Particle methods and the pricing of American options

Transcription

1 Particle methods and the pricing of American options Peng HU Oxford-Man Institute April 29, 2013 Joint works with P. Del Moral, N. Oudjane & B. Rémillard P. HU (OMI) University of Oxford 1 / 46

2 Outline 1 Snell envelope 2 Broadie-Glasserman methods 3 Genealogical/Ancestral tree based method 4 Snell envelope with small probability criteria P. HU (OMI) University of Oxford 2 / 46

3 Summary 1 Snell envelope Some notation Path space models Snell envelope Robustness lemma Examples Small probability criteria Exponential concentration inequalities 2 Broadie-Glasserman methods 3 Genealogical/Ancestral tree based method 4 Snell envelope with small probability criteria P. HU (OMI) University of Oxford 3 / 46

4 Some notation E state space, P(E) proba. on E & B(E) bounded functions (µ, f ) P(E) B(E) µ(f ) = µ(dx) f (x) M(x, dy) integral operator over E M(f )(x) = M(x, dy)f (y) [µm](dy) = µ(dx)m(x, dy) (= [µm](f ) = µ[m(f )] ) Markov chain X n with transitions M n (x n 1, dx n ) from E n 1 to E n E Pη0 {f n (X n ) X 0,..., X k } = M k,n (f n )(X k ) := M k,n (X k, dx n ) f n (x n ) E n with M k,n (x k, dx n ) = (M k+1 M k+2... M n )(x k, dx n ) = P (X n dx n X k = x k ) P. HU (OMI) University of Oxford 4 / 46

5 Path space models Path space notations Given a elementary X k Markov chain with transitions M k (x k 1, dx k ) from E k 1 into E k. The historical process X k = (X 0,..., X k ) E k = (E 0 E k ) can be seen as a Markov chain with transitions M k (x k 1, dx k ) P. HU (OMI) University of Oxford 5 / 46

6 Snell envelope Description For 0 k n, some process Z k (gain) with F k available information on k, T k set of stopping times taking value in(k,k+1... n) Purpose: find sup τ Tk E(Z τ F k ) Y k the Snell envelope of Z k : Y n = Z n Y k = Z k E(Y k+1 F k ) Main property of the Snell envelope: Y k = sup τ T k E(Z τ F k ) = E(Z τ k F k ) τ k = min {k j n : Y j = Z j } T k P. HU (OMI) University of Oxford 6 / 46

7 Snell envelope Assumption Some Markov chain (X k ) 0 k n, with η 0 P(E 0 ), M n (x n 1, dx n ) from E n 1 to E n on filtered space (Ω, F, P η0 ), F k associated natural filtration. For f k B(E k ), assume Z k = f k (X k ) (payoff) Then Y k = u k (X k ) Snell envelope recursion: u k = f k M k+1 (u k+1 ) with u n = f n A NSC for the existence of the Snell envelope M k,l f l (x) < for any 1 k l n, and any state x E k. To check this claim, we simply notice that f k u k f k + M k+1 u k+1 = f k u k M k,l f l k l n P. HU (OMI) University of Oxford 7 / 46

8 Preliminary Numerical solution Replacing (f k, M k ) 0 k n by some approximation model ( f k, M k ) 0 k n on some possibly reduced measurable subsets Êk E k. û k = f k M k+1 (û k+1 ) with terminal condition û n = f n for 0 k n A robustness/continuity lemma For any 0 k < n, on the state space Êk, we have that u k û k n l=k M k,l f l f n 1 l + l=k M k,l (M l+1 M l+1 )u l+1 Proof: By inequality (a b) (a b ) a a + b b and induction. P. HU (OMI) University of Oxford 8 / 46

9 Application examples of the lemma Deterministic methods Cut-off type methods Euler approximation methods Interpolation type methods Quantization tree methods Monte Carlo methods ( stoch. N-grid approximation) Broadie-Glasserman methods [N 2 ] BG type adapted mean-field particle method [N 2 ] Importance sampling method for path dependent case [N 2 ] Genealogical tree based method [N] P. HU (OMI) University of Oxford 9 / 46

