Lower bounds for the Stock Price density in a Local-Stochastic - PowerPoint PPT Presentation

Lower bounds for the Stock Price density in a Local-Stochastic Volatility Model V. Bally and S. De Marco Vlad Bally Université de Marne-la-Vallée and équipe Math… October 2010

Local - Stochastic Volatility models We consider the model p dX t = � 1 2 � 2 ( t; X t ) V t dt + � ( t; X t ) V t ( �dW 1 t + � � dW 2 t ) ; p V t dW 1 dV t = b ( t; V t ) V t dt + � ( t; X t ) t q 1 � � 2 : where � � = Remark : S t = S 0 e X t is the stock price which satis…es p V t ( �dW 1 t + � � dW 2 dS t = S t � ( t; ln S t =S 0 ) t ) : For � � 1 , b ( v ) = a � bv and constant � , this is the Heston model. Problem : In the Heston model, it is known that the moments blow up : For some p > 1 there exists T > 0 such that E ( S p T ) = 1 :

Consequences . Implied volatility : � ( T; k ) de…ned as the solution of the equation E ( S 0 e X T � S 0 e k ) + = C BS ( S 0 e k ; T; � ( T; k )) with k = log( K=S 0 ) = log-forward moneyness. Critical exponents : T ( X ) = sup f p : E ( S p p � T ) = E ( e pX T ) < 1g ; T ( X ) = sup f p : E ( S � p q � T ) = E ( e � pX T ) < 1g : Lee’s moment formula (model free) : T� 2 ( T; k ) T� 2 ( T; k ) = g ( p � = g ( q � lim sup T ( X ) � 1) lim sup T ( X )) k k k !1 k !�1 q p 2 + p � p ) ; g ( p ) = 2 � 4( g ( 1 ) = 0 :

Calibration : In Heston model one may compute explicitly, for each p 2 N T p ( a; b; � ) = sup f t : E ( S p t ) = E ( e pX t ) < 1g < 1 p � T ( X ) = p � T ( a; b; � ) ! One employs the market data to compute T� 2 ( T; k ) T� 2 ( T; k ) lim sup = s + ; lim sup = s � k k k !1 k !�1 and obtains parameters guesses from g ( p � g ( q � T ( a; b; � ) � 1) = s + ; T ( a; b; � )) = s � Our aim : Proving that in the previous local-stochastic volatility models we have moment explosion : p � T ( X ) � C T < 1 8 T Drawback : C T is a rough constant.

Theorem . (B - S. De Marco) A . Suppose that i ) ( t; v ) ! � ( t; v ) ; ( t; x ) ! � ( t; x ) Lipschitz continuous and bounded ii ) 0 < � � � ( t; v ) ; 0 < � � � ( t; x ) iii ) v ! b ( t; v ) sub-linear growth Then P ( X T > x ) � e � c T x In particular, for each x > 0 E ( e pX T ) � E ( e pX T 1 f X T >x g ) � e px � c T x = e ( p � c T ) x ! 1 for p > c T : B . Suppose moreover that is in C 3 x ! � ( t; x ) b : Then p T ( x ) � e � c 0 T x : P ( X T 2 dx ) = p T ( x ) dx and

Tubes estimates (B, Fernandez, Meda, 2008) We consider a general Itô process Y t 2 R n Z t Z t d X 0 � j ( s; Y s ) dW j Y t = Y 0 + s + 0 � 0 ( s; Y s ) ds j =1 and a deterministic curve y t 2 R n : We want to give a lower bound of the form Z T P ( j Y t � y t j � R t ; 0 � t � T ) � exp( � C (1 + 0 F ( t ) dt )) where R t is a time depending radius and F ( t ) is a rate function which is explicit. Remark . i) The coe¢cients may depend on the trajectory in an adapted way : � j ( s; y ) = � j ( s; !; y ) which is � ( W u ; u � s ) measurable. In particular, if � j ( s; y ) = � j ( s; ! ) we get a general Itô process. ii) Y may be some Non - Markov process . EX : If X t is a di¤usion process on R m and � : R m ! R n is a twice di¤erentiable function then Y t = �( X t ) is no more a Markov process (options on a basket).

Hypothesis . Consider the exit time from the tube � R = inf f t : j Y t � y t j � R t g : We assume that d � � X 2 + j � 0 ( t; Y t ) j � c t � � i ) (Bounded) � � j ( t; Y t ) 0 � t � � R ; � j =1 d � � X 2 1 f s _ t<� R g j F s ^ t ) � L t j s � t j ; � � ii ) (Lip) E ( � � j ( s; Y s ) � � j ( t; Y t � j =1 d D E X iii ) (Ellipticity) inf � j ( t; Y t ) ; � � � t 0 � t � � R j � j =1 j =1 We also assume that the deterministic curves f t = y t ; R t ; � t ; c t ; L t satisfy : There exists � � 1 and h > 0 such that iv ) (Growth) f t � �f s for j t � s j � h or put it otherwise j ln f t � ln f s j � ln � for j t � s j � h:

And we de…ne the rate function h + j @ t y t j 2 F ( t ) = 1 t )( 1 + 1 + ( c 2 t + L 2 ) : R 2 � t � t t Theorem . A . (B-F-M 2008) Under the above hypothesis Z T P ( j Y t � y t j � R t ; 0 � t � T ) � exp( � C ( n ) � p ( n ) (1 + 0 F ( t ) dt ))) : B . (B 2005) Suppose moreover that � j ( t; Y t ) 2 D n +3 ;p ; t � 0 ; j = 0 ; :::; d: Then Z T p T ( x ) � exp( � S ( n ) � p ( n ) (1 + P ( Y T 2 dx ) = p T ( x ) dx and 0 F ( t ) dt )) : Here C ( n ) is an universal constant depending on the dimension n only. And S ( n ) is a constant which depends on n but also on the Sobolev norms (in Malliavin sense) of � j ( t; Y t ) :

