Model Free Hedging. David Hobson. Bachelier World Congress Brussels, June University of Warwick

Transcription

1 Model Free Hedging David Hobson University of Warwick Bachelier World Congress Brussels, June 2014

2 Overview The classical model-based approach Robust or model-independent pricing and hedging Primal and dual versions of the problem Martingale inequalities The Skorokhod embedding problem Martingale optimal transport PCOCs and fake diffusions David Hobson (Warwick) Model Free Hedging June / 32

3 A partial and incomplete list of contributors to the subject Acciaio, Beiglböck, Bouchard, Brown, Carr, Cox, Davis, Dolinsky, Dupire, Henry-Labordère, Huang, Kahale, Kardaras, Klimmek, Laurance, Lee, Oberhauser, Obloj, Neuberger, Nutz, Penkner, Rogers, Schachermayer, Spoida, Soner, Tan, Temme, Touzi, Tsuzuki, Wang, Wang. David Hobson (Warwick) Model Free Hedging June / 32

4 The classical approach to derivative pricing The standard approach is to postulate a model M = (Ω, F, P, F = (F t ) t 0, S = (S t ) t 0 ) (or a parametric family of models M = {M λ ; λ Λ}). Then (working in a discounted universe) we price a contingent claim F T = F (S t ; 0 t T ) as a expectation under an equivalent martingale measure Q: E Q [F T ] Example The Black-Scholes-Merton model For the family of exponential Brownian motion models ds t /S t = µdt + σdw t, for a call with payoff (S T K) +, the price is C BS where C BS = C BS (T, K; 0, S 0 ; µ, σ) David Hobson (Warwick) Model Free Hedging June / 32

5 The classical approach to derivative pricing The standard approach is to postulate a model M = (Ω, F, P, F = (F t ) t 0, S = (S t ) t 0 ) (or a parametric family of models M = {M λ ; λ Λ}). Then (working in a discounted universe) we price a contingent claim F T = F (S t ; 0 t T ) as a expectation under an equivalent martingale measure Q: E Q [F T ] Example The Black-Scholes-Merton model For the family of exponential Brownian motion models ds t /S t = µdt + σdw t, for a call with payoff (S T K) +, the price is C BS where C BS = C BS (T, K; 0, S 0 ; µ, σ) David Hobson (Warwick) Model Free Hedging June / 32

6 Underpinning the theory is the notion of complete markets and replication: T F T = E Q [F T ] + θ t ds t 0 P/Q a.s. In a complete market the price and hedge are uniquely specified and replication is perfect provided the model is a perfect description of the real world. For convex payoffs (eg puts and calls) C BS (σ) is increasing in σ. Hence, in practice the Black-Scholes-Merton model is calibrated using a liquid option. C BS (T, K) is an extrapolation device. David Hobson (Warwick) Model Free Hedging June / 32

7 Underpinning the theory is the notion of complete markets and replication: T F T = E Q [F T ] + θ t ds t 0 P/Q a.s. In a complete market the price and hedge are uniquely specified and replication is perfect provided the model is a perfect description of the real world. For convex payoffs (eg puts and calls) C BS (σ) is increasing in σ. Hence, in practice the Black-Scholes-Merton model is calibrated using a liquid option. C BS (T, K) is an extrapolation device. David Hobson (Warwick) Model Free Hedging June / 32

8 Underpinning the theory is the notion of complete markets and replication: T F T = E Q [F T ] + θ t ds t 0 P/Q a.s. In a complete market the price and hedge are uniquely specified and replication is perfect provided the model is a perfect description of the real world. For convex payoffs (eg puts and calls) C BS (σ) is increasing in σ. Hence, in practice the Black-Scholes-Merton model is calibrated using a liquid option. C BS (T, K) is an extrapolation device. David Hobson (Warwick) Model Free Hedging June / 32

9 Underpinning the theory is the notion of complete markets and replication: T F T = E Q [F T ] + θ t ds t 0 P/Q a.s. In a complete market the price and hedge are uniquely specified and replication is perfect provided the model is a perfect description of the real world. For convex payoffs (eg puts and calls) C BS (σ) is increasing in σ. Hence, in practice the Black-Scholes-Merton model is calibrated using a liquid option. C BS (T, K) is an extrapolation device. David Hobson (Warwick) Model Free Hedging June / 32

10 A reverse approach: constructing models to match vanilla option prices Typically we are not given M = (Ω, F, P, F, S) but rather we observe a collection of prices for traded securities, including derivatives. Rather than starting with the model the issue is to find a model M to match the prices of traded assets. In the base case we assume that in addition to the asset itself, vanilla puts and calls are liquidly traded and are used as inputs. Other derivatives are treated as exotics. David Hobson (Warwick) Model Free Hedging June / 32

