MARTINGALES BASED ON IID: ADDITIVE MG Y 1,..., Y t,... : IID EY = 0 X t = Y 1 +... + Y t is MG MULTIPLICATIVE MG Y 1,..., Y t,... : IID EY = 1 X t = Y 1... Y t : X t+1 = X t Y t+1 E(X t+1 F t ) = E(X t Y t+1 F t ) = X t E(Y t+1 F t ) = X t 1 = X t MORE REALISTIC FOR STOCKS, FOR EXAMPLE BINOMIAL MODEL: X t = S t, Y = e r u or e r d under π WInter 2006 1 Per A. Mykland
MARKOV PROCESSES X t, t = 0,..., T is process adapted to filtration (F t ) (X t ) is a Markov process if for every function f there is a function g so that E(f(X t+1 ) F t ) = g(x t ) In other words: all current info relevant to the future of (X t ) is contained in the current value of (X t ). FOR EXAMPLE: (S t ) IS MARKOV IN THE BINOMIAL TREE. S t (ω t 1, H) = us t 1 (ω t 1 ) S t 1 (ω t 1 ). S t (ω t 1, T ) = ds t 1 (ω t 1 ) For S t 1 (ω t 1 ) = s t 1, and if Z = u with probability p(h) and = d with probability p(t ) E(f(S t ) F t 1 )(ω t 1 ) = E(f(ZS t 1 ) F t 1 )(ω t 1 ) where: = E(f(Zs t 1 ) F t 1 )(ω t 1 ) = E(f(Zs t 1 )) (since Z is independent of F t 1 ) = g(s t 1 ) g(s) = E(f(Zs)) = f(us)p(h) + f(ds)p(t ) This works for both actual and risk neutral probability WInter 2006 2 Per A. Mykland
General principle for taking out what is known Lemma: Let X = (X 1,..., X K ) and Y = (Y 1,..., X L ) be random variables, and let f(x, y) be a function. If X = (X 1,..., X p ) is G-measurable, then on the set ω {X = x} E(f(X, Y ) G)(ω) = E(f(x, Y ) G)(ω). Proof of Lemma: Let P be the partition corresponding to G. For each ω {X = x}, there is a B P, so that ω B {X = x}. (This is since X is G-measurable.) Then E(f(X, Y ) G)(ω) = E(f(X, Y ) B) = E(f(x, Y ) B) since X(ω) = x on B = E(f(x, Y ) G)(ω) When Y is independent of G QED consequently E(f(x, Y ) G)(ω) = E(f(x, Y )) E(f(X, Y ) G)(ω) = E(f(x, Y )) WInter 2006 3 Per A. Mykland
THE STRUCTURE OF EUROPEAN OPTIONS IN THE BINOMIAL TREE Payoff V = v T (S T ). Induction: assume value of option at t+1 is v t+1 (S t+1 ). Then where value of option at t = E π (e r v t+1 (S t+1 ) F t ) = E π (e r v t+1 (ZS t ) F t ) where Z = u or = d = v t (S t ) v t (s) = E π v t+1 (Zs) = π(h)v t+1 (us) + π(t )v t+1 (ds) WInter 2006 4 Per A. Mykland
A NON-MARKOV PROCESS M t = max 0 u t S u (from Shreve p. 48) WInter 2006 5 Per A. Mykland
CREATING A MARKOV PROCESS BY ADDING STATE VARIABLES M t = max 0 u t S u CLAIM: (S t, M t ) is a Markov process Proof: If S t = s and M t = m: with Z = u or d: S t+1 = sz and M t+1 = M t S t+1 = m sz (x y = max(x, y) and x y = min(x, y)) E(f(S t+1, M t+1 ) F t ) = E(f(sZ, m (sz))) = g(s, m) by appropriate definition of g QED It follows, by induction, that, for payoff v T (S T, M T ): value of option at t = v t (S t, M t ) General Theorem (Feynman-Kac): If X 0, X 1,..., X T is a Markov process under π (e.g. X t = (S t, M t )). For payoff v T (X T ), value of option at t = v t (X t ) for some function v t (x). WInter 2006 6 Per A. Mykland