Back to our problem : p dX t = � 1 2 � 2 ( t; X t ) V t dt + � ( t; X t ) V t ( �dW 1 t + � � dW 2 t ) ; p V t dW 1 dV t = b ( t; V t ) V t dt + � ( t; X t ) t : Remark 1 . P ( X T > x ) � P ( X T 2 B R ( x + R )) � P ( j X t � x t j � R; t � � R ) : so we need ball estimates for X T : Remark 2 . The ellipticity condition is ( � ^ � � ) � ( � ^ � ) � V t � � t ; 0 � t � � R so we need tubes estimates for V t : We take two deterministic curves x t and v t and a deterministic time dependent radius R t and we want to lower bound P ( j X t � x t j + j V t � v t j � R t ; t � � R ) :

Rate function h + j @ t x t j 2 + j @ t v t j 2 F x;v ( t ) = 1 t )( 1 + 1 + ( c 2 t + L 2 ) R 2 � t � t t with � t = c ( �; �; � ) � v t ; c t = C � (1 + v t ) ; L t = C � (1 + v t ) so, up to a constant h + j @ t x t j 2 + j @ t v t j 2 F x;v ( t ) = 1 + (1 + v t ) 2 ( 1 + 1 ) : R 2 v t v t t Optimization v t = R 2 j @ t x t j = j @ t v t j t : So we take R t = p v t x t = v t + ( X 0 � V 0 ) and and we get h + j @ t v t j 2 h + j @ t v t j 2 + (1 + v t ) 2 F x;v ( t ) = 1 � 1 + v t : v t v t v t

We look for v t which minimizes Z T 0 ( j @ t v t j 2 + v t ) dt v t under the constrained x = x T = v T + ( X 0 � V 0 ) ! v T = x � ( X 0 � V 0 ) : Solution q q x + V 0 � sinh t 2 v t = V 0 � : sinh T 2 Final result. A. p � ( X ) � c 0 q � ( X ) � c 00 P ( X T � x ) � exp( � c T ( � ) x ) ; T ( � ) ; T ( � ) : The constant c T ( � ) blows up as j � j ! 1 : Problem : get estimates which are uniform in � ? Andersen & Piterbag ’07 - for Heston : � ! � 1 ) T ( � ) ! 1 ) c T ( � ) ! 18 T B. Density : under regularity assumptions for the coe¢cients : p T ( x ) � exp( � c 0 T ( � ) x ) : Malliavin calculus with respect to W 2 ; conditionally with respect to W 1 :

Back to Itô processes : Idea of the proof. We want to give "tubes estimates" around y t for Z t Z t d X 0 � j ( s; Y s ) dW j Y t = Y 0 + s + 0 � 0 ( s; Y s ) ds: j =1 Step 1 . We choose a time grid 0 = t 0 < t 1 < ::: < t N = T and we write Y t k +1 = Y t k + I k + R k with d X � j ( t k ; Y t k )( W j t k +1 � W j I k = t k ) j =1 Z t k +1 Z t k +1 d X ( � j ( s; Y s ) � � j ( t k ; Y t k )) dW j R k = s + � 0 ( s; Y s ) ds: t k t k j =1 Remark 1 . Let � k = t k +1 � t k : Then q I k � � k and R k � � k

Remark 2 . Y t k + I k is Gaussian conditionally to � ( W s ; s � t k ) and � � 2 � � Z � y � Y t k � P ( Y t k + I k 2 B r k ( y t k +1 )) � 1 B rk ( y tk +1 ) exp( � ) dy: � d 2 � t k t k q So if we take r k � � t k � � q P ( Y t k + I k 2 B r k ( y t k +1 )) � r d � � � e � c � e � c : k � y t k +1 � Y t k ) � + r k � � t k ! � d t k Consequence : then P ( \ N � 1 k =0 f Y t k +1 2 B r k ( y t k +1 ) g ) � P ( \ N � 1 k =0 f Y t k + I k 2 B r k ( y t k +1 ) g ) � e � cN : Problems : R T 1. Compute N (this gives 0 F ( t ) dt ) ( Do not N ! 1 )

2. How to det rid of R k ??? Taylor expansion : E ( � " ( Y t k + I k + R k � y t k +1 )) = E ( � " ( Y t k � y t k +1 + I k )) Z 1 0 E ( � 0 + " ( Y t k � y t k +1 + I k + �R k ) R k ) d� 1. Tubes - stochastic calculus. If " � � n= 4 then k � � Z 1 � � � � 1 � � 0 E ( � 0 " ( Y t k � y t k +1 + I k + �R k ) R k ) d� 2 E ( � " ( Y t k � y t k +1 + I k )) � � � 2. Density : We need to let " ! 0 : We take � " s.t. � 0 " = � " and we write Z 1 Z 1 0 E ( � 0 0 E (� 00 " ( Y t k � y t k +1 + I k + �R k ) R k ) d� = " ( Y t k � y t k +1 + I k + �R k ) R k ) d� Z 1 = 0 E (� " ( Y t k � y t k +1 + I k + �R k ) H (2) ) d�: