inequality, we have E τ,t [ S s ] ≤ (3 s + 1) e s . In view of (A.33) to demon-strate iii) it suffices to show that E τ,t "Z V ( t − ε )0 S s ds < ∞ . However, Z V ( t − ε )0 S s ds ! = S V ( t − ε ) − S − W tV ( t − ε ) + Z V ( t − ε )0 η t ( V − ( s ) , S s ) S s ds ! , which obviously has a finite expectation due to earlier results. (cid:4)(cid:4)