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Absolute continuity for some one-dimensional processes

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Absolute continuity for some one-dimensional processes
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    a  r   X   i  v  :   0   8   0   4 .   3   0   3   7  v   2   [  m  a   t   h .   P   R   ]   8   J  u   l   2   0   0   8 Absolute continuity for some one-dimensional processes Nicolas  Fournier ∗ , Jacques  Printems ∗ July 8, 2008 Abstract We introduce an elementary method for proving the absolute continuity of the time marginals of one-dimensional processes. It is based on a comparison between the Fourier transform of such timemarginals with those of the one-step Euler approximation of the underlying process. We obtain someabsolute continuity results for stochastic differential equations with H¨older continuous coefficients.Furthermore, we allow such coefficients to be random and to depend on the whole path of the solution.We also show how it can be extended to some stochastic partial differential equations, and to someL´evy-driven stochastic differential equations. In the cases under study, the Malliavin calculus cannotbe used, because the solution in generally not Malliavin-differentiable. Key words   : Absolute continuity, Stochastic differential equations, L´evy processes, Stochastic partialdifferential equations, H¨older coefficients, Random coefficients. MSC 2000   : 60H10, 60H15, 60J75. 1 Introduction In this paper, we introduce a new method for proving the absolute continuity of the time marginals of some one-dimensional processes. The main idea is elementary, and quite rough. It is based on the explicitlaw of the associated one-step Euler scheme, and to an estimate saying that the process and its Eulerscheme remain very close to each other during one step.As we will see, this method is quite robust, and applies to many processesfor which the use of the Malliavincalculus (see Nualart [20], Malliavin [18]) is not possible, because the processes do not have Malliavinderivatives: examples for this are S.D.E.s with H¨older coefficients, S.D.E.s with random coefficients,...However, we are not able, for the moment, to extend it to multi-dimensional processes. The difficultyseems to be that we use some integrability properties of some Fourier transforms, which heavily dependson the dimension.To illustrate this method, we will consider four types of one-dimensional processes. Let us summarizeroughly the results we obtain, and compare them to existing results. Brownian S.D.E.s with H¨older coefficients.  To introduce our method in a simple way, we considera process satisfying a S.D.E. of the form  dX  t  =  σ ( X  t ) dB t  +  b ( X  t ) dt . We assume that  b  is measurablewith at most linear growth, and that  σ  is H¨older continuous with exponent  θ >  1 / 2. We show that  X  t has a density on  { σ   = 0 }  as soon as  t >  0. The proof is very short.Such a result is probably not far from being alreadyknown. In the case where σ  is bounded below, Aronson[1] obtains some absolute continuity result assuming only that  σ  and  b  are measurable (together withsome growth conditions) by analytical methods. Our result might be deduced