10 Path dependent case Problematic Given gain functions (f k ) 0 k n and obstacle functions (G k ) 0 k n Snell envelope of f k (X k ) k 1 p=0 G p(x p ) = F k (X 0,..., X k )? Impossible to compute if G k too small on typical trajectories. New recursion Original Snell envelope : u k (X 0,..., X k ) = F k (X 0,..., X k ) E(u k+1 (X 0,..., X k+1 ) F k ) with u n (X 0,..., X n ) = F n (X 0,..., X n ) We provide a new recursion v k = f k (G k M k+1 (v k+1 )) with v n = f n u k (x 0,..., x k ) = v k (x k ) k 1 p=0 G p(x p ) P. HU (OMI) University of Oxford 10 / 46

11 Exponential concentration inequalities Important constants p 0 a(2p) 2p = (2p) p 2 p and a(2p + 1) 2p+1 = (2p+1)p+1 p+1/2 2 (p+1/2) Proposition If we have a Khinchine s type L p -mean error bounds in the following form: integer p 1 and constant c N sup u k (x) û k (x) Lp a(p) c x E k then we have the following exponential concentration inequality ( sup P u k (x k ) û k (x k ) > c ) + ɛ exp ( Nɛ 2 /c 2) x E k N P. HU (OMI) University of Oxford 11 / 46

12 Summary 1 Snell envelope 2 Broadie-Glasserman methods Original Broadie-Glasserman BG mean-field particle method 3 Genealogical/Ancestral tree based method 4 Snell envelope with small probability criteria P. HU (OMI) University of Oxford 12 / 46

13 Broadie-Glasserman methods M. Broadie and P. Glasserman. A Stochastic Mesh Method for Pricing High- Dimensional American Options Journal of Computational Finance (04) Original Broadie-Glasserman methods (hyp : M k η k) η k η k = 1 N N i=1 δ ξ i k where ξ k := (ξ i k ) 1 i N i.i.d. N-grid η k on Ê k = E k M k+1(x k, dx k+1) M k+1(x k, dx k+1) = η k+1 (dx k+1) R k+1 (x k, x k+1) }{{} = η k+1 (dx k+1) dm k+1 (x k,.) (x dη k+1) k+1 P. HU (OMI) University of Oxford 13 / 46

14 Broadie-Glasserman methods By Khintchine s inequality we notice: [ N M l+1 M ] l+1 (f )(x l ) 2 a(p) η l+1 [(R l+1 (x l,.)f ) p ] 1 p Lp We provide the following non asymptotic convergence estimate Theorem For any integer p 1, we denote by p the smallest even integer greater than p. Then for any time horizon 0 k n, and any x k E k, we have N u k (x k) û k(x k) Lp 2a(p) k l<n { [ 1 M k,l(x k, dx l )η l+1 (R l+1 (x l,.)u l+1 ) p ]} p P. HU (OMI) University of Oxford 14 / 46

15 BG adapted mean-field particle method (N 2 ) algorithm with the choice η k = Law(X k ) = η k 1M k Description (hyp. : M k λ k) η k η k = 1 N N i=1 δ ξ i k with i.i.d. copies ξi k of X k M k+1(x k, dx k+1) M k+1(x k, dx k+1) = η k+1 (dx k+1) H k+1 (x k, x k+1 ) η k (H k+1 (., x k+1 )) with (H) 0 H n (x n 1, x n) = dm n(x n 1,.) dλ n (x n) P. HU (OMI) University of Oxford 15 / 46

16 BG mean-field particle method Snell envelope ( Set by recursion û k (x k ) = f k (x k ) Ê k+1 with terminal condition û n = f n Theorem M k+1(x k, dx k+1) û k+1(x k+1) ) (H) 1 M l+1(h 2p l+1 M l+1(u 2p l+1 ) < ) < with sup x l,y l E l H l+1 (x l, x l+1 ) H l+1 (y l, x l+1 ) h l+1(x l+1) Under the conditions (H) 0 and (H) 1 stated above, for any even integer p > 1, any 0 k n, and x k E k, we have N u k (x k ) û k (x k ) L p 2a(p) k l<n ( ) M l+1(h 2p l+1 ) M l+1(u 2p 1 l+1 ) 2p P. HU (OMI) University of Oxford 16 / 46

17 Summary 1 Snell envelope 2 Broadie-Glasserman methods 3 Genealogical/Ancestral tree based method 4 Snell envelope with small probability criteria P. HU (OMI) University of Oxford 17 / 46