11 Lemma Suppose, for fixed T, call prices are known for every strike K (0, ). Then, assuming C(T, K) = E Q [(S T K) + ] we have Q(S T > K) = Corollary (Breeden and Litzenberger) K C(T, K) Q(S T dk) = 2 2 C(T, K) K Any co-maturing European claim can be priced and hedged perfectly. F (k)(s k) dk + F (k)c(t, k)dk + x f (y) = f (x) + (y x)f (x) + f (k)(y k) + dk + f (k)(k y) + dk x 0 S0 F (S T ) = F (S 0 ) + (S T S 0 )F (S 0 ) + F (k)(k s) + dk S 0 0 S0 E Q [F (S T )] = F (S 0 )+ 0 + F (k)p(t, k)dk S 0 0 David Hobson (Warwick) Model Free Hedging June / 32

12 Lemma Suppose, for fixed T, call prices are known for every strike K (0, ). Then, assuming C(T, K) = E Q [(S T K) + ] we have Q(S T > K) = Corollary (Breeden and Litzenberger) K C(T, K) Q(S T dk) = 2 2 C(T, K) K Any co-maturing European claim can be priced and hedged perfectly. F (k)(s k) dk + F (k)c(t, k)dk + x f (y) = f (x) + (y x)f (x) + f (k)(y k) + dk + f (k)(k y) + dk x 0 S0 F (S T ) = F (S 0 ) + (S T S 0 )F (S 0 ) + F (k)(k s) + dk S 0 0 S0 E Q [F (S T )] = F (S 0 )+ 0 + F (k)p(t, k)dk S 0 0 David Hobson (Warwick) Model Free Hedging June / 32

13 Corollary (Neuberger - Dupire) Given the prices of vanilla puts and calls, and assuming that the underlying is continuous semi-martingale the the fixed leg of a T -maturity variance swap is model independent and given by S 0 2C(T, k) k 2 dk + S0 0 2P(T, k) k 2 dk Proof. Then if ds t = S tσ tdw t, d( 2 ln S t) = 2 dst S t + d[s]t S 2 t T 0 σ 2 tdt= T 0 d[s] t S 2 t = = T 0 T 0 2 ds t 2 ln(s T /S 0) S t [ 2 2 ] ds t + S t S 0 S 0 2 k 2(S T k) + dk + S0 0 2 k 2(k S T ) + dk David Hobson (Warwick) Model Free Hedging June / 32

14 Corollary (Neuberger - Dupire) Given the prices of vanilla puts and calls, and assuming that the underlying is continuous semi-martingale the the fixed leg of a T -maturity variance swap is model independent and given by S 0 2C(T, k) k 2 dk + S0 0 2P(T, k) k 2 dk Proof. Then if ds t = S tσ tdw t, d( 2 ln S t) = 2 dst S t + d[s]t S 2 t T 0 σ 2 tdt= T 0 d[s] t S 2 t = = T 0 T 0 2 ds t 2 ln(s T /S 0) S t [ 2 2 ] ds t + S t S 0 S 0 2 k 2(S T k) + dk + S0 0 2 k 2(k S T ) + dk David Hobson (Warwick) Model Free Hedging June / 32

15 Model free Hedging Suppose the underlying is a forward price (ie suppose we are working in discounted units). Suppose calls and puts of a single fixed maturity are liquidly traded. The objective is to price and hedge a co-maturing (path-dependent) exotic option. David Hobson (Warwick) Model Free Hedging June / 32

16 Definition A perfectly calibrated model is a quintuple (Ω, F, Q, F, S) such that S is a (Q, F)-martingale and E Q [(S T K) + ] = C(T, K) for each traded vanilla call. More generally, given a set of option prices is there a perfectly calibrated model consistent with those option prices? is there a model with continuous paths? is there a model with paths of finite (quadratic) variation? if there is, is it unique? If not one ambition is to choose a parsimonious model which captures essential features of the problem is realistic another ambition is to search for the class M of all consistent models if there is not, is there an arbitrage? is there a model-independent (strong) arbitrage? is there a model-dependent (weak) arbitrage?? A perfectly calibrated model can be used for pricing and hedging. David Hobson (Warwick) Model Free Hedging June / 32

17 Definition A perfectly calibrated model is a quintuple (Ω, F, Q, F, S) such that S is a (Q, F)-martingale and E Q [(S T K) + ] = C(T, K) for each traded vanilla call. More generally, given a set of option prices is there a perfectly calibrated model consistent with those option prices? is there a model with continuous paths? is there a model with paths of finite (quadratic) variation? if there is, is it unique? If not one ambition is to choose a parsimonious model which captures essential features of the problem is realistic another ambition is to search for the class M of all consistent models if there is not, is there an arbitrage? is there a model-independent (strong) arbitrage? is there a model-dependent (weak) arbitrage?? A perfectly calibrated model can be used for pricing and hedging. David Hobson (Warwick) Model Free Hedging June / 32