from [1] by a localizationargument, however, we did not succeed in this direction. Anyway, our proof is much more simple. ∗ Universit´e Paris Est, LAMA, Facult´e des Sciences et Technologies, 61 avenue du G´en´eral de Gaulle, 94010 Cr´eteilCedex, France. E-mail:  nicolas.fournier@univ-paris12.fr, printems@univ-paris12.fr 1  Let us observe that to our knowledge, all the probabilistic papers about this topic assume at least that σ,b  are Lipschitz-continuous, see the paper of Bouleau-Hirsch [8] (the case where  b  is measurable can alsobe treated by using the Girsanov Theorem).Finally, let us mention that in [8], one gets the absolute continuity of the law of   X  t  for all  t >  0 assoon as  σ ( x 0 )   = 0, if   X  0  =  x 0 . Such a result cannot hold in full generality for H¨older continuouscoefficients: choose  x 0  >  0,  σ ( x ) =  x , and  b ( x ) =  − sign ( x ) | x | α , for some  α  ∈  (0 , 1). Denote by τ  ǫ  = inf  { t  ≥  0 ,X  t  =  ǫ } , for  ǫ  ∈  R + . One can check, using the Itˆo formula, that for  ǫ  ∈  (0 ,x 0 ), E [ X  1 − αt ∧ τ  ǫ ] =  x 1 − α 0  − E [   t ∧ τ  ǫ 0  ( α (1 − α )2  X  1 − αs  +(1 − α )) ds ]  ≤  x 1 − α 0  − (1 − α ) E [ τ  ǫ ∧ t ], whence E [ τ  ǫ ]  ≤  x 1 − α 0  / (1 − α ).As a consequence,  E [ τ  0 ]  ≤  x 1 − α 0  / (1  −  α ). But it also holds that  X  τ  0 + t  = 0 a.s. for all  t  ≥  0. ThusPr[ X  t  = 0]  >  0, at least for  t  large enough. Brownian S.D.E.s with random coefficients depending on the paths.  We consider here a processsolving a S.D.E. of the form  dX  t  =  σ ( X  t ) κ ( t, ( X  u ) u ≤ t ,H  t ) dB t  +  b ( t, ( X  u ) u ≤ t ,H  t ) dt , for some auxiliaryadapted process  H  . We assume some H¨older conditions on  σκ , some growth conditions, and that  κ  isbounded below. We prove the absolute continuity of the law of   X  t  on the set  { σ   = 0 }  for all  t >  0.Observe that we do not assume that  H   is Malliavin-differentiable, which would of course be needed if wewanted to use the Malliavin calculus.S.D.E.s with random coefficients arise for example in finance. Indeed, stochastic volatility models are nowwidely used, see e.g. Heston [14], Fouque-Papanicolaou-Sircar [11],... S.D.E.s with coefficients depending on the paths of the solutions arise in random mechanics: if one writes a S.D.E. satisfied by the velocityof a particle, the coefficients will often depend on its position, which is nothing but the integral of itsvelocity. One can also imagine a particle with position  X  t  whose diffusion and drift coefficients dependon the length covered by the particle at time  t , that is sup [0 ,t ] X  s  − inf  [0 ,t ] X  s .Here again, the result is not far from being known: if   σκ  is bounded below, one may use the resultof Gyongy [13], which says that the solution of a S.D.E. (with random coefficients depending on thewhole paths of the solution) has the same time marginals as the solution of a S.D.E. with deterministiccoefficients depending only on time and position. These coefficients being measurable and uniformlyelliptic, one may then use the result of Aronson [1]. However, our method is extremely simple, and wedo not have to assume that  σ  is bounded below. Stochastic heat equation.  