18 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

19 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

20 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

21 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

22 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

23 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

24 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

25 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

26 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

27 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

28 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

29 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

30 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

31 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

32 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

33 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

34 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

35 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

36 Genealogical/Ancestral tree based method Evolution example of genealogical tree : P. HU (OMI) University of Oxford 18 / 46

37 Genealogical tree based method notation The k-th coordinate mapping π k : x n = (x 0,..., x n) E n = (E 0... E n) π k (x n ) = x k E k 0 k < n, x k E k and any function f B(E k+1 ), we have η n = Law(X 0,..., X n) and M k+1(f )(x) := η n((1 x π k ) (f π k+1 )) η n ((1 x π k )) Remark η n = η 0 M 1 M n = η n 1 M n P. HU (OMI) University of Oxford 19 / 46

38 Genealogical tree based method Particle system = Neutral genetic particle algorithm Markov chain taking values in the product state spaces E N k. Initial system X 0 = ( X i 0 ) 1 i N i.i.d. random copies of X 0 Evolution X k E N k Selection X ( ) k := X i k Ek N 1 i N Mutation X k+1 E N k+1 P. HU (OMI) University of Oxford 20 / 46

39 Genealogical tree based method Structure = Ancestral lines X 1 ( X k 0,k 1, X 1,k 1,..., X k,k 1 ).. X k = X k i = ( X 0,k i, X 1,k i,..., X k,k i ).. X k N ( X 0,k N, X 1,k N,..., X k,k N ) Remark X i k+1 = = ( ( X i 0,k+1, X 1,k+1, i..., X k,k+1) i, X ) k+1,k+1 i }{{} ( { ( }} ) { ) ( ) X i i i 0,k, X 1,k,..., X k,k, X k+1,k+1 i = X i k, X k+1,k+1 i P. HU (OMI) University of Oxford 21 / 46

40 Genealogical tree based method Occupation measures η N k := 1 N 1 i N δ X i k and η N k := 1 N 1 i N δ X i k η N k : empirical meas. of X i k c iid η N k : empirical meas. of X i k c iid η N k η N k 1 M k P. HU (OMI) University of Oxford 22 / 46

41 Genealogical tree based method With elementary decomposition [η N n η N k 1M k,n ] = n [ηl N (ηl 1M N l )]M l,n and Khintchine s inequality, by induction we have following estimates Lemma l=k For any p 1, p the smallest even integer greater than p. In this notation, for any k 0 and any function f, we have the almost sure estimate N E ( [η N n η N k 1M k 1,n ](f ) p F N k 1 ) 1 p 2a(p) n l=k [ ] 1 ηk 1M N k 1,l ( M l,n (f ) p p ) P. HU (OMI) University of Oxford 23 / 46

42 Genealogical tree based method Approximation of the Markov transitions M k+1 M k+1(f )(x) := ηn n ((1 x π k ) (f π k+1 )) η N n ((1 x π k )) Construction of Model := û k (x) = { fk (x) M k+1 (û k+1)(x) x Êk,n 0 otherwise 1 i N 1 x( X i k,n ) f ( X i k+1,n ) 1 i N 1 x( X i k,n ) In terms of the ancestors at level k, this recursion takes the following form 1 i N û k ( X i k,n ) = fk ( X i k,n ) M k+1 (û k+1 ) ( X i k,n ) P. HU (OMI) University of Oxford 24 / 46

43 Genealogical tree based method Applying the local error given by precedent lemma and the robustness lemma, we finally get Theorem For any p 1, and 1 i N we have the following uniform estimate (uk û k )( X k,n) i Lp c p (n)/ N sup 0 k n with some collection of finite constants c p (n) < whose values only depend on the parameters p and n. P. HU (OMI) University of Oxford 25 / 46

44 Numerical examples Asset modeling dx (i) t X (i) t = rdt + σ i dz i t, i = 1,..., d = 6. z i independent standard Brownian motions r =5% annually X 0 (i) = 1 and σ i = 20% annually Bermudan options Maturity T = 1 year and 11 equally distributed exercise opportunities: geometric average put with payoff (K d i=1 X (i) T ) +, K = 1 arithmetic average put with payoff (K 1 d d i=1 X (i) T ) +, K = 1 P. HU (OMI) University of Oxford 26 / 46