CHANGE OF MEASURE Q, R: PROBABILITIES, G: σ-field R IS ABSOLUTELY CONTINUOUS UNDER Q (on G): R Q: A G : Q(A) = 0 P (A) = 0 R IS EQUIVALENT TO Q (on G): R Q : R Q AND Q R BECAUSE ACTUAL AND RISK NEUTRAL: NO ARBITRAGE P π (i) If π(a) > 0, P (A) = 0 { S T = I A = 1 if A sop (S T = 0) = 1 0 if not ( ) 1 But π(s T > 0) > 0 π B T S T > 0 > 0 S 0 = E π 1 B T S T > 0 (ii) If π(a) = 0, P (A) > 0: P (S T > 0) > 0 S 0 = E π 1 B T S T = 0 ARBITRAGE ARBITRAGE WInter 2006 7 Per A. Mykland
BOND WIH DEFAULT NUMERAIRE: B t = e rt t = 0, 1 BOND: V t = 1 for t = 0 { } e = m no default 0 default Ṽ t = V t = 1 for t = 0 B t { } e m r no default = 0 default for t = 1 for t = 1 1 = Ṽ 0 = E π Ṽ 1 = e m r π( no default) (+0 π( default)) Condition: m > r or m r? or?? P (default) = 0 π(default) = 0 (π P ) π(no default) = 1 m = r 1 > P (default) > 0 1 > π(default) > 0 m > r WInter 2006 8 Per A. Mykland
RADON-NIKODYM DERIVATIVES Q, R: PROBABILITIES, Q R (on G) P is partition associated with G DEFINITION OF R-N DERIVATIVE dr R(B) (ω) = dq Q(B) when ω B P CAVEAT: dr dq DEPENDS ON G: dr dq = dr dq G Q( dr dq 0) = 1 If R Q: Q( dr dq > 0) = 1 ( ) dr E Q dq = 1 PROPERTIES For all G-measurable Y : E R (Y ) = E Q (Y dr dq ( ) 1 dq If R Q: dr = dr dq ) WInter 2006 9 Per A. Mykland
Since Y is G- measurable: EXAMPLE OF PROOF for every B P: Y (ω) is constant for ω B: Y (ω) = Y (B). Hence: E R (Y ) = B P Y (B)R(B) = Y (B) dr dq (B)Q(B) B P ( = E Q Y dr ) dq WInter 2006 10 Per A. Mykland
BINOMIAL TREES. S 2 = u 2 S 0 H, then H H, then T. S 2 = uds 0 T, then H. S 2 = d 2 S 0 T, then T ACTUAL MEASURE: P (H), P (T ) RISK NEUTRAL MEASURE: π(h), π(t ) σ-field F n OR PARTITION P n : BASED ON n FIRST COIN TOSSES: (ω 1,...ω n ) WInter 2006 11 Per A. Mykland
CALCULATION OF R-N DERIVATIVE IN BINOMIAL CASE For every sequence ω = (ω 1,..., ω n ): {ω} = {(ω 1,..., ω n )} is a set in P n Value of derivative for this set: dπ dp (ω 1,...ω n ) = π(ω 1,...ω n ) P (ω 1,..., ω n ) = π(ω 1)... π(ω n ) P (ω 1 )... P (ω n ) ( ) #H ( π(h) π(t ) = P (H) P (T ) ) #T where H = H(ω) = # of heads among ω 1,..., ω n AMBIGUITY ABOUT ω Could consider ω = (ω 1,..., ω n,...) (the outcome of infinitely many coin tosses). Then dπ dp F n (ω) = dπ dp (ω 1,..., ω n ) Each set in P n : one (ω 1,..., ω n ), many (ω 1,..., ω n,...) WInter 2006 12 Per A. Mykland
RELATIONSHIP BETWEEN R-N DERIVATIVE AND STOCK PRICE IN BINOMIAL CASE since #H + #T = n. Thus: S n = S 0 u #H d #T ( u ) #H = S 0 d n d #H = log S n log S 0 n log d log u log d ON THE OTHER HAND: dπ dp (ω) = ( ) #H ( ) #T π(h) π(t ) P (H) P (T ) = f(s n, n) dπ dp IS A FUNCTION OF S n WInter 2006 13 Per A. Mykland
For payoffs at time t: Price at time 0 for payoff STATE PRICE DENSITY rt dπ ξ t (ω) = e dp F t (ω) I A (ω) = { 1 if ω A 0 otherwise at time t: e rt E π (I A ) = e rt E P ( I A dπ dp = E P (I A ξ t ) ) State price corresponding to ω: A = {ω}: E P ( I{ω} ξ t ) = ξt (ω)p {ω} General valuation formula: price for payoff V at time t = e rt E π (V ) = E P (V ξ t ) WInter 2006 14 Per A. Mykland
THE RADON-NIKODYM DERIVATIVE IS A MARTINGALE DEFINE Z t = dπ dp F t IN THE BINOMIAL CASE: Z t+1 = Z t Y t+1 WHERE Y 1, Y 2,... ARE IID, Y t = { π(h) P (H) π(t ) P (T ) if outcome H at time t if outcome T at time t AND E P (Y ) = π(h) π(t ) P (H) + P (T ) = π(h) + π(t ) = 1 P (H) P (T ) Z 0 = 1 A TYPICAL MULTIPLICATIVE MARTINGALE WInter 2006 15 Per A. Mykland