We also study the heat equation  ∂  t U   =  ∂  xx U  + b ( U  )+ σ ( U  ) ˙ W   on R + × [0 , 1],with Neumann boundary conditions, where  W   is a space-time white noise, see Walsh [23]. We prove that U  ( t,x ) has a density on  { σ   = 0 }  for all  t >  0, all  x  ∈  [0 , 1], as soon as  σ  is H¨older continuous withexponent  θ >  1 / 2, and  b  is measurable and has at most linear growth.This result shows the robustness of our method: the best absolute continuity result was due to Pardoux-Zhang [21], who assume that  b  and  σ  are Lipschitz continuous. Let us however mention that theirnondegeneracy condition is very sharp, since they obtain the absolute continuity of   U  ( t,x ) for all  t >  0,all  x  ∈  [0 , 1] assuming only that  σ ( U  (0 ,x 0 ))   = 0 for some  x 0  ∈  [0 , 1] (if   U  (0 ,. ) is continuous). L´evy-driven S.D.E.s.  We finally consider the S.D.E.  dX  t  =  σ ( X  t ) dZ  t  +  b ( X  t ) dt , where ( Z  t ) t ≥ 0  is aL´evy process without drift and without Brownian part, and with L´evy measure  ν  . Roughly, we assumethat   | z |≤ ǫ z 2 ν  ( dz )  ≃  ǫ 2 − λ , for all  ǫ  ∈  (0 , 1], for some  λ  ∈  (3 / 4 , 2). We obtain that the law of   X  t  has adensity on  { σ   = 0 }  for all  t >  0, under the following assumption:(a) if   λ  ∈  (3 / 2 , 2),  b  is measurable and has at most linear growth, and  σ  is H¨older continuous withexponent  θ >  1 / 2;(b) if   λ  ∈  [1 , 3 / 2],  b  and  σ  are H¨older continuous with exponents  α >  3 / 2 − λ  and  θ >  1 / 2;(c) if   λ  ∈  (3 / 4 , 1),  b,σ  are H¨older continuous with exponent  θ >  3 / (2 λ ) − 1.This result seems to be the first absolute continuity result for jumping S.D.E.s with non Lipschitz coef-ficients. Observe that in some cases we allow the drift coefficient to be only measurable, even when thedriving L´evy process has no Brownian part. Such a result cannot be obtained using a trick like Girsanov’sTheorem (because even the law of such a L´evy process ( Z  t ) t ∈ [0 , 1]  and that of ( Z  t  +  t ) t ∈ [0 , 1]  are clearlynot equivalent). To our knowledge, this gives the first absolute continuity result for L´evy-driven S.D.E.swith measurable drift.2  Observe also that we allow the intensity measure of the Poissonian part to be singular: even withoutBrownian part and without drift, our result yields some absolute continuity for L´evy-driven S.D.E.s, evenwhen the L´evy measure of the driving process is completely singular. Such cases are not included inthe famous works of Bichteler-Jacod [7], Bichteler-Gravereaux-Jacod [6]. Picard [22] obtained some very complete results in that direction, for S.D.E.s with smooth coefficients. Notice that Picard obtains hisresults for any  λ  ∈  (0 , 2): our assumption is quite heavy, since we have to restrict our study to the casewhere  λ >  3 / 4.Let us finally mention a completely different approach developped by Denis [10], Nourdin-Simon [19], Bally [2] and others, where singular L´evy measures are allowed when the drift coefficient is sufficientlynon constant.We will frequently use the following classical Lemma. Lemma 1.1  For   µ  a nonnegative finite measure on   R , we denote by    µ ( ξ  ) =   R e iξx µ ( dx )  its Fourier transform (for all   ξ   ∈ R ). If    R |   µ ( ξ  ) | 2 dξ <  ∞ , then   µ  has a density