45 Numerical examples Benchmark Nb. assets Geometric Arithmetic Figure : Benchmark values for the geometric and arithmetic put options (taken from B. Bouchard and X. Warin, Monte-Carlo valorisation of American options: facts and new algorithms to improve existing methods, Numerical Methods in Finance, eds. R. Carmona, P. Del Moral, P. Hu and N. Oudjane, Springer-Verlag (2012). P. HU (OMI) University of Oxford 27 / 46

46 State space discretization Methods : random tree, stochastic mesh, Binomial tree, quantization partitioning or quantization-like approach: State space partitioning 1 Simulate N i.i.d. paths according to asset dynamic 2 At each time step, partition the particles into M subsets (V j k ) 1 j M,1 k n 3 For each subset, compute the representative state (S j k ) 1 j M,1 k n as average of particles Finite state space Markov chain 1 Define Ẽk = {S 1 k,..., S M k } 2 The dynamic of new Markov chain X k : ( P Xk = S j k X ) ( ) k 1 = Sk 1 i = P X k V j k X k 1 = Sk 1 i P. HU (OMI) University of Oxford 28 / 46

47 Numerical examples Complexity and errors Complexity : Forward step O(MN), Backward step O(N) State discretization error bounded by c M 1 d G.T algorithm error bounded by c Mβ N, for β > 0 Optimization Set M = O(N d 2βd+2 ) Global complexity of order N (1+2β)d+2 2βd+2 Approximation error bounded by c N 1 2βd+2 In following example, we set β = 1/2 so that the complexity grows with the dimension from N 4/3, N 3/2, N 8/5,, N 2 for dimensions d = 1, 2, 3,,. P. HU (OMI) University of Oxford 29 / 46

48 Numerical examples Normalized estimated values Normalized estimated values Normalized estimated values x10^3 10^4 2.5x10^4 5x10^4 10^5 2x10^5 4x10^5 10^6 2x10^6 Number of particles (log scale) 5x10^3 10^4 2.5x10^4 5x10^4 10^5 2x10^5 4x10^5 10^6 2x10^6 Number of particles (log scale) x10^3 10^4 2.5x10^4 5x10^4 10^5 2x10^5 4x10^5 10^6 2x10^6 Number of particles (log scale) 1.25 Normalized estimated values Normalized estimated values Normalized estimated values x10^3 10^4 2.5x10^4 5x10^4 10^5 2x10^5 4x10^5 10^6 2x10^6 Number of particles (log scale) 1 5x10^3 10^4 2.5x10^4 5x10^4 10^5 2x10^5 4x10^5 10^6 2x10^6 Number of particles (log scale) 1 5x10^3 10^4 2.5x10^4 5x10^4 10^5 2x10^5 4x10^5 10^6 2x10^6 Number of particles (log scale) Boxplots for estimated option values (divided by the benchmark values) as a function of the number of particles for the geometric put-payoff. The box stretches from the 25th percentile to the 75th percentile, the median is shown as a line across the box, the whiskers extend from the box out to the most extreme data value within 1.5 IQR (Interquartile Range) and red crosses indicates outliers. P. HU (OMI) University of Oxford 30 / 46

49 Numerical examples Normalized estimated values Normalized estimated values Normalized estimated values x10^3 10^4 2.5x10^4 5x10^4 10^5 2x10^5 4x10^5 10^6 2x10^6 Number of particles (log scale) 5x10^3 10^4 2.5x10^4 5x10^4 10^5 2x10^5 4x10^5 10^6 2x10^6 Number of particles (log scale) 5x10^3 10^4 2.5x10^4 5x10^4 10^5 2x10^5 4x10^5 10^6 2x10^6 Number of particles (log scale) Normalized estimated values Normalized estimated values Normalized estimated values x10^3 10^4 2.5x10^4 5x10^4 10^5 2x10^5 4x10^5 10^6 2x10^6 Number of particles (log scale) 1 5x10^3 10^4 2.5x10^4 5x10^4 10^5 2x10^5 4x10^5 10^6 2x10^6 Number of particles (log scale) 1 5x10^3 10^4 2.5x10^4 5x10^4 10^5 2x10^5 4x10^5 10^6 2x10^6 Number of particles (log scale) Boxplots for estimated option values (divided by the benchmark values) as a function of the number of particles for the arithmetic put-payoff. P. HU (OMI) University of Oxford 31 / 46