18 Definition A perfectly calibrated model is a quintuple (Ω, F, Q, F, S) such that S is a (Q, F)-martingale and E Q [(S T K) + ] = C(T, K) for each traded vanilla call. More generally, given a set of option prices is there a perfectly calibrated model consistent with those option prices? is there a model with continuous paths? is there a model with paths of finite (quadratic) variation? if there is, is it unique? If not one ambition is to choose a parsimonious model which captures essential features of the problem is realistic another ambition is to search for the class M of all consistent models if there is not, is there an arbitrage? is there a model-independent (strong) arbitrage? is there a model-dependent (weak) arbitrage?? A perfectly calibrated model can be used for pricing and hedging. David Hobson (Warwick) Model Free Hedging June / 32

19 Definition A perfectly calibrated model is a quintuple (Ω, F, Q, F, S) such that S is a (Q, F)-martingale and E Q [(S T K) + ] = C(T, K) for each traded vanilla call. More generally, given a set of option prices is there a perfectly calibrated model consistent with those option prices? is there a model with continuous paths? is there a model with paths of finite (quadratic) variation? if there is, is it unique? If not one ambition is to choose a parsimonious model which captures essential features of the problem is realistic another ambition is to search for the class M of all consistent models if there is not, is there an arbitrage? is there a model-independent (strong) arbitrage? is there a model-dependent (weak) arbitrage?? A perfectly calibrated model can be used for pricing and hedging. David Hobson (Warwick) Model Free Hedging June / 32

20 Classical approach Parametric Family of Models { Prices of vanillas } { Calibrated models } Prices and hedges of exotics Model-free approach Prices of underlyings and vanillas Perfectly calibrated Models/ Classes of models Prices and hedges for exotics David Hobson (Warwick) Model Free Hedging June / 32

21 Classical approach Parametric Family of Models { Prices of vanillas } { Calibrated models } Prices and hedges of exotics Model-free approach Prices of underlyings and vanillas Perfectly calibrated Models/ Classes of models Prices and hedges for exotics David Hobson (Warwick) Model Free Hedging June / 32

22 Classical approach Parametric Family of Models { Prices of vanillas } { Calibrated models } Prices and hedges of exotics Model-free approach Prices of underlyings and vanillas Perfectly calibrated Models/ Classes of models Prices and hedges for exotics David Hobson (Warwick) Model Free Hedging June / 32

23 Classical approach Parametric Family of Models { Prices of vanillas } { Calibrated models } Prices and hedges of exotics Model-free approach Prices of underlyings and vanillas Perfectly calibrated Models/ Classes of models Prices and hedges for exotics David Hobson (Warwick) Model Free Hedging June / 32

24 Classical approach Parametric Family of Models { Prices of vanillas } { Calibrated models } Prices and hedges of exotics Model-free approach Prices of underlyings and vanillas Perfectly calibrated Models/ Classes of models Prices and hedges for exotics David Hobson (Warwick) Model Free Hedging June / 32

25 The pricing (primal) problem Suppose we are given a family {(S Tα K α ) + } α A of traded calls with associated prices C(T α, K α ) α A. The class M of perfectly calibrated models is the class of models M such that E Q [(S Tα K α ) + ] = C(T α, K α ) α A. Now consider adding an extra security F T = F (((S t ) 0 t T )) (an exotic). Assuming M is non-empty the primal problem is to find: P = sup E Q [F T ] M M David Hobson (Warwick) Model Free Hedging June / 32

26 The pricing (primal) problem Suppose we are given a family {(S Tα K α ) + } α A of traded calls with associated prices C(T α, K α ) α A. The class M of perfectly calibrated models is the class of models M such that E Q [(S Tα K α ) + ] = C(T α, K α ) α A. Now consider adding an extra security F T = F (((S t ) 0 t T )) (an exotic). Assuming M is non-empty the primal problem is to find: P = sup E Q [F T ] M M David Hobson (Warwick) Model Free Hedging June / 32

27 The dual (hedging) problem In the setting as above, suppose we can find π = {(π i ) i=1,...,n }, α = {(α i ) 1 i N }, θ = (θ t ) 0 t T such that (π, α, θ) is a semi-static superhedge, i.e. F T N i=1 T π i (S Tαi K αi ) + + θ t ds t 0 Remarks: We want to be able to give a pathwise interpretation to the integral. For this we may need to decide which paths of S are to be considered as feasible (càdlàg? continuous?), and to put restrictions on θ, (elementary? bounded variation?) Then the price of the super-hedging portfolio is N π i C(T αi, K αi ) i=1 David Hobson (Warwick) Model Free Hedging June / 32

28 The dual (hedging) problem In the setting as above, suppose we can find π = {(π i ) i=1,...,n }, α = {(α i ) 1 i N }, θ = (θ t ) 0 t T such that (π, α, θ) is a semi-static superhedge, i.e. F T N i=1 T π i (S Tαi K αi ) + + θ t ds t 0 Remarks: We want to be able to give a pathwise interpretation to the integral. For this we may need to decide which paths of S are to be considered as feasible (càdlàg? continuous?), and to put restrictions on θ, (elementary? bounded variation?) Then the price of the super-hedging portfolio is N π i C(T αi, K αi ) i=1 David Hobson (Warwick) Model Free Hedging June / 32