with respect to the Lebesgue measure. Proof   For  n  ≥  1, consider  µ n  =  µ ⋆ g n , where  g n  is the centered Gaussian distribution with variance1 /n . Then of course,  |   µ n ( ξ  ) | ≤ |   µ ( ξ  ) | . Furthermore,  µ n  has a density  f  n  ∈  L 1 ∩ L ∞ ( R ,dx ) (for each fixed n  ≥  1), so that we may apply the Plancherel equality, which yields   R f  2 n ( x ) dx  = (2 π ) − 1    R |   µ n ( ξ  ) | 2 dξ   ≤ (2 π ) − 1    R |   µ ( ξ  ) | 2 dξ   =:  C <  ∞ . Due to the weak compactness of the balls of   L 2 ( R ,dx ), we may extract asubsequence  n k  and find a function  f   ∈  L 2 ( R ,dx ) such that  f  n k  goes weakly in  L 2 ( R ,dx ) to  f  . But onthe other hand,  µ n ( dx ) =  f  n ( x ) dx  tends weakly (in the sense of measures) to  µ . As a consequence,  µ  isnothing but  f  ( x ) dx .   Observe here that this Lemma is optimal. Indeed, the fact that   µ  belongs to  L  p , with  p >  2, doesnot imply that  µ  has a density, see counter-examples in Kahane-Salem [16]. The following localizationargument will also be of constant use. Lemma 1.2  For   δ >  0 , we introduce a function   f  δ  : R +  →  [0 , 1] , vanishing on   [0 ,δ  ] , positive on   ( δ, ∞ ) ,and globally Lipschitz continuous (with Lipschitz constant   1 ).Consider a probability measure   µ  on   R  and a function   σ  :  R  →  R + . Assume that for each   δ >  0 , the measure   µ δ ( dx ) =  f  δ ( σ ( x )) µ ( dx )  has a density. Thus   µ  has a density on   { x  ∈ R , σ ( x )  >  0 } . Proof   Let  A  ⊂ R be a Borel set with Lebesgue measure 0. We have to prove that  µ ( A ∩{ σ >  0 } ) = 0. Foreach  δ >  0, the measures  1 { σ ( x ) >δ } µ ( dx ) and  µ δ ( dx ) are clearly equivalent. By assumption,  µ δ ( A ) = 0for each  δ >  0, whence  µ ( A ∩{ σ > δ  } ) = 0. Hence,  µ ( A ∩{ σ >  0 } ) = lim δ → 0  µ ( A ∩{ σ > δ  } ) = 0.   The different sections of this paper are almost independent. In Section 2, we consider the case of simpleBrownian S.D.E.s. Section 3 is devoted to Brownian S.D.E.s with random coefficients depending on thewhole path of the solution. The stochastic heat equation is treated in Section 4. Finally, we considersome L´evy-driven S.D.E.s in Section 5. 2 Simple Brownian S.D.E.s We consider a filtered probability space (Ω , F  , ( F  t ) t ≥ 0 ,P  ) and a ( F  t ) t ≥ 0 -Brownian motion ( B t ) t ≥ 0 . For x  ∈ R  and  σ,b  : R → R , we consider the one-dimensional S.D.E. X  t  =  x +    t 0 σ ( X  s ) dB s  +    t 0 b ( X  s ) ds.  (2.1)Our aim in this section is to prove the following result. Theorem 2.1  Assume that   σ  is H¨ older continuous with exponent   θ  ∈  (1 / 2 , 1] , and that   b  is measurable and has at most linear growth. Consider a continuous   ( F  t ) t ≥ 0 -adapted solution   ( X  t ) t ≥ 0  to (2.1). Then  for all   t >  0 , the law of   X  t  has a density on the set   { x  ∈ R ,σ ( x )   = 0 } . 3  Observe that the (weak or strong) existence of solutions to (2.1) does not hold under the sole assumptionsof Theorem 2.1. However, at least weak existence holds if one assumes additionally that  b  is continuousor that  σ  is bounded below, see Karatzas-Shreve [17]. Proof   By a scaling argument, it suffices to consider the case  t  = 1. We divide the proof into three parts. Step 1.  