50 Summary 1 Snell envelope 2 Broadie-Glasserman methods 3 Genealogical/Ancestral tree based method 4 Snell envelope with small probability criteria P. HU (OMI) University of Oxford 32 / 46

51 Change of measure Problematic Markov chain (X k ) 0 k n on (E k, E k ) 0 k n with (M k ) 0 k n (P k ) 0 k n (X 0,..., X k ) 0 k n To calculate, with v n = f n : v k = f k (G k M k+1 (v k+1 )) 0 G k 1 (Barrier options.) v k = f k (M k+1 (v k+1 )) but f k are localized in a small region (Deep out of money options.) P. HU (OMI) University of Oxford 33 / 46

52 Change of measure Change of measure Natural and optimal choice to reduce variance: [ dq n = 1 n 1 ] G k (X k ) dp n, with Z n = E Z n ( n 1 ) n 1 G k (X k ) = η k (G k ) k=0 k=0 k=0 where η k (f ) := ( E f (X k ) ) k 1 p=0 G p(x p ) ( k 1 ) E p=0 G p(x p ) Remark η k (f ) = η k 1(G k 1 M k (f )) η k 1 (G k 1 ) and define η k = Φ k (η k 1 ) P. HU (OMI) University of Oxford 34 / 46

53 With small probability criteria Notation & Hyp. Q k (f )(x k 1 ) := G k 1 (x k 1 )M k (x k 1, dx k )f (x k ) M k (x k 1, dx k ) = H k (x k 1, x k )λ k (dx k ) Lemma on the change of measure For any measure η on E k, recursion of v k can be rewritten: ( ) dqk+1 (x k, ) v k (x k ) = f k (x k ) Q k+1 (v k+1 )(x k ) = f k (x k ) Φ k+1 (η) dφ k+1 (η) v k+1, for any x k E k, where dq k+1 (x k, ) dφ k+1 (η) (x k+1) = G k(x k )H k+1 (x k, x k+1 )η(g k ) η(g k H k+1 (, x k+1 )) for any (x k, x k+1 ) E k E k+1. P. HU (OMI) University of Oxford 35 / 46

54 With small probability criteria Algorithm ξ 0 = ( ) ξ0 i 1 i N i.i.d. random copies of X 0 Evolution ξ k Ek N Selection ) ξk := ( ξi k S k,η N k 1 i N E N k Mutation M k+1 ξ k+1 E N k+1 P. HU (OMI) University of Oxford 36 / 46

55 With small probability criteria Selection S k,η N k First step ξ i k stays on ξi k with proba ɛg k(ξ i k ) Go to 2 nd step with proba 1 ɛg k (ξ i k ) Second step ξ i k ξ i k with proba G k (ξ i k ) N j=1 G k (ξ j k ) Occupation measure η N k = 1 N N i=1 δ ξ i k P. HU (OMI) University of Oxford 37 / 46

56 With small probability criteria Proposition (Khintchine s inequality) For any integer p 1, we denote by p the smallest even integer greater than p. In this notation, for any 0 k n and any integrable function f on space E k+1, we have: E ( ηk+1(f N ) Fk N ) = Φk+1 (ηk N )(f ) and N E ( [ η N k+1 Φ k+1 (η N k ) ] (f ) p F N k ) 1 p 2 a(p) [ ] 1 Φ k+1 (ηk N )( f p p ) P. HU (OMI) University of Oxford 38 / 46

57 With small probability criteria Approximation of the transition operator Q k+1 Q k+1 (f )(x k ) := = = E N k+1 E N k+1 1 i N η N k+1(dx k+1 ) dq k+1(x k, ) dφ k+1 (η N k ) (x k+1)f (x k+1 ) ηk+1(dx N k+1 ) G k(x k )H k+1 (x k, x k+1 )ηk N(G k) ηk N(G f (x k+1 ) kh k+1 (, x k+1 )) G k (x k )H k+1 (x k, ξ i k+1 ) 1 j N G k(ξ j k ) 1 l N G k(ξ l k )H k+1(ξ l k, ξi k+1 ) f (ξ i k+1) Remark Q k+1 (x k, dx k+1 ):= Φ k+1 (η N k )(dx k+1) dq k+1(x k, ) dφ k+1 (η N k ) (x k+1) No Error! Q k+1 (x k, dx k+1 ) := η N k+1 (dx k+1) dq k+1(x k, ) dφ k+1 (η N k ) (x k+1) Error of order 1 N P. HU (OMI) University of Oxford 39 / 46