29 The dual problem is to find D = inf N π i C(T αi, K αi ) i=1 where the infimum is taken over semi-static superhedges. Taking expectations in any perfectly calibrated model and for any semi-static superhedge E Q F T N π i C(T αi, K αi ) i=1 we get weak duality P D For specific examples, and under some general assumptions, there is strong duality P = D. David Hobson (Warwick) Model Free Hedging June / 32

30 The dual problem is to find D = inf N π i C(T αi, K αi ) i=1 where the infimum is taken over semi-static superhedges. Taking expectations in any perfectly calibrated model and for any semi-static superhedge E Q F T N π i C(T αi, K αi ) i=1 we get weak duality P D For specific examples, and under some general assumptions, there is strong duality P = D. David Hobson (Warwick) Model Free Hedging June / 32

31 The dual problem is to find D = inf N π i C(T αi, K αi ) i=1 where the infimum is taken over semi-static superhedges. Taking expectations in any perfectly calibrated model and for any semi-static superhedge E Q F T N π i C(T αi, K αi ) i=1 we get weak duality P D For specific examples, and under some general assumptions, there is strong duality P = D. David Hobson (Warwick) Model Free Hedging June / 32

32 Martingale inequalities Let (x 0, x 1,... x n ) be a sequence of real numbers. Let f = f (x 0, x 1,..., x n ) be given. Suppose we can find (c i, h i ) 0 i n where c i = c i (x i ) and h i = h i (x 0, x 1,..., x i ) such that for all sequences (x 0, x 1,..., x n ) f (x 0, x 1,..., x n ) n n 1 c i (x i ) + (x i+1 x i )h i (x 0, x 1,..., x i ) i=0 Then if M = (M k ) 0 k n is a discrete-parameter martingale then f (M 0, M 1..., M n ) and then i=0 n n 1 c(m i ) + (M i+1 M i )h i (M 0, M 1,..., M i ) i=0 i=0 E[f (M 0, M 1..., M n )] n E[c(M i )] This inequality relates the path-dependent functional to the marginals. David Hobson (Warwick) Model Free Hedging June / 32 i=0

33 Martingale inequalities Let (x 0, x 1,... x n ) be a sequence of real numbers. Let f = f (x 0, x 1,..., x n ) be given. Suppose we can find (c i, h i ) 0 i n where c i = c i (x i ) and h i = h i (x 0, x 1,..., x i ) such that for all sequences (x 0, x 1,..., x n ) f (x 0, x 1,..., x n ) n n 1 c i (x i ) + (x i+1 x i )h i (x 0, x 1,..., x i ) i=0 Then if M = (M k ) 0 k n is a discrete-parameter martingale then f (M 0, M 1..., M n ) and then i=0 n n 1 c(m i ) + (M i+1 M i )h i (M 0, M 1,..., M i ) i=0 i=0 E[f (M 0, M 1..., M n )] n E[c(M i )] This inequality relates the path-dependent functional to the marginals. David Hobson (Warwick) Model Free Hedging June / 32 i=0

34 For any sequence (x 0, x 1,... x n), let y k = max{x 0, x 1,... x k }. Then n 1 n 1 h(y i )(x i+1 x i ) = h(y i )(y i+1 y i ) + h(y n)(x n y n) i=0 i=0 Then with h(x) = 4x, and using 4u(v u) 2(v 2 u 2 ), and then n 1 4y i (x i+1 x i ) = i=0 For any sequence n 1 4y i (y i+1 y i ) + 4y n(x n y n) i=0 n 1 2(yi+1 2 y i 2 ) + 4yn(xn yn) = 2y n 2 2y xnyn i=0 n 1 4y i (x i+1 x i ) + 2x0 2 + y n 2 4xn 2 4x ny n yn 2 4xn 2 0 i=0 and it follows that, for any martingale [ n 1 yn 2 4x2 n 2x y i (x i+1 x i ) E i=0 ] ( sup M k ) 2 4E[Mn 2 ] 2E[M2 0 ] 0 k n David Hobson (Warwick) Model Free Hedging June / 32

35 x σr = y σr = y n x n x 3 = y 3 x 0 = y 0 = y 1 = y 2 σ 1 = 3 σ r n n 1 h(y i )(x i+1 x i ) = i=0 = = r 1 σ i+1 1 s=0 i=σ i h(y i )(x i+1 x i ) + n 1 r 1 h(y s)(y s+1 y s) + h(y n)(x n y n) s=0 n 1 h(y i )(y i+1 y i ) + h(y n)(x n y n) i=0 i=σ r h(y i )(x i+1 x i ) David Hobson (Warwick) Model Free Hedging June / 32