For  ǫ  ∈  (0 , 1), we consider the random variable Z  ǫ  :=  X  1 − ǫ  +    11 − ǫ σ ( X  1 − ǫ ) dB s  =  X  1 − ǫ  + σ ( X  1 − ǫ )( B 1  − B 1 − ǫ ) . Conditioning with respect to  F  1 − ǫ , we get, for all  ξ   ∈ R , | E  e iξZ  ǫ |F  1 − ǫ  |  =  | exp  iξX  1 − ǫ  − ǫσ 2 ( X  1 − ǫ ) ξ  2 / 2  |  = exp  − ǫσ 2 ( X  1 − ǫ ) ξ  2 / 2  . Step 2.  Using classical arguments (Doob’s inequality and Gronwall Lemma), and the fact that  σ  and  b have at most linear growth one may show that there is a constant  C   such that for all 0  ≤  s  ≤  t  ≤  1, E  sup [0 , 1] X  2 t   ≤  C,  E  ( X  t  − X  s ) 2   ≤  C  ( t − s ) .  (2.2)Next, since  σ  is H¨older continuous with index  θ  ∈  (1 / 2 , 1] and since  b  has at most linear growth, we get,for all  ǫ  ∈  (0 , 1), E [( X  1  − Z  ǫ ) 2 ]  ≤  2    11 − ǫ E  ( σ ( X  s ) − σ ( X  1 − ǫ )) 2  ds  + 2 E    11 − ǫ b ( X  s ) ds  2  ≤  C     11 − ǫ E  | X  s − X  1 − ǫ | 2 θ  ds  + 2 ǫ    11 − ǫ E [ b 2 ( X  s )] ds ≤  C     11 − ǫ E  | X  s − X  1 − ǫ | 2  θ ds  + Cǫ    11 − ǫ E [1 + X  2 s ] ds ≤  Cǫ 1+ θ + Cǫ 2 ≤  Cǫ 1+ θ , where we used (2.2). Step 3.  Let  δ >  0 be fixed, consider the function  f  δ  defined in Lemma 1.2, and the measure  µ δ,X 1 ( dx ) = f  δ ( | σ ( x ) | ) µ X 1 ( dx ), where  µ X 1  is the law of   X  1 . Then for all  ξ   ∈ R , all  ǫ  ∈  (0 , 1), we may write |    µ δ,X 1 ( ξ  ) |  =  | E [ e iξX 1 f  δ ( | σ ( X  1 ) | )] |≤ | E [ e iξX 1 f  δ ( | σ ( X  1 − ǫ ) | )] | + E [ | f  δ ( | σ ( X  1 ) | ) − f  δ ( | σ ( X  1 − ǫ ) | ) | ] ≤ | E [ e iξZ  ǫ f  δ ( | σ ( X  1 − ǫ ) | )] | + | ξ  | E [ | X  1  − Z  ǫ | ] + E [ | f  δ ( | σ ( X  1 ) | ) − f  δ ( | σ ( X  1 − ǫ ) | ) | ] , where we used the inequality  | e iξx − e iξz | ≤ | ξ  | . | x − z |  and the fact that  f  δ  is bounded by 1. First, Step1 implies that  E  e iξZ  ǫ f  δ ( | σ ( X  1 − ǫ ) | )   ≤  E  E  e iξZ  ǫ f  δ ( | σ ( X  1 − ǫ ) | ) |F  1 − ǫ  ≤  E  f  δ ( | σ ( X  1 − ǫ ) | ) e − ǫσ 2 ( X 1 − ǫ ) ξ 2 / 2   ≤  exp( − ǫδ  2 ξ  2 / 2) , since  f  δ  is bounded by 1 and vanishes on [0 ,δ  ]. Step 2 implies that  | ξ  | E [ | X  1  − Z  ǫ | ]  ≤  C  | ξ  | ǫ (1+ θ ) / 2 . Since f  δ  is Lipschitz continuous and  σ  is H¨older continuous with index  θ  ∈  (1 / 2 , 1], we deduce from (2.2) that E [ | f  δ ( | σ ( X  1 ) | ) − f  δ ( | σ ( X  1 − ǫ ) | ) | ]  ≤  C  E [ | X  1  − X  1 − ǫ | θ ]  ≤  Cǫ θ/ 2 .As a conclusion, we deduce that for all  ξ   ∈ R , for all  ǫ  ∈  (0 , 1), |    µ δ,X 1 ( ξ  ) | ≤  exp( − ǫδ  2 ξ  2 / 2) + C  | ξ  | ǫ (1+ θ ) / 2 + Cǫ θ/ 2 . 4  For each  | ξ  | ≥  1 fixed, we apply this formula with the choice  ǫ  := (log | ξ  | ) 2 /ξ  2 ∈  (0 , 1). This gives |    µ δ,X 1 ( ξ  ) | ≤  exp( − δ  2 (log | ξ  | ) 2 / 2) + C  (log | ξ  | ) 1+ θ / | ξ  | θ + C  (log | ξ  | ) θ / | ξ  | θ . This holding for all  | ξ  | ≥  1, and    µ δ,X 1  being bounded by 1, we get that   R |    µ δ,X 1 ( ξ  ) | 2 dξ <  ∞ , since θ >  1 / 2 by assumption. Lemma 1.1 implies that the measure  µ δ,X 1  has a density, for each  δ >  0. Lemma1.2 allows us to conclude that  µ X 1  has a density on  {| σ |  >  0 } .   