THE GENERAL CASE Z t = dπ dp F t Filtration on partition form: P t IF A P t AND A = q B q WHERE B q P t+1 : E(Z t+1 A) = q = q = q Z t+1 (B q )P (B q A) π(b q ) P (B q ) P (B q A) π(b q ) P (B q ) P (B q A) P (A) IF ω A: = q = 1 P (A) = π(a) P (A) = Z t (A) π(b q ) P (B q ) P (B q ) P (A) since B q A π(b q ) q E(Z t+1 P t )(ω) = E(Z t+1 A) = Z t (A) = Z t (ω) THUS: (Z t ) IS A MARTINGALE WInter 2006 16 Per A. Mykland
THE R-N DERIVATIVE REPRESENTATION BY FINAL VALUE IF T IS THE FINAL TIME, AND dπ dp TIVE ON F T : ( ) dπ Z t = E (Z T F t ) = E P dp F t IS R-N DERIVA- for t T WInter 2006 17 Per A. Mykland
THE R-N DERIVATIVE: CONDITIONAL EXPECTATIONS Theorem: Suppose Q P on F. Let dq/dp m F and G F. Then ( ) E P X dq dp G E Q (X G) = ( ) E dq. P dp G IN THE TIME DEPENDENT SYSTEM: G = F ( ) t E P XZ T Ft E π (X F t ) = E P (Z T F t ) ( ) E P XZ T F t = PRICE AT TIME t OF PAYOFF V AT TIME ) T : E P (Ṽ ZT F t Ṽ t = E π (Ṽ F t ) = Z ( t ) E P e rt V Z T F t V t = e rt Z ( t ) E P V ξ T F t = ξ t Z t WInter 2006 18 Per A. Mykland
ANOTHER APPLICATION OF THE CONDITIONAL EXPECTATION FORMULA: MODIFYING PATH DEPENDENT OPTIONS IN THE BINOMIAL MODEL V = payoff of option (at T ) V = E π (V S T ) ( ) E V dπ dp S T = ( ) dπ (G = σ(s T )) E dp S T ) dπ dp (V E S T = dπ dp (since dπ dp is a function of S T ) = E(V S T ) WE HAVE HERE USED G = σ(s T ) WInter 2006 19 Per A. Mykland
ARE PATH DEPENDENT OPTIONS OPTIMAL? V = payoff V = E π (V S T ) = E(V S T ) Price: Price of V = e rt E π V = e rt E π V tower property = price of V Expected payoff of V e rt EV = e rt EV tower property = expected payoff of V The Rao-Blackwell inequality: Var (V ) = E (Var (V S T )) + Var (E(V S T )) = E (Var (V S T )) + Var (V ) > Var (V ) unless V = V. WInter 2006 20 Per A. Mykland
UTILITY: TYPICAL ASSUMPTIONS more is better: V 1 V 2 => U(V 1 ) U(V 2 ) risk aversion: strict concavity: U(αV 1 + (1 α)v 2 ) > αu(v 1 ) + (1 α)u(v 2 ) non-strict concavity: replace < by Jensen: U strictly concave: unless V is G-measurable: E(U(V ) G) < U(E(V G)) P -a.s. PATH DEPENDENT OPTIONS AGAIN V = E π (V S T ) = E(V S T ) Price of V = price of V EU(V ) = EE(U(V ) S T ) < EU(E(V S T )) unless V m σ(s T ) = EU(V ) WInter 2006 21 Per A. Mykland
OPTIMAL INVESTMENT IN BIN. MODEL MAXIMIZATION OF UTILITY SUBJECT TO INITIAL CAPITAL: max EU(V T ) subject to constraints: capital constraint: V 0 = v replicability: Ṽ T = V 0 + T 1 t=0 t S t BY COMPLETENESS: CONSTRAINTS EQUIVALENT TO (with ξ T = e rt (dπ/dp )) E π Ṽ T = v or EV T ξ T = v BY ARGUMENT ON PREVIOUS PAGE: V T IF FUNCTION OF S T, OR, EQUIVALENTLY, ξ T : V T = f(ξ T ) WInter 2006 22 Per A. Mykland
REFORMULATION OF PROBLEM ξ T can take values x 1,..., x T +1 Problem becomes: max f U(f(xi ))P (ξ = x i ) subject to: f(xi )x i P (ξ = x i ) = v Lagrangian: L = ( ) U(f(x i ))P (ξ = x i ) λ f(xi )x i P (ξ = x i ) v Want: 0 = L f(x i ) = U (f(x i ))P (ξ = x i ) λx i P (ξ = x i ) In other words: or: U (f(x i )) = λx i U (V T ) = λξ T Finally: V T = (U ) ( 1) (λξ T ) WInter 2006 23 Per A. Mykland