36 One-touch digitals Suppose S 0 < B and consider the first time H B that S gets to level B or above. Consider a one-touch digital which pays a unit amount on this event. Then, for K < B I { max 0 u T S u B} Then = T 1 B K (S T K) I {t H B } 0 (B K) ds t 1 B K (S T K) + + I {T H B } S H B S T B K sup Q{ max S u B} = P D = M M 0 u T C(T, K) inf K<B B K Note C(T, K) is a decreasing convex function with C(0) = S 0 < B, and we can find the infimum = minimum over K by calculus, or by picture. We can prove equality throughout by finding a perfectly calibrated model, and a K for which Q{max 0 u T S u B} = C(T, K )/(B K ). David Hobson (Warwick) Model Free Hedging June / 32

37 One-touch digitals Suppose S 0 < B and consider the first time H B that S gets to level B or above. Consider a one-touch digital which pays a unit amount on this event. Then, for K < B I { max 0 u T S u B} Then = T 1 B K (S T K) I {t H B } 0 (B K) ds t 1 B K (S T K) + + I {T H B } S H B S T B K sup Q{ max S u B} = P D = M M 0 u T C(T, K) inf K<B B K Note C(T, K) is a decreasing convex function with C(0) = S 0 < B, and we can find the infimum = minimum over K by calculus, or by picture. We can prove equality throughout by finding a perfectly calibrated model, and a K for which Q{max 0 u T S u B} = C(T, K )/(B K ). David Hobson (Warwick) Model Free Hedging June / 32

38 One-touch digitals Suppose S 0 < B and consider the first time H B that S gets to level B or above. Consider a one-touch digital which pays a unit amount on this event. Then, for K < B I { max 0 u T S u B} Then = T 1 B K (S T K) I {t H B } 0 (B K) ds t 1 B K (S T K) + + I {T H B } S H B S T B K sup Q{ max S u B} = P D = M M 0 u T C(T, K) inf K<B B K Note C(T, K) is a decreasing convex function with C(0) = S 0 < B, and we can find the infimum = minimum over K by calculus, or by picture. We can prove equality throughout by finding a perfectly calibrated model, and a K for which Q{max 0 u T S u B} = C(T, K )/(B K ). David Hobson (Warwick) Model Free Hedging June / 32

39 One-touch digitals Suppose S 0 < B and consider the first time H B that S gets to level B or above. Consider a one-touch digital which pays a unit amount on this event. Then, for K < B I { max 0 u T S u B} Then = T 1 B K (S T K) I {t H B } 0 (B K) ds t 1 B K (S T K) + + I {T H B } S H B S T B K sup Q{ max S u B} = P D = M M 0 u T C(T, K) inf K<B B K Note C(T, K) is a decreasing convex function with C(0) = S 0 < B, and we can find the infimum = minimum over K by calculus, or by picture. We can prove equality throughout by finding a perfectly calibrated model, and a K for which Q{max 0 u T S u B} = C(T, K )/(B K ). David Hobson (Warwick) Model Free Hedging June / 32

40 S0 C(k) K K S0 B k David Hobson (Warwick) Model Free Hedging June / 32

41 Skorokhod Embedding Problems Definition Let W be Brownian motion started at 0. Let η be a centred probability measure. The Skorokhod embedding problem (SEP) is to find a stopping time τ such that W τ η. We can switch to W 0 = S 0 and η has mean S 0 by a translation. In the base problem, knowledge of a continuum of calls for a single maturity T is equivalent to knowledge of the law of M T. But, by a theorem of Monroe, any right continuous martingale can be written as a time-change M t = W τt of Brownian motion. Conversely, given W a solution of the SEP such that W τ µ we can find a perfectly calibrated model by setting S t = W (t/t t) τ. David Hobson (Warwick) Model Free Hedging June / 32

42 If F is invariant under time change, so that F ((S t ) 0 t T ) = F ((W t ) 0 t τ ) we have sup E[F ] = sup E[F ((W t ) 0 t τ )] M M τ:w τ µ and the search over models becomes a search over solutions of the SEP. For increasing functions φ, the solution of the SEP which maximises E[φ(sup u τ W s )] is the Azéma-Yor solution. Hence we can prove optimality (and no duality gap) for the one-touch digital. Renewed interest in the Skorokhod Embedding Problem: New proofs of existing results based on pathwise inequalities New (financial) interpretations of existing results New embeddings based on maximising functionals Forward starting versions (non-trivial initial law) and non-centred versions David Hobson (Warwick) Model Free Hedging June / 32

43 If F is invariant under time change, so that F ((S t ) 0 t T ) = F ((W t ) 0 t τ ) we have sup E[F ] = sup E[F ((W t ) 0 t τ )] M M τ:w τ µ and the search over models becomes a search over solutions of the SEP. For increasing functions φ, the solution of the SEP which maximises E[φ(sup u τ W s )] is the Azéma-Yor solution. Hence we can prove optimality (and no duality gap) for the one-touch digital. Renewed interest in the Skorokhod Embedding Problem: New proofs of existing results based on pathwise inequalities New (financial) interpretations of existing results New embeddings based on maximising functionals Forward starting versions (non-trivial initial law) and non-centred versions David Hobson (Warwick) Model Free Hedging June / 32