3 Brownian S.D.E.s with random coefficients We start again with a filtered probability space (Ω , F  , ( F  t ) t ≥ 0 ,P  ) a ( F  t ) t ≥ 0 -Brownian motion ( B t ) t ≥ 0 .To model the randomness of the coefficients, we consider an auxiliary predictable process ( H  t ) t ≥ 0 , withvalues in some normed space ( S  , ||  .  || ). Then we consider  σ  :  R  →  R  and two measurable maps κ,b  :  A → R , where A  :=  { ( s, ( x u ) u ≤ s ,h ) , s  ≥  0 , ( x u ) u ≥ 0  ∈  C  ( R + , R ) ,h  ∈ S} , and the following one-dimensional S.D.E. X  t  =  x +    t 0 σ ( X  s ) κ ( s, ( X  u ) u ≤ s ,H  s ) dB s  +    t 0 b ( s, ( X  u ) u ≤ s ,H  s ) ds.  (3.1)Here again, the existence of solutions to such a general equation does of course not always hold, even underthe assumptions below. However, there are many particular cases for which the (weak or strong) existencecan be proved by classical methods (Picard iteration, martingale problems, change of probability, changeof time, ...) Theorem 3.1  Assume that the auxiliary process   H   satisfies, for some   η >  1 / 2 , for all   0  ≤  s  ≤  t  ≤  T  , E  || H  t || 2   ≤  C  T   and   E  || H  t − H  s || 2   ≤  C  T  ( t − s ) η .  (3.2) Assume also that   κσ  and   b  have at most linear growth, that is for all   0  ≤  t  ≤  T  , all   ( x u ) u ≥ 0  ∈  C  ( R + , R ) ,all   h  ∈ S  , | σ ( x t ) κ ( t, ( x u ) u ≤ t ,h ) | + | b ( t, ( x u ) u ≤ t ,h ) | ≤  C  T  (1 + sup [0 ,t ] | x u | + || h || ) ,  (3.3) that   σ  is H¨ older continuous with index   α  ∈  (1 / 2 , 1] , and that for some   θ 1  ∈  (1 / 4 , 1] ,  θ 2  ∈  (1 / 2 , 1] , and  θ 3  ∈  (1 / 2 η, 1] , for all   0  ≤  s  ≤  t  ≤  T  , all   ( x u ) u ≥ 0  ∈  C  ( R + , R ) , all   h,h ′  ∈ S  , | σ ( x t ) κ ( t, ( x u ) u ≤ t ,h ) − σ ( x s ) κ ( s, ( x u ) u ≤ s ,h ′ ) | ≤  C  T   ( t − s ) θ 1 + sup u ∈ [ s,t ] | x u  − x s | θ 2 + || h − h ′ || θ 3  .  (3.4) Finally, assume that   κ  is bounded below by some constant   κ 0  >  0 . Consider a continuous   ( F  t ) t ≥ 0 -adapted solution   ( X  t ) t ≥ 0  to (3.1). Then the law of   X  t  has a density on   { x  ∈ R , σ ( x )   = 0 }  as soon as   t >  0 . Notice that (3.2) does not imply that  H   is a.s. continuous: it is just a sort of   L 2 -continuity. Observe alsothat we assume no regularity about the drift coefficient  b . This is not so surprising, thinking about theGirsanov Theorem. However, the Girsanov Theorem might be difficult to use in such a context, due tothe randomness of the coefficients (a change of probability also changes the law of the auxiliary process).Let us briefly illustrate (3.4). Example 3.2  (a) Let   σ ( x s ) κ ( s, ( x u ) u ≤ s ,h ) =  φ ( s,x s , sup [0 ,s ] ϕ ( x u ) ,h ) , with   φ  :  R +  × R × R ×S →  R satisfying   | φ ( s,x,m,h ) − φ ( s ′ ,x ′ ,m ′ ,h ′ ) | ≤  C  ( | s − s ′ | θ 1 + | x − x ′ | θ 2 + | m − m ′ | ζ  + || h − h ′ || θ 3 )  and   ϕ  : R → R satisfying   | ϕ ( x ) − ϕ ( x ′ ) | ≤  C  | x − x ′ | r , with   ζr  ≥  θ 2 . Then   σκ  satisfies (3.4).(b) Let   σ ( x s ) κ ( s, ( x u ) u ≤ s ,h ) =  φ ( s,x s ,   s 0  ϕ ( x u ) du,h )  with   φ  :  R +  ×  R  ×  R  × S →  R  satisfying the condition   | φ ( s,x,m,h ) − φ ( s ′ ,x ′ ,m ′ ,h ′ ) | ≤  C  ( | s − s ′ | θ 1 + | x − x ′ | θ 2 + | m − m ′ | θ 1 + || h − h ′ || θ 3 )  and with  ϕ  : R → R  bounded. Then   σκ  satisfies (3.4). 5
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