58 With small probability criteria Construction of model v k (x) = { fk (x) Q k+1 ( v k+1 )(x) x E N k 0 otherwise Theorem For any 0 k n and any integer p 1, we have sup ( v k v k )(x) Lp 2 a(p) x E k N k<l<n p 1 p q k,l [ ] 1 Q k,l+1 (h p 1 l+1 v p l+1 )(x) p with a collection of constants q k,l and functions h k defined as q k,l := G l h k+1 l 1 m=k G m and h k (x k ) := sup x,y E k 1 H k (x, x k ) H k (y, x k ) P. HU (OMI) University of Oxford 40 / 46

59 Numerical results Prices dynamics ds t (i) = S t (i)(rdt + σdz i t), with r = 10%, σ = 20%, T = 1, n = 11, and S t0 (i) = 1, for i = 1, 2, 3. Options Model 1 Geometric average put option with payoff (K d i=1 S T (i)) +, 2 Arithmetic average put option with payoff (K 1 d d i=1 S T (i)) +, P. HU (OMI) University of Oxford 41 / 46

60 Numerical results Choice of potential functions (sequential importance sampling) G 0 (x 1 ) = (f 1 (x 1 ) ε) α, G k (x k, x k+1 ) = (f k+1(x k+1 ) ε) α (f k (x k ) ε) α, for all k = 1,, n 1, where f k are the payoff functions and α (0, 1] and ε > 0 are parameters fixed in our simulations to the values α = 1/5 and ε = P. HU (OMI) University of Oxford 42 / 46

61 Numerical results SM vs. SMCM Payoff K d = 1 d = 2 d = 3 d = 4 d = 5 Geometric (1%) 1 (3%) 1 (6%) 1 (9%) 1 (10%) Put (2%) 8 (6%) 6 (11%) 4 (14%) 3 (14%) (6%) 28 (11%) 18 (17%) 16 (18%) 11 (16%) Arithmetic (1%) 3 (2%) 3 (7%) 4 (13%) 5 (18%) Put (2%) 13 (6%) 24 (19%) 56 (24%) 100 (20%) (6%) 71 (15%) 363 (14%) 866 (16%) ( ) Table : Variance ratio ( Var(ˆv SM ) Var(ˆv SMCM ) ) and Bias ratio ( E(ˆv SM ) E(ˆv SMCM ) E(ˆv SM ) ) (within parentheses) computed over 1000 runs for N = 3200 mesh points. (For the arithmetic put, when d = 5 and K = 0.75, the 1000 estimates provided by the standard SM algorithm were all equal to zero, hence the associated variance ratio has not been reported). P. HU (OMI) University of Oxford 43 / 46

62 Numerical results SM PB SM CM PB SM NB SM CM NB SM PB SM CM PB SM NB SM CM NB Option value estimates Option value estimates Number of particles (in logarithmic scale) Number of particles (in logarithmic scale) (a) Geometric Put with d = 3 (b) Arithmetic Put with d = 3 x SM PB SM CM PB SM NB SM CM NB 9 SM PB SM CM PB SM NB SM CM NB Option value estimates Option value estimates Number of particles (in logarithmic scale) Number of particles (in logarithmic scale) (c) Geometric Put with d = 4 (d) Arithmetic Put with d = 4 P. HU (OMI) University of Oxford 44 / 46

63 Numerical results x SM PB SM CM PB SM NB SM CM NB SM PB SM CM PB SM NB SM CM NB Option value estimates Option value estimates Number of particles (in logarithmic scale) (e) Geometric Put with d = Number of particles (in logarithmic scale) (f) Arithmetic Put with d = 5 Figure : Positively-biased option values estimates (average estimates with 95% confidence interval computed over 1000 runs) and Negatively-biased option values estimates (average estimates over the 1000 runs each forward estimate being evaluated over forward Monte Carlo simulations), computed by the SM algorithm (in blue line) and the SMCM algorithm (in red line), as a function of the number of mesh points for geometric (on the left column) and arithmetic (on the right column) put options with strike K = P. HU (OMI) University of Oxford 45 / 46

64 END Thank you! Questions? P. HU (OMI) University of Oxford 46 / 46