44 Optimal transport Let µ, ν be probability measures on a space E and let c : E E R be a cost function. The optimal transport problem, (Monge, Kantorovich,..., Villani... ) is to find inf E[c(X, Y )] X µ,y ν If E = R and c(x, y) = γ(y x) with γ convex, and if µ is atom free, then the solution is trivial; set Y = F 1 ν (F µ (X )). Suppose there exists φ, ψ s.t. c(x, y) φ(x) + ψ(y). Then for any X, Y P(µ, ν) = We want inf E[c(X, Y )] sup X µ,y ν E[φ(X ) + ψ(y )] = D(µ, ν) φ,ψ:c φ+ψ conditions for no duality gap characterisations of the optimal solution (eg Brenier s theorem) David Hobson (Warwick) Model Free Hedging June / 32

45 Optimal transport Let µ, ν be probability measures on a space E and let c : E E R be a cost function. The optimal transport problem, (Monge, Kantorovich,..., Villani... ) is to find inf E[c(X, Y )] X µ,y ν If E = R and c(x, y) = γ(y x) with γ convex, and if µ is atom free, then the solution is trivial; set Y = F 1 ν (F µ (X )). Suppose there exists φ, ψ s.t. c(x, y) φ(x) + ψ(y). Then for any X, Y P(µ, ν) = We want inf E[c(X, Y )] sup X µ,y ν E[φ(X ) + ψ(y )] = D(µ, ν) φ,ψ:c φ+ψ conditions for no duality gap characterisations of the optimal solution (eg Brenier s theorem) David Hobson (Warwick) Model Free Hedging June / 32

46 Optimal transport Let µ, ν be probability measures on a space E and let c : E E R be a cost function. The optimal transport problem, (Monge, Kantorovich,..., Villani... ) is to find inf E[c(X, Y )] X µ,y ν If E = R and c(x, y) = γ(y x) with γ convex, and if µ is atom free, then the solution is trivial; set Y = F 1 ν (F µ (X )). Suppose there exists φ, ψ s.t. c(x, y) φ(x) + ψ(y). Then for any X, Y P(µ, ν) = We want inf E[c(X, Y )] sup X µ,y ν E[φ(X ) + ψ(y )] = D(µ, ν) φ,ψ:c φ+ψ conditions for no duality gap characterisations of the optimal solution (eg Brenier s theorem) David Hobson (Warwick) Model Free Hedging June / 32

47 Optimal transport Let µ, ν be probability measures on a space E and let c : E E R be a cost function. The optimal transport problem, (Monge, Kantorovich,..., Villani... ) is to find inf E[c(X, Y )] X µ,y ν If E = R and c(x, y) = γ(y x) with γ convex, and if µ is atom free, then the solution is trivial; set Y = F 1 ν (F µ (X )). Suppose there exists φ, ψ s.t. c(x, y) φ(x) + ψ(y). Then for any X, Y P(µ, ν) = We want inf E[c(X, Y )] sup X µ,y ν E[φ(X ) + ψ(y )] = D(µ, ν) φ,ψ:c φ+ψ conditions for no duality gap characterisations of the optimal solution (eg Brenier s theorem) David Hobson (Warwick) Model Free Hedging June / 32

48 Martingale Optimal Transport Suppose E = R and we add a condition that E[Y X ] = X to represent a martingale condition. The martingale optimal transport problem is to find P mg = Suppose there exists φ, ψ, h such that c(x, y) φ(x) + ψ(y) + h(x)(y x). inf E[c(X, Y )] X µ,y ν,e[y X ]=X P mg (µ, ν) = inf E[c(X, Y )] D mg (µ, ν) = sup E[φ(X ) + ψ(y )] ρ A φ,ψ,h where ρ is a joint law of (X, Y ) and A = A(µ, ν) = {ρ : y ρ(dx, dy) µ(dx); x ρ(dx, dy) ν(dy); y (y x)ρ(dx, dy) = 0}, and the supremum is taken over φ, ψ, h such that c(x, y) φ(x) + ψ(y) + h(x)(y x). David Hobson (Warwick) Model Free Hedging June / 32

49 Martingale Optimal Transport Suppose E = R and we add a condition that E[Y X ] = X to represent a martingale condition. The martingale optimal transport problem is to find P mg = Suppose there exists φ, ψ, h such that c(x, y) φ(x) + ψ(y) + h(x)(y x). inf E[c(X, Y )] X µ,y ν,e[y X ]=X P mg (µ, ν) = inf E[c(X, Y )] D mg (µ, ν) = sup E[φ(X ) + ψ(y )] ρ A φ,ψ,h where ρ is a joint law of (X, Y ) and A = A(µ, ν) = {ρ : y ρ(dx, dy) µ(dx); x ρ(dx, dy) ν(dy); y (y x)ρ(dx, dy) = 0}, and the supremum is taken over φ, ψ, h such that c(x, y) φ(x) + ψ(y) + h(x)(y x). David Hobson (Warwick) Model Free Hedging June / 32

50 Martingale Optimal Transport Suppose E = R and we add a condition that E[Y X ] = X to represent a martingale condition. The martingale optimal transport problem is to find P mg = Suppose there exists φ, ψ, h such that c(x, y) φ(x) + ψ(y) + h(x)(y x). inf E[c(X, Y )] X µ,y ν,e[y X ]=X P mg (µ, ν) = inf E[c(X, Y )] D mg (µ, ν) = sup E[φ(X ) + ψ(y )] ρ A φ,ψ,h where ρ is a joint law of (X, Y ) and A = A(µ, ν) = {ρ : y ρ(dx, dy) µ(dx); x ρ(dx, dy) ν(dy); y (y x)ρ(dx, dy) = 0}, and the supremum is taken over φ, ψ, h such that c(x, y) φ(x) + ψ(y) + h(x)(y x). David Hobson (Warwick) Model Free Hedging June / 32

51 An Example Suppose µ U[ 1, 1] and ν U[ 2, 2]. Suppose c(x, y) = y x. Primal problem: Let Z = ±1 with P(Z = 1) = 1 2 = P(Z = 1). Let Y = X + Z. Then, if X µ we have Y ν and E[Y X ] = X. Then (X, Y ) is a solution for the problem, and in this case E[ Y X ] = 1. Hence P 1. Dual problem: Since b 1 2 (b2 + 1) and (y x) 2 = y 2 x 2 2x(y x), Then Y X Y X X (Y X ) 1 2 E[ Y X ] 1 2 E[Y 2 ] E[X 2 ] 1 2 Hence D 1 Finally 1 P D 1 and there is equality throughout. David Hobson (Warwick) Model Free Hedging June / 32

52 An Example Suppose µ U[ 1, 1] and ν U[ 2, 2]. Suppose c(x, y) = y x. Primal problem: Let Z = ±1 with P(Z = 1) = 1 2 = P(Z = 1). Let Y = X + Z. Then, if X µ we have Y ν and E[Y X ] = X. Then (X, Y ) is a solution for the problem, and in this case E[ Y X ] = 1. Hence P 1. Dual problem: Since b 1 2 (b2 + 1) and (y x) 2 = y 2 x 2 2x(y x), Then Y X Y X X (Y X ) 1 2 E[ Y X ] 1 2 E[Y 2 ] E[X 2 ] 1 2 Hence D 1 Finally 1 P D 1 and there is equality throughout. David Hobson (Warwick) Model Free Hedging June / 32

53 An Example Suppose µ U[ 1, 1] and ν U[ 2, 2]. Suppose c(x, y) = y x. Primal problem: Let Z = ±1 with P(Z = 1) = 1 2 = P(Z = 1). Let Y = X + Z. Then, if X µ we have Y ν and E[Y X ] = X. Then (X, Y ) is a solution for the problem, and in this case E[ Y X ] = 1. Hence P 1. Dual problem: Since b 1 2 (b2 + 1) and (y x) 2 = y 2 x 2 2x(y x), Then Y X Y X X (Y X ) 1 2 E[ Y X ] 1 2 E[Y 2 ] E[X 2 ] 1 2 Hence D 1 Finally 1 P D 1 and there is equality throughout. David Hobson (Warwick) Model Free Hedging June / 32

54 An Example Suppose µ U[ 1, 1] and ν U[ 2, 2]. Suppose c(x, y) = y x. Primal problem: Let Z = ±1 with P(Z = 1) = 1 2 = P(Z = 1). Let Y = X + Z. Then, if X µ we have Y ν and E[Y X ] = X. Then (X, Y ) is a solution for the problem, and in this case E[ Y X ] = 1. Hence P 1. Dual problem: Since b 1 2 (b2 + 1) and (y x) 2 = y 2 x 2 2x(y x), Then Y X Y X X (Y X ) 1 2 E[ Y X ] 1 2 E[Y 2 ] E[X 2 ] 1 2 Hence D 1 Finally 1 P D 1 and there is equality throughout. David Hobson (Warwick) Model Free Hedging June / 32

55 Martingale Optimal Transport: Open Issues Solve the problem for explicit cost functionals eg c(x, y) = y x or c(x, y) = (ln y ln x) 2. Characterisations of the optimal solution (eg analogues of Brenier s theorem). Prove duality in a general setting. Extend to multiple time-steps by concatenation. David Hobson (Warwick) Model Free Hedging June / 32

56 PCOCs and fake diffusions Definition A fake Brownian motion is a martingale X = (X t ) t 0 with N(0, t) marginals which is not Brownian motion. (Note (X s, X t ) is not jointly normal.) There are several constructions of fake Brownian motion, typically involving dropping either the path-continuity property, or the Markov property. Definition A fake diffusion (fake version of a martingale diffusion Y ) is a martingale X = (X t ) t 0 such that X t Y t. David Hobson (Warwick) Model Free Hedging June / 32

57 PCOCs and fake diffusions Definition A fake Brownian motion is a martingale X = (X t ) t 0 with N(0, t) marginals which is not Brownian motion. (Note (X s, X t ) is not jointly normal.) There are several constructions of fake Brownian motion, typically involving dropping either the path-continuity property, or the Markov property. Definition A fake diffusion (fake version of a martingale diffusion Y ) is a martingale X = (X t ) t 0 such that X t Y t. David Hobson (Warwick) Model Free Hedging June / 32

58 Definition A PCOC is a Processus Croissant d Ordre Convexe: given a set of marginals which is increasing in convex order, a PCOC is a martingale which matches those marginals. David Hobson (Warwick) Model Free Hedging June / 32

59 The mathematical finance interpretation given a set of call option prices {C(T, K)} T 0,K 0 satisfying no-arbitrage, construct a perfectly calibrated model elegantly to maximise some path-dependent functional given an exotic derivative, find the cheapest superhedging strategy. duality gap? Note, the Dupire construction gives a perfectly calibrated model. David Hobson (Warwick) Model Free Hedging June / 32

60 Local volatility models Lemma (Dupire) Suppose call prices are known for every strike K (0, ) and every T (0, T ]. Assuming C(T, K) is sufficiently differentiable, there exists a unique diffusion of the form such that In particular σ(t, s) solves ds t = S t σ(t, S t )db t C(T, K) = E[(S T K) + ] σ(t, K) 2 = 2 T C(T, K) C(T, K) K K Are there other models which calibrate perfectly to the data? If so, why choose the local-vol model? David Hobson (Warwick) Model Free Hedging June / 32

61 Summary and conclusions The model-free framework provides an alternative to the classical approach. The idea is to use liquidly traded instruments (including vanilla derivatives) to restrict attention to the class of perfectly calibrated models (primal approach) or as hedging instruments (dual approach). Implications for trading of financial derivatives: Upper and lower price bounds; sub and superhedges. But bounds are often wide. Connections with mathematics: Martingale inequalities, Skorokhod embeddings, martingale optimal transport, fake diffusions,.... David Hobson (Warwick) Model Free Hedging June / 32

62 Summary and conclusions The model-free framework provides an alternative to the classical approach. The idea is to use liquidly traded instruments (including vanilla derivatives) to restrict attention to the class of perfectly calibrated models (primal approach) or as hedging instruments (dual approach). Implications for trading of financial derivatives: Upper and lower price bounds; sub and superhedges. But bounds are often wide. Connections with mathematics: Martingale inequalities, Skorokhod embeddings, martingale optimal transport, fake diffusions,.... David Hobson (Warwick) Model Free Hedging June / 32

63 Summary and conclusions The model-free framework provides an alternative to the classical approach. The idea is to use liquidly traded instruments (including vanilla derivatives) to restrict attention to the class of perfectly calibrated models (primal approach) or as hedging instruments (dual approach). Implications for trading of financial derivatives: Upper and lower price bounds; sub and superhedges. But bounds are often wide. Connections with mathematics: Martingale inequalities, Skorokhod embeddings, martingale optimal transport, fake diffusions,.... David Hobson (Warwick) Model Free Hedging June / 32

64 Thankyou for your attention! David Hobson (Warwick) Model Free Hedging June / 32

65 References Model free pricing: D.G. Hobson; Robust Hedging of the Lookback Option; Fin. Stoch.2(4) , D.G. Hobson; The Skorokhod Embedding Problem and Model-Independent Bounds for Option Prices; Paris-Princeton Lectures on Mathematical Finance Springer. Duality Gaps M. Beiglböck, P. Henry-Labordère and F. Penkner; Model independent bounds for option prices - a mass transport approach; Fin. Stoch. 17(3) , J. Dolinsky and H. Mete Soner; Martingale optimal transport and robust hedging in continuous time; Prob. Th. Rel. Fields (forthcoming). Martingale inequalities B. Acciaio, M. Beiglböck, W. Schachermayer, J. Temme; A trajectorial interpretation of Doob s martingale inequalities; Ann. Appl. Prob. 23(4) , 2013 Fake diffusions K. Hamza and F.C. Klebaner; A family of non-gaussian martingales with Gaussian marginals; J. Appl. Math. Stoch. Anal P. Henry-Labordere and X. Tan and N. Touzi; An explicit martingale version of the one-dimensional Brenier s theorem with full martingales constraint (preprint). David Hobson (Warwick) Model Free